Galois TheoryTom Leinster, University of Edinburgh

Version of 13 May 2021

Note to the reader 21 Overview of Galois theory 4

1.1 The view of C from Q . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Every polynomial has a symmetry group. . . . . . . . . . . . . . . 91.3 . . . which determines whether it can be solved . . . . . . . . . . . 11

2 Rings and fields 142.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Polynomials 243.1 The ring of polynomials . . . . . . . . . . . . . . . . . . . . . . . 243.2 Factorizing polynomials . . . . . . . . . . . . . . . . . . . . . . 283.3 Irreducible polynomials . . . . . . . . . . . . . . . . . . . . . . . 33

4 Field extensions 384.1 Definition and examples . . . . . . . . . . . . . . . . . . . . . . 384.2 Algebraic and transcendental elements . . . . . . . . . . . . . . . 434.3 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5 Degree 515.1 Degrees of extensions and polynomials . . . . . . . . . . . . . . . 515.2 The tower law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . 585.4 Ruler and compass constructions . . . . . . . . . . . . . . . . . . 60

6 Splitting fields 676.1 Extending homomorphisms . . . . . . . . . . . . . . . . . . . . . 686.2 Existence and uniqueness of splitting fields . . . . . . . . . . . . 706.3 The Galois group . . . . . . . . . . . . . . . . . . . . . . . . . . 76

7 Preparation for the fundamental theorem 827.1 Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.2 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917.3 Fixed fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

8 The fundamental theorem of Galois theory 1008.1 Introducing the Galois correspondence . . . . . . . . . . . . . . . 1008.2 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048.3 A specific example . . . . . . . . . . . . . . . . . . . . . . . . . 109

9 Solvability by radicals 1159.1 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1169.2 Solvable polynomials have solvable groups . . . . . . . . . . . . . 1199.3 An unsolvable polynomial . . . . . . . . . . . . . . . . . . . . . 126

10 Finite fields 13010.1 ?th roots in characteristic ? . . . . . . . . . . . . . . . . . . . . . 13110.2 Classification of finite fields . . . . . . . . . . . . . . . . . . . . 13310.3 Multiplicative structure . . . . . . . . . . . . . . . . . . . . . . . 13510.4 Galois groups for finite fields . . . . . . . . . . . . . . . . . . . . 136

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Note to the reader

These are the 2020–21 course notes for Galois Theory (MATH10080).

Structure Each chapter corresponds to one week of the semester. You shouldread Chapter 1 in Week 1, Chapter 2 in Week 2, and so on. I’m writing the notesas we go along, so the chapters will appear one by one: keep your eye on Learn.

Exercises looking like this are sprinkled through the notes. The ideais that you try them immediately, before you continue reading.Most of them are meant to be quick and easy, much easier than as-signment or workshop questions. If you can do them, you can take itas a sign that you’re following. For those that defeat you, talk withyour group or ask on Piazza; or if you’re really stuck, ask me.I promise you that if you make a habit of trying every exercise, you’llenjoy the course more and understand it better than if you don’t bother.

Digressions like this are optional and not examinable, but might interestyou. They’re usually on points that I find interesting, and often describeconnections between Galois theory and other parts of mathematics.

Here you’ll seetitles of relevant

videosReferences to theorem numbers, page numbers, etc., are clickable links.

What to prioritize You know by now that the most important things in almostany course are the definitions and the results called Theorem. But I also want toemphasize the proofs. This course presents a wonderful body of theory, and theidea is that you learn it all, including the proofs that are its beating heart.

A closed-book exam would test that by asking you to reproduce some proofs.Your exam will be open book, so it can’t ask you to reproduce proofs, but it willtest something arguably harder: that you understand them. So the proofs will needyour attention and energy.

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Prerequisites You are required to have taken these two courses:

• Honours Algebra: We’ll need some linear algebra, corresponding to Chap-ter 1 of that course. For example, you should be able to convince yourselfthat an endomorphism of a finite-dimensional vector space is injective if andonly if it is surjective.We’ll also need everything from Honours Algebra about rings and polyno-mials (Chapter 3 there), including ideals, quotient rings (factor rings), theuniversal property of quotient rings, and the first isomorphism theorem forrings.

• Group Theory: From that course, we’ll need fundamentals such as normalsubgroups, quotient groups, the universal property of quotient groups, andthe first isomorphism theorem for groups. You should know lots about thesymmetric groups (=, alternating groups �=, cyclic groups �= and dihedralgroups �=, and I hope you can list all of the groups of order < 8 withouthaving to think too hard.Chapter 8 of Group Theory, on solvable groups, will be crucial for us! Ifyou skipped it, you’ll need to go back and fix that. For example, you’ll needto understand the statement that (4 is solvable but �5 is not.We won’t need anything on free groups, the Sylow theorems, or the Jordan–Hölder theorem.

If you’re a visiting student and didn’t take those courses, please get in touch so wecan decide whether your background is suitable.

Mistakes I’ll be grateful to hear of any mistakes (Tom.Leinster@ed.ac.uk), evenif it’s something very small and even if you’re not sure.

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Chapter 1

Overview of Galois theory

This chapter stands apart from all the others,Modern treatments of Galois theory take advantage of several well-developed

branches of algebra: the theories of groups, rings, fields, and vector spaces. Thisis as it should be! However, assembling all the algebraic apparatus will take ussome time, during which it’s easy to lose sight of what it’s all for.

Introduction toWeek 1

Galois theory came from two basic insights:

• every polynomial has a symmetry group;

• this group determines whether the polynomial can be solved by radicals (ina sense I’ll define).

In this chapter, I’ll explain these two ideas in as short and low-tech a way as I canmanage. In Chapter 2 we’ll start again, beginning the modern approach that willtake up the rest of the course. But I hope that all through that long build-up, you’llkeep in mind the fundamental ideas that you learn in this chapter.

1.1 The view of C from QImagine you lived several centuries ago, before the discovery of complex numbers.Your whole mathematical world is the real numbers, and there is no square root of−1. This situation frustrates you, and you decide to do something about it.

So, you invent a new symbol 8 (for ‘imaginary’) and decree that 82 = −1. Youstill want to be able to do all the usual arithmetic operations (+, ×, etc.), and youwant to keep all the rules that govern them (associativity, commutativity, etc.). Soyou’re also forced to introduce new numbers such as 2+ 3× 8, and you end up withwhat today we call the complex numbers.

So far, so good. But then you notice something strange. When you inventedthe complex numbers, you only intended to introduce one square root of −1. But

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accidentally, you introduced a second one at the same time: −8. (Youwait centuriesfor a square root of −1, then two arrive at once.) Maybe that’s not so strange initself; after all, positive reals have two square roots too. But then you realizesomething genuinely weird:

There’s nothing you can do to distinguish 8 from −8.

Try as you might, you can’t find any reasonable statement that’s true for 8 but not−8. For example, you notice that 8 is a solution of

I3 − 3I2 − 16I − 3 =17I,

but then you realize that −8 is too.Of course, there are unreasonable statements that are true for 8 but not −8, such

as ‘I = 8’. We should restrict to statements that only refer to the known world ofreal numbers. More precisely, let’s consider statements of the form

?1(I)?2(I)

=?3(I)?4(I)

, (1.1)

where ?1, ?2, ?3, ?4 are polynomials with real coefficients. Any such equationcan be rearranged to give

?(I) = 0,

where again ? is a polynomial with real coefficients, so we might as well justconsider statements of that form. The point is that if ?(8) = 0 then ?(−8) = 0.

Let’s make this formal.

Definition 1.1.1 Two complex numbers I and I′ are indistinguishable when seenfrom R, or conjugate over R, if for all polynomials ? with coefficients in R,

?(I) = 0 ⇐⇒ ?(I′) = 0.

Warning 1.1.2 The standard term is ‘conjugate over R’. ‘Indistin-guishable’ is a term I invented for this chapter only, to express the ideaof not being able to tell the two numbers apart.

For example, 8 and −8 are indistinguishable when seen from R. This followsfrom a more general result:

Lemma 1.1.3 Let I, I′ ∈ C. Then I and I′ are indistinguishable when seen fromR if and only if I′ = I or I′ = I.

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Proof ‘Only if’: suppose that I and I′ are indistinguishable when seen from R.Write I = G + 8H with G, H ∈ R. Then (I − G)2 = −H2. Since G and H are real,indistinguishability implies that (I′ − G)2 = −H2, so I′ − G = ±8H, so I′ = G ± 8H.

‘If’: obviously I is indistinguishable from itself, so it’s enough to prove that Iis indinguishable from I. I’ll give two proofs. Each one teaches us a lesson thatwill be valuable later.

First proof: recall that complex conjugation satisfies

F1 + F2 = F1 + F2, F1 · F2 = F1 · F2

for all F1, F2 ∈ C, and 0 = 0 for all 0 ∈ R. It follows by induction that for anypolynomial ? over R,

?(F) = ?(F)for all F ∈ C. So

?(I) = 0 ⇐⇒ ?(I) = 0 ⇐⇒ ?(I) = 0.

Second proof: write I = G + 8H with G, H ∈ R. Let ? be a polynomial overR such that ?(I) = 0. We will prove that ?(I) = 0. This is trivial if H = 0, sosuppose that H ≠ 0.

Consider the polynomial<(C) = (C−G)2 + H2. Then<(I) = 0. You know fromHonours Algebra that

?(C) = <(C)@(C) + A (C) (1.2)

for some polynomials @ and A with deg(A) < deg(<) = 2 (so A is either a constantor of degree 1). Putting C = I in (1.2) gives A (I) = 0. But it’s easy to see that thisis impossible unless A is the zero polynomial (using the assumption that H ≠ 0).So ?(C) = <(C)@(C). But <(I) = 0, so ?(I) = 0, as required.

We have just shown that for all polynomials ? over R, if ?(I) = 0 then?(I) = 0. Exchanging the roles of I and I proves the converse. Hence I and I areindistinguishable when seen from R. �

Exercise 1.1.4 Both proofs of ‘if’ contain little gaps: ‘It follows byinduction’ in the first proof, and ‘it’s easy to see’ in the second. Fillthem.

Digression 1.1.5 With complex analysis in mind, we could imagine a stricterdefinition of indistinguishability in which polynomials are replaced by arbi-trary convergent power series (still with coefficients in R). This would allowfunctions such as exp, cos and sin, and equations such as exp(8c) = −1.

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But this apparently different definition of indistinguishability is, in fact,equivalent. A complex number is still indistinguishable from exactly itselfand its complex conjugate. (For example, exp(−8c) = −1 too.) Do you seewhy?

Lemma 1.1.3 tells us that indistinguishability over R is rather simple. Butthe same idea becomes much more interesting if we replace R by Q. And in thiscourse, we will mainly focus on polynomials over Q.

Define indistinguishability seen from Q, or officially conjugacy over Q,by replacing R by Q in Definition 1.1.1. From now on, I will usually just say‘indistinguishable’, dropping the ‘seen from Q’.

Example 1.1.6 I claim that√

2 and −√

2 are indistinguishable. And I’ll giveyou two different proofs, closely analogous to the two proofs of the ‘if’ part ofLemma 1.1.3.

First proof: write

Q(√

2) = {0 + 1√

2 : 0, 1 ∈ Q}.

For F ∈ Q(√

2), there are unique 0, 1 ∈ Q such that F = 0 + 1√

2, because√

2 isirrational. So it is logically valid to define

Example 1.1.6F = 0 − 1

√2 ∈ Q(

√2).

It is straightforward to check that�F1 + F2 = F1 + F2, �F1 · F2 = F1 · F2

for all F1, F2 ∈ Q(√

2), and that 0 = 0 for all 0 ∈ Q. Just as in the proof ofLemma 1.1.3, it follows that F and F are indistinguishable for every F ∈ Q(

√2).

In particular,√

2 is indistinguishable from −√

2.Second proof: let ? = ?(C) be a polynomial with coefficients in Q such that

?(√

2) = 0. You know from Honours Algebra that

?(C) = (C2 − 2)@(C) + A (C)

for some polynomials @(C) and A (C) over Q with deg A < 2. Putting C =√

2 givesA (√

2) = 0. But√

2 is irrational and A (C) is of the form 0C + 1 with 0, 1 ∈ Q, so Amust be the zero polynomial. Hence ?(C) = (C2 − 2)@(C), giving ?(−

√2) = 0.

We have just shown that for all polynomials ? over Q, if ?(√

2) = 0 then?(−√

2) = 0. The same argument with the roles of√

2 and −√

2 reversed provesthe converse. Hence ±

√2 are indistinguishable.

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1

l

l2

l3

l4

Figure 1.1: The 5th roots of unity.

Exercise 1.1.7 Let I ∈ Q. Show that I is distinguishable from I′ forany complex number I′ ≠ I.

One thing that makes indistinguishability more subtle over Q than over R isthat over Q, more than two numbers can be indistinguishable:

Example 1.1.8 The 5th roots of unity are

1, l, l2, l3, l4,

where l = 42c8/5 (Figure 1.1). Now 1 is distinguishable from the rest, since it is aroot of the polynomial C − 1 and the others are not. (See also Exercise 1.1.7.) Butit turns out that l, l2, l3, l4 are all indistinguishable from each other.

Complex conjugates are indistinguishable when seen from R, so they’re cer-tainly indistinguishable when seen from Q. Since l4 = 1/l = l, it follows thatl and l4 are indistinguishable when seen from Q. By the same argument, l2

and l3 are indistinguishable. What’s not so obvious is that l and l2 are indistin-guishable. I know two proofs, which are like the two proofs of Lemma 1.1.3 andExample 1.1.6. But we’re not equipped to do either yet.

Example 1.1.9 More generally, let ? be any prime and put l = 42c8/?. Thenl, l2, . . . , l?−1 are indistinguishable.

So far, we have asked when one complex number can be distinguished fromanother, using only polynomials over Q. But what about more than one?

Definition 1.1.10 Let : ≥ 0 and let (I1, . . . , I: ) and (I′1, . . . , I′:) be :-tuples of

complex numbers. We say that (I1, . . . , I: ) and (I′1, . . . , I′:) are indistinguishable

if for all polynomials ?(C1, . . . , C: ) over Q in : variables,

?(I1, . . . , I: ) = 0 ⇐⇒ ?(I′1, . . . , I′: ) = 0.

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When : = 1, this is just the previous definition of indistinguishability.

Exercise 1.1.11 Suppose that (I1, . . . , I: ) and (I′1, . . . , I′:) are in-

distinguishable. Show that I8 and I′8 are indistinguishable, for each8 ∈ {1, . . . , :}.

Example 1.1.12 For any I1, . . . , I: ∈ C, the :-tuples (I1, . . . , I: ) and (I1, . . . , I: )are indistinguishable. For let ?(C1, . . . , C: ) be a polynomial over Q. Then

?(I1, . . . , I: ) = ?(I1, . . . , I: )

since the coefficients of ? are real, by a similar argument to the one in the firstproof of Lemma 1.1.3. Hence

?(I1, . . . , I: ) = 0 ⇐⇒ ?(I1, . . . , I: ) = 0,

which is what we had to prove.

Example 1.1.13 Let l = 42c8/5, as in Example 1.1.8. Then

(l, l2, l3, l4) and (l4, l3, l2, l)

are indistinguishable, by Example 1.1.12. It can also be shown that

(l, l2, l3, l4) and (l2, l4, l, l3)

are indistinguishable, although the proof is beyond us for now. But

(l, l2, l3, l4) and (l2, l, l3, l4)

are distinguishable, since if we put ?(C1, C2, C3, C4) = C2 − C21 then

?(l, l2, l3, l4) = 0 ≠ ?(l2, l, l3, l4).

So the converse of Exercise 1.1.11 is false: just because I8 and I′8 are indistinguish-able for all 8, it doesn’t follow that (I1, . . . , I: ) and (I′1, . . . , I

′:) are indistinguish-

able.

1.2 Every polynomial has a symmetry group. . .Weare now ready to describe the first main idea of Galois theory: every polynomialhas a symmetry group.

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Definition 1.2.1 Let 5 be a polynomial with coefficients in Q. Write U1, . . . , U:for its distinct roots in C. The Galois group of 5 is

Gal( 5 ) = {f ∈ (: : (U1, . . . , U: ) and (Uf(1) , . . . , Uf(:)) are indistinguishable}.

‘Distinct roots’ means that we ignore any repetition of roots: e.g. if 5 (C) =C5(C − 1)9 then : = 2 and {U1, U2} = {0, 1}.

Exercise 1.2.2 Show that Gal( 5 ) is a subgroup of (: . (This one ismaybe a bit tricky notationally. Stick to : = 3 if you like.)

Exercise 1.2.2Digression 1.2.3 I brushed something under the carpet. The definition ofGal( 5 ) depends on the order in which the roots are listed. Different orderingsgives different subgroups of (: . However, these subgroups are all conjugateto each other, and therefore isomorphic as abstract groups. So Gal( 5 ) iswell-defined as an abstract group, independently of the choice of ordering.

Example 1.2.4 Let 5 be a polynomial over Q all of whose complex rootsU1, . . . , U: are rational. If f ∈ Gal( 5 ) then Uf(8) and U8 are indistinguishablefor each 8, by Exercise 1.1.11. But since they are rational, that forces Uf(8) = U8(by Exercise 1.1.7), and since U1, . . . , U: are distinct, f(8) = 8. Hence f = id. Sothe Galois group of 5 is trivial.

Example 1.2.5 Let 5 be a quadratic over Q. If 5 has rational roots then as wehave just seen, Gal( 5 ) is trivial. If 5 has two non-real roots then they are complexconjugate, so Gal( 5 ) = (2 by Example 1.1.12. The remaining case is where 5 hastwo distinct roots that are real but not rational, and it can be shown that in that casetoo, Gal( 5 ) = (2.

Warning 1.2.6 On terminology: note that just now I said ‘non-real’.Sometimes people casually say ‘complex’ to mean ‘not real’. But trynot to do this yourself. It makes as little sense as saying ‘real’ to mean‘irrational’, or ‘rational’ to mean ‘not an integer’.

Example 1.2.7 Let 5 (C) = C4 + C3 + C2 + C + 1. Then (C − 1) 5 (C) = C5 − 1, so 5 hasroots l, l2, l3, l4 where l = 42c8/5. We saw in Example 1.1.13 that(

1 2 3 44 3 2 1

),

(1 2 3 42 4 1 3

)∈ Gal( 5 ),

(1 2 3 42 1 3 4

)∉ Gal( 5 ).

In fact, it can be shown that

Gal( 5 ) =⟨(

1 2 3 42 4 1 3

)⟩� �4.

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Example 1.2.8 Let 5 (C) = C3 + 1C2 + 2C + 3 be a cubic over Q with no rationalroots. Then

Gal( 5 ) �{�3 if

√−2732 + 18123 − 423 − 4133 + 1222 ∈ Q,

(3 otherwise.

This appears as Proposition 22.4 in Stewart, but is way beyond us for now: calcu-lating Galois groups is hard.

Galois groups,informally

1.3 . . . which determines whether it can be solvedHere we meet the second main idea of Galois theory: the Galois group of apolynomial determines whether it can be solved. More exactly, it determineswhether the polynomial can be ‘solved by radicals’.

To explain what this means, let’s begin with the quadratic formula. The rootsof a quadratic 0C2 + 1C + 2 are

−1 ±√12 − 40220

.

After much struggling, it was discovered that there is a similar formula for cubics0C3 + 1C2 + 2C + 3: the roots are given by

3√−27023+9012−213+30

√3(270232−180123+4023+4133−1222) + 3

√−27023+9012−213−30

√3(270232−180123+4023+4133−1222)

3 3√20.

(No, you don’t need to memorize that!) This is a complicated formula, and there’salso something strange about it. Any nonzero complex number has three cube roots,and there are two 3

√ signs in the formula (ignoring the 3√2 in the denominator), soit looks as if the formula gives nine roots for the cubic. But a cubic can only havethree roots. What’s going on?

It turns out that some of the nine aren’t roots of the cubic at all. You have tochoose your cube roots carefully. Section 1.4 of Stewart’s book has much more onthis point, as well as an explanation of how the cubic formula was obtained. (Wewon’t be going into this ourselves.)

As Stewart also explains, there is a similar but even more complicated formulafor quartics (polynomials of degree 4).

Digression 1.3.1 Stewart doesn’t actually write out the explicit formula forthe cubic, let alone the much worse one for the quartic. He just describesalgorithms by which they can be solved. But if you unwind the algorithm forthe cubic, you get the formula above. I have done this exercise once and donot recommend it.

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Once mathematicians discovered how to solve quartics, they naturally lookedfor a formula for quintics (polynomials of degree 5). But it was eventually provedby Abel and Ruffini, in the early 19th century, that there is no formula like thequadratic, cubic or quartic formula for polynomials of degree ≥ 5. A bit moreprecisely, there is no formula for the roots in terms of the coefficients that usesonly the usual arithmetic operations (+, −, ×, ÷) and :th roots (for integers :).

Spectacular as this result was, Galois went further, and so will we.Informally, let us say that a complex number is radical if it can be obtained

from the rationals using only the usual arithmetic operations and :th roots. Forexample,

12 +

3√

7√2 − 2√7

4

√6 + 5

√23

is radical, whichever square root, cube root, etc., we choose. A polynomial overQ is solvable (or soluble) by radicals if all of its complex roots are radical.

Example 1.3.2 Every quadratic over Q is solvable by radicals. This follows fromthe quadratic formula: (−1 ±

√12 − 402)/20 is visibly a radical number.

Example 1.3.3 Similarly, the cubic formula shows that every cubic over Q issolvable by radicals. The same goes for quartics.

Example 1.3.4 Some quintics are solvable by radicals. For instance,

(C − 1) (C − 2) (C − 3) (C − 4) (C − 5)

is solvable by radicals, since all its roots are rational and, therefore, radical. A bitless trivially, (C − 123)5 + 456 is solvable by radicals, since its roots are the fivecomplex numbers 123 + 5√−456, which are all radical.

What determines whether a polynomial is solvable by radicals? Galois’samazing achievement was to answer this question completely:

Theorem 1.3.5 (Galois) Let 5 be a polynomial over Q. Then

5 is solvable by radicals ⇐⇒ Gal( 5 ) is a solvable group.

Example 1.3.6 Definition 1.2.1 implies that if 5 has degree = then Gal( 5 ) isisomorphic to a subgroup of (=. You saw in Group Theory that (4 is solvable,and that every subgroup of a solvable group is solvable. Hence the Galois groupof any polynomial of degree ≤ 4 is solvable. It follows from Theorem 1.3.5 thatevery polynomial of degree ≤ 4 is solvable by radicals.

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Example 1.3.7 Put 5 (C) = C5−6C+3. Later we’ll show that Gal( 5 ) = (5. You sawin Group Theory that (5 is not solvable (or at least, you saw that �5 isn’t solvable,which implies that (5 isn’t either, as (5 contains �5 as a subgroup). Hence 5 isnot solvable by radicals.

If there was a quintic formula then all quintics would be solvable by radicals,for the same reason as in Examples 1.3.2 and 1.3.3. But since this is not the case,there is no quintic formula.

Galois’s result is much sharper than Abel and Ruffini’s. They proved that thereis no formula providing a solution by radicals of every quintic, whereas Galoisfound a way of determining which quintics (and higher) can be solved by radicalsand which cannot.

Digression 1.3.8 From the point of view of modern numerical computation,this is all a bit odd. Computationally speaking, there is probably not muchdifference between solving C5 + 3 = 0 to 100 decimal places (that is, finding5√−3) and solving C5 − 6C + 3 = 0 to 100 decimal places (that is, solvinga polynomial that isn’t solvable by radicals). Numerical computation andabstract algebra have different ideas about what is easy and what is hard!

∗ ∗ ∗

This completes our overview of Galois theory. What’s next?Mathematics increasingly emphasizes abstraction over calculation. Individual

mathematicians’ tastes vary, but the historical trend is clear. In the case of Galoistheory, this means dealing with abstract algebraic structures, principally fields,instead of manipulating explicit algebraic expressions such as polynomials. Thecubic formula already gave you a taste of how hairy that can get.

Developing Galois theory using abstract algebraic structures helps us to see itsconnections to other parts of mathematics, and also has some fringe benefits. Forexample, we’ll solve some notorious geometry problems that perplexed the ancientGreeks and remained unsolved for millennia. For that and many other things, we’llneed some ring theory and some field theory—and that’s what’s next.

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Chapter 2

Rings and fields

We now start again. This chapter is a mixture of revision and material that islikely to be new to you. The revision is from Honours Algebra and Introductionto Number Theory (if you took it, which I won’t assume).

Introduction toWeek 2

2.1 RingsWe’ll begin with some stuff you know—but with a twist.

In this course, the word ring means commutative ring with 1 (multiplicativeidentity). Noncommutative rings and rings without 1 are important in some partsof mathematics, but since we’ll be focusing on commutative rings with 1, it willbe easier to just call them ‘rings’.

Given rings ' and (, a homomorphism from ' to ( is a function i : ' → (

satisfying the equations

i(A + A′) = i(A) + i(A′), i(0) = 0, i(−A) = −i(A),i(AA′) = i(A)i(A′), i(1) = 1 (note this!)

for all A, A′ ∈ '. For example, complex conjugation is a homomorphism C → C.It is a very useful lemma that if

i(A + A′) = i(A) + i(A′), i(AA′) = i(A)i(A′), i(1) = 1

for all A, A′ ∈ ' then i is a homomorphism. In other words, to show that i is ahomomorphism, you only need to check it preserves +, · and 1; preservation of 0and negatives then comes for free. But you do need to check it preserves 1. Thatdoesn’t follow from the other conditions.

A subring of a ring ' is a subset ( ⊆ ' that contains 0 and 1 and is closedunder addition, multiplication and negatives. Whenever ( is a subring of ', theinclusion ] : ( → ' (defined by ](B) = B) is a homomorphism.

14

Warning 2.1.1 In Honours Algebra, rings had 1s but homomorphismswere not required to preserve 1. Similarly, subrings of ' had to havea 1, but it was not required to be the same as the 1 of '.For example, take the ringC of complex numbers, the noncommutativering " of 2 × 2 matrices over C, and the function i : C→ " definedby

i(I) =(I 00 0

).

In the terminology of Honours Algebra, i is a homomorphism andits image im i is a subring of " . But in our terminology, i is nota homomorphism (as i(1) ≠ �) and im i is not a subring of " (as� ∉ im i).

Exercise 2.1.2 Let ' be a ring and let S be any set (perhaps infinite)of subrings of '. Prove that their intersection

⋂(∈S ( is also a subring

of '.(In contrast, in the Honours Algebra setup, even the intersection oftwo subrings need not be a subring.)

Example 2.1.3 For any ring ', there is exactly one homomorphism Z→ '. Hereis a sketch of the proof.

To show there is at least one homomorphism j : Z→ ', we will construct one.Define j on nonnegative integers = inductively by j(0) = 0 and j(=+1) = j(=)+1.(Thus, j(=) = 1' + · · · + 1'.) Define j on negative integers = by j(=) = −j(−=).A series of tedious checks shows that j is indeed a ring homomorphism.

To show there is only one homomorphism Z→ ', let i be any homomorphismZ → '; we have to prove that i = j. Certainly i(0) = 0 = j(0). Next proveby induction on = that i(=) = j(=) for nonnegative integers =. I leave the detailsto you, but the crucial point is that because homomorphisms preserve 1, we musthave

i(= + 1) = i(=) + i(1) = i(=) + 1

for all = ≥ 0. Once we have shown that i and j agree on the nonnegative integers,it follows that for negative =,

i(=) = −i(−=) = −j(−=) = j(=).

Hence i(=) = j(=) for all = ∈ Z; that is, i = j.Usually we write j(=) as = · 1', or simply as = if it is clear from the contextThe meaning of

‘= · 1’, andExercise 2.2.7

that = is to be interpreted as an element of '.

15

Every ring homomorphism i : ' → ( has an image im i, which is a subringof (, and a kernel ker i, which is an ideal of '.

Warning 2.1.4 Subrings in ring theory are analogous to subgroupsin group theory, and ideals in ring theory are analogous to normalsubgroups in group theory. But whereas normal subgroups are aspecial kind of subgroup, ideals are not a special kind of subring!Subrings must contain 1, but most ideals don’t.

Exercise 2.1.5 Prove that the only subring of a ring ' that is also anideal is ' itself.

Given a ring ' and an ideal � P ', we obtain the quotient ring or factor ring'/� and the canonical homomorphism c� : ' → '/�, which is surjective and haskernel �.

Quotient rings As explained in Honours Algebra, the quotient ring together with the canonicalhomomorphism has a ‘universal property’: given any ring ( and any homomor-phism i : ' → ( satisfying ker i ⊇ �, there is exactly one homomorphismi : '/� → ( such that this diagram commutes:

'c� //

i

'/�i

��(.

(For a diagram to commute means that whenever there are two different pathsfrom one object to another, the composites along the two paths are equal. Here, itmeans that i = i ◦ c� .) The first isomorphism theorem says that if i is surjectiveand has kernel equal to � then i is an isomorphism. So c� : ' → '/� is essentiallythe only surjective homomorphism out of ' with kernel �.

Digression 2.1.6 Loosely, the ideals of a ring ' correspond one-to-one withthe surjective homomorphisms out of '. This means four things:

• given an ideal � P ', we get a surjective homomorphism out of '(namely, c� : ' → '/�);

• given a surjective homomorphism i out of ', we get an ideal of '(namely, ker i);

• if we start with an ideal � of ', take its associated surjective homomor-phism c� : ' → '/�, then take its associated ideal, we end up wherewe started (that is, ker(c� ) = �);

16

• if we start with a surjective homomorphism i : ' → (, take its asso-ciated ideal ker i, then take its associated surjective homomorphismcker i : ' → '/ker i, we end up where we started (at least ‘up to iso-morphism’, in that we have the isomorphism i : '/ker i→ ( makingthe triangle commute). This is the first isomorphism theorem.

Analogous stories can be told for groups and for modules.

An integral domain is a ring ' such that 0' ≠ 1' and for A, A′ ∈ ',

AA′ = 0⇒ A = 0 or A′ = 0.

Exercise 2.1.7 The trivial ring or zero ring is the one-element setwith its only possible ring structure. Show that the only ring in which0 = 1 is the trivial ring.

Equivalently, an integral domain is a nontrivial ring in which cancellation isvalid: AB = A′B implies A = A′ or B = 0.

Digression 2.1.8 Why is the condition 0 ≠ 1 in the definition of integraldomain?

My answer begins with a useful general point: the sum of no things shouldalways be interpreted as 0. (The amount you pay in a shop is the sum of theprices of the individual things. If you buy no things, you pay £0.) Similarly,the product of no things should be interpreted as 1.

Now consider the following condition on a ring ': for all = ≥ 0 andA1, . . . , A= ∈ ',

A1A2 · · · A= = 0⇒ there exists 8 ∈ {1, . . . , =} such that A8 = 0. (2.1)

In the case = = 0, this says: if 1 = 0 then there exists 8 ∈ ∅ such that A8 = 0.But any statement beginning ‘there exists 8 ∈ ∅’ is false! So in the case = = 0,condition (2.1) states that 1 ≠ 0. And for = = 2, it’s the main condition in thedefinition of integral domain. So ‘1 ≠ 0’ is the 0-fold analogue of the maincondition.

On the other hand, if (2.1) holds for = = 0 and = = 2 then a simple inductionshows that it holds for all = ≥ 0. Conclusion: an integral domain canequivalently be defined as a ring in which (2.1) holds for all = ≥ 0.

Let ) be a subset of a ring '. The ideal 〈)〉 generated by ) is the intersectionof all the ideals of ' containing ) . You can show that any intersection of ideals isan ideal (much as you did for subrings in Exercise 2.1.2), so 〈)〉 is an ideal. It is

17

the smallest ideal of ' containing ) . That is, 〈)〉 is an ideal containing ) , and if �is another ideal containing ) then 〈)〉 ⊆ �. When ) is a finite set {A1, . . . , A=}, wewrite 〈)〉 as 〈A1, . . . , A=〉, and it satisfies

〈A1, . . . , A=〉 = {01A1 + · · · + 0=A= : 01, . . . , 0= ∈ '}. (2.2)

In particular, when = = 1,〈A〉 = {0A : 0 ∈ '}.

Ideals of the form 〈A〉 are called principal ideals. A principal ideal domain isan integral domain in which every ideal is principal.Example 2.1.9 Z is a principal ideal domain. Indeed, if �PZ then either � = {0},in which case � = 〈0〉, or � contains some positive integer, in which case we candefine = to be the least positive integer in � and use the division algorithm to showthat � = 〈=〉.

Exercise 2.1.10 Fill in the details of Example 2.1.9.

Let A and B be elements of a ring '. We say that A divides B, and write A | B, ifthere exists 0 ∈ ' such that B = 0A . This condition is equivalent to B ∈ 〈A〉, and to〈B〉 ⊆ 〈A〉.

An element D ∈ ' is a unit if it has a multiplicative inverse, or equivalentlyif 〈D〉 = '. The units form a group '× under multiplication. For instance,Z× = {1,−1}.

Exercise 2.1.11 Let A and B be elements of an integral domain. Showthat A | B | A ⇐⇒ 〈A〉 = 〈B〉 ⇐⇒ B = DA for some unit D.

Elements A and B of a ring are coprime if for 0 ∈ ',0 | A and 0 | B⇒ 0 is a unit.

Proposition 2.1.12 Let ' be a principal ideal domain and A, B ∈ '. ThenA and B are coprime ⇐⇒ 0A + 1B = 1 for some 0, 1 ∈ '.

Proof ⇒: suppose that A and B are coprime. Since ' is a principal ideal domain,〈A, B〉 = 〈D〉 for some D ∈ '. Since A ∈ 〈A, B〉 = 〈D〉, we must have D | A, andsimilarly D | B. But A and B are coprime, so D is a unit. Hence 1 ∈ 〈D〉 = 〈A, B〉.But by equation (2.2),

〈A, B〉 = {0A + 1B : 0, 1 ∈ '},and the result follows.⇐: suppose that 0A + 1B = 1 for some 0, 1 ∈ '. If D ∈ ' with D | A and D | B

then D | (0A + 1B) = 1, so D is a unit. Hence A and B are coprime. �

18

2.2 FieldsA field is a ring in which 0 ≠ 1 and every nonzero element is a unit. Equivalently,it is a ring such that × = \ {0}. Every field is an integral domain.

Exercise 2.2.1 Write down all the examples of fields that you know.

A field has exactly two ideals: {0} and . For if {0} ≠ � P then D ∈ � forsome D ≠ 0; but then D is a unit, so 〈D〉 = , so � = .

Lemma 2.2.2 Every homomorphism between fields is injective.

Proof Let i : → ! be a homomorphism between fields. Then ker i P , soker i is either {0} or . If ker i = then i(1) = 0; but i(1) = 1 by definitionof homomorphism, so 0 = 1 in !, contradicting the assumption that ! is a field.Hence ker i = {0}, that is, i is injective. �

Warning 2.2.3 With the Honours Algebra definition of homomor-phism, Lemma 2.2.2 would be false, since the map with constant value0 would be a homomorphism.

Let ' be any ring. By Example 2.1.3, there is a unique homomorphismj : Z → '. Its kernel is an ideal of the principal ideal domain Z. Henceker j = 〈=〉 for a unique integer = ≥ 0. This = is called the characteristic of ',and written as char '. Explicitly,

char ' =

{the least = > 0 such that = · 1' = 0', if such an = exists;0, otherwise.

(2.3)

Another way to say it: for < ∈ Z, we have < · 1' = 0 if and only if < is a multipleof char '.

The concept of characteristic is mostly used in the case of fields.

Examples 2.2.4 i. Q, R and C all have characteristic 0.

ii. For a prime number ?, we write F? for the field Z/〈?〉 of integers modulo?. Then char F? = ?.

Lemma 2.2.5 The characteristic of an integral domain is 0 or a prime number.

19

Proof Let ' be an integral domain and write = = char '. Suppose that = > 0; wemust prove that = is prime.

Since 1 ≠ 0 in an integral domain, = ≠ 1. (Remember that 1 is not a prime!So that step was necessary.) Now let :, < > 0 with :< = =. Writing j for theunique homomorphism Z→ ', we have

j(:)j(<) = j(:<) = j(=) = 0,

and ' is an integral domain, so j(:) = 0 or j(<) = 0. WLOG, j(:) = 0. Butker j = 〈=〉, so = | : , so : = =. Hence = is prime. �

Examples 2.2.4 show that there exist fields of every possible characteristic.But there is no way of mapping between fields of different characteristics:

Lemma 2.2.6 Let i : → ! be a homomorphism of fields. Then char = char !.

Proof Write j and j! for the unique homomorphisms from Z to and !,respectively. Since j! is the unique homomorphism Z→ !, the triangle

Zj

��

j!

�� i

// !

commutes. (Concretely, this says that i(= · 1 ) = = · 1! for all = ∈ Z.) Henceker(i◦ j ) = ker j! . But i is injective by Lemma 2.2.2, so ker(i◦ j ) = ker j .Hence ker j = ker j! , or equivalently, char = char !. �

For example, the inclusionQ→ R is a homomorphism of fields, and both havecharacteristic 0.

Exercise 2.2.7 This proof of Lemma 2.2.6 is quite abstract. Finda more concrete proof, taking equation (2.3) as your definition ofcharacteristic. (You will still need the fact that i is injective.)

The meaning of‘= · 1’, and

Exercise 2.2.7A subfield of a field is a subring that is a field. The prime subfield of

is the intersection of all the subfields of . It is straightforward to show that anyintersection of subfields is a subfield (just as you showed in Exercise 2.1.2 that anyintersection of subrings is a subring). Hence the prime subfield is a subfield. It isthe smallest subfield of , in the sense that any other subfield of contains it.

Concretely, the prime subfield of is{< · 1 = · 1

: <, = ∈ Z with = · 1 ≠ 0}.

20

To see this, first note that this set is a subfield of . It is the smallest subfield of :for if ! is a subfield of then 1 ∈ ! by definition of subfield, so < · 1 ∈ ! forall integers<, so (< ·1 )/(= ·1 ) ∈ ! for all integers< and = such that = ·1 ≠ 0.

Examples 2.2.8 i. The field Q has no proper subfields, so the prime subfieldof Q is Q itself.

ii. Let ? be a prime. The field F? has no proper subfields, so the prime subfieldof F? is F? itself.

Exercise 2.2.9 What is the prime subfield of R? Of C?

The prime subfields appearing in Examples 2.2.8 wereQ and F?. In fact, theseare the only prime subfields of anything:

Lemma 2.2.10 Let be a field.

i. If char = 0 then the prime subfield of is Q.

ii. If char = ? > 0 then the prime subfield of is F?.

In the statement of this lemma, as so often in mathematics, the word ‘is’ means‘is isomorphic to’. I hope you’re comfortable with that by now.

Proof For (i), suppose that char = 0. By definition of characteristic, = · 1 ≠ 0for all integers = ≥ 0. One can check that there is a well-defined homomorphismi : Q → defined by </= ↦→ (< · 1 )/(= · 1 ). (The check uses the fact thatj : =→ = · 1 is a homomorphism.) Now i is injective, being a homomorphismof fields, so im i � Q. But im i is a subring of , and in fact a subfield since it isisomorphic to Q. It is the prime subfield, since Q has no proper subfields.

For (ii), suppose that char = ? > 0. By Lemma 2.2.5, ? is prime. Theunique homomorphism j : Z → has kernel 〈?〉, by definition. By the firstisomorphism theorem, im j � Z/〈?〉 = F?. But im j is a subring of , and infact a subfield since is it isomorphic to F?. It is the prime subfield, since F? hasno proper subfields. �

Lemma 2.2.11 Every finite field has positive characteristic.

Proof By Lemma 2.2.10, a field of characteristic 0 contains a copy of Q and istherefore infinite. �

21

Warning 2.2.12 There are also infinite fields of positive characteris-tic! We haven’t met one yet, but we will soon.

So far, we have rather few examples of fields. The following construction willallow us to manufacture many, many more.

An element A of a ring ' is irreducible if A is not 0 or a unit, and if for 0, 1 ∈ ',Building blocks

A = 01 ⇒ 0 or 1 is a unit.

For example, the irreducibles in Z are ±2,±3,±5, . . .. An element of a ring isreducible if it is not 0, a unit, or irreducible. So 0 and units count as neitherreducible nor irreducible.

Exercise 2.2.13 What are the irreducible elements of a field?

Proposition 2.2.14 Let ' be a principal ideal domain and 0 ≠ A ∈ '. Then

A is irreducible ⇐⇒ '/〈A〉 is a field.

Proof Write c for the canonical homomorphism ' → '/〈A〉.⇒: suppose that A is irreducible. To show that 1'/〈A〉 ≠ 0'/〈A〉, note that since

A is not a unit, 1' ∉ 〈A〉 = ker c, so

1'/〈A〉 = c(1') ≠ 0'/〈A〉 .

Next we have to show that every nonzero element of '/〈A〉 is a unit, orequivalently that c(B) is a unit whenever B ∈ ' with B ∉ 〈A〉. We have A - B, andA is irreducible, so A and B are coprime. Hence by Proposition 2.1.12 (and theassumption that ' is a principal ideal domain), we can choose 0, 1 ∈ ' such that

0A + 1B = 1' .

Applying c to each side gives

c(0)c(A) + c(1)c(B) = 1'/〈A〉 .

But c(A) = 0, so c(1)c(B) = 1, so c(B) is a unit.⇐: suppose that '/〈A〉 is a field. Then 1'/〈A〉 ≠ 0'/〈A〉, that is, c(1') ∉ ker c =

〈A〉, that is, A - 1'. Hence A is not a unit.Next we have to show that if 0, 1 ∈ ' with A = 01 then 0 or 1 is a unit. We

have0 = c(A) = c(0)c(1)

22

and '/〈A〉 is an integral domain, so WLOG c(0) = 0. Then 0 ∈ ker c = 〈A〉, so0 = A1′ for some 1′ ∈ '. This gives

A = 01 = A1′1.

But A ≠ 0 by hypothesis, and ' is an integral domain, so 1′1 = 1. Hence 1 is aunit. �

Example 2.2.15 When = is an integer, Z/〈=〉 is a field if and only if = is irreducible(that is, ± a prime number).

Proposition 2.2.14 enables us to construct fields from irreducible ele-ments. . . but irreducible elements of a principal ideal domain. Right now that’snot much help, because we don’t have many examples of principal ideal domains.But we will soon.

23

Chapter 3

Polynomials

This chapter revisits and develops some themes you met in Honours Algebra.Although it’s long, it contains material you’ve seen before. Before you begin, itmay help you to reread Section 3.3 (Polynomials) of the Honours Algebra notes.

Introduction toWeek 3

3.1 The ring of polynomialsYou already know the definition of polynomial, but I want to make a point byphrasing it in an unfamiliar way.

Definition 3.1.1 Let ' be a ring. A polynomial over ' is an infinite sequence(00, 01, 02, . . .) of elements of ' such that {8 : 08 ≠ 0} is finite.

The set of polynomials over ' forms a ring as follows:

(00, 01, . . .) + (10, 11, . . .) = (00 + 10, 01 + 11, . . .), (3.1)(00, 01, . . .) · (10, 11, . . .) = (20, 21, . . .) (3.2)

where 2: =∑

8, 9 : 8+ 9=:081 9 , (3.3)

the zero of the ring is (0, 0, . . .), and the multiplicative identity is (1, 0, 0, . . .).Of course, we almost always write (00, 01, 02, . . .) as 00 + 01C + 02C

2 + · · · , orthe same with some other symbol in place of C. In that notation, formulas (3.1)and (3.2) look like the usual formulas for addition and multiplication of polyno-mials. Nevertheless:

Warning 3.1.2 A polynomial is not a function!A polynomial gives rise to a function, as we’ll recall in a moment. Buta polynomial itself is a purely formal object.

24

The set of polynomials over ' is written as '[C] (or '[D], '[G], etc.). Since( = '[C] is itself a ring, we can consider the ring ([D] = ('[C]) [D], usually writtenas '[C, D], and similarly '[C, D, E] = ('[C, D]) [E], etc.

Why studypolynomials?

We use 5 , 6, ℎ, . . . and 5 (C), 6(C), ℎ(C), . . . interchangeably to denote elementsof '[C]. A polynomial 5 = (00, 01, . . .) over ' gives rise to a function

' → '

A ↦→ 00 + 01A + 02A2 + · · · .

(The sum on the right-hand side makes sense because only finitely many 08s arenonzero.) This function is usually called 5 too. But calling it that is slightlydangerous, because:

Warning 3.1.3 Different polynomials can give rise to the same func-tion. For example, consider C, C2 ∈ F2 [C]. They are different polyno-mials: going back to Definition 3.1.1, they’re alternative notation forthe sequences

(0, 1, 0, 0, . . .) and (0, 0, 1, 0, . . .),

which are plainly not the same. On the other hand, they induce thesame function F2 → F2, because 0 = 02 for all (both) 0 ∈ F2.

Exercise 3.1.4 Show that whenever ' is a finite nontrivial ring, it ispossible to find distinct polynomials over ' that induce the same func-tion ' → '. (Hint: are there finitely or infinitely many polynomialsover '? Functions ' → '?)

The ring of polynomials has a universal property: a homomorphism from '[C]to some other ring � is determined by its effect on constant polynomials and on Citself, in the following sense.

The universalproperty of '[C] Lemma 3.1.5 (Universal property of the polynomial ring) Let ' and � be

rings. For every homomorphism i : ' → � and every 1 ∈ �, there is exactly onehomomorphism \ : '[C] → � such that

\ (0) = i(0) for all 0 ∈ ', (3.4)\ (C) = 1. (3.5)

On the left-hand side of (3.4), the ‘0’ means the polynomial 0 + 0C + 0C2 + · · · .

25

Proof To show there is at most one such \, take any homomorphism \ : '[C] → �

satisfying (3.4) and (3.5). Then for every polynomial∑8 08C

8 over ',

\

(∑8

08C8)=

∑8

\ (08)\ (C)8 since \ is a homomorphism

=∑8

i(08)18 by (3.4) and (3.5).

So \ is uniquely determined.To show there is at least one such \, define a function \ : '[C] → � by

\

(∑8

08C8)=

∑8

i(08)18

(∑8 08C

8 ∈ '[C]). Then \ clearly satisfies conditions (3.4) and (3.5). It remains tocheck that \ is a homomorphism. I will do the worst part of this, which is to checkthat \ preserves multiplication, and leave the rest to you.

So, take polynomials 5 (C) = ∑8 08C

8 and 6(C) = ∑9 1 9 C

9 . Then 5 (C)6(C) =∑: 2: C

: , where 2: is as defined in equation (3.3). We have

\ ( 5 6) = \(∑:

2: C:)

=∑:

i(2: )1: by definition of \

=∑:

i

( ∑8, 9 : 8+ 9=:

081 9

)1: by definition of 2:

=∑:

∑8, 9 : 8+ 9=:

i(08)i(1 9 )1: since i is a homomorphism

=∑8, 9

i(08)i(1 9 )18+ 9

=

(∑8

i(08)18) (∑

9

i(1 9 )1 9)

= \ ( 5 )\ (6) by definition of \. �

Here are three uses for the universal property of the ring of polynomials. First:

Definition 3.1.6 Let i : ' → ( be a ring homomorphism. The induced homo-morphism

i∗ : '[C] → ([C]is the unique homomorphism '[C] → ([C] such that i∗(0) = i(0) for all 0 ∈ 'and i∗(C) = C.

26

The universal property guarantees that there is one and only one homomor-phism i∗ with these properties. Concretely,

i∗(∑8

08C8)=

∑8

i(08)C8

for all∑8 08C

8 ∈ '[C].Second, let ' be a ring and A ∈ '. By the universal property, there is a unique

homomorphism evA : '[C] → ' such that evA (0) = 0 for all 0 ∈ ' and evA (C) = A.Concretely,

evA(∑8

08C8)=

∑8

08A8

for all∑8 08C

8 ∈ '[C]. This map evA is called evaluation at A.(The notation

∑08C

8 for what is officially (00, 01, . . .)makes it look obvious thatwe can evaluate a polynomial at an element and that this gives a homomorphism:of course ( 5 · 6) (A) = 5 (A)6(A), for instance! But that’s only because of thenotation: there was actually something to prove here.)

Third, let ' be a ring and 2 ∈ '. For any 5 (C) ∈ '[C], we can ‘substituteC = D + 2’ to get a polynomial in D. What exactly does this mean? Formally, thereis a unique homomorphism \ : '[C] → '[D] such that \ (0) = 0 for all 0 ∈ ' and\ (C) = D + 2. Concretely,

\

(∑8

08C8)=

∑8

08 (D + 2)8 .

This particular substitution is invertible. Informally, the inverse is ‘substituteD = C − 2’. Formally, there is a unique homomorphism \′ : '[D] → '[C] suchthat \′(0) = 0 for all 0 ∈ ' and \′(D) = C − 2. These maps \ and \′ carrying outthe substitutions are inverse to each other, as you can deduce this from either theuniversal property or the concrete descriptions. So, the substitution maps

'[C]\ //

'[D]\ ′oo (3.6)

define an isomorphism between '[C] and '[D]. For example, since isomorphismpreserve irreducibility (and everything else that matters!), 5 (C) is irreducible ifand only if 5 (C − 2) is irreducible.

Exercise 3.1.7 What happens to everything in the previous paragraphif we substitute C = D2 + 2 instead?

The rest of this section is about degree.

27

Definition 3.1.8 The degree deg( 5 ) of a nonzero polynomial 5 (C) = ∑08C

8 is thelargest = ≥ 0 such that 0= ≠ 0. By convention, deg(0) = −∞, where −∞ is aformal symbol which we give the properties

−∞ < =, (−∞) + = = −∞, (−∞) + (−∞) = −∞

for all integers =.

Digression 3.1.9 Defining deg(0) like this is helpful because it allows usto make statements about all polynomials without having to make annoyingexceptions for the zero polynomial (e.g. Lemma 3.1.10(i)).

But putting deg(0) = −∞ alsomakes intuitive sense. At least for polynomialsover R, the degree of a nonzero polynomial tells us how fast it grows: when Cis large, 5 (C) behaves roughly like Cdeg( 5 ) . What about the zero polynomial?Well, whether or not C is large, 0(C) = 0, and C−∞ can sensibly be interpretedas limA→−∞ CA = 0. So it makes sense to put deg(0) = −∞.

Lemma 3.1.10 Let ' be an integral domain. Then:

i. deg( 5 6) = deg( 5 ) + deg(6) for all 5 , 6 ∈ '[C];

ii. '[C] is an integral domain;

iii. when ' is a field, the units in '[C] are the polynomials of degree 0 (that is,the nonzero constants);

iv. when ' is a field, 5 (C) ∈ '[C] is irreducible if and only if deg( 5 ) > 0 and5 cannot be expressed as a product of two polynomials of degree > 0.

Proof At least parts (i)–(iii) were in Honours Algebra (Section 3.3). Part (iv)follows from the general definition of irreducible element of a ring. �

3.2 Factorizing polynomialsEvery nonzero integer can be expressed as a product of primes in an essentiallyunique way. But the analogous statement is not true in all rings, or even allintegral domains. Some rings have elements that can’t be expressed as a productof irreducibles at all. In other rings, factorizations into irreducibles exist but arenot unique. (By ‘not unique’ I mean more than just changing the order of thefactors or multiplying them by units.)

The big theorem of this section is that, happily, every polynomial over a fieldcan be factorized, essentially uniquely, into irreducibles.

We begin with a result on division of polynomials from Section 3.3 of HonoursAlgebra.

28

Proposition 3.2.1 Let be a field and 5 , 6 ∈ [C]. Then there is exactly one pairof polynomials @, A ∈ [C] such that 5 = @6 + A and deg(A) < deg(6). �

We use this to prove an extremely useful fact:

Proposition 3.2.2 Let be a field. Then [C] is a principal ideal domain.

Proof First, [C] is an integral domain, by Lemma 3.1.10(ii).Now let � P [C]. If � = {0} then � = 〈0〉. Otherwise, put 3 = min{deg( 5 ) :

0 ≠ 5 ∈ �} and choose 6 ∈ � such that deg(6) = 3.I claim that � = 〈6〉. To prove this, let 5 ∈ �; we must show that 6 | 5 .

By Proposition 3.2.1, 5 = @6 + A for some @, A ∈ [C] with deg(A) < 3. NowA = 5 − @6 ∈ � since 5 , 6 ∈ �, so the minimality of 3 implies that A = 0. Hence5 = @6, as required. �

If you struggled with Exercise 2.1.10, that proof should give you a clue.

Warning 3.2.3 Lemma 3.1.10(ii) implies that [C1, . . . , C=] is anintegral domain for all = ≥ 2, but it is not a principal ideal domain.For example, the ideal

〈C1, C2〉 = { 5 (C1, C2) ∈ Q[C1, C2] : 5 has constant term 0}

is not principal.Also, Proposition 3.2.2 really needed the hypothesis that is a field;it’s not enough for it to be a principal ideal domain. For example, Z isa principal ideal domain, but in Z[C], the ideal

〈2, C〉 = { 5 (C) ∈ Z[C] : the constant term of 5 is even}

is not principal.

Exercise 3.2.4 Prove that the ideals in Warning 3.2.3 are indeed notprincipal.

Exercise 3.2.4: anon-principal ideal

At the end of Chapter 2, I promised I’d give you a way of manufacturing lotsof new fields. Here it is!

Corollary 3.2.5 Let be a field and let 0 ≠ 5 ∈ [C]. Then

5 is irreducible ⇐⇒ [C]/〈 5 〉 is a field.

Proof This follows from Propositions 2.2.14 and 3.2.2. �

29

To manufacture new fields using Corollary 3.2.5, we’ll need a way of knowingwhich polynomials are irreducible. That’s the topic of Section 3.3, but for now let’sstick to our mission: proving that every polynomial factorizes into irreducibles inan essentially unique way.

To achieve this mission, we’ll need two more lemmas.

Lemma 3.2.6 Let be a field and let 5 (C) ∈ [C] be a nonconstant polynomial.Then 5 (C) is divisible by some irreducible in [C].

The word nonconstant just means ‘of degree > 0’.

Proof Let 6 be a nonconstant polynomial of smallest possible degree such that6 | 5 . (For this to make sense, there must be at least one nonconstant polynomialdividing 5 , and there is: 5 .) I claim that 6 is irreducible. Proof: if 6 = 6162 theneach 68 divides 5 , so by the minimality of deg(6), each 68 has degree 0 or deg(6).They cannot both have degree deg(6), since deg(61) + deg(62) = deg(6) > 0. Soat least one has degree 0, i.e., is a unit. �

Lemma 3.2.7 Let be a field and 5 , 6, ℎ ∈ [C]. Suppose that 5 is irreducibleand 5 | 6ℎ. Then 5 | 6 or 5 | ℎ.

This behaviour is familiar in the integers: if a prime ? divides some product01, then ? | 0 or ? | 1.

Proof Suppose that 5 - 6. Since 5 is irreducible, 5 and 6 are coprime. Since [C]is a principal ideal domain, Proposition 2.1.12 implies that there are ?, @ ∈ [C]such that

? 5 + @6 = 1.

Multiplying both sides by ℎ gives

? 5 ℎ + @6ℎ = ℎ.

But 5 | ? 5 ℎ and 5 | 6ℎ, so 5 | ℎ. �

Theorem 3.2.8 Let be a field and 0 ≠ 5 ∈ [C]. Then

5 = 0 51 52 · · · 5=

for some = ≥ 0, 0 ∈ and monic irreducibles 51, . . . , 5= ∈ [C]. Moreover, =and 0 are uniquely determined by 5 , and 51, . . . , 5= are uniquely determinedup to reordering.

30

In the case = = 0, the product 51 · · · 5= should be interpreted as 1 (as inDigression 2.1.8). Monic means that the leading coefficient is 1.

Proof First we prove that such a factorization exists, by induction on deg( 5 ). Ifdeg( 5 ) = 0 then 5 is a constant 0 and we take = = 0. Now suppose that deg( 5 ) > 0and assume the result for polynomials of smaller degree. By Lemma 3.2.6, thereis an irreducible 6 dividing 5 , and we can assume that 6 is monic by dividing bya constant if necessary. Then 5 /6 is a nonzero polynomial of smaller degree than5 , so by inductive hypothesis,

5 /6 = 0ℎ1 · · · ℎ<for some 0 ∈ and monic irreducibles ℎ1, . . . , ℎ<. Rearranging gives

5 = 0ℎ1 · · · ℎ<6,

completing the induction.Now we prove uniqueness, again by induction on deg( 5 ). If deg( 5 ) = 0 then

5 is a constant 0 and the only possible factorization is the one with = = 0. Nowsuppose that deg( 5 ) > 0, and take two factorizations

0 51 · · · 5= = 5 = 161 · · · 6< (3.7)

where 0, 1 ∈ and 58, 6 9 are monic irreducible. Since deg( 5 ) > 0, we have=, < ≥ 1. Now 5= | 161 · · · 6<, so by Lemma 3.2.7, 5= | 6 9 for some 9 . Byrearranging, we can assume that 9 = <. But 6< is also irreducible, so 5= = 26<for some nonzero 2 ∈ , and both 5= and 6< are monic, so 2 = 1. Hence 5= = 6<.Cancelling in (3.7) (which we can do as [C] is an integral domain) gives

0 51 · · · 5=−1 = 161 · · · 6<−1.

By inductive hypothesis, = = <, 0 = 1, and the lists of irreducibles 51, . . . , 5=−1and 61, . . . , 6<−1 are the same up to reordering. This completes the induction. �

One way to find an irreducible factor of a polynomial 5 (C) ∈ [C] is to find aroot (an element 0 ∈ such that 5 (0) = 0):

Lemma 3.2.9 Let be a field, 5 (C) ∈ [C] and 0 ∈ . Then

5 (0) = 0 ⇐⇒ (C − 0) | 5 (C).

Proof ⇒: suppose that 5 (0) = 0. By Proposition 3.2.1,

5 (C) = (C − 0)@(C) + A (C) (3.8)

for some @, A ∈ [C] with deg(A) < 1. Then A is a constant, so putting C = 0

in (3.8) gives A = 0.⇐: if 5 (C) = (C − 0)@(C) for some polynomial @ then 5 (0) = 0. �

31

Definition 3.2.10 Let be a field, let 0 ≠ 5 ∈ [C], and let 0 ∈ be a root of5 . Themultiplicity of 0 is the unique integer < ≥ 1 such that (C − 0)< | 5 (C) but(C − 0)<+1 - 5 (C).

Exercise 3.2.11 This definition assumes that there is a unique < withthese properties (that is, there is one and only one such <). Prove it.

Example 3.2.12 Over any field , the polynomial C has a root at 0withmultiplicity1, and the polynomial C2 has a root at 0 with multiplicity 2. This is true even when = F2, in which case C and C2 induce the same function → (Warning 3.1.2).

Proposition 3.2.13 Let be a field and let 0 ≠ 5 ∈ [C]. Write 01, . . . , 0: forthe distinct roots of 5 in , and <1, . . . , <: for their multiplicities. Then

5 (C) = (C − 01)<1 · · · (C − 0: )<:6(C)for some 6(C) ∈ [C] that has no roots.

Proof By induction on : . If : = 0, this is immediate (again interpreting an emptyproduct as 1). Now suppose that : ≥ 1. By definition, (C − 0: )<: | 5 (C), so wecan put

5 (C) = 5 (C)(C − 0: )<:

∈ [C] .

Any root of 5 is a root of 5 , so it is one of 01, . . . , 0: . But 5 (0: ) ≠ 0: forif 5 (0: ) = 0 then (C − 0: ) | 5 (C) by Lemma 3.2.9, so (C − 0: )<:+1 | 5 (C), acontradiction. Hence any root of 5 is one of 01, . . . , 0:−1. These are indeed rootsof 5 , with multiplicities <1, . . . , <:−1. So by inductive hypothesis,

5 (C) = (C − 01)<1 · · · (C − 0:−1)<:−16(C)for some 6(C) ∈ [C] with no roots, completing the induction. �

Corollary 3.2.14 Let be a field and 5 ∈ [C]. Suppose that 5 has distinct roots01, . . . , 0: ∈ with multiplicities <1, . . . , <: . Then <1 + · · · + <: ≤ deg( 5 ). �

In other words, a polynomial of degree = has no more than = roots, even whenyou count the roots with multiplicities (e.g. count a double root twice).

A field is algebraically closed if every nonconstant polynomial has at least oneroot. For example, C is algebraically closed (the fundamental theorem of algebra).Proposition 3.2.13 implies:Corollary 3.2.15 Let be an algebraically closed field and let 0 ≠ 5 ∈ [C].Write 01, . . . , 0: for the distinct roots of 5 in , and <1, . . . , <: for their multi-plicities. Then

5 (C) = 2(C − 01)<1 · · · (C − 0: )<: ,where 2 is the leading coefficient of 5 . �

32

3.3 Irreducible polynomialsDetermining whether an integer is prime is generally hard, and determiningwhether a polynomial is irreducible is hard too. This section presents a fewtechniques for doing so.

Let’s begin with the simplest cases. Recall Lemma 3.1.10(iv): a polynomial isirreducible if and only if it is nonconstant (has degree > 0) and cannot be expressedas a product of two nonconstant polynomials.

Lemma 3.3.1 Let be a field and 5 ∈ [C].

i. If 5 is constant then 5 is not irreducible.

ii. If deg( 5 ) = 1 then 5 is irreducible.

iii. If deg( 5 ) ≥ 2 and 5 has a root then 5 is reducible.

iv. If deg( 5 ) ∈ {2, 3} and 5 has no root then 5 is irreducible.

Proof Parts (i) and (ii) follow from what we just recalled, and (iii) followsfrom Lemma 3.2.9. For (iv), suppose for a contradiction that 5 = 6ℎ withdeg(6), deg(ℎ) ≥ 1. We have deg(6) + deg(ℎ) ∈ {2, 3}, so without loss of gener-ality, deg(6) = 1. Also without loss of generality, 6 is monic, say 6(C) = C + 0; butthen 5 (−0) = 0, a contradiction. �

Warning 3.3.2 The converse of (iii) is false! To show a polynomial isirreducible, it’s not enough to show it has no root (unless it has degree2 or 3). For instance, (C2 + 1)2 ∈ Q[C] has no root but is reducible.

Examples 3.3.3 i. Let ? be a prime. Then 5 (C) = 1 + C + · · · + C ?−1 ∈ F? [C] isreducible, since 5 (1) = 0.

ii. Let 5 (C) = C3 − 10 ∈ Q[C]. Then deg( 5 ) = 3 and 5 has no root in Q, so 5 isirreducible by part (iv) of the lemma.

iii. Over C or any other algebraically closed field, the irreducibles are exactlythe polynomials of degree 1.

Exercise 3.3.4 If I gave you a quadratic overQ, howwould you decidewhether it is reducible or irreducible?

From now on we focus on = Q. Any polynomial over Q can be multipliedby a nonzero rational constant to get a polynomial over Z, and that’s often a helpfulmove, so we’ll look at Z[C] too.

33

Definition 3.3.5 A polynomial over Z is primitive if its coefficients have nocommon divisor except for ±1.

For example, 15 + 6C + 10C2 is primitive but 15 + 6C + 30C2 is not.Lemma 3.3.6 Let 5 (C) ∈ Q[C]. Then there exist a primitive polynomial � (C) ∈Z[C] and U ∈ Q such that 5 = U�.

Proof Write 5 (C) = ∑8 (08/18)C8, where 08 ∈ Z and 0 ≠ 18 ∈ Z. Take any common

multiple 1 of the 18s; then writing 28 = 081/18 ∈ Z, we have 5 (C) = (1/1)∑28C

8.Now let 2 be the greatest common divisor of the 28s, put 38 = 28/2 ∈ Z, and put� (C) = ∑

38C8. Then � (C) is primitive and 5 (C) = (2/1)� (C). �

If the coefficients of a polynomial 5 (C) ∈ Q[C] happen to all be integers, theword ‘irreducible’ could mean two things: irreducibility in the ring Q[C] or in thering Z[C]. We say that 5 is irreducible over Q or Z to distinguish between the two.

Suppose we have a polynomial over Z that’s irreducible over Z. In principle itcould still be reducible over Q: although there’s no nontrivial way of factorizingit over Z, perhaps it can be factorized when you give yourself the freedom ofnon-integer coefficients. But the next result tells us that you can’t.Lemma 3.3.7 (Gauss) i. The product of two primitive polynomials over Z is

primitive.

ii. If a polynomial over Z is irreducible over Z, it is irreducible over Q.

Proof For (i), let 5 and 6 be primitive polynomials over Z. Let ? be a primenumber. (We’re going to show that ? doesn’t divide all the coefficients of 5 6.)Write c : Z → Z/?Z = F? for the canonical homomorphism, which induces ahomomorphism c∗ : Z[C] → F? [C] as in Definition 3.1.6.

Since 5 is primitive, ? does not divide all the coefficients of 5 . Equivalently,c∗( 5 ) ≠ 0. Similarly, c∗(6) ≠ 0. But F? [C] is an integral domain, so

c∗( 5 6) = c∗( 5 )c∗(6) ≠ 0,

so ? does not divide all the coefficients of 5 6. This holds for all primes ?, so 5 6is primitive.

For (ii), let 5 ∈ Z[C] be a polynomial irreducible over Z. Let 6, ℎ ∈ Q[C] with5 = 6ℎ. By Lemma 3.3.6, 6 = U� and ℎ = V� for some U, V ∈ Q and primitive�, � ∈ Z[C]. Then UV = </= for some coprime integers < and =, giving

= 5 = <��.

(All three of these polynomials are over Z.) Now = divides every coefficient of= 5 , hence every coefficient of <��. Since < and = are coprime, = divides everycoefficient of ��. But �� is primitive by (i), so = = ±1, so 5 = ±<��. Since 5is irreducible over Z, either � or � is constant, so 6 or ℎ is constant, as required.�

34

Gauss’s lemma quickly leads to a test for irreducibility. It involves taking apolynomial over Z and reducing it mod ?, for some prime ?. This means applyingthe map c∗ : Z[C] → F? [C] from the last proof. As we saw after Definition 3.1.6,if 5 (C) = ∑

08C8 then c∗( 5 ) (C) =

∑c(08)C8, where c(08) is the congruence class of

08 mod ?. I’ll write 0 = c(0) and 5 = c∗( 5 ). That is, 5 is ‘ 5 mod ?’.

Proposition 3.3.8 (Mod p method) Let 5 (C) = 00 + 01C + · · · + 0=C= ∈ Z[C]. Ifthere is some prime ? such that ? - 0= and 5 ∈ F? [C] is irreducible, then 5 isirreducible over Q.

I’ll give some examples first, then the proof.

Examples 3.3.9 i. Let’s use themod ?method to show that 5 (C) = 9+14C−8C3is irreducible over Q. Take ? = 7: then 5 (C) = 2− C3 ∈ F7 [C], so it’s enoughto show that 2− C3 is irreducible over F7. Since this has degree 3, it’s enoughto show that C3 = 2 has no solution in F7 (by Lemma 3.3.1(iv)). And youcan easily check this by computing 03, (±1)3, (±2)3 and (±3)3 mod 7.

ii. The condition in Proposition 3.3.8 that ? - 0= can’t be dropped. For instance,consider 5 (C) = 6C2 + C and ? = 2.

Warning 3.3.10 Take 5 (C) as in Example 3.3.9(i), but this time take? = 3. Then 5 (C) = −C + C3 ∈ F3 [C], which is reducible. But thatdoesn’t mean 5 is reducible! The mod ? method only ever lets youshow that a polynomial is irreducible over Q, not reducible.

Proof of Proposition 3.3.8 Take a prime ? satisfying the conditions in the Propo-sition. By Gauss’s lemma, it is enough to prove that 5 is irreducible over Z.

Since 5 is irreducible, deg( 5 ) > 0, so deg( 5 ) > 0.Now let 5 = 6ℎ in Z[C]. We have 5 = 6ℎ and 5 is irreducible, so without

loss of generality, 6 is constant. The leading coefficient of 5 is the product of theleading coefficients of 6 and ℎ, and is not divisible by ?, so the leading coefficientof 6 is not divisible by ?. Hence deg(6) = deg(6) = 0. �

We finish with an irreducibility test that turns out to be surprisingly powerful.

Proposition 3.3.11 (Eisenstein’s criterion) Let 5 (C) = 00 + · · · + 0=C= ∈ Z[C],with = ≥ 1. Suppose there exists a prime ? such that:

• ? - 0=;

• ? | 08 for all 8 ∈ {0, . . . , = − 1};

• ?2 - 00.

35

Then 5 is irreducible over Q.

To prove this, wewill use the concept of the codegree codeg( 5 ) of a polynomial5 (C) = ∑

8 08C8, which is defined to be the least 8 such that 08 ≠ 0 (if 5 ≠ 0), or as

the formal symbol∞ if 5 = 0. For polynomials 5 and 6 over an integral domain,

codeg( 5 6) = codeg( 5 ) + codeg(6).

Clearly codeg( 5 ) ≤ deg( 5 ) unless 5 = 0.

Proof By Gauss’s lemma, it is enough to show 5 is irreducible over Z. Let6, ℎ ∈ Z[C] with 5 = 6ℎ. Continue to write 5 (C) ∈ F? [C] for 5 reduced mod ?;then 5 = 6ℎ. Since

?2 - 00 = 5 (0) = 6(0)ℎ(0),we may assume without loss of generality that ? - 6(0). Hence codeg(6) = 0.Also, codeg( 5 ) = =, since ? divides each of 00, . . . , 0=−1 but not 0=. So

= = codeg( 5 ) = codeg(6) + codeg(ℎ) = codeg(ℎ) ≤ deg(ℎ) ≤ deg(ℎ), (3.9)

giving = ≤ deg(ℎ). But 5 = 6ℎ with deg( 5 ) = =, so deg(ℎ) = = and deg(6) = 0.�

Exercise 3.3.12 The last step in (3.9) was ‘deg(ℎ) ≤ deg(ℎ)’. Whyis that true? And when does equality hold?

Example 3.3.13 Let

6(C) = 29C5 − 5

3C4 + C3 + 1

3∈ Q[C] .

Then 6 is irreducible over Q if and only if

96(C) = 2C5 − 15C4 + 9C3 + 3

is irreducible over Q, which it is by Eisenstein’s criterion with ? = 3.Testing for

irreducibilityExercise 3.3.14 Use Eisenstein’s criterion to show that for every= ≥ 1, there is an irreducible polynomial over Q of degree =.

I’ll give you one more example, and it’s not just any old polynomial: it’s animportant one that we’ll need when we come to think about solvability by radicals.It needs a lemma.

Lemma 3.3.15 Let ? be a prime and 0 < 8 < ?. Then ? |(?8

).

36

For example, the 7th row of Pascal’s triangle is 1, 7, 21, 35, 35, 21, 7, 1, and 7divides all of these numbers apart from the first and last.

Proof We have 8!(? − 8)!(?8

)= ?!, and ? divides ?! but not 8! or (? − 8)! (since ?

is prime and 0 < 8 < ?), so ? must divide(?8

). �

Example 3.3.16 Let ? be a prime. The ?th cyclotomic polynomial is

Φ? (C) = 1 + C + · · · + C ?−1 =C ? − 1C − 1

. (3.10)

I claim thatΦ? is irreducible. We can’t apply Eisenstein toΦ? as it stands, becausewhichever prime we choose (whether it’s ? or another one) doesn’t divide any ofthe coefficients. However, we saw on p. 27 that Φ? (C) is irreducible if and only ifΦ? (C − 2) is irreducible, for any 2 ∈ Q. We’ll take 2 = −1. We have

Φ? (C + 1) = (C + 1)? − 1(C + 1) − 1

=1C

?∑8=1

(C

8

)C8

= ? +(?

2

)C + · · · +

(?

? − 1

)C ?−2 + C ?−1.

So Φ? (C + 1) is irreducible by Eisenstein’s criterion and Lemma 3.3.15, henceΦ? (C) is irreducible too.

Digression 3.3.17 I defined the ?th cyclotomic polynomial Φ? only when? is prime. The definition of Φ= for general = ≥ 1 is not the obviousgeneralization of (3.10). It’s this:

Φ= (C) =∏Z

(C − Z),

where the product runs over all primitive =th roots of unity Z . (In thiscontext, ‘primitive’ means that = is the smallest number satisfying Z= = 1;it’s a different usage from ‘primitive polynomial’.)

Many surprising things are true. It’s not obvious that the coefficients of Φ=are real, but they are. Even given that they’re real, it’s not obvious that they’rerational, but they are. Even given that they’re rational, it’s not obvious thatthey’re integers, but they are. The degree of Φ= is i(=), the number ofintegers between 1 and = that are coprime with = (Euler’s function). It’s alsotrue that the polynomial Φ= is irreducible for all =, not just primes.

Some of these things are quite hard to prove, and results from Galois theoryhelp. We probably won’t get into all of this, but you can read more here.

37

Chapter 4

Field extensions

Roughly speaking, an ‘extension’ of a field is a field " that contains as asubfield. It’s not much of an exaggeration to say that field extensions are the centralobjects of Galois theory, in much the same way that vector spaces are the centralobjects of linear algebra.

Introduction toWeek 4

It will be a while before it becomes truly clear why field extensions are soimportant, but here are a couple of indications:

• For any polynomial 5 over Q, we can take the smallest subfield " of C thatcontains all the complex roots of 5 , and that’s an extension of Q.

• For any irreducible polynomial 5 over a field , the quotient ring " =

[C]/〈 5 〉 is a field. The constant polynomials form a subfield of " isomor-phic to , so " is an extension of .

It’s important to distinguish between these two types of example. The first extendsQ by all the roots of 5 , whereas the second extends by just one root of 5—aswe’ll see.

But let’s begin at the beginning.

4.1 Definition and examplesDefinition 4.1.1 A field extension consists of a field , a field " , and a homo-morphism ] : → " .

Since homomorphisms between fields are injective (Lemma 2.2.2), is iso-morphic to the subfield im(]) of " . It is usually safe to identify ](0) with 0for each 0 ∈ —in other words, pretend that is actually a subfield of " and](0) = 0. We then say that " is an extension of and write " : .

38

It’s worth taking aminute tomake sure you understand the relationship betweensubsets and injections. This is a fundamental point about sets, not fields. Givena set � and a subset � ⊆ �, there’s an inclusion function ] : � → �, defined by](1) = 1 for all 1 ∈ �. (Remember that any function has a specified domain andcodomain, so this isn’t the same thing as the identity on �.) This inclusion functionis an injective. On the other hand, given any injective function between sets, sayi : - → �, the image im i is a subset of �, and there’s a bijection i′ : - → im i

given by i′(G) = i(G) (G ∈ -). Hence - is ‘isomorphic to’ (in bijection with) thesubset im i of �. This back-and-forth process means that subsets and injectionsare more or less the same thing.

Digression 4.1.2 If a field extension is basically the same thing as a field "together with a subfield , you might wonder why we bother with the moregeneral Definition 4.1.1, involving an arbitrary homomorphism ]. It turnsout that in the long run, it makes things easier.

There are two factors at play. The first is purely set-theoretic. The conceptof subset isn’t actually as simple as it appears, at least when you look at whatmathematicians do rather than what we claim we do. For example, (1) it’scommon to define the set C as R2, (2) everyone treats R as a subset of C,but (3) almost no one would say that R is a subset of R2 (if you wrote ‘thepoint c of R2’ on an exam, you’d be marked wrong). In truth, the commonconventions are inconsistent. A good way to make everything respectable isto do everything in terms of injections rather than subsets. It would take uptoo much space to go into this here, but Definition 4.1.1 is one example ofthis approach in action.

The second factor is algebraic. According to Definition 4.1.1, a field exten-sion is simply a homomorphism between fields, so it includes examples suchas the conjugation map : C → C. You might feel this example obeys theletter but not the spirit of Definition 4.1.1. But again, it turns out to be usefulto include such examples. When we come to count isomorphisms betweenfields in a few weeks, you’ll see why.

Examples 4.1.3 i. The inclusion ] : Q → C is a field extension, usually justwritten as C : Q. Similarly, there are field extensions C : R and R : Q.

ii. LetQ(√

2) = {0 + 1√

2 : 0, 1 ∈ Q}.Then Q(

√2) is a subring of C (easily), and in fact it’s a subfield: for if

(0, 1) ≠ (0, 0) then1

0 + 1√

2=0 − 1

√2

02 − 212

39

(noting that the denominators are not 0 because√

2 is irrational). So wehave an extension C : Q(

√2). Also, because Q ⊆ Q(

√2), we have another

extension Q(√

2) : Q.

iii. By direct calculation or later theory (which will make it much easier),

Q(√

2, 8) = {0 + 1√

2 + 28 + 3√

28 : 0, 1, 2, 3 ∈ Q}

is also a subfield of C, so we have an extension Q(√

2, 8) : Q.

iv. Let be a field. A rational expression over is a ratio of two polynomials

5 (C)6(C) ,

where 5 (C), 6(C) ∈ [C] with 6 ≠ 0. Two such expressions, 51/61 and52/62, are regarded as equal if 5162 = 5261 in [C]. So formally, a rationalexpression is an equivalence class of pairs ( 5 , 6) under the equivalencerelation in the last sentence. The set of rational expressions over isdenoted by (C).Rational expressions are added, subtracted and multiplied in the ways you’dexpect, making (C) into a field. There is a homomorphism ] : → (C)given by ](0) = 0/1 (0 ∈ ). In other words, (C) contains a copy of asthe constant rational expressions. So, we have a field extension (C) : .

v. In particular, when = F? for some prime ?, we have the field extensionF? (C) : F?. Note that F? (C) is an infinite field of characteristic ?! Fieldsof positive characteristic don’t have to be finite. So I’ve now fulfilled thepromise I made in Warning 2.2.12.

Warning 4.1.4 People sometimes say ‘rational function’ to mean‘rational expression’. But just as for polynomials (Warnings 3.1.2and 3.1.3), I want to emphasize that rational expressions are not func-tions. For instance, 1/(C − 1) is a totally respectable element of (C).You don’t have to—and shouldn’t—worry about what happens whenC = 1, because C is just a formal symbol (a mark on a piece of paper)rather than a variable, and 1/(C − 1) is just a formal expression, not afunction.If this puzzles you, I suggest going back to those warnings aboutpolynomials, which make the same point in a simpler setting.

40

Exercise 4.1.5 Find two examples of fields such that Q ( (Q(√

2, 8). (The symbol ( means proper subset.)

For any kind of algebraic structure, there is a notion of the ‘substructuregenerated by’ a given subset. For example, when - is a subset of a group �,you know what the subgroup generated by - is: it’s the intersection of all thesubgroups containing - . Similarly, when - is a subset of a vector space + , thelinear subspace generated by - (or ‘spanned by -’, as one usually says) is theintersection of all the linear subspaces containing - . We now make a similardefinition for fields.

Definition 4.1.6 Let be a field and - a subset of . The subfield of generatedby - is the intersection of all subfields of containing - .

You can check that any intersection of subfields of is a subfield of (even ifit’s an uncountably infinite intersection). In fact, you already did most of the workfor this in Exercise 2.1.2. So, the subfield � of generated by - really is a subfieldof . It contains - itself. By definition of intersection, � is the smallest subfieldof containing - , in the sense that any subfield of containing - contains �.

Exercise 4.1.7 Check the truth of all the statements in the previousparagraph.

Examples 4.1.8 i. The subfield of generated by ∅ is the prime subfieldof .

ii. Let ! be the subfield of C generated by {8}. I claim that

! = {0 + 18 : 0, 1 ∈ Q}.

To prove this, we have to show that ! is the smallest subfield of C containing8. First, it is a subfield of C (by an argument similar to Example 4.1.3(ii))and it contains 0+18 = 8. Now let !′ be any subfield of C containing 8. Then!′ contains the prime subfield of C (by definition of prime subfield), whichis Q. So whenever 0, 1 ∈ Q, we have 0, 1, 8 ∈ !′ and so 0 + 18 ∈ !′. Hence! ⊆ !′, as required.

iii. A very similar argument shows that the subfield of C generated by√

2 iswhat we have been calling Q(

√2).

41

Exercise 4.1.9 What is the subfield of C generated by {7/8}? By{2 + 38}? By R ∪ {8}?

We will be very interested in chains of fields

⊆ ! ⊆ "

in which and " are regarded as fixed and ! as variable. You can think of asthe floor, " as the ceiling, and ! as varying in between.

Definition 4.1.10 Let " : be a field extension and . ⊆ " . We write (. ) forthe subfield of " generated by ∪ . , and call it with . adjoined.

So, (. ) is the smallest subfield of " containing both and . .When . is a finite set {U1, . . . , U=}, we write ({U1, . . . , U=}) as

(U1, . . . , U=).

Examples 4.1.11 i. Take " : to be C : Q and . = {√

2}. Then (. ) is thesmallest subfield ofC containingQ∪{

√2}. But every subfield ofC contains

Q: that’s what it means for Q to be the prime subfield of C. So, (. ) is thesmallest subfield of C containing

√2. By Example 4.1.8(iii), that’s exactly

what we’ve been calling Q(√

2) all along. We refer to Q(√

2) as ‘Q with√

2adjoined’.

ii. Similarly, Q with 8 adjoined is

Q(8) = {0 + 18 : 0, 1 ∈ Q}

(Example 4.1.8(ii)), and Q with {√

2, 8} adjoined is the subfield denoted byQ(√

2, 8) in Example 4.1.3(iii).

iii. Let " be a field and - ⊆ " . Write for the prime subfield of " . Then (-) is the smallest subfield of " containing and - . But every subfieldof " contains , by definition of prime subfield. So (-) is the smallestsubfield of " containing -; that is, it’s the subfield of " generated by - .We already saw this argument in (i), in the case " = C and - = {

√2}.

iv. Let be any field and let " be the field (C) of rational expressions over ,which is an extension of . You might worry that there’s some ambiguityin the notation: (C) could either mean the field of rational expressionsover (as defined in Example 4.1.3(iv)) or the subfield of (C) obtained byadjoining the element C of (C) to (as in Definition 4.1.10).

42

In fact, they’re the same. In other words, the smallest subfield of (C)containing and C is (C) itself. Or equivalently, the only subfield of (C)containing and C is (C) itself. To see this, let ! be any such subfield. Forany polynomial 5 (C) = ∑

08C8 over , we have 5 (C) ∈ !, since 08, C ∈ ! and

! is closed under multiplication and addition. Hence for any polynomials5 (C), 6(C) over with 6(C) ≠ 0, we have 5 (C), 6(C) ∈ !, so 5 (C)/6(C) ∈ !as ! is closed under division by nonzero elements. So ! = (C).

Warning 4.1.12 It is not true in general that

(U) = {0 + 1U : 0, 1 ∈ } (false!) (4.1)

Examples like Q(√

2) and Q(8) do satisfy this, but that’s only because√2 and 8 satisfy quadratic equations. Certainly the right-hand side is

a subset of (U), but in general it’s much smaller, and isn’t a subfield.You’ve just seen an example: the field (C) of rational expressions ismuch bigger than the set {0 + 1C : 0, 1 ∈ } of polynomials of degree≤ 1. And that set of polynomials isn’t closed under multiplication.Another example: let U be the real cube root of 2. You can show thatU2 cannot be expressed as 0 + 1U for any 0, 1 ∈ Q (a fact we’ll comeback to in Example 4.2.9(ii)). But U ∈ Q(U), so U2 ∈ Q(U), so (4.1)fails in this case. In fact,

Q(U) = {0 + 1U + 2U2 : 0, 1, 2 ∈ Q}.

We’ll see why next week.

Exercise 4.1.13 Let" : be a field extension. Show that (.∪/) =( (. )) (/) whenever., / ⊆ " . (For example, (U, V) = ( (U)) (V)whenever U, V ∈ " .)

4.2 Algebraic and transcendental elementsA complex number U is said to be ‘algebraic’ if

00 + 01U + · · · + 0=U= = 0

for some rational numbers 08, not all zero. (You may have seen this definition with‘integer’ instead of ‘rational number’; it makes no difference, as you can alwaysclear the denominators.) This concept generalizes to arbitrary field extensions:

43

Definition 4.2.1 Let " : be a field extension and U ∈ " . Then U is algebraicover if there exists 5 ∈ [C] such that 5 (U) = 0 but 5 ≠ 0, and transcendentalotherwise.

Exercise 4.2.2 Show that every element of is algebraic over .

Examples 4.2.3 i. Let = ≥ 1. Then 42c8/= ∈ C is algebraic over Q, since5 (C) = C= − 1 is a nonzero polynomial such that 5 (42c8/=) = 0.

ii. The numbers c and 4 are both transcendental over Q. Both statements arehard to prove (and we won’t). By Exercise 4.2.2, any complex numbertranscendental over Q is irrational. Proving the irrationality of c and 4 isalready a challenge; proving they’re transcendental is even harder.

iii. Although c is transcendental over Q, it is algebraic over R, since it’s anelement of R. (Again, we’re using Exercise 4.2.2.) Moral: you shouldn’t sayan element of a field is just ‘algebraic’ or ‘transcendental’; you should sayit’s ‘algebraic/transcendental over ’, specifying your . Or at least, youshould do this when there’s any danger of confusion.

iv. Take the field (C) of rational expressions over a field . Then C ∈ (C) istranscendental over , since 5 (C) = 0 ⇐⇒ 5 = 0.

The set of complex numbers algebraic over Q is written as Q. It’s a fact that Qis a subfield of C, but this is extremely hard to prove by elementary means. Nextweek I’ll show you that with a surprisingly small amount of abstract algebra, youcan transform this from a very hard problem into an easy one.

So that you appreciate the miracle later, I give you this unusual exercise now.

Exercise 4.2.4 Attempt to prove any part of the statement that Q is asubfield ofC. For example, try to show thatQ is closed under addition,or multiplication, or reciprocals. I have no idea how to do any of theseusing only our current tools, but it’s definitely worth a few minutes ofdoomed effort to get a sense of the difficulties.

Let " : be a field extension and U ∈ " . An annihilating polynomial of Uis a polynomial 5 ∈ [C] such that 5 (U) = 0. So, U is algebraic if and only if ithas some nonzero annihilating polynomial.

It is natural to ask not only whether U is annihilated by some nonzero polyno-mial, but which polynomials annihilate it. The situation is pleasantly simple:

44

Lemma 4.2.5 Let " : be a field extension and U ∈ " . Then there is apolynomial <(C) ∈ [C] such that

〈<〉 = {annihilating polynomials of U over }. (4.2)

If U is transcendental over then < = 0. If U is algebraic over then there is aunique monic polynomial < satisfying (4.2).

Proof By the universal property of polynomial rings (Lemma 3.1.5), there is aunique homomorphism

\ : [C] → "

such that \ (0) = 0 for all 0 ∈ and \ (C) = U. (Here we’re taking the ‘i’ ofLemma 3.1.5 to be the inclusion → " .) Then

\

(∑08C

8)=

∑08U

8

for all∑08C

8 ∈ [C], so

ker \ = {annihilating polynomials of U over }.

Now ker \ is an ideal of the principal ideal domain [C] (using Proposition 3.2.2),so ker \ = 〈<〉 for some < ∈ [C].

If U is transcendental then ker \ = {0}, so < = 0.If U is algebraic then < ≠ 0. Multiplying a polynomial by a nonzero constant

does not change the ideal it generates (by Exercise 2.1.11 and Lemma 3.1.10(iii)),so we can assume that < is monic. It remains to prove that < is the only monicpolynomial such that 〈<〉 = ker \. But if < is another monic polynomial such that〈<〉 = ker \ then < = 2< for some nonzero constant 2 (again by Exercise 2.1.11and Lemma 3.1.10(iii)), and both are monic, so 2 = 1 and < = <. �

Definition 4.2.6 Let" : be a field extension and let U ∈ " be algebraic over .Theminimal polynomial of U is the unique monic polynomial < satisfying (4.2).

Exercise 4.2.7 What is the minimal polynomial of an element 0 of ?

This is an important definition, so we give some equivalent ways of stating it.

Lemma 4.2.8 Let " : be a field extension, let U ∈ " be algebraic over , andlet < ∈ [C] be a monic polynomial. The following are equivalent:

i. < is the minimal polynomial of U over ;

45

ii. <(U) = 0, and < | 5 for all annihilating polynomials 5 of U over ;

iii. <(U) = 0, and deg(<) ≤ deg( 5 ) for all nonzero annihilating polynomials5 of U over ;

iv. <(U) = 0 and < is irreducible over .

Part (iii) says the minimal polynomial is a monic annihilating polynomial ofleast degree.Proof (i)⇒(ii) follows from the definition of minimal polynomial.

(ii)⇒(iii) because if < | 5 ≠ 0 then deg(<) ≤ deg( 5 ).(iii)⇒(iv): assume (iii). First,< is not constant: for if< is constant then< = 1

(since < is monic); but <(U) = 0, so 1 = 0 in , a contradiction. Next, supposethat < = 5 6 for some 5 , 6 ∈ [C]. Then 0 = <(U) = 5 (U)6(U), so without lossof generality, 5 (U) = 0. By (iii), deg( 5 ) ≥ deg(<), so deg( 5 ) = deg(<) anddeg(6) = 0. This proves (iv).

(iv)⇒(i): assume (iv), and write <U for the minimal polynomial of U. Wehave <U | < by definition of <U and since <(U) = 0. But < is irreducible and <Uis not constant, so < is a nonzero constant multiple of <U. Since both are monic,< = <U, proving (i). �

Examples 4.2.9 i. The minimal polynomial of√

2 over Q is C2 − 2. There areseveral ways to see this.Firstmethod: C2−2 is amonic annihilating polynomial of

√2, and no nonzero

polynomial of degree ≤ 1 over Q annihilates√

2 since it is irrational. Thenuse Lemma 4.2.8(iii).Second method: C2 − 2 is an irreducible monic annihilating polynomial. Itis irreducible either because C2 − 2 has degree 2 and has no rational roots(using Lemma 3.3.1(iv)), or by Eisenstein’s criterion with prime 2. Thenuse Lemma 4.2.8(iv).

ii. The minimal polynomial of 3√2 over Q is C3 − 2. This will follow fromLemma 4.2.8(iv) as long as C3 − 2 is irreducible, which you can show usingeither Lemma 3.3.1(iv) or Eisenstein.But unlike in (i), it’s not so easy to show directly that C3−2 is the annihilatingpolynomial of least degree. Try proving with your bare hands that 3√2satisfies no quadratic equation over Q, i.e. that the equation

3√22= 0

3√2 + 1has no solution for 0, 1 ∈ Q. It’s not impossible, but it’s a mess. (Younaturally begin by cubing both sides, but look what happens next. . . ) Sothe theory really gets us something here.

Two traps46

iii. Let ? be a prime number, and put l = 42c8/? ∈ C. Then l is a root of C ? −1,but that is not the minimal polynomial of l, since it is reducible:

C ? − 1 = (C − 1)<(C)

where<(C) = C ?−1 + · · · + C + 1.

Sincel?−1 = 0 butl−1 ≠ 0, wemust have<(l) = 0. By Example 3.3.16,< is irreducible over Q. Hence < is the minimal polynomial of l over Q.

4.3 Simple extensionsWe can say a lot about extensions generated by a single element.

Definition 4.3.1 A field extension " : is simple if there exists U ∈ " such that" = (U).

Examples 4.3.2 i. Surprisingly many algebraic extensions are simple. Forinstance,Q(

√2,√

3) : Q is a simple extension (despite appearances), becausein fact Q(

√2,√

3) = Q(√

2 +√

3).

ii. (C) : is simple, where (C) is the field of rational expressions over .

Exercise 4.3.3 Prove that Q(√

2,√

3) = Q(√

2 +√

3). Hint: begin byfinding (

√2 +√

3)3.

Given a simple extension " : and an element U that generates it, we cantake the minimal polynomial of U, which is irreducible over . But in the oppositedirection, if you hand me a field and an irreducible polynomial < over , I cancook up for you a simple extension " : and an element U ∈ " whose minimalpolynomial is the < you gave me.

This works as follows. Whenever< is an irreducible polynomial over a field ,the quotient [C]/〈<〉 is a field (Corollary 3.2.5). We have ring homomorphisms

→ [C] c−→ [C]/〈<〉,

where the first map sends 0 ∈ to the constant polynomial 0 and c is the canonicalhomomorphism. Their composite is a homomorphism of fields → [C]/〈<〉.So, we have a field extension

( [C]/〈<〉

): . And one of the elements of

[C]/〈<〉 is c(C), which I will call U.

47

Lemma 4.3.4 Let < be a monic irreducible polynomial over a field . Then( [C]/〈<〉

): is a simple extension generated by U (in the notation above), and

the minimal polynomial of U over is <.

Proof We have ker c = 〈<〉 by definition of c, and ker c is the set of annihilatingpolynomials of U over , so< is the minimal polynomial of U over (by definitionof minimal polynomial).

To see that U generates [C]/〈<〉 over , let ! be a subfield of [C]/〈<〉containing and U. Then c−1! is a subring of [C] containing and C (sincec(C) = U), which forces c−1! = [C] and so ! = [C]/〈<〉. �

We’ll show that [C]/〈<〉 is the only simple extension of by an element withminimal polynomial <. But the word ‘only’ is going to have to be interpreted inan up-to-isomorphism sense (as in ‘there’s only one group of order 2’). Here it isformally.

Definition 4.3.5 Let be a field, and let ] : → " and ]′ : → "′ be extensionsof . A homomorphism i : " → "′ is said to be a homomorphism over if

"i // "′

]

``

]′

>>

commutes. If i is invertible then we call i an isomorphism over .

Exercise 4.3.6 In this definition, show that if i is invertible then i−1

is also a homomorphism over .

The next result not only classifies the simple extensions by an algebraic elementHow to understandsimple algebraic

extensions(part (i)), but also those by a transcendental element (part (ii)).

Theorem 4.3.7 (Classification of simple extensions) Let be a field.

i. Let < ∈ [C] be a monic irreducible polynomial. Then there exist anextension " : and an algebraic element U ∈ " such that " = (U)and U has minimal polynomial < over .Moreover, if (", U) and ("′, U′) are two such pairs, there is an isomor-phism i : " → "′ over such that i(U) = U′.

ii. There exist an extension " : and a transcendental element U ∈ "such that " = (U).Moreover, if (", U) and ("′, U′) are two such pairs, there is an isomor-

48

phism i : " → "′ over such that i(U) = U′.

Proof The first part of (i) follows from Lemma 4.3.4. For ‘Moreover’, we may aswell take " = [C]/〈<〉 and U as in Lemma 4.3.4. The homomorphism

\ : [C] → "′∑08C

8 ↦→ ∑08U′8 (4.3)

has kernel 〈<〉, so [C]/〈<〉 � im \ by the first isomorphism theorem, so im \ isa field. Also, \ (0) = 0 for all 0 ∈ . Hence im \ is a subfield of "′ containing\ (C) = U′ and . But we are assuming that "′ = (U′), which means thatthe only subfield of "′ containing and U′ is "′ itself. So im \ = "′, giving [C]/〈<〉 � "′. Diagram:

[C] c //

\

%% [C]/〈<〉 \

�// "′

ee OO 99

(If you’re losing the thread, it may help to go back to the review of the universalproperty of quotients and the first isomorphism theorem on p. 16.) The isomor-phism \ : [C]/〈<〉 → "′ that we have constructed is an isomorphism over ,and \ (U) = \ (c(C)) = U′. So i = \ satisfies the conditions of the theorem.

To prove the first part of (ii), we simply take " to be the field (C) of ra-tional expressions over and U = C. Then C ∈ " is transcendental over (Example 4.2.3(iv)).

For ‘Moreover’, take any simple extension"′ of by a transcendental elementU′. Any 5 , 6 ∈ [C] with 6 ≠ 0 give rise to an element 5 (U′)/6(U′) ∈ "′, where6(U′) ≠ 0 becauseU′ is transcendental. One can check that this gives awell-definedhomomorphism

\ : (C) → "′

5 (C)6(C) ↦→

5 (U′)6(U′)

( 5 , 6 ∈ [C], 6 ≠ 0). Now \ is injective (being a homomorphism of fields), so (C) � im \, so im \ is a subfield of "′. Also, im \ contains \ (C) = U′, and\ (0) = 0 for each 0 ∈ . So im \ is a subfield of "′ containing U′ and , whichsince "′ = (U′) implies that im \ = "′. So \ : (C) → "′ is an isomorphism,and it is an isomorphism over satisfying \ (C) = U′. Hence i = \ satisfies theconditions of the theorem. �

49

Conclusion: given any field (not necessarily Q!) and any monic irreducible<(C) ∈ [C], we can say the words ‘adjoin to a root U of<’, and this unambigu-ously defines an extension (U) : . (At least, unambiguously up to isomorphismover —but who could want more?) Similarly, we can unambiguously adjoin to a transcendental element.

Examples 4.3.8 i. Let be any field not containing a square root of 2. ThenC2 − 2 is irreducible over . So we can adjoin to a root of C2 − 2, givingan extension (

√2) : .

We have already seen this example many times when = Q, in which case (√

2) can be seen as a subfield of C. But the construction works for any . For instance, 2 has no square root in F3, so there is an extension F3(

√2)

of F3. It can be constructed as F3 [C]/〈C2 − 2〉.

ii. The polynomial <(C) = 1 + C + C2 is irreducible over F2, so we may adjoin toF2 a root U of <. Then F2(U) = F2 [C]/〈1 + C + C2〉.

Exercise 4.3.9 Howmany elements does the field F3(√

2) have? Whatabout F2(U)?

Warning 4.3.10 Take = Q and <(C) = C3 − 2, which is irreducible.Write U1, U2, U3 for the roots of < in C. Then Q(U1), Q(U2) andQ(U3) are all different as subsets of C. For example, one of the U8is the real cube root of 2 (say U1), which implies that Q(U1) ⊆ R,whereas the other two are not real, so Q(U8) * R for 8 ≠ 1. However,Q(U1) : Q, Q(U2) : Q and Q(U3) : Q are all isomorphic as abstractfield extensions of Q. This follows from Theorem 4.3.7, since all theU8 have the same minimal polynomial, <.You’re already very familiar with this kind of situation in otherbranches of algebra, whether you realize it or not. For instance, in lin-ear algebra, take three vectors v1, v2, v3 in R2, none a scalar multipleof any other. Then span(v1), span(v2) and span(v3) are all differentas subsets of R2, but they are all isomorphic as abstract vector spaces(since they’re all 1-dimensional). A similar example could be givenwith a group containing several subgroups that are all isomorphic.

You’ve seen that Galois theory involves aspects of group theory and ring theory.In the next chapter, you’ll see how linear algebra enters the picture too.

50

Chapter 5

Degree

We’ve already seen that if you adjoin to Q a square root of 2, then each element ofthe resulting field can be specified using two rational numbers, 0 and 1:

Q(√

2) ={0 + 1

√2 : 0, 1 ∈ Q

}.

We’ve also seen that if you adjoin to Q a cube root of 2, then it takes three rationalnumbers to specify each element of the resulting field:

Introduction toWeek 5

Q( 3√2) ={0 + 1 3√2 + 2 3√2

2: 0, 1, 2 ∈ Q

}(Warning 4.1.12). This might lead us to suspect that Q( 3√2) : Q is in some sensea ‘bigger’ extension than Q(

√2) : Q.

The first thing we’ll do in this chapter is to make this intuition rigorous. We’lldefine the ‘degree’ of an extension and see that Q(

√2) : Q and Q( 3√2) : Q have

degrees 2 and 3, respectively.The concept of degree is incredibly useful, and not only in Galois theory.

In fact, I’ll show you how it can be used to solve three problems that remainedunsolved for literally millennia, since the time of the ancient Greeks.

5.1 Degrees of extensions and polynomialsThe concept of degree is an excellent illustration of a powerful mathematicaltechnique: forgetting.

Let " : be a field extension. What happens if we forget how to multiplytogether elements of " that aren’t in ?

We still have the field . What remains of " is its underlying additive abeliangroup (", +, 0), and because we haven’t forgotten how to multiply elements of

51

with elements of " , we still have the multiplication function × " → " . So,we have a field , an abelian group " , and an action of on " .

In other words, whenever " : is a field extension, " is a vector space over in a natural way.

Definition 5.1.1 The degree [" : ] of a field extension " : is the dimensionof " as a vector space over .

If " is a finite-dimensional vector space over , it’s clear what this means. If" is infinite-dimensional over , we write [" : ] = ∞, where ∞ is a formalsymbol which we give the properties

= < ∞, = · ∞ = ∞ (= ≥ 1), ∞ · ∞ = ∞

(where = is an integer). An extension " : is finite if [" : ] < ∞.

Digression 5.1.2 You know that whenever + is a finite-dimensional vectorspace, (i) there exists a basis of + , and (ii) there is a bijection between anytwo bases. This makes it possible to define the dimension of a vector spaceas the number of elements in a basis. In fact, both (i) and (ii) are true forevery vector space, not just the finite-dimensional ones. So we can definethe dimension of an arbitrary vector space as the ‘number’ of elements in abasis, where now ‘number’ means cardinal, i.e. isomorphism class of sets.

We could interpret Definition 5.1.1 using this general definition of dimension.For instance, suppose we had one field extension " : such that " had acountably infinite basis over , and another, " ′ : , such that " ′ had anuncountably infinite basis over . Then [" : ] and [" ′ : ] would bedifferent.

However, we’ll lump all the infinite-dimensional extensions together and saythat their degrees are all ∞. We’ll mostly be dealing with finite extensionsanyway, and won’t need to distinguish between sizes of∞. It’s a bit like thedifference between a house that costs a million pounds and a house that coststen million: although the difference in cost is huge, most of us would lumpthem together in a single category called ‘unaffordable’.

Examples 5.1.3 i. Every field " contains at least one nonzero element,namely, 1. So [" : ] ≥ 1 for every field extension " : .If " = then {1} is a basis, so [" : ] = 1. On the other hand, if[" : ] = 1 then the one-element linearly independent set {1} must be abasis, which implies that every element of " is equal to 0 · 1 = 0 for some0 ∈ , and so " = . Hence

[" : ] = 1 ⇐⇒ " = .

52

ii. Every element of C is equal to G + H8 for a unique pair (G, H) of elements ofR. That is, {1, 8} is a basis of C over R. Hence [C : R] = 2.

iii. Let be a field and (C) the field of rational expressions over . Then1, C, C2, . . . are linearly independent over , so [ (C) : ] = ∞.

Warning 5.1.4 The degree [ : ] of over itself is 1, not 0.Degrees of extensions are never 0. See Example 5.1.3(i).

Theorem 5.1.5 Let (U) : be a simple extension, with U algebraic over .Write < ∈ [C] for the minimal polynomial of U and = = deg(<). Then

1, U, . . . , U=−1

is a basis of (U) over . In particular, [ (U) : ] = deg(<).

Proof By Lemma 4.3.4 and Theorem 4.3.7(i), we might as well take (U) = [C]/〈<〉 and U = c(C), where c : [C] → [C]/〈<〉 is the canonical homomor-phism.

Since c is surjective, every element of (U) is equal to c( 5 ) for some 5 ∈ [C].By Proposition 3.2.1, there are unique @, A ∈ [C] such that 5 = @< + A anddeg(A) < =. In particular, there is a unique polynomial A ∈ [C] such that5 − A ∈ 〈<〉 and deg(A) < =. Equivalently, there are unique 00, . . . , 0=−1 ∈ suchthat

5 (C) −(00 + 01C + · · · + 0=−1C

=−1) ∈ 〈<〉.Equivalently, there are unique 00, . . . , 0=−1 ∈ such that

c( 5 ) = c(00 + 01C + · · · + 0=−1C

=−1) .Equivalently (since c(C) = U), there are unique 00, . . . , 0=−1 ∈ such that

c( 5 ) = 00 + 01U + · · · 0=−1U=−1.

We have now shown that every element of (U) can be expressed as a -linearcombination of 1, U, . . . , U=−1 in a unique way. In other words, 1, U, . . . , U=−1 is abasis of (U) over . �

Exercise 5.1.6 For a prime ?, find [Q(42c8/?) : Q].

53

Theorem 5.1.5 implies that when U ∈ " is algebraic over , with minimalpolynomial of degree =, the subset {00 + 01U + · · · + 0=−1U

=−1 : 08 ∈ } is asubfield of " . This isn’t particularly obvious: for instance, why is it closed undertaking reciprocals? But it’s true.

Corollary 5.1.7 Let " : be a field extension and U ∈ " . Then

(U) : is finite ⇐⇒ U is algebraic over .

Proof For ⇒, we prove the contrapositive. If U is transcendental over then (U) is isomorphic to (C) over (by Theorem 4.3.7(ii)), and [ (C) : ] = ∞by Example 5.1.3(iii).

Theorem 5.1.5 gives⇐. �

For a field extension " : and U ∈ " , the degree of U over is [ (U) : ].We write it as deg (U). So, deg (U) < ∞ if and only if U is algebraic over , andin that case, deg (U) is the degree of the minimal polynomial of U over .

Examples 5.1.8 i. Let U ∈ C be an algebraic number over Q whose minimalpolynomial is quadratic. Then by Theorem 5.1.5,

Q(U) = {0 + 1U : 0, 1 ∈ Q}.

We’ve already seen this in many examples, such as U =√

2 and U = 8.

ii. Let U be the real cube root of 2. By Example 4.2.9(ii), the minimalpolynomial of U over Q is C3 − 2, so degQ(U) = 3. It follows thatQ(U) ≠ {0 + 1U : 0, 1 ∈ Q}, since otherwise the two-element set {1, U}would span the three-dimensional vector space Q(U). So we have anotherproof that 22/3 cannot be written as a Q-linear combination of 1 and 21/3.As observed in Example 4.2.9(ii), this is messy to prove directly.

Theorem 5.1.5 is quite powerful. Here are two more corollaries of it.

Corollary 5.1.9 i. Let " : ! : be field extensions and V ∈ " . Then[! (V) : !] ≤ [ (V) : ].

ii. Let " : be a field extension and U, V ∈ " . Then [ (U, V) : (U)] ≤[ (V) : ].

Proof For (i): if [ (V) : ] = ∞ then the inequality is clear. Otherwise, V isalgebraic over (by Corollary 5.1.7), with minimal polynomial < ∈ [C], say.Then < is an annihilating polynomial for V over !, so the minimal polynomial ofV over ! has degree ≤ deg(<). The result follows from Theorem 5.1.5.

Part (ii) follows by taking ! = (U). �

54

"

!

V

[! (V):!][ (V): ]

Figure 5.1: Visualization of Corollary 5.1.9(i) (not to be taken too seriously).

Informally, I think of part (i) as in Figure 5.1. The degree of V over measureshow far V is from being in . Since ! contains , it might be that V is closer to !than to (i.e. [! (V) : !] < [ (V) : ]), and it’s certainly no further away.

Exercise 5.1.10 Give an example to show that the inequality in Corol-lary 5.1.9(ii) can be strict. Your example can be as trivial as you like.

Corollary 5.1.11 Let " : be a field extension and U1, . . . , U= ∈ " . Sup-pose that each U8 is algebraic over of degree 38. Then every elementU ∈ (U1, . . . , U=) can be represented as a polynomial in U1, . . . , U= over :

U =∑

A1,...,A=

2A1,...,A=UA11 · · · U

A==

for some 2A1,...,A= ∈ , where A8 ranges over 0, . . . , 38 − 1.

Proof By induction on =. When = = 0, this is trivial. Now let = ≥ 1 and supposethe result holds for = − 1. Let

U ∈ (U1, . . . , U=) =( (U1, . . . , U=−1)

)(U=).

By Theorem 5.1.5 applied to the extension ( (U1, . . . , U=−1)) (U=) : (U1, . . . , U=−1), noting that deg (U1,...,U=−1) (U=) ≤ deg (U=) = 3=, we have

U =

3=−1∑A=0

2AUA= (5.1)

for some 20, . . . , 23=−1 ∈ (U1, . . . , U=−1). By inductive hypothesis, for each A wehave

2A =∑

A1,...,A=−1

2A1,...,A=−1,AUA11 · · · U

A=−1=−1 (5.2)

for some 2A1,...,A=−1,A ∈ , where A8 ranges over 0, . . . , 38 − 1. Substituting (5.2)into (5.1) completes the induction. �

55

Example 5.1.12 Back in Example 4.1.11(ii), I claimed that

Q(√

2, 8) = {0 + 1√

2 + 28 + 3√

28 : 0, 1, 2, 3 ∈ Q}.

Corollary 5.1.11 applied to Q(√

2, 8) : Q proves this, since degQ(√

2) = degQ(8) =2.

Exercise 5.1.13 Let " : be a field extension and U a transcen-dental element of " . Can every element of (U) be represented as apolynomial in U over ?

5.2 The tower lawWenowknow about the degrees of simple extensions—those obtained by adjoininga single element. What about extensions obtained by adjoining several elements?The following result is invaluable.

Theorem 5.2.1 (Tower law) Let " : ! : be field extensions.

i. If (U8)8∈� is a basis of ! over and (V 9 ) 9∈� is a basis of " over !, then(U8V 9 )(8, 9)∈�×� is a basis of " over .

ii. " : is finite ⇐⇒ " : ! and ! : are finite.

iii. [" : ] = [" : !] [! : ].

The sets � and � here could be infinite. I’ll say that a family (08)8∈� of elementsof a field is finitely supported if the set {8 ∈ � : 08 ≠ 0} is finite.

Proof To prove (i), we show that (U8V 9 )(8, 9)∈�×� is a linearly independent spanningset of " over .

For linear independence, let (28 9 )(8, 9)∈�×� be a finitely supported family of ele-ments of such that

∑8, 9 28 9U8V 9 = 0. Then

∑9 (∑8 28 9U8)V 9 = 0, with

∑8 28 9U8 ∈ !

for each 9 ∈ �. Since (V 9 ) 9∈� is linearly independent over !, we have∑8 28 9U8 = 0

for each 9 ∈ �. But (U8)8∈� is linearly independent over , so 28 9 = 0 for each 8 ∈ �and 9 ∈ �.

To show (U8V 9 )(8, 9)∈�×� spans" over , let 4 ∈ " . Since (V 9 ) 9∈� spans" over!, we have 4 =

∑9 3 9 V 9 for some finitely supported family (3 9 ) 9∈� of elements of

!. Since (U8)8∈� spans ! over , for each 9 ∈ � we have 3 9 =∑8 28 9U8 for some

finitely supported family (28 9 )8∈� of . Hence 4 =∑8, 9 28 9U8V 9 , as required.

Parts (ii) and (iii) follow. �

56

Example 5.2.2 What is [Q(√

2,√

3) : Q]? The tower law gives[Q(√

2,√

3) : Q]=

[Q(√

2,√

3) : Q(√

2)] [Q(√

2) : Q]

= 2[Q(√

2,√

3) : Q(√

2)].

We have [Q(√

2,√

3) : Q(√

2)]≤

[Q(√

3) : Q]= 2

by Corollary 5.1.9(ii). On the other hand,√

3 ∉ Q(√

2), so Q(√

2,√

3) ≠ Q(√

2),so [Q(

√2,√

3) : Q(√

2)] > 1 by Example 5.1.3(i). So [Q(√

2,√

3) : Q(√

2)] = 2,giving the answer: [Q(

√2,√

3) : Q] = 4.By the same argument as in Example 5.1.12, {1,

√2,√

3,√

6} spansQ(√

2,√

3)over Q. But we have just shown that Q(

√2,√

3) has dimension 4 over Q. Hencethis spanning set is a basis. That is, for every element U ∈ Q(

√2,√

3), there is oneand only one 4-tuple (0, 1, 2, 3) of rational numbers such that

U = 0 + 1√

2 + 2√

3 + 3√

6.

Exercise 5.2.3 In that example, I claimed that√

3 ∉ Q(√

2). Prove it.

Corollary 5.2.4 Let " : !′ : ! : be field extensions. If " : is finite then[!′ : !] divides [" : ].

Proof Apply the tower law to " : !′ : ! then " : ! : . �

That result might remind you of Lagrange’s theorem on group orders. Theresemblance is no coincidence, as we’ll see.

Exercise 5.2.5 Show that a field extension whose degree is a primenumber must be simple.

That result might remind you of the fact that a group of prime order must becyclic, and that’s no coincidence either!

A second corollary of the tower law:

Corollary 5.2.6 Let " : be a field extension and U1, . . . , U= ∈ " . Then

[ (U1, . . . , U=) : ] ≤ [ (U1) : ] · · · [ (U=) : ] .

Proof By the tower law and then Corollary 5.1.9(ii),

[ (U1, . . . , U=) : ]= [ (U1, . . . , U=) : (U1, . . . , U=−1)] · · · [ (U1, U2) : (U1)] [ (U1) : ]≤ [ (U=) : ] · · · [ (U2) : ] [ (U1) : ] . �

57

algebraic finitelygenerated

finite

Figure 5.2: Finiteness conditions on a field extension

Example 5.2.7 What is [Q(121/4, 61/15) : Q]? You can check (hint, hint) thatdegQ(121/4) = 4 and degQ(61/15) = 15. So by Corollary 5.2.4, [Q(121/4, 61/15) :Q] is divisible by 4 and 15. But also, Corollary 5.2.6 implies that [Q(121/4, 61/15) :Q] ≤ 4 × 15 = 60. Since 4 and 15 are coprime, the answer is 60.

Exercise 5.2.8 Generalize Example 5.2.7. In other words, whatgeneral result does the argument of Example 5.2.7 prove, not involvingthe particular numbers chosen there?

5.3 Algebraic extensionsWe defined a field extension " : to be finite if [" : ] < ∞, that is, " isfinite-dimensional as a vector space over . Here are two related conditions.

Definition 5.3.1 A field extension " : is finitely generated if " = (. ) forsome finite subset . ⊆ " .

Definition 5.3.2 A field extension " : is algebraic if every element of " isalgebraic over .

Recall from Corollary 5.1.7 that U is algebraic over if and only if (U) : isfinite. So for a field extension to be algebraic is also a kind of finiteness condition.

Examples 5.3.3 i. For any field , the extension (C) : is finitely generated(take the ‘. ’ above to be {C}) but not finite, by Corollary 5.1.7.

ii. In Section 4.2 youmet the setQ of complex numbers algebraic overQ. We’llvery soon prove that it’s a subfield of C. It is algebraic over Q, by definition.But you’ll show in Workshop 3 that it is not finite over Q.

Our three finiteness conditions are related as follows (Figure 5.2).

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Proposition 5.3.4 The following conditions on a field extension " : are equiv-alent:

i. " : is finite;

ii. " : is finitely generated and algebraic;

iii. " = (U1, . . . , U=) for some finite set {U1, . . . , U=} of elements of " alge-braic over .

Proof (i)⇒(ii): suppose that " : is finite.To show that " : is finitely generated, take a basis U1, . . . , U= of " over

. Every subfield ! of " containing is a -linear subspace of " , so ifU1, . . . , U= ∈ ! then ! = " . This proves that the only subfield of " containing ∪ {U1, . . . , U=} is " itself; that is, " = (U1, . . . , U=). So " : is finitelygenerated.

To show that " : is algebraic, let U ∈ " . Then by part (ii) of the tower law(Theorem 5.2.1), (U) : is finite, so by Corollary 5.1.7, U is algebraic over .

(ii)⇒(iii) is immediate from the definitions.(iii)⇒(i): suppose that " = (U1, . . . , U=) for some U8 ∈ " algebraic over .

Then[" : ] ≤ [ (U1) : ] · · · [ (U=) : ]

by Corollary 5.2.6. For each 8, we have [ (U8) : ] < ∞ since U8 is algebraicover (using Corollary 5.1.7 again). So [" : ] < ∞. �

We already saw that when " = (U1, . . . , U=) with each U8 algebraic, everyelement of " is a polynomial in U1, . . . , U= (Corollary 5.1.11). So for any finiteextension " : , there is some finite set of elements such that everything in "can be expressed as a polynomial over in these elements.

Exercise 5.3.5 Let " : be a field extension and ⊆ ! ⊆ " . Inthe proof of Proposition 5.3.4, I said that if ! is a subfield of " then! is a -linear subspace of " . Why is that true? And is the conversealso true? Give a proof or a counterexample.

Corollary 5.3.6 Let (U) : be a simple extension. The following are equivalent:

i. (U) : is finite;

ii. (U) : is algebraic;

iii. U is algebraic over .

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Proof (i)⇒(ii) follows from (i)⇒(ii) of Proposition 5.3.4.(ii)⇒(iii) is immediate from the definitions.(iii)⇒(i) follows from (iii)⇒(i) of Proposition 5.3.4. �

Here’s a spectacular application of Corollary 5.3.6.

Proposition 5.3.7 Q is a subfield of C.

Proof By Corollary 5.3.6,

Q = {U ∈ C : [Q(U) : Q] < ∞}.

For all U, V ∈ Q,

[Q(U, V) : Q] ≤ [Q(U) : Q] [Q(V) : Q] < ∞

by Corollary 5.2.6. Hence

[Q(U + V) : Q] ≤ [Q(U, V) : Q] < ∞,

giving U + V ∈ Q. Similarly, U · V ∈ Q. For all U ∈ Q,

[Q(−U) : Q] = [Q(U) : Q] < ∞,

giving −U ∈ Q. Similarly, 1/U ∈ Q (if U ≠ 0). And clearly 0, 1 ∈ Q. �

If you did Exercise 4.2.4, you’ll appreciate how hard that result is to prove fromfirst principles, and how amazing it is that the proof above is so clean and simple.

5.4 Ruler and compass constructionsThe ancient Greeks developed planar geometry to an extraordinary degree, dis-covering how to perform a very wide range of constructions using only ruler andcompasses. But there were three particular constructions that they couldn’t figureout how to do using only these instruments:

• Trisect the angle: given an angle \, construct the angle \/3.

• Duplicate the cube: given a length, construct a new length whose cube istwice the cube of the original. That is, given two points distance ! apart,construct two points distance 3√2! apart.

• Square the circle: given a circle, construct a square with the same area.That is, given two points distance ! apart, construct two points distance√c! apart.

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The challenge of finding constructions lay unanswered for thousands of years. Andit wasn’t for lack of attention: mathematicians kept on looking. My Galois theorylecture notes from when I was an undergraduate contain the following words:

Thomas Hobbes claimed to have solved these. John Wallis disagreed.A 17th century pamphlet war ensued.

Twitter users may conclude that human nature has not changed.It turns out that the reasonwhy no one could find away to do these constructions

is that they’re impossible. We’ll prove it using field theory.In order to prove that you can’t do these things using ruler and compasses, it’s

necessary to know that you can do certain other things using ruler and compasses.I’ve made a video showing how to do the various constructions we’ll need.

Ruler and compassconstructions Digression 5.4.1 The standard phrase is ‘ruler and compass constructions’,

but it’s slightly misleading. A ruler has distance markings on it, whereasfor the problems of ancient Greece, you’re supposed to use only a ‘straightedge’: a ruler without markings (and no, you’re not allowed to mark it). AsStewart explains (Section 7.1), with a marked or markable straight edge, youcan solve all three problems. Also, for what it’s worth, an instrument fordrawing circles is strictly speaking a pair of compasses. But like everyoneelse, we’ll say ‘ruler and compass’—

—when we really mean ‘straight edge and compasses’—

The problems as stated above are maybe not quite precise; let’s formalize them.Starting from a subset Σ of the plane, our instruments allow the following

constructions:

• given two distinct points �, � of Σ, draw the (infinite) line through � and �;

• given two distinct points �, � of Σ, draw the circle with centre � passingthrough �.

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A point in the plane is immediately constructible from Σ if it is a point ofintersection between two distinct lines, or two distinct circles, or a line and acircle, of the form above. A point � in the plane is constructible from Σ ifthere is a finite sequence �1, . . . , �= = � of points such that �8 is immediatelyconstructible from Σ ∪ {�1, . . . , �8−1} for each 8.

So far I have written in the Greek spirit by saying ‘the plane’ rather than R2.But now fix a coordinate system. For Σ ⊆ R2, write

Σ = Q({U ∈ R : U is a coordinate of some point in Σ}

),

which is a subfield of R. The condition on U means that (U, H) ∈ Σ for some H ∈ Ror (G, U) ∈ Σ for some G ∈ R.

The key to the impossibility proofs is the following definition. For subfields ⊆ " ⊆ R, let us say that " : is an iterated quadratic extension if there issome finite sequence of subfields

= 0 ⊆ 1 ⊆ · · · ⊆ = = "

such that [ 8 : 8−1] = 2 for all 8 ∈ {1, . . . , =}.

Theorem 5.4.2 Let Σ ⊆ R2 and (G, H) ∈ R2. If (G, H) is constructible from Σthen there is an iterated quadratic extension of Σ containing G and H.

Before I show you the proof, I’ll state a corollary that reveals how this theoremleads to proofs of impossibility.

Corollary 5.4.3 Let Σ ⊆ R2 and (G, H) ∈ R2. If (G, H) is constructible from Σ thenG and H are algebraic over Σ, and their degrees over Σ are powers of 2.

Proof Take an iterated quadratic extension " of Σ with G ∈ " . Then [" : Σ] = 2= for some = ≥ 0, by the tower law. But then deg Σ (G) = [ Σ(G) : Σ]divides 2= by Corollary 5.2.4, and is therefore a power of 2. And similarly for H.�

We will show that if (for instance) we could trisect angles, we would be ableto construct a point whose coordinates do not have degree a power of 2, giving acontradiction.

To prove Theorem 5.4.2, we need a lemma.

Lemma 5.4.4 Let be a subfield of R and U, V ∈ R. Suppose that U and V areeach contained in some iterated quadratic extension of . Then there is someiterated quadratic extension of containing both U and V.

62

Proof Take subfields

= 0 ⊆ 1 ⊆ · · · ⊆ = ⊆ R, = !0 ⊆ !1 ⊆ · · · ⊆ !< ⊆ R

with U ∈ =, V ∈ !<, and [ 8 : 8−1] = 2 = [! 9 : ! 9−1] for all 8, 9 .For each 9 ∈ {1, . . . , <}, choose some V 9 ∈ ! 9 \ ! 9−1; then ! 9 = ! 9−1(V 9 ).

Hence ! 9 = (V1, . . . , V 9 ) for each 9 .Now consider the chain of subfields

= 0 ⊆ 1 ⊆ · · · ⊆ = ⊆ = (V1) ⊆ = (V1, V2) ⊆ · · · ⊆ = (V1, . . . , V<).(5.3)

For each 9 ∈ {1, . . . , <}, Corollary 5.1.9(i) gives the inequality

[ = (V1, . . . , V 9 ) : = (V1, . . . , V 9−1)] ≤ [ (V1, . . . , V 9 ) : (V1, . . . , V 9−1)]= [! 9 : ! 9−1] = 2.

So in the chain of subfields (5.3), each successive extension has degree 1 or2. An extension of degree 1 is an equality, so by ignoring repeats, we see that = (V1, . . . , V<) is an iterated quadratic extension of . Finally,

U ∈ = ⊆ = (V1, . . . , V<), V ∈ !< = (V1, . . . , V<) ⊆ = (V1, . . . , V<),

so U, V ∈ = (V1, . . . , V<), as required. �

Exercise 5.4.5 In the second paragraph of the proof, I claimed that! 9 = ! 9−1(V 9 ). The general principle here is that if " : is a fieldextension of degree 2 and W ∈ " \ then " = (W). Prove this.

Proof of Theorem 5.4.2 Suppose that (G, H) is constructible from Σ in = steps.If = = 0 then (G, H) ∈ Σ, so G, H ∈ Σ, which is trivially an iterated quadraticextension of Σ.

Now let = ≥ 1, and suppose inductively that each coordinate of each pointof R2 constructible from Σ in < = steps lies in some iterated quadratic extensionof Σ. By definition, (G, H) is an intersection point of two distinct lines/circlesthrough points constructible in < = steps. By inductive hypothesis, each coordinateof each of those points lies in some iterated quadratic extension of Σ, so byLemma 5.4.4, there is an iterated quadratic extension ! of Σ containing all thepoints’ coordinates. The coefficients in the equations of the lines/circles then alsolie in !.

We now show that deg! (G) ∈ {1, 2}.

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If (G, H) is the intersection point of two distinct lines, then G and H satisfy twolinearly independent equations

0G + 1H + 2 = 0,0′G + 1′H + 2′ = 0

with 0, 1, 2, 0′, 1′, 2′ ∈ !. Solving gives G ∈ !.If (G, H) is an intersection point of a line and a circle, then

0G + 1H + 2 = 0,G2 + H2 + 3G + 4H + 5 = 0

with 0, 1, 2, 3, 4, 5 ∈ !. If 1 = 0 then 0 ≠ 0 and G = −2/0 ∈ !. Otherwise, wecan eliminate H to give a quadratic over ! satisfied by G, so that deg! (G) ∈ {1, 2}.

If (G, H) is an intersection point of two circles, then

G2 + H2 + 3G + 4H + 5 = 0,G2 + H2 + 3′G + 4′H + 5 ′ = 0

with 3, 4, 5 , 3′, 4′, 5 ′ ∈ !. Subtracting, we reduce to the case of a line and a circle,again giving deg! (G) ∈ {1, 2}.

Hence deg! (G) ∈ {1, 2}. If deg! (G) = 1 then G ∈ !, which is an iteratedquadratic extension of Σ. If deg! (G) = 2, i.e. [! (G) : !] = 2, then ! (G) is aniterated quadratic extension of Σ. In either case, G lies in some iterated quadraticextension of Σ. The same is true of H. Hence by Lemma 5.4.4, there is an iteratedquadratic extension of Σ containing G and H. This completes the induction. �

Now we solve the problems of ancient Greece. In all cases, we take Σ =

{(0, 0), (1, 0)}. Then Σ = Q.Proposition 5.4.6 The angle cannot be trisected by ruler and compass.

Proof Suppose it can be. Construct an equilateral triangle with (0, 0) and (1, 0)as two of its vertices (which can be done by ruler and compass; Figure 5.3). Trisectthe angle of the triangle at (0, 0). Plot the point (G, H) where the trisector meets thecircle with centre (0, 0) through (1, 0). Then G = cos(c/9), so by Corollary 5.4.3,degQ(cos(c/9)) is a power of 2.

Now we use the trigonometric formula

cos 3\ = 4(cos \)3 − 3 cos \.

Taking \ = c/9 and using cos(c/3) = 1/2, we get 8G3 − 6G − 1 = 0. Reduced mod5, this cubic has no roots and is therefore irreducible (by Lemma 3.3.1(iv)). So bythe mod ? test, 8C3 − 6C − 1 is irreducible over Q. Hence (8C3 − 6C − 1)/8 is theminimal polynomial of G over Q, giving degQ(G) = 3. Since 3 is not a power of 2,this is a contradiction. �

64

(0, 0) (1, 0)

(G, H)

Figure 5.3: The impossibility of trisecting 60◦.

Proposition 5.4.7 The cube cannot be duplicated by ruler and compass.

Proof Suppose it can be. Since (0, 0) and (1, 0) are distance 1 apart, we canconstruct from them two points � and � distance 3√2 apart. From � and � we canconstruct, using ruler and compass, the point ( 3√2, 0). (The video shows how to‘transport distances’ like this.) So degQ(

3√2) is a power of 2, by Corollary 5.4.3.But degQ(

3√2) = 3 by Example 5.1.8(ii), a contradiction. �

Proposition 5.4.8 The circle cannot be squared by ruler and compass.

This one is the most outrageously false, yet the hardest to prove.

Proof Suppose it can be. Since the circle with centre (0, 0) through (1, 0) hasarea c, we can construct by ruler and compass a square with side-length

√c,

and from that, we can construct by ruler and compass the point (√c, 0). So by

Corollary 5.4.3,√c is algebraic over Q with degree a power of 2. Since Q is a

subfield of C, it follows that c is algebraic over Q. But it is a (hard) theorem thatc is transcendental over Q. �

Digression 5.4.9 Stewart has a nice alternative approach to all this, in hisChapter 7. He treats the plane as the complex plane, and he shows that theset of all points in C constructible from 0 and 1 is a subfield. In fact, it is thesmallest subfield of C closed under taking square roots. He calls it Qpy, the‘Pythagorean closure’ of Q. It can also be described as the set of complexnumbers contained in some iterated quadratic extension of Q.

There is one more famous ruler and compass problem: for which integers = isthe regular =-sided polygon constructible, starting from just a pair of points in theplane?

The answer has to do with Fermat primes, which are prime numbers of theform 2D + 1 for some D ≥ 1. A little exercise in number theory shows that if 2D + 1is prime then D must itself be a power of 2. The only known Fermat primes are

220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537.

65

Whether there are any others is a longstanding open question. In any case, it canbe shown that the regular =-sided polygon is constructible if and only if

= = 2A ?1 · · · ?:

for some A, : ≥ 0 and distinct Fermat primes ?1, . . . , ?: .We will not do the proof, but it involves cyclotomic polynomials. A glimpse

of the connection: let ? be a prime such that the regular ?-sided polygon isconstructible. Consider the regular ?-sided polygon inscribed in the unit circlein C, with one of its vertices at 1. Then another vertex is at 42c8/?, and fromconstructibility, one can deduce that degQ(42c8/?) is a power of 2. But we saw inExample 3.3.16 that degQ(42c8/?) = ? − 1. So ? − 1 is a power of 2, that is, ?is a Fermat prime. Galois theory, number theory and Euclidean geometry cometogether!

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Chapter 6

Splitting fields

In Chapter 1, we met a definition of the symmetry group of a polynomial overQ. It was phrased in terms of indistinguishable tuples, it was possibly a littlemysterious, and it was definitely difficult to work with (e.g. we couldn’t computethe symmetry group of 1 + C + C2 + C3 + C4).

Introduction toWeek 6

In this chapter, we’re going to give a different but equivalent definition of thesymmetry group of a polynomial. It’s a two-step process:

1. We show how every polynomial 5 over gives rise to an extension of ,called the ‘splitting field’ of 5 .

2. We show how every field extension has a symmetry group.

The symmetry group, or ‘Galois group’, of a polynomial is then defined to be thesymmetry group of its splitting field extension.

How does these two steps work?

1. When = Q, the splitting field of 5 is the smallest subfield of C containingall the complex roots of 5 . For a general field , it’s constructed by addingthe roots of 5 one at a time, using simple extensions, until we obtain anextension of in which 5 splits into linear factors.

2. The symmetry group of a field extension " : is defined as the group ofautomorphisms of " over . This is the same idea you’ve seen many timesbefore, for symmetry groups of other mathematical objects.

Why bother? Why not define the symmetry group of 5 directly, as in Chapter 1?

• Because this strategy works over every field , not just Q.

• Because there are field extensions that do not arise from a polynomial, andtheir symmetry groups are sometimes important. For example, an important

67

structure in number theory, somewhatmysterious to this day, is the symmetrygroup of the algebraic numbers Q over Q.

• Because using abstract algebra means you can cut down on explicit calcu-lations with polynomials. (By way of analogy, you’ve seen how abstractlinear algebra with vector spaces and linear maps allows you to cut down oncalculations with matrices.) It also makes connections with other parts ofmathematics more apparent.

6.1 Extending homomorphismsIn your degree so far, you’ll have picked up the general principle that for manykinds of mathematical object (such as groups, rings, fields, vector spaces, modules,metric spaces, topological spaces, measure spaces, . . . ), it’s important to considerthe appropriate notion of mapping between them (such as homomorphisms, linearmaps, continuousmaps, . . . ). And since Chapter 4, you know that the basic objectsof Galois theory are field extensions.

So it’s no surprise that sooner or later, we’ll have to think about mappings fromone field extension to another. That moment is now: we’ll need what’s in thissection in order to establish the basic facts about splitting fields.

When we think about a field extension " : , we generally regard the field as our starting point and " as a field that extends it. (This doesn’t meananything rigorous.) Similarly, we might start with a homomorphism k : → ′

between fields, together with extensions " of and "′ of ′, and look for ahomomorphism " → "′ that extends k. The language is as follows.

Definition 6.1.1 Let ] : → " and ]′ : ′ → "′ be field extensions. Letk : → ′ be a homomorphism of fields. A homomorphism i : " → "′

extends k if the squareExtension problems "

i // "′

]

OO

k// ′

]′

OO

commutes (that is, i ◦ ] = ]′ ◦ k).

Here I’ve used the definition of a field extension as a homomorphism ] of fields(Definition 4.1.1). Most of the time we view as a subset of " and ′ as a subsetof "′, with ] and ]′ being the inclusions. In that case, for i to extend k just meansthat

i(0) = k(0) for all 0 ∈ .

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The basic questions are: given the two field extensions and the homomorphism k,is there some i that extends k? If so, how many?

We’ll consider these questions later. This section simply gathers together threegeneral results about extensions of field homomorphisms.

Recall that any ring homomorphism k : ' → ( induces a homomorphismk∗ : '[C] → ([C] (Definition 3.1.6).

Lemma 6.1.2 Let " : and "′ : ′ be field extensions, let k : → ′ bea homomorphism, and let i : " → "′ be a homomorphism extending k. Let

Explanation ofLemma 6.1.2

U ∈ " and 5 (C) ∈ [C]. Then

5 (U) = 0 ⇐⇒(k∗( 5 )

) (i(U)

)= 0.

Proof Write 5 (C) = ∑8 08C

8, where 08 ∈ . Then (k∗( 5 )) (C) =∑8 k(08)C8 ∈

′[C], so

(k∗( 5 )) (i(U)) =∑8

k(08)i(U)8 =∑8

i(08)i(U)8 = i( 5 (U)),

where the second equality holds because i extends k. Since i is injective(Lemma 2.2.2), the result follows. �

Exercise 6.1.3 Show that if k is injective then so is k∗, and if k is anisomorphism then so is k∗.

Lemma 6.1.4 Let " : and "′ : ′ be field extensions, let k : → ′ be anisomorphism, and let i : " → "′ be a homomorphism extending k. Let U ∈ "be algebraic over with minimal polynomial <. Then i(U) is algebraic over ′with minimal polynomial k∗(<).

Proof By Lemma 6.1.2, k∗(<) is an annihilating polynomial of i(U) over ′.Also k∗(<) ≠ 0, since < ≠ 0 and k∗ is injective. So i(U) is algebraic over ′.(Recall that an element is algebraic if it has a nonzero annihilating polynomial.)

Since k∗ : [C] → ′[C] is an isomorphism and < ∈ [C] is irreducible,k∗(<) ∈ ′[C] is irreducible. It is also monic. Hence k∗(<) is a monic irreducibleannihilating polynomial of i(U), so it is the minimal polynomial of i(U). �

An isomorphism between fields, rings, groups, vector spaces, etc., can beunderstood as simply a renaming of the elements. For example, if I tell you thatthe ring ' is left Noetherian but not right Artinian, and that ( is isomorphic to ',then you can deduce that ( is left Noetherian but not right Artinian without havingthe slightest idea what those terms mean. Just as long as they don’t depend on thenames of the elements of the ring concerned (which such definitions never do),you’re fine.

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Proposition 6.1.5 Let k : → ′ be an isomorphism of fields, let (U) : be asimple extension where U has minimal polynomial < over , and let ′(U′) : ′be a simple extension where U′ has minimal polynomial k∗(<) over ′. Thenthere is a unique isomorphism i : (U) → ′(U′) that extends k and satisfiesi(U) = U′.

Diagram: (U) i

�// ′(U′)

OO

k

� // ′

OO

We often use a dotted arrow to denote a map whose existence is part of theconclusion of a theorem, rather than a hypothesis.

Proof View ′(U′) as an extension of via the composite homomorphism k−→

′ → ′(U′). Then the minimal polynomial of U′ over is <. (If this isn’tintuitively clear to you, think of the isomorphism k as renaming.) Hence bythe classification of simple extensions, Theorem 4.3.7, there is an isomorphismi : (U) → ′(U′) over such that i(U) = U′. Then i extends k.

It only remains to prove uniqueness. Let i be any homomorphism (U) → ′(U′) that extends k and satisfies i(U) = U′. Then i(0) = k(0) = i(0) for all0 ∈ and i(U) = U′ = i(U). Since every element of (U) is a polynomial in Uwith coefficients in , it follows that i = i. �

6.2 Existence and uniqueness of splitting fieldsLet 5 be a polynomial over a field . Informally, a splitting field for 5 is anextension of where 5 has all its roots, and which is no bigger than it needs to be.

Warning 6.2.1 If 5 is irreducible, we know how to create an extensionof where 5 has at least one root: take the simple extension [C]/〈 5 〉,in which the equivalence class of C is a root of 5 (Lemma 4.3.4).But [C]/〈 5 〉 is not usually a splitting field for 5 . For example, take = Q and 5 (C) = C3 − 2, as in Warning 4.3.10. Write U for the realcube root of 2. (Half the counterexamples in Galois theory involve thereal cube root of 2.) ThenQ[C]/〈 5 〉 is isomorphic to the subfieldQ(U)of R, which only contains one root of 5 : the other two are non-real,hence not in Q(U).

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Definition 6.2.2 Let 5 be a polynomial over a field " . Then 5 splits in " if

5 (C) = V(C − U1) · · · (C − U=)

for some = ≥ 0 and V, U1, . . . , U= ∈ " .

Equivalently, 5 splits in " if all its irreducible factors in " [C] are linear.

Examples 6.2.3 i. A field " is algebraically closed if and only if every poly-nomial over " splits in " .

ii. Let 5 (C) = C4 − 4C2 − 5. Then 5 splits in Q(8,√

5), since

5 (C) = (C2 + 1) (C2 − 5)= (C − 8) (C + 8) (C −

√5) (C +

√5).

But 5 does not split in Q(8), as its factorization into irreducibles in Q(8) [C]is

5 (C) = (C − 8) (C + 8) (C2 − 5),which contains a nonlinear factor. Moral: 5 may have one root or evenseveral roots in " , but still not split in " .

iii. Let" = F2(U), where U is a root of 5 (C) = 1+ C+ C2, as in Example 4.3.8(ii).We have

5 (1 + U) = 1 + (1 + U) + (1 + 2U + U2) = 1 + U + U2 = 0,

so 5 has two distinct roots in " , giving

5 (C) = (C − U) (C − (1 + U))

in " [C]. Hence 5 splits in " .In this example, adjoining one root of 5 gave us a second root for free. Butthis doesn’t typically happen (Warning 6.2.1).

Definition 6.2.4 Let 5 be a polynomial over a field . A splitting field of 5 over is an extension " of such that:

i. 5 splits in ";

ii. " = (U1, . . . , U=), where U1, . . . , U= are the roots of 5 in " .

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Exercise 6.2.5 Show that (ii) can equivalently be replaced by: ‘if !is a subfield of " containing , and 5 splits in !, then ! = "’.

Examples 6.2.6 i. Let 5 ∈ Q[C]. Write U1, . . . , U= for the complex roots of5 . Then Q(U1, . . . , U=) (the smallest subfield of C containing U1, . . . , U=) isa splitting field of 5 over Q.

ii. Let 5 (C) = C3 − 2 ∈ Q[C]. Its complex roots are U, lU and l2U, where U isthe real cube root of 2 and l = 42c8/3. Hence a splitting field of 5 over Q is

Q(U, lU, l2U) = Q(U, l).

Now degQ(U) = 3 as 5 is irreducible, and degQ(l) = 2 as l has minimalpolynomial 1 + C + C2. By an argument like that in Example 5.2.7, it followsthat [Q(U, l) : Q] = 6. On the other hand, [Q(U) : Q] = 3. So again, theextension we get by adjoining all the roots of 5 is bigger than the one we getby adjoining just one root of 5 .

iii. Take 5 (C) = 1 + C + C2 ∈ F2 [C], as in Example 6.2.3(iii). By Theorem 5.1.5,{1, U} is a basis of F2(U) over F2, so

F2(U) = {0, 1, U, 1 + U}= F2 ∪ {the roots of 5 in F2(U)}.

Hence F2(U) is a splitting field of 5 over F2.

Exercise 6.2.7 In Example 6.2.6(ii), I said that Q(U, lU, l2U) =Q(U, l). Why is that true?

Digression 6.2.8 Splitting fields overQ are easy, as Example 6.2.6(i) shows.That’s because we have a ready-made algebraically closed field containingQ, namely, C.

In fact, for any field , it’s possible to build an algebraically closed fieldcontaining . And there is a unique ‘smallest’ algebraically closed fieldcontaining , called its algebraic closure . For example, the algebraicclosure ofQ isQ, the subfield of C consisting of the algebraic numbers. (Fora proof that Q is indeed algebraically closed, see Workshop 3.)

Many texts on Galois theory include constructions of the algebraic closureof a field, but we won’t do it.

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Our mission for the rest of this section is to show that every polynomial 5 hasexactly one splitting field. So that’s actually two tasks: first, show that 5 has atleast one splitting field, then, show that 5 has only one splitting field. The firsttask is easy, and in fact we prove a little bit more:

Lemma 6.2.9 Let 5 be a polynomial over a field . Then there exists a splittingfield " of 5 over such that [" : ] ≤ deg( 5 )!.

(So that this lemma holds for 5 = 0, we had better define (−∞)! = 1.)

Proof If 5 is constant then is a splitting field of 5 over , and the result holdstrivially.

Now suppose inductively that deg( 5 ) ≥ 1. We may choose an irreduciblefactor < of 5 . By Theorem 4.3.7, there is an extension (U) of with <(U) = 0.Then (C − U) | 5 (C) in (U) [C], giving a polynomial 6(C) = 5 (C)/(C − U) over (U).

We have deg(6) = deg( 5 ) − 1, so by inductive hypothesis, there is a splittingfield " of 6 over (U) with [" : (U)] ≤ deg(6)!. Then " is a splitting field of5 over . (Check that you understand why.) Also, by the tower law,

[" : ] = [" : (U)] [ (U) : ] ≤ (deg( 5 ) − 1)! · deg(<) ≤ deg( 5 )!,

completing the induction. �

Proving that every polynomial has only one splitting field is harder. As ever,‘only one’ has to be understood up to isomorphism: after all, if you’re given asplitting field, you can always rename its elements to get an isomorphic copy that’snot literally identical to the original one. But isomorphism is all that matters.

Our proof of the uniqueness of splitting fields depends on the following result,which will also be useful for other purposes as we head towards the fundamentaltheorem of Galois theory.

Proposition 6.2.10 Let k : → ′ be an isomorphism of fields, let 5 ∈ [C], let" be a splitting field of 5 over , and let "′ be a splitting field of k∗( 5 ) over ′.Then:

i. there exists an isomorphism i : " → "′ extending k;

ii. there are at most [" : ] such extensions i.

Diagram:"

i // "′

OO

k// ′

OO

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Proof We prove both statements by induction on deg( 5 ). If 5 is constant thenboth field extensions are trivial, so there is exactly one isomorphism i extendingk.

Now suppose that deg( 5 ) ≥ 1. We can choose a monic irreducible factor < of5 . Then< splits in " since 5 does and< | 5 ; choose a root U ∈ " of<. We have5 (U) = 0, so (C − U) | 5 (C) in (U) [C], giving a polynomial 6(C) = 5 (C)/(C − U)over (U). Then " is a splitting field of 6 over (U), and deg(6) = deg( 5 ) − 1.

Also, k∗(<) splits in "′ since k∗( 5 ) does and k∗(<) | k∗( 5 ). WriteU′1, . . . , U

′B for the distinct roots of k∗(<) in "′. Note that

1 ≤ B ≤ deg(k∗(<)) = deg(<). (6.1)

Since k∗ is an isomorphism, k∗(<) is monic and irreducible, and is therefore theCountingisomorphisms: the

proof ofProposition 6.2.10

minimal polynomial of U′9for each 9 ∈ {1, . . . , B}. Hence by Proposition 6.1.5,

for each 9 , there is a unique isomorphism \ 9 : (U) → ′(U′9) that extends k and

satisfies \ 9 (U) = U′9 . (See diagram below.)For each 9 ∈ {1, . . . , B}, we have a polynomial

\ 9∗(6) =\ 9∗( 5 )

\ 9∗(C − U)=k∗( 5 )C − U′

9

over ′(U′9), and "′ is a splitting field of k∗( 5 ) over ′, so "′ is also a splitting

field of \ 9∗(6) over ′(U′9 ).To prove that there is at least one isomorphism i extending k, choose any

9 ∈ {1, . . . , B} (as we may since B ≥ 1). By applying the inductive hypothesis to6 and \ 9 , there is an isomorphism i extending \ 9 :

"i // "′

(U)

OO

\ 9 // ′(U′9)

OO

OO

k// ′

OO

But then i also extends k, as required.To prove there are at most [" : ] isomorphisms i : " → "′ extending

k, first note that any such i satisfies (k∗( 5 )) (i(U)) = 0 (by Lemma 6.1.2), soi(U) = U′

9for some 9 ∈ {1, . . . , B}. Every element of (U) is a polynomial

in U over , and i(0) = k(0) ∈ ′ for all 0 ∈ , so i maps (U) into ′(U′

9). Now i( (U)) contains ′ (since k is an isomorphism) and U′

9, so

i( (U)) = ′(U′9). Since homomorphisms of fields are injective, i restricts to

an isomorphism (U) → ′(U′9) satisfying U ↦→ U′

9. By the uniqueness part of

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Proposition 6.1.5, this restricted isomorphism must be \ 9 . Thus, i extends \ 9 fora unique 9 ∈ {1, . . . , B}, giving

(number of isos i extending k) =B∑9=1(number of isos i extending \ 9 ).

For each 9 , the number of isomorphisms i extending \ 9 is ≤ [" : (U)], byinductive hypothesis. So, using the tower law and (6.1),

(number of isos i extending k) ≤ B · [" : (U)] = B · [" : ]deg(<) ≤ [" : ],

completing the induction. �

Exercise 6.2.11 Why does the proof of Proposition 6.2.10 not showthat there are exactly [" : ] isomorphisms i extending k? Howcould you strengthen the hypotheses in order to obtain that conclusion?(The second question is a bit harder, and we’ll see the answer nextweek.)

This brings us to the foundational result on splitting fields. Recall that anautomorphism of an object - is an isomorphism - → - .

Theorem 6.2.12 Let 5 be a polynomial over a field . Then:

i. there exists a splitting field of 5 over ;

ii. any two splitting fields of 5 are isomorphic over ;

iii. when " is a splitting field of 5 over ,

(number of automorphisms of " over ) ≤ [" : ] ≤ deg( 5 )!.

Proof Part (i) is immediate from Lemma 6.2.9, and part (ii) follows from Propo-sition 6.2.10 by taking ′ = and k = id . The first inequality in (iii) followsfrom Proposition 6.2.10 by taking ′ = , "′ = " and k = id , and the secondfollows from Lemma 6.2.9. �

Up to now we have been saying ‘a’ splitting field. Theorem 6.2.12 gives us theright to speak of the splitting field of a given polynomial 5 over a given field .We write it as SF ( 5 ).

We finish with a left over lemma that will be useful later.

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Lemma 6.2.13 i. Let " : ( : be field extensions, 5 ∈ [C], and . ⊆ " .Suppose that ( is the splitting field of 5 over . Then ((. ) is the splittingfield of 5 over (. ).

ii. Let 5 be a polynomial over a field , and let ! be a subfield of SF ( 5 )containing (so that SF ( 5 ) : ! : ). Then SF ( 5 ) is the splitting field of5 over !.

Proof For (i), 5 splits in (, hence in ((. ). Writing - for the set of roots of 5in (, we have ( = (-) and so ((. ) = (-) (. ) = (- ∪ . ) = (. ) (-); thatis, ((. ) is generated over (. ) by - . This proves (i), and (ii) follows by taking" = SF ( 5 ) and . = !. �

6.3 The Galois groupWhat gives Galois theory its special flavour is the use of groups to study fields andpolynomials. Here is the central definition.

Definition 6.3.1 The Galois group Gal(" : ) of a field extension is the groupof automorphisms of " over , with composition as the group operation.

Exercise 6.3.2 Check that this really does define a group. You’ll needthe result of Exercise 4.3.6, for instance.

In other words, an element of Gal(" : ) is an isomorphism \ : " → " suchthat \ (0) = 0 for all 0 ∈ .

Examples 6.3.3 i. What is Gal(C : R)? Certainly the identity is an auto-morphism of C over R. So is complex conjugation ^, as implicitly shownin the first proof of Lemma 1.1.3. So {id, ^} ⊆ Gal(C : R). I claim thatGal(C : R) has no other elements. For let \ ∈ Gal(C : R). Then

(\ (8))2 = \ (82) = \ (−1) = −\ (1) = −1

as \ is a homomorphism, so \ (8) = ±8. If \ (8) = 8 then

\ (0 + 18) = \ (0) + \ (1)\ (8) = 0 + 18

for all 0, 1 ∈ R (since \ is an automorphism overR), giving \ = id. Similarly,if \ (8) = −8 then \ = ^. So Gal(C : R) = {id, ^} � �2.

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ii. Let U be the real cube root of 2. For each \ ∈ Gal(Q(U) : Q), we have

(\ (U))3 = \ (U3) = \ (2) = 2

and \ (U) ∈ Q(U) ⊆ R, so \ (U) = U. Every element of Q(U) can beexpressed as a polynomial in U overQ (by Theorem 5.1.5), so \ = id. HenceGal(Q(U) : Q) is trivial.

Exercise 6.3.4 Prove that Gal(Q(42c8/3) : Q) = {id, ^}, where ^(I) =I. (Hint: imitate Example 6.3.3(i).)

The Galois group of a polynomial is defined to be the Galois group of itssplitting field extension:

Definition 6.3.5 Let 5 be a polynomial over a field . TheGalois group Gal ( 5 )of 5 over is Gal(SF ( 5 ) : ).

So the definitions fit together like this:

polynomial ↦−→ field extension ↦−→ group.

We will soon prove that Definition 6.3.5 is equivalent to the definition of Galoisgroup in Chapter 1, where we went straight from polynomials to groups.

Theorem 6.2.12(iii) says that

|Gal ( 5 ) | ≤ [SF ( 5 ) : ] ≤ deg( 5 )!. (6.2)

In particular, Gal ( 5 ) is always a finite group.

Examples 6.3.6 i. GalQ(C2 + 1) = Gal(Q(8) : Q) = {id, ^} � �2, where ^ iscomplex conjugation on Q(8). The second equality is proved by the sameargument as in Example 6.3.3(i), replacing C : R by Q(8) : Q.

Calculating theGalois group withbare hands, part 1

Calculating theGalois group withbare hands, part 2

ii. Let 5 (C) = (C2 + 1) (C2 − 2). Then GalQ( 5 ) is the group of automorphismsof Q(8,

√2) over Q. Similar arguments to those in Examples 6.3.3 show that

every \ ∈ GalQ( 5 ) must satisfy \ (8) = ±8 and \ (√

2) = ±√

2, and that thetwo choices of sign determine \ completely. And one can show that all fourchoices are possible, so that |GalQ( 5 ) | = 4. There are two groups of orderfour, �4 and �2 × �2. But each element of GalQ( 5 ) has order 1 or 2, soGalQ( 5 ) is not �4, so GalQ( 5 ) � �2 × �2.I’ve been sketchy with the details here, because it’s not really sensible to tryto calculate Galois groups until we have a few more tools at our disposal.We start to assemble them now.

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Figure 6.1: The Galois group � = Gal ( 5 ) permutes the roots U8 of 5 . (Imageadapted from @rowvector.)

In the examples so far, we’ve seen that if U is a root of 5 then so is \ (U) forevery \ ∈ Gal ( 5 ). This is true in general, and is best expressed in terms of groupactions (Figure 6.1). In a slogan: the Galois group permutes the roots.

Lemma 6.3.7 Let 5 be a polynomial over a field . Let - be the set of roots of 5in SF ( 5 ). Then there is an action of Gal ( 5 ) on - defined by

Gal ( 5 ) × - → -

(\, U) ↦→ \ (U). (6.3)

Proof First, if \ ∈ Gal ( 5 ) and U ∈ - then \ (U) ∈ - , by Lemma 6.1.2 with = ′, " = "′ and k = id. For the function (6.3) to be an action means that(i ◦ \) (U) = i(\ (U)) and id(U) = U for all i, \ ∈ Gal ( 5 ) and U ∈ - , which aretrue by definition. �

The action of theGalois group

An action of a group � on a set - is essentially the same thing as a homomor-phism from � to the group Sym(-) of bijections from - to - . (If we write 6G asf6 (G), then the homomorphism � → Sym(-) is 6 ↦→ f6.) We now adopt thisviewpoint in the case of Galois groups.

Let 5 be a polynomial over a field , and write U1, . . . , U: for the distinct rootsof 5 in SF ( 5 ). For each 8 ∈ {1, . . . , :}, Lemma 6.3.7 implies that \ (U8) = U 9 fora unique 9 . Write 9 as f\ (8), so that

\ (U8) = Uf\ (8) .

Then we have a function

Γ : Gal ( 5 ) → (:\ ↦→ f\ ,

(6.4)

and it is straightforward to check that Γ is a homomorphism.

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Exercise 6.3.8 What is the kernel of Γ, in concrete terms?

If you remember the definition of Galois group in Chapter 1 (Definition 1.2.1),the mention of (: should have set your antennae tingling. There, we defined theGalois group as a certain subgroup of (: , namely, the one consisting of thosepermutations f for which the tuples

(U1, . . . , U: ), (Uf(1) , . . . , Uf(:))

are indistinguishable. Let’s nowmake the definition of indistinguishability official,switch to the standard name (recall Warning 1.1.2), and generalize from Q to anarbitrary field.

Definition 6.3.9 Let" : be a field extension, let : ≥ 0, and let (U1, . . . , U: ) and(U′1, . . . , U

′:) be :-tuples of elements of " . Then (U1, . . . , U: ) and (U′1, . . . , U

′:)

are conjugate over if for all ? ∈ [C1, . . . , C: ],

?(U1, . . . , U: ) = 0 ⇐⇒ ?(U′1, . . . , U′: ) = 0.

In the case : = 1, we omit the brackets and say that U and U′ are conjugate to meanthat (U) and (U′) are.

We show now that the two definitions of the Galois group of 5 are equivalent.

Proposition 6.3.10 Let 5 be a polynomial over a field , with distinct rootsU1, . . . , U: in SF ( 5 ). Define the group homomorphism Γ : Gal ( 5 ) → (: asin (6.4). Then Γ is injective, and its image is

{f ∈ (: : (U1, . . . , U: ) and (Uf(1) , . . . , Uf(:)) are conjugate over }. (6.5)

In particular, (6.5) is a subgroup of (: isomorphic to Gal ( 5 ).

Proof To show that Γ is injective, let \ ∈ kerΓ. Then \ (U8) = U8 for all 8.Now SF ( 5 ) = (U1, . . . , U: ), with each U8 algebraic over , so every elementof SF ( 5 ) can be expressed as a polynomial in U1, . . . , U: over (by Corol-lary 5.1.11). Since \ : SF ( 5 ) → SF ( 5 ) fixes each element of and each U8, itis the identity on all of SF ( 5 ). Thus, kerΓ is trivial, so Γ is injective.

Now we prove that imΓ is the set (6.5). In one direction, let f ∈ imΓ.Then f = f\ for some \ ∈ Gal ( 5 ) (writing Γ(\) = f\ , as before). For every? ∈ [C1, . . . , C: ],

?(Uf\ (1) , . . . , Uf\ (:)) = ?(\ (U1), . . . , \ (U: )) = \ (?(U1, . . . , U: )),

79

where the first equality is by definition of f\ and the second is because \ is ahomomorphism over . But \ is an isomorphism, so it follows that

?(Uf\ (1) , . . . , Uf\ (:)) = 0 ⇐⇒ ?(U1, . . . , U: ) = 0.

Hence f belongs to the set (6.5).In the other direction, let f be a permutation in (6.5). By Corollary 5.1.11,

every element of SF ( 5 ) can be expressed as ?(U1, . . . , U: ) for some ? ∈ [C1, . . . , C: ]. Now for ?, @ ∈ [C1, . . . , C: ], we have

?(U1, . . . , U: ) = @(U1, . . . , U: ) ⇐⇒ ?(Uf(1) , . . . , Uf(:)) = @(Uf(1) , . . . , Uf(:))

(by applying Definition 6.3.9 of conjugacy with ? − @ as the ‘?’). So there is awell-defined, injective function \ : SF ( 5 ) → SF ( 5 ) satisfying

\ (?(U1, . . . , U: )) = ?(Uf(1) , . . . , Uf(:)) (6.6)

for all ? ∈ [C1, . . . , C: ]. Moreover, \ is surjective because f is a permutation,\ (0) = 0 for all 0 ∈ (by taking ? = 0 in (6.6)), and \ (U8) = Uf(8) for all 8 (bytaking ? = C8 in (6.6)). You can check that \ is a homomorphism of fields. Hence\ ∈ Gal ( 5 ) with Γ(\) = f, proving that f ∈ imΓ.

The final sentence of the proposition follows because every injective grouphomomorphism W : � → � induces an isomorphism between� and the subgroupim W of �. �

Exercise 6.3.11 I skipped two small bits in that proof: ‘\ is surjectivebecause f is a permutation’ (why?), and ‘You can check that \ is ahomomorphism of fields’. Fill the gaps.

Digression 6.3.12 As you may know, an action of a group � on a set -is called faithful if the corresponding homomorphism � → Sym(-) isinjective. A more concrete way to say that is that the only element 6 ∈ �that fixes everything (6G = G for all G ∈ -) is the identity. Equivalently, if6, ℎ ∈ � and 6G = ℎG for all G then 6 = ℎ. Most actions that one meets inpractice are faithful; those that aren’t involve a kind of redundancy.

It’s important in Galois theory to be able to move easily between fields. Forexample, you might start with a polynomial whose coefficients belong to one field , but later decide to consider the coefficients as belonging to some larger field !.Here’s what happens to the Galois group when you do that.

Corollary 6.3.13 Let ! : be a field extension and 5 ∈ [C]. Then Gal! ( 5 )embeds naturally as a subgroup of Gal ( 5 ).

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The phrasing here is slightly vague: it means there is an injective homomor-phismGal! ( 5 ) → Gal ( 5 ), and there is such an obvious choice of homomorphismthat we tend to regard Gal! ( 5 ) as being a subgroup of Gal ( 5 ).

Proof This follows from Proposition 6.3.10 together with the observation that iftwo :-tuples are conjugate over !, they are conjugate over . �

Example 6.3.14 Consider the Galois group of 5 (C) = (C2 + 1) (C2 − 2) over Q, Rand C. In Example 6.3.6(ii), we saw that GalQ( 5 ) � �2 × �2.

Over R, the Galois group of 5 is the same as that of C2 + 1, since bothroots of C2 − 2 lie in R and are therefore preserved by elements of GalR( 5 ). SoGalR( 5 ) = GalR(C2 + 1) � �2.

Finally, GalC( 5 ) is trivial. Indeed, every polynomial 6 ∈ C[C] has trivialGalois group over C: for 6 splits in C, so SFC(6) = C, so SFC(6) : C is a trivialextension and so has trivial Galois group.

Corollary 6.3.15 Let 5 be a polynomial over a field , with : distinct roots inSF ( 5 ). Then |Gal ( 5 ) | divides :!.

Proof By Proposition 6.3.10, Gal ( 5 ) is isomorphic to a subgroup of (: , whichhas :! elements. The result follows from Lagrange’s theorem. �

The inequalities (6.2) already gave us |Gal ( 5 ) | ≤ deg( 5 )!. Corollary 6.3.15improves on this in two respects. First, it gives us |Gal ( 5 ) | ≤ :!, where: ≤ deg( 5 ) in all cases, and : < deg( 5 ) if 5 has repeated roots in its splittingfield. A trivial example: if 5 (C) = C2 then : = 1 and deg( 5 ) = 2. Second, it tellsus that |Gal ( 5 ) | is not only less than or equal to :!, but a factor of it.

Galois theory is about the interplay between field extensions and groups.In the next chapter, we’ll see that just as every field extension giving rise toa group of automorphisms (its Galois group), every group of automorphismsgives rise to a field extension. We’ll also go deeper into the different types offield extension: normal extensions (the mirror image of normal subgroups) andseparable extensions (which have to do with repeated roots). All that will lead ustowards the fundamental theorem of Galois theory.

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Chapter 7

Preparation for the fundamentaltheorem

Very roughly, the fundamental theorem of Galois theory says that you can tell alot about a field extension by looking at its Galois group. A bit more specifically,

Introduction toWeek 7

it says that the subgroups and quotients of Gal(" : ), and their orders, giveus information about the subfields of " containing , and their degrees. Forexample, one part of the fundamental theorem is that

[" : ] = |Gal(" : ) |.

The theorem doesn’t hold for all extensions, just those that are ‘nice enough’.Crucially, this includes splitting field extensions SFQ( 5 ) : Q of polynomials 5over Q—the starting point of classical Galois theory.

Let’s dip our toes into the water by thinking about why it might be true that[" : ] = |Gal(" : ) |, at least for extensions that are nice enough.

The simplest nontrivial extensions are the simple algebraic extensions, " =

(U). Write < for the minimal polynomial of U over and U1, U2, . . . , UB forthe distinct roots of < in " . For every element i of Gal(" : ), we have<(i(U)) = 0 by Lemma 6.1.2, and so i(U) = U 9 for some 9 ∈ {1, . . . , B}. On theother hand, for each 9 ∈ {1, . . . , B}, there is exactly one i ∈ Gal(" : ) such thati(U) = U 9 , by Proposition 6.1.5. So |Gal(" : ) | = B.

We know that [" : ] = deg(<). So [" : ] = |Gal(" : ) | if and only ifdeg(<) is equal to B, the number of distinct roots of< in" . Certainly B ≤ deg(<).But are B and deg(<) equal?

There are two reasons why they might not be. First, < might not split in " .For instance, if = Q and U = 3√2 then <(C) = C3 − 2, which has only one root inQ( 3√2), so |Gal(Q( 3√2) : Q) | = 1 < 3 = deg(<). An algebraic extension is called‘normal’ if this problem doesn’t occur, that is, if the minimal polynomial of everyelement does split. That’s what Section 7.1 is about.

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Second, we might have B < deg(<) because some of the roots of < in " arerepeated. If they are, the number B of distinct roots will be less then deg(<). Analgebraic extension is called ‘separable’ if this problem doesn’t occur, that is, if theminimal polynomial of every element has no repeated roots in its splitting field.That’s what Section 7.2 is about.

If we take any finite extension " : (not necessarily simple) that is bothnormal and separable, then it is indeed true that |Gal(" : ) | = [" : ]. And infact, these conditions are enough to make the whole fundamental theorem work,as we’ll see next week.

I hesitated before putting normality and separability into the same chapter,because you should think of them in quite different ways:

• Normality has a clear conceptual meaning (as I explain in a video), and itsimportance was recognized by Galois himself. Despite the name, most fieldextensions aren’t normal; normality isn’t something to be taken for granted.

• In contrast, Galois never considered separability, because it holds automat-ically over Q (his focus), and in fact over any field of characteristic 0, aswell as any finite field. It takes some work to find an extension that isn’tseparable. You can view separability as more of a technicality.

There’s one more concept in this chapter: the ‘fixed field’ of a group ofautomorphisms (Section 7.3). Every Galois theory text I’ve seen contains at leastone proof that makes you ask ‘how did anyone think of that?’ I would argue thatthe proof of Theorem 7.3.6 is the one and only truly ingenious argument in thiscourse: maybe not the hardest, but the most ingenious. Your opinion may differ!

7.1 NormalityDefinition 7.1.1 An algebraic field extension " : is normal if for all U ∈ " ,the minimal polynomial of U splits in " .

We also say " is normal over to mean that " : is normal.

Lemma 7.1.2 Let " : be an algebraic extension. Then " : is normal if andonly if every irreducible polynomial over either has no roots in " or splits in " .

Put another way, normality means that any irreducible polynomial over withat least one root in " has all its roots in " .

Proof Suppose that " : is normal, and let 5 be an irreducible polynomial over . If 5 has a root U in " then the minimal polynomial of U is 5 /2, where 2 ∈

83

is the leading coefficient of 5 . Since " : is normal, 5 /2 splits in " , so 5 doestoo.

Conversely, suppose that every irreducible polynomial over either has noroots in " or splits in " . Let U ∈ " . Then the minimal polynomial of U has atleast one root in " (namely, U), so it splits in " . �

Examples 7.1.3 i. Consider Q( 3√2) : Q. The minimal polynomial of 3√2 overQ is C3 − 2, whose roots in C are 3√2 ∈ R and l 3√2, l2 3√2 ∈ C \ R, wherel = 42c8/3. Since Q( 3√2) ⊆ R, the minimal polynomial C3 − 2 does not splitin Q( 3√2). Hence Q( 3√2) is not normal over Q.Alternatively, using the equivalent condition in Lemma 7.1.2, Q( 3√2) : Q isnot normal because C3−2 is an irreducible polynomial overQ that has a rootin Q( 3√2) but does not split there.One way to think about the non-normality of Q( 3√2) : Q is as follows. Thethree roots of C3−2 are indistinguishable (conjugate) overQ, since they havethe same minimal polynomial. But if they’re indistinguishable, it would

What does it meanto be normal?

be strange for an extension to contain some but not all of them, since thatwould be making a distinction between elements that are supposed to beindistinguishable. In that sense, Q( 3√2) is ‘abnormal’.

ii. Let 5 be a polynomial over a field . Then SF ( 5 ) : is always normal,as we shall see (Theorem 7.1.5).

iii. Every extension of degree 2 is normal (just as, in group theory, everysubgroup of index 2 is normal). You’ll be asked to show this in Workshop 4,but you also know enough to do it now.

Exercise 7.1.4 What happens if you drop the word ‘irreducible’ fromLemma 7.1.2? Is it still true?

Normality of field extensions is intimately related to normality of subgroups,and conjugacy in field extensions is also related to conjugacy in groups. The video‘What does it mean to be normal?’ explains both kinds of normality and conjugacyin intuitive terms.

Here’s the first of our two theorems about normal extensions. It describeswhich extensions arise as splitting field extensions.

Theorem 7.1.5 Let " : be a field extension. Then

" = SF ( 5 ) for some 5 ∈ [C] ⇐⇒ " : is finite and normal.

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SF" (<)

" = (U1, . . . , U=)i

�// (U1, . . . , U=, Y)

OO

(X) �

\//

OO

(Y)

OO

gg 77

Figure 7.1: Maps used in the proof that splitting field extensions are normal.

Proof For⇐, suppose that " : is finite and normal. By finiteness, there is abasisU1, . . . , U= of" over , and eachU8 is algebraic over (by Proposition 5.3.4).For each 8, let <8 be the minimal polynomial of U8 over ; then by normality, <8splits in " . Hence 5 = <1<2 · · ·<= ∈ [C] splits in " . The set of roots of 5 in" contains {U1, . . . , U=}, and " = (U1, . . . , U=), so " is generated over bythe set of roots of 5 in " . Thus, " is a splitting field of 5 over .

For ⇒, take 5 ∈ [C] such that " = SF ( 5 ). We may assume that 5 ≠ 0,since if 5 = 0 then " = , which is certainly finite and normal over .

Write U1, . . . , U= for the roots of 5 in " . Then " = (U1, . . . , U=). Each U8is algebraic over (since 5 ≠ 0), so by Proposition 5.3.4, " : is finite.

We now show that " : is normal, which is the most substantial part of theproof (Figure 7.1). Let X ∈ " , with minimal polynomial < ∈ [C]. Certainly <splits in SF" (<), so to show that < splits in " , it is enough to show that everyroot Y of < in SF" (<) lies in " .Splitting field

extensions arenormal

Since < is a monic irreducible annihilating polynomial of Y over , it is theminimal polynomial of Y over . Hence byTheorem4.3.7, there is an isomorphism\ : (X) → (Y) over . Now observe that:

• " = SF (X) ( 5 ), by Lemma 6.2.13(ii);

• (U1, . . . , U=, Y) is a splitting field of 5 over (Y), by Lemma 6.2.13(i)(taking the ‘"’, ‘(’ and ‘. ’ there to be SF" ( 5 ), " and {Y});

• \∗( 5 ) = 5 , since 5 ∈ [C] and \ is a homomorphism over .

So by Proposition 6.2.10, there is some isomorphism i : " → (U1, . . . , U=, Y)extending \. Then i is an isomorphism over , since \ is.

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Since X is in (U1, . . . , U=), it can be expressed as a polynomial in U1, . . . , U=over (by Corollary 5.1.11). Since i is a map over , it follows that i(X) is a poly-nomial in i(U1), . . . , i(U=) over . But i(X) = \ (X) = Y; moreover, for each 8 wehave 5 (i(U8)) = 0 (by Lemma 6.1.2 with k = id ) and so i(U8) ∈ {U1, . . . , U=}.Hence Y is a polynomial in U1, . . . , U= over , giving Y ∈ (U1, . . . , U=) = " , asrequired. �

Corollary 7.1.6 Let " : ! : be field extensions. If " : is finite and normalthen so is " : !.

Proof Follows from Theorem 7.1.5 and Lemma 6.2.13(ii). �

Warning 7.1.7 It does not follow that ! : is normal. For instance,consider Q( 3√2, 42c8/3) : Q( 3√2) : Q. The first field is the splittingfield of C3 − 2 over Q, and therefore normal over Q, but Q( 3√2) is not(Example 7.1.3(i)).

Theorem 7.1.5 is the first of two theorems about normality. The second is todo with the action of the Galois group of an extension.

Warning 7.1.8 By definition, the Galois group Gal(" : ) of anextension " : acts on " . But if " is the splitting field of somepolynomial 5 over then the action of Gal(" : ) on " restricts toan action on the roots of 5 (a finite set), as we saw in Section 6.3. Sothere are two actions of the Galois group under consideration, one therestriction of the other. Both are important.

When a group acts on a set, a basic question is: what are the orbits? ForGal(" : ) acting on " , the answer is: the conjugacy classes of " over . Or atleast, that’s the case when " : is finite and normal:

Proposition 7.1.9 Let " : be a finite normal extension and U, U′ ∈ " . Then

U and U′ are conjugate over ⇐⇒ U′ = i(U) for some i ∈ Gal(" : ).

Proof For⇐, let i ∈ Gal(" : ) with U′ = i(U). Then U and U′ are conjugateover , by Lemma 6.1.2 (with "′ = " , ′ = and k = id ).

For ⇒, suppose that U and U′ are conjugate over . Since " : is finite,both are algebraic over , and since they are conjugate over , they have thesame minimal polynomial < ∈ [C]. By Theorem 4.3.7, there is an isomorphism\ : (U) → (U′) over such that \ (U) = U′ (see diagram below).

86

By Theorem 7.1.5," is the splitting field of some polynomial 5 over . Hence" is also the splitting field of 5 over both (U) and (U′), by Lemma 6.2.13(ii).Moreover, \∗( 5 ) = 5 since \ is a homomorphism over and 5 is a polynomialover . So by Proposition 6.2.10(i), there is an automorphism i of " extending\:

"i

�// "

(U) �

\//

OO

(U′)

OO

bb <<

Then i ∈ Gal(" : ) with i(U) = \ (U) = U′, as required. �

That result was about the action of Gal(" : ) on the field " , but it hasa powerful corollary involving the action of the Galois group on the roots of anirreducible polynomial 5 , when " = SF ( 5 ):

Corollary 7.1.10 Let 5 be an irreducible polynomial over a field . Then theaction of Gal ( 5 ) on the roots of 5 in SF ( 5 ) is transitive.

Recall what transitive means, for an action of a group � on a set -: for allG, G′ ∈ - , there exists 6 ∈ � such that 6G = G′.

Proof Since 5 is irreducible, the roots of 5 in SF ( 5 ) all have the same minimalpolynomial, namely, 5 divided by its leading coefficient. So they are all conjugateover . Since SF ( 5 ) : is finite and normal (by Theorem 7.1.5), the resultfollows from Proposition 7.1.9. �

Exercise 7.1.11 Show by example that Corollary 7.1.10 becomes falseif you drop the word ‘irreducible’.

Example 7.1.12 Let 5 (C) = 1 + C + · · · + C ?−1 ∈ Q[C], where ? is prime. Since(1 − C) 5 (C) = 1 − C ?, the roots of 5 in C are l, l2, . . . , l?−1, where l = 42c8/?.By Example 3.3.16, 5 is irreducible over Q. Hence by Corollary 7.1.10, for each8 ∈ {1, . . . , ? − 1}, there is some i ∈ GalQ( 5 ) such that i(l) = l8.

This is spectacular! Until now, we’ve been unable to prove such things withouta huge amount of explicit checking, which, moreover, only works on a case-by-casebasis. For example, if you watched the video ‘Calculating Galois groups with barehands, part 2’, you’ll have seen how much tedious calculation went into the single

87

case ? = 5, 8 = 2:

But the theorems we’ve proved make all this unnecessary.In fact, for each 8 ∈ {1, . . . , ? − 1}, there’s exactly one element i8 of GalQ( 5 )

such that i8 (l) = l8. For since SFQ( 5 ) = Q(l, . . . , l?−1) = Q(l), two elementsof GalQ( 5 ) that take the same value on l must be equal. Hence

GalQ( 5 ) = {i1, . . . , i?−1}.

We’ll see later that GalQ( 5 ) � �?−1.

Example 7.1.13 Let’s calculate � = GalQ(C3 − 2). Since C3 − 2 has 3 distinctroots in C, it has 3 distinct roots in its splitting field. By Proposition 6.3.10, � isisomorphic to a subgroup of (3. Now� acts transitively on the 3 roots, so it has atleast 3 elements, so it is isomorphic to either �3 or (3. Since two of the roots arenon-real complex conjugates, one of the elements of � is complex conjugation,which has order 2. Hence 2 divides |� |, forcing � � (3.

We now show how a normal field extension gives rise to a normal subgroup.Whenever you meet a normal subgroup, you should immediately want to form theresulting quotient, so we do that too.

Theorem 7.1.14 Let " : ! : be field extensions with " : finite andnormal.

i. ! : is a normal extension ⇐⇒ i! = ! for all i ∈ Gal(" : ).

ii. If ! : is a normal extension then Gal(" : !) is a normal subgroup ofGal(" : ) and

Gal(" : )Gal(" : !) � Gal(! : ).

88

Before the proof, here’s some context and explanation.Part (i) answers the question implicit in Warning 7.1.7: we know from Corol-

lary 7.1.6 that " : ! is normal, but when is ! : normal? In part (i), i! means{i(U) : U ∈ !}. For i! to be equal to ! means that i fixes ! as a set (in otherwords, permutes it within itself), not that i fixes each element of !.

In part (ii), it’s true for all" : ! : thatGal(" : !) is a subset ofGal(" : ),since

Gal(" : !) = {automorphisms i of " such that i(U) = U for all U ∈ !}⊆ {automorphisms i of " such that i(U) = U for all U ∈ }= Gal(" : ).

It’s also always true that Gal(" : !) is a subgroup of Gal(" : ), as you caneasily check. But part (ii) tells us something much more substantial: it’s a normalsubgroup when ! : is a normal extension.

Proof of Theorem 7.1.14 For (i), first suppose that ! is normal over , and leti ∈ Gal(" : ). For all U ∈ !, Proposition 7.1.9 implies that U and i(U) areconjugate over , so they have the same minimal polynomial, so i(U) ∈ ! bynormality. Hence i! ⊆ !. The same argument with i−1 in place of i givesi−1! ⊆ !, and applying i to each side then gives ! ⊆ i!. So i! = !.

Conversely, suppose that i! = ! for all i ∈ Gal(" : ). Let U ∈ ! withminimal polynomial <. Since " : is normal, < splits in " . Each root U′ of< in " is conjugate to U over , so by Proposition 7.1.9, U′ = i(U) for somei ∈ Gal(" : ), giving U′ ∈ i! = !. Hence < splits in ! and ! : is normal.

For (ii), suppose that ! : is normal. To prove that Gal(" : !) is a normalsubgroup of Gal(" : ), let i ∈ Gal(" : ) and \ ∈ Gal(" : !). We show thati−1\i ∈ Gal(" : !), or equivalently,

i−1\i(U) = U for all U ∈ !,

or equivalently,\i(U) = i(U) for all U ∈ !.

But by (i), i(U) ∈ ! for all U ∈ !, so \ (i(U)) = i(U) since \ ∈ Gal(" : !).This completes the proof that Gal(" : !) P Gal(" : ).

Finally, we prove the statement on quotients (still supposing that ! : is anormal extension). Every automorphism i of " over satisfies i! = ! (by (i)),and therefore restricts to an automorphism i of !. The function

a : Gal(" : ) → Gal(! : )i ↦→ i

89

is a group homomorphism, since it preserves composition. Its kernel is Gal(" :!), by definition. If we can prove that a is surjective then the last part of thetheorem will follow from the first isomorphism theorem.

To prove that a is surjective, we must show that each automorphism k of !over extends to an automorphism i of ":

"i

�// "

!�

k//

OO

!

OO

`` >>

The argument is similar to the second half of the proof of Proposition 7.1.9. ByTheorem 7.1.5, " is the splitting field of some 5 ∈ [C]. Then " is also thesplittting field of 5 over !. Also,k∗( 5 ) = 5 sincek is a homomorphismover and5 is a polynomial over . So by Proposition 6.2.10(i), there is an automorphismi of " extending k, as required. �

Example 7.1.15 Take " : ! : to be

Q( 3√2, l

): Q(l) : Q,

where l = 42c8/3. As you will recognize by now, Q( 3√2, l) is the splitting field ofC3 − 2 over Q, so it is a finite normal extension of Q by Theorem 7.1.5.

Also,Q(l) is the splitting field of C2+C+1 overQ, so it too is a normal extensionofQ. Part (i) of Theorem 7.1.14 implies that every element ofGalQ(C3−2) restrictsto an automorphism of Q(l).

Part (ii) implies that

Gal(Q

( 3√2, l)

: Q(l))P Gal

(Q

( 3√2, l)

: Q)

and that

Gal(Q

( 3√2, l)

: Q)

Gal(Q

( 3√2, l)

: Q(l)) � Gal(Q(l) : Q). (7.1)

What does this say explicitly? We showed in Example 7.1.13 that Gal(Q( 3√2, l) :Q) � (3. That is, each element of the Galois group permutes the three roots

3√2, l 3√2, l2 3√2

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of C3 − 2, and all six permutations are realized by some element of the Galoisgroup. An element of Gal(Q( 3√2, l) : Q) that fixes l is determined by whichof the three roots 3√2 is mapped to, so Gal(Q( 3√2, l) : Q(l)) � �3. Finally,Gal(Q(l) : Q) � �2 by Example 7.1.12. So in this case, the isomorphism (7.1)states that

(3�3� �2.

Exercise 7.1.16 Draw a diagram showing the three roots of C3−2 andthe elements of � = Gal(Q( 3√2, l) : Q(l)) acting on them. There isa simple geometric description of the elements of Gal(Q( 3√2, l) : Q)that belong to the subgroup �. What is it?

7.2 SeparabilityTheorem 6.2.12 implies that |Gal(" : ) | ≤ [" : ] whenever " : is asplitting field extension. Why is this an inequality, not an equality? The answer canbe traced back to the proof of Proposition 6.2.10 on extension of isomorphisms.There, we had an irreducible polynomial called k∗(<), and we wrote B for thenumber of distinct roots of k∗(<) in its splitting field. Ultimately, the source ofthe inequality |Gal(" : ) | ≤ [" : ] was the fact that B ≤ deg(k∗(<)).

But is this last inequality actually an equality? That is, does an irreduciblepolynomial of degree 3 always have 3 distinct roots in its splitting field? Certainlyit has 3 roots when counted with multiplicity. But there will be fewer than 3distinct roots if any of the roots are repeated (have multiplicity ≥ 2, as defined inDefinition 3.2.10). The question is whether this can ever happen.

Exercise 7.2.1 Try to find an example of an irreducible polynomialof degree 3 with fewer than 3 distinct roots in its splitting field. Orif you can’t, see if you can prove that this is impossible over Q—thatis, an irreducible over Q has no repeated roots in C. Both are quitehard, but ten minutes spent trying may help you to appreciate what’sto come.

Definition 7.2.2 An irreducible polynomial over a field is separable if it has norepeated roots in its splitting field.

Equivalently, an irreducible polynomial 5 ∈ [C] is separable if it splits intodistinct linear factors in SF ( 5 ):

5 (C) = 0(C − U1) · · · (C − U=)

91

for some 0 ∈ and distinct U1, . . . , U= ∈ SF ( 5 ). Put another way, an irreducible5 is separable if and only if it has deg( 5 ) distinct roots in its splitting field.

Example 7.2.3 C3 − 2 ∈ Q[C] is separable, since it has 3 distinct roots in C, hencein its splitting field.

Example 7.2.4 This is an example of an irreducible polynomial that’s inseparable.It’s a little bit complicated, although in fact it’s the simplest example there is.

Let ? be a prime, let be the field F? (D) of rational expressions over F?(where D is the variable symbol), and let 5 (C) = C ? − D. By definition, 5 has atleast one root U in its splitting field. We have

(C − U)? =?∑8=0

(?

8

)C8 (−U)?−8 = C ? − U? = 5 (C),

where the second equality follows from Lemma 3.3.15. So 5 (C) = (C − U)? inSF ( 5 ), which means that U is the only root of 5 in its splitting field—despite 5having degree ? > 1.

We now show that 5 is irreducible over . The unique factorization of 5into irreducible polynomials over SF ( 5 ) is 5 (C) = (C − U)?, so any nontrivialfactorization of 5 in [C] is of the form

5 (C) = (C − U)@ (C − U)?−@

where 0 < @ < ? and both factors belong to [C]. The coefficient of C@−1 in(C − U)@ is −@U, so −@U ∈ . But @ is invertible in , so U ∈ , contradictingExercise 7.2.5.

Exercise 7.2.5 Show that D has no ?th root in F? (D); that is, there is noU ∈ F? (D) with U? = D. (Hint: consider the degree of polynomials.)

Warning 7.2.6 Definition 7.2.2 is only a definition of separabilityfor irreducible polynomials. There is a definition of separability forarbitrary polynomials, but it’s not simply Definition 7.2.2 with theword ‘irreducible’ deleted. We won’t need it, but here it is: anarbitrary polynomial is called separable if each of its irreducible factorsis separable. So C2 is separable, even though it has a repeated root.

In real analysis, we can test whether a root is repeated by asking whether the

92

derivative is 0 there:

Over an arbitrary field, there’s no general definition of differentiation, as there isno sense of what a ‘limit’ might be. But even without limits, we can differentiatepolynomials in the following sense.

Definition 7.2.7 Let be a field and let 5 (C) = ∑=8=0 08C

8 ∈ [C]. The formalderivative of 5 is

(� 5 ) (C) ==∑8=1808C

8−1 ∈ [C] .

We use � 5 rather than 5 ′ to remind ourselves not to take the familiar propertiesof differentiation for granted. Nevertheless, the usual basic laws hold:

Lemma 7.2.8 Let be a field. Then

� ( 5 + 6) = � 5 + �6, � ( 5 6) = 5 · �6 + � 5 · 6, �0 = 0

for all 5 , 6 ∈ [C] and 0 ∈ . �

Exercise 7.2.9 Check a couple of the properties in Lemma 7.2.8.

The real analysis test for repetition of roots has an algebraic analogue:

Lemma 7.2.10 Let 5 be a nonzero polynomial over a field . The following areequivalent:

i. 5 has a repeated root in SF ( 5 );

ii. 5 and � 5 have a common root in SF ( 5 );

iii. 5 and � 5 have a nonconstant common factor in [C].

Proof (i)⇒(ii): suppose that 5 has a repeated root U in SF ( 5 ). Then 5 (C) =(C − U)26(C) for some 6(C) ∈ (SF ( 5 )) [C]. Hence

(� 5 ) (C) = (C − U){26(C) + (C − U) · (�6) (C)

},

93

so U is a common root of 5 and � 5 in SF ( 5 ).(ii)⇒(iii): suppose that 5 and � 5 have a common root U in SF ( 5 ). Then U

is algebraic over (since 5 ≠ 0), and the minimal polynomial of U over is thena nonconstant common factor of 5 and � 5 .

(iii)⇒(ii): if 5 and � 5 have a nonconstant common factor 6 then 6 splits inSF ( 5 ), and any root of 6 in SF ( 5 ) is a common root of 5 and � 5 .

(ii)⇒(i): suppose that 5 and � 5 have a common root U ∈ SF ( 5 ). Then5 (C) = (C − U)6(C) for some 6 ∈ (SF ( 5 )) [C], giving

(� 5 ) (C) = 6(C) + (C − U) · (�6) (C).

But (� 5 ) (U) = 0, so 6(U) = 0, so 6(C) = (C − U)ℎ(C) for some ℎ ∈ (SF ( 5 )) [C].Hence 5 (C) = (C − U)2ℎ(C), and U is a repeated root of 5 in its splitting field. �

The point of Lemma 7.2.10 is that condition (iii) allows us to test for repetitionof roots in SF ( 5 ) without ever leaving [C], or even knowing what SF ( 5 ) is.

Proposition 7.2.11 Let 5 be an irreducible polynomial over a field. Then 5 isinseparable if and only if � 5 = 0.

Proof This follows from (i) ⇐⇒ (iii) in Lemma 7.2.10. Since 5 is irreducible,5 and � 5 have a nonconstant common factor if and only if 5 divides � 5 ; butdeg(� 5 ) < deg( 5 ), so 5 | � 5 if and only if � 5 = 0. �

Corollary 7.2.12 Let be a field.

i. If char = 0 then every irreducible polynomial over is separable.

ii. If char = ? > 0 then an irreducible polynomial 5 ∈ [C] is inseparable ifand only if

5 (C) = 10 + 11C? + · · · + 1A CA ?

for some 10, . . . , 1A ∈ .

In other words, the only irreducible polynomials that are inseparable are thepolynomials in C ? in characteristic ?. Inevitably, Example 7.2.4 is of this form.

Proof Let 5 (C) = ∑08C

8 be an irreducible polynomial. Then 5 is inseparable ifand only if � 5 = 0, if and only if 808 = 0 for all 8 ≥ 1. If char = 0, this impliesthat 08 = 0 for all 8 ≥ 1, so 5 is constant, which contradicts 5 being irreducible. Ifchar = ?, then 808 = 0 for all 8 ≥ 1 is equivalent to 08 = 0 whenever ? - 8. �

Remark 7.2.13 In the final chapter wewill show that every irreducible polynomialover a finite field is separable. So, it is only over infinite fields of characteristic ?that you have to worry about inseparability.

94

We now build up to showing that |Gal(" : ) | = [" : ] whenever " : is a finite normal extension in which the minimal polynomial of every element of" is separable. First, some terminology:

Definition 7.2.14 Let " : be an algebraic extension. An element of " isseparable over if its minimal polynomial over is separable. The extension" : is separable if every element of " is separable over .

Examples 7.2.15 i. Every algebraic extension of fields of characteristic 0 isseparable, by Corollary 7.2.12.

ii. Every algebraic extension of a finite field is separable, by Remark 7.2.13.

iii. The splitting field of C ? − D over F? (D) is inseparable: indeed, the elementdenoted by U in Example 7.2.4 is inseparable over F? (D), since its minimalpolynomial is the inseparable polynomial C ? − D.

Exercise 7.2.16 Let " : ! : be field extensions. Show that if" : is algebraic then so are " : ! and ! : .

Lemma 7.2.17 Let " : ! : be field extensions, with " : algebraic. If " : is separable then so are " : ! and ! : .

Proof Both " : ! and ! : are algebraic by Exercise 7.2.16, so it does makesense to ask whether they are separable. (We only defined what it means for analgebraic extension to be separable.) That ! : is separable is immediate from thedefinition. To show that " : ! is separable, let U ∈ " . Write <! and < for theminimal polynomials of U over ! and , respectively. Then < is an annihilatingpolynomial of U over !, so <! | < in ! [C]. Since " : is separable, < splits into distinct linear factors in SF (< ). Since <! | < , so does <! . Hence<! ∈ ! [C] is separable, so U is separable over !. �

As hinted in the introduction to this section, we will prove that |Gal(" : ) | =[" : ] by refining Proposition 6.2.10.

Proposition 7.2.18 Let k : → ′ be an isomorphism of fields, let 5 ∈ [C], let" be a splitting field of 5 over , and let "′ be a splitting field of k∗( 5 ) over ′.Suppose that the extension "′ : ′ is separable. Then there are exactly [" : ]isomorphisms i : " → "′ extending k.

Proof This is almost the same as the proof of Proposition 6.2.10, but with theinequality B ≤ deg(k∗(<)) replaced by an equality, which holds by separability.For the inductive hypothesis to go through, we need the extension "′ : ′(U′

9) to

be separable, and this follows from the separability of "′ : ′ by Lemma 7.2.17.�

95

Theorem 7.2.19 |Gal(" : ) | = [" : ] for every finite normal separableextension " : .

Proof By Theorem 7.1.5, " = SF ( 5 ) for some 5 ∈ [C]. The result followsfrom Proposition 7.2.18, taking "′ = " , ′ = , and k = id . �

Examples 7.2.20 i. |Gal ( 5 ) | = [SF ( 5 ) : ] for any polynomial 5 over afield of characteristic 0.For instance, [SFQ(C3 −2) : Q] = 6 by a similar argument to Example 5.2.7,using that SFQ(C3 − 2) contains elements of degree 2 and 3 over Q. Hence|GalQ( 5 ) | = 6. But GalQ( 5 ) embeds into (3 by Proposition 6.3.10, soGalQ( 5 ) � (3. We already proved this in Example 7.1.15, using a differentargument.

ii. Consider = F? (D) and " = SF (C ? − D). With notation as in Exam-ple 7.2.4, we have " = (U), so [" : ] = deg (U) = ?. On the otherhand, |Gal(" : ) | = 1 by Corollary 6.3.15. So Theorem 7.2.19 fails if wedrop the separability hypothesis.

Digression 7.2.21 With some effort, one can show that in any algebraicextension " : , the separable elements form a subfield of " . (See Stewart,Theorem 17.22.) It follows that a finite extension (U1, . . . , U=) : isseparable if and only if each U8 is. Hence a splitting field extension SF ( 5 ) : is separable if and only if every root of 5 is separable in SF ( 5 ), whichitself is equivalent to 5 being separable in the sense of Warning 7.2.6.

So: SF ( 5 ) is separable over if and only if 5 is separable over . Thus,the different meanings of ‘separable’ interact nicely.

Digression 7.2.22 It’s a stunning fact that every finite separable extensionis simple. This is called the theorem of the primitive element. For instance,whenever U1, . . . , U= are complex numbers algebraic over Q, there is someU ∈ C (a ‘primitive element’) such that Q(U1, . . . , U=) = Q(U). We saw onecase of this in Example 4.3.2(i): Q(

√2,√

3) = Q(√

2 +√

3).The theorem of the primitive element was at the heart of most early accountsof Galois theory, and appears in many modern treatments too, but we willnot prove it.

7.3 Fixed fieldsWhen a group � acts on a set - , we usually want to know which elements of- are fixed by every element of �. Here, we’ll ask this question for groups of

96

automorphisms of a field.The following preliminary definition will come in handy.

Definition 7.3.1 Let - and . be sets, and let � ⊆ {functions - → . }. Theequalizer of � is

Eq(�) = {G ∈ - : 5 (G) = 6(G) for all 5 , 6 ∈ �}.

So the equalizer of � is the part of - on which all the functions in � are equal.

Lemma 7.3.2 Let " and "′ be fields, and let � ⊆ {homomorphisms " → "′}.Then Eq(�) is a subfield of " .

Proof We must show that 0, 1 ∈ Eq(�), that if U ∈ Eq(�) then −U ∈ Eq(�) and1/U ∈ Eq(�) (when U ≠ 0), and that if U, V ∈ Eq(�) then U + V, UV ∈ Eq(�). Iwill show just the last of these, leaving the rest to you. Suppose that U, V ∈ Eq(�).For all i, \ ∈ �, we have

i(UV) = i(U)i(V) = \ (U)\ (V) = \ (UV),

so UV ∈ Eq(�). �

Write Aut(") for the group of automorphisms of a field " .

Definition 7.3.3 Let " be a field and let � a subgroup of Aut("). The fixedfield of � is

Fix(�) = {U ∈ " : i(U) = U for all i ∈ �}.

Lemma 7.3.4 Fix(�) is a subfield of " .

Proof In fact, Fix(�) = Eq(�), since id" is an element of �. The result followsfrom Lemma 7.3.2. �

Exercise 7.3.5 Using Lemma 7.3.4, show that every automorphismof a field is an automorphism over its prime subfield. In other words,Aut(") = Gal(" : ) whenever " is a field with prime subfield .

Here’s the big, ingenious, result about fixed fields. It will play a crucial part inthe proof of the fundamental theorem of Galois theory.

97

Theorem 7.3.6 Let " be a field and � a finite subgroup of Aut("). Then[" : Fix(�)] ≤ |� |.

As Fix(�) gets bigger, [" : Fix(�)] gets smaller. Theorem 7.3.6 tells us thatthe smaller |� | is, the more of " must be fixed by �. For some intuition aboutboth the statement and the proof, watch the video ‘The size of fixed fields’.

Proof Write = = |� |. It is enough to prove that any = + 1 elements U0, . . . , U= of" are linearly dependent over Fix(�).

Let

, ={(G0, . . . , G=) ∈ "=+1 : G0\ (U0) + · · · + G=\ (U=) = 0 for all \ ∈ �

}.

Then, is defined by = homogeneous linear equations in = + 1 variables, so it is aThe size of fixed

fieldsnontrivial "-linear subspace of "=+1.

Claim: let (G0, . . . , G=) ∈ , and i ∈ �. Then (i(G0), . . . , i(G=)) ∈ , .Proof: For all \ ∈ �, we have

G0(i−1 ◦ \) (U0) + · · · + G= (i−1 ◦ \) (U=) = 0

since i−1 ◦ \ ∈ �. Applying i to both sides gives that for all \ ∈ �,

i(G0)\ (U0) + · · · + i(G=)\ (U=) = 0,

proving the claim.Since, is nontrivial, there is a least : ≥ 0 such that, contains some nonzero

vector x = (G0, . . . , G=) with G8 = 0 for all 8 > : . Then G: ≠ 0, and since , isclosed under scalar multiplication by " , we may assume that G: = 1.

Remarks on theproof of

Theorem 7.3.6(actually a PDF,not a video)

We now show that G8 ∈ Fix(�) for all 8. Let i ∈ �. By the claim,(i(G0), . . . , i(G=)) ∈ , . Since, is a linear subspace,

(i(G0) − G0, . . . , i(G=) − G=) ∈ ,.

But i(G: ) − G: = i(1) − 1 = 0 and i(G8) − G8 = 0 − 0 = 0 for all 8 > : , so byminimality of : , also i(G8) − G8 = 0 for all 8 < : . Hence G8 ∈ Fix(�) for all 8.

We have shown that , contains a nonzero vector x ∈ Fix(�)=+1. But taking\ = id in the definition of , gives

∑G8U8 = 0. Hence U0, . . . , U= are linearly

dependent over Fix(�). �

Example 7.3.7 Let ^ : C → C be complex conjugation. Then � = {id, ^} is asubgroup of Aut(C), and Theorem 7.3.6 predicts that [C : Fix(�)] ≤ 2. SinceFix(�) = R, this is true.

98

Exercise 7.3.8 Find another example of Theorem 7.3.6.

Digression 7.3.9 In fact, Theorem 7.3.6 is an equality: [" : Fix(�)] =|� |. This is proved directly in many Galois theory books (e.g. Stewart,Theorem 10.5). In our approach it will be a consequence of the fundamentaltheorem of Galois theory (rather than a step on the way to proving it), at leastunder certain hypotheses on " and �.

The reverse inequality, [" : Fix(�)] ≥ |� |, is closely related to the resultcalled ‘linear independence of characters’. (A good reference is Lang, Alge-bra, 3rd edition, Theorem 4.1.) Another instance of linear independence ofcharacters is that the functions I ↦→ 42c8=I (= ∈ Z) are linearly independent,a fundamental fact in the theory of Fourier series.

Wefinish by adding a further connecting strand between the concepts of normalextension and normal subgroup, complementary to the strands in Theorem 7.1.14.It needs a lemma that fundamentally has nothing to do with fields, and is really ageneral result about groups acting on sets, as the proof shows.

Lemma 7.3.10 Let " be a field, � a subgroup of Aut("), and i ∈ Aut(").Then Fix(i�i−1) = i Fix(�).

Proof Let U ∈ " . Then

U ∈ Fix(i�i−1) ⇐⇒ i\i−1(U) = U for all \ ∈ �⇐⇒ \i−1(U) = i−1(U) for all \ ∈ �⇐⇒ i−1(U) ∈ Fix(�)⇐⇒ U ∈ i Fix(�). �

Proposition 7.3.11 Let " : be a finite normal extension and � a normalsubgroup of Gal(" : ). Then Fix(�) is a normal extension of .

Proof Since every element of � is an automorphism over , the subfield Fix(�)of " contains . For each i ∈ Gal(" : ), we have

i Fix(�) = Fix(i�i−1) = Fix(�),

where the first equality holds by Lemma 7.3.10 and the second because� is normalin Gal(" : ). Hence by Theorem 7.1.14(i), Fix(�) : is a normal extension.�

The stage is now set for the central result of the course: the fundamentaltheorem of Galois theory.

99

Chapter 8

The fundamental theorem of Galoistheory

We’ve been building up to this moment all semester. Let’s do it!Introduction to

Week 8

8.1 Introducing the Galois correspondenceLet " : be a field extension, with viewed as a subfield of " , as usual.

An intermediate field of " : is a subfield of " containing . Write

F = {intermediate fields of " : }.For ! ∈ F , we draw diagrams like this:

"

!

,

with the bigger fields higher up.Also write

G = {subgroups of Gal(" : )}.For � ∈ G , we draw diagrams like this:

1

�

Gal(" : ),

100

where 1 denotes the trivial subgroup and the bigger groups are lower down. It willbecome clear soon why we’re using opposite conventions for the field and groupdiagrams.

For ! ∈ F , the group Gal(" : !) consists of all automorphisms i of " thatfix each element of !. Since ⊆ !, any such i certainly fixes each element of .Hence Gal(" : !) is a subgroup of Gal(" : ). This process defines a function

Gal(" : −) : F → G! ↦→ Gal(" : !).

In the expression Gal(" : −), the symbol − should be seen as a blank space intowhich arguments can be inserted.

Warning 8.1.1 The group we’re associating with ! is Gal(" : !),not Gal(! : )! Both groups matter, but only one is a subgroup ofGal(" : ), which is what we’re interested in here.We showed just now that Gal(" : !) is a subgroup of Gal(" : ).If you wanted to show that Gal(! : ) is (isomorphic to) a subgroupof Gal(" : )—which it isn’t—then you’d probably do it by tryingto prove that every automorphism of ! over extends uniquely to " .And that’s false. For instance, when ! = , the identity on ! typicallyhas many extensions to ": they’re the elements of Gal(" : ).Although Gal(! : ) isn’t a subgroup of Gal(" : ), it is a quotientof it, at least when both extensions are finite and normal. We saw thisin Theorem 7.1.14, and we’ll come back to it in Section 8.2.

In the other direction, for� ∈ G , the subfield Fix(�) of" contains . Indeed,� ⊆ Gal(" : ), and by definition, every element of Gal(" : ) fixes everyelement of , so Fix(�) ⊇ . Hence Fix(�) is an intermediate field of " : .This process defines a function

Fix : G → F� ↦→ Fix(�).

We have now defined functions

FGal(":−) // G .

Fixoo

The fundamental theorem of Galois theory tells us how these functions behave:how the concepts of Galois group and fixed field interact. The proof will bringtogether most of the big results we’ve proved so far, and assumes that the extensionis finite, normal and separable. But first, let’s say the simple things that are truefor all extensions:

101

"

!2

!1

1

Gal(" : !2)

Gal(" : !1)

Gal(" : )

Figure 8.1: The function ! ↦→ Gal(" : !) is order-reversing (Lemma 8.1.2(i)).

Lemma 8.1.2 Let " : be a field extension, and define F and G as above.

i. For !1, !2 ∈ F ,

!1 ⊆ !2 ⇒ Gal(" : !1) ⊇ Gal(" : !2)

(Figure 8.1). For �1, �2 ∈ G ,

�1 ⊆ �2 ⇒ Fix(�1) ⊇ Fix(�2).

ii. For ! ∈ F and � ∈ G ,

! ⊆ Fix(�) ⇐⇒ � ⊆ Gal(" : !).

iii. For all ! ∈ F ,! ⊆ Fix(Gal(" : !)).

For all � ∈ G ,� ⊆ Gal(" : Fix(�)).

Warning 8.1.3 In part (i), the functions Gal(" : −) and Fix reverseinclusions. The bigger youmake !, the smaller youmakeGal(" : !),because it gets harder for an automorphism to fix everything in !. Andthe bigger you make �, the smaller you make Fix(�), because it getsharder for an element of " to be fixed by everything in �. That’s whythe field and group diagrams are opposite ways up.

102

Proof (i): I leave the first half as an exercise. For the second, suppose that�1 ⊆ �2, and let U ∈ Fix(�2). Then \ (U) = U for all \ ∈ �2, so \ (U) = U for all\ ∈ �1, so U ∈ Fix(�1).

(ii): both sides are equivalent to the statement that \ (U) = U for all \ ∈ � andU ∈ !.

(iii): the first statement follows from the ⇐ direction of (ii) by taking � =

Gal(" : !), and the second follows from the ⇒ direction of (ii) by taking! = Fix(�). (Or, they can be proved directly.) �

Exercise 8.1.4 Prove the first half of Lemma 8.1.2(i).

Exercise 8.1.5 Draw a diagram like Figure 8.1 for the second half ofLemma 8.1.2(i).

Digression 8.1.6 If you’ve done some algebraic geometry, the formal struc-ture of Lemma 8.1.2 might seem familiar. Given a field and a naturalnumber =, we can form the set F of subsets of = and the set G of ideals of [C1, . . . , C=], and there are functions F � G defined by taking the annihi-lating ideal of a subset of = and the zero-set of an ideal of [C1, . . . , C=].The analogue of Lemma 8.1.2 holds.

In general, a pair of ordered sets F and G equipped with functions F � Gsatisfying the properties above is called aGalois connection. This in turn isa special case of the category-theoretic notion of adjoint functors.

The functions

FGal(":−) // G .

Fixoo

are called theGalois correspondence for " : . This terminology is mostly usedin the case where the functions are mutually inverse, meaning that

! = Fix(Gal(" : !)), � = Gal(" : Fix(�))

for all ! ∈ F and � ∈ G . We saw in Lemma 8.1.2(iii) that in both cases, theleft-hand side is a subset of the right-hand side. But they are not always equal:

Example 8.1.7 Let " : be Q( 3√2) : Q. Since [" : ] is 3, which is aprime number, the tower law implies that there are no nontrivial intermediatefields: F = {", }. We saw in Example 6.3.3(ii) that Gal(" : ) is trivial,so G = {Gal(" : )}. Hence F has two elements and G has only one. This

103

makes it impossible for there to be mutually inverse functions between F and G .Specifically, what goes wrong is that

Fix(Gal

(Q

( 3√2)

: Q) )= Fix

({idQ( 3√2)

})= Q

( 3√2)≠ Q.

Exercise 8.1.8 Let ? be a prime number, let = F? (D), and let " bethe splitting field of C ? −D over , as in Examples 7.2.4 and 7.2.20(ii).Prove that Gal(" : −) and Fix are not mutually inverse.

If Gal(" : −) and Fix are mutually inverse then they set up a one-to-onecorrespondence between the set F of intermediate fields of " : and the set Gof subgroups of Gal(" : ). The fundamental theorem of Galois theory tells usthat this dream comes true when " : is finite, normal and separable. And ittells us more besides.

8.2 The theoremThe moment has come.

Theorem 8.2.1 (Fundamental theorem of Galois theory) Let " : be afinite normal separable extension. Write

F = {intermediate fields of " : },G = {subgroups of Gal(" : )}.

i. The functions FGal(":−) //G

Fixoo are mutually inverse.

ii. |Gal(" : !) | = [" : !] for all ! ∈ F , and [" : Fix(�)] = |� | forall � ∈ G .

iii. Let ! ∈ F . Then

! is a normal extension of ⇐⇒ Gal(" : !) is a normal subgroup of Gal(" : ),

and in that case,

Gal(" : )Gal(" : !) � Gal(! : ).

104

Proof First note that for each ! ∈ F , the extension " : ! is finite and normal (byCorollary 7.1.6) and separable (by Lemma 7.2.17). Also, the group Gal(" : )is finite (by Theorem 7.2.19), so every subgroup is finite too.

We prove (i) and (ii) together. First let � ∈ G . We have

|� | ≤ |Gal(" : Fix(�)) | = [" : Fix(�)] ≤ |� |, (8.1)

where the first inequality holds because� ⊆ Gal(" : Fix(�)) (Lemma 8.1.2(iii)),the equality follows from Theorem 7.2.19 (since " : Fix(�) is finite, normal andseparable), and the second inequality follows from Theorem 7.3.6 (since � isfinite). So equality holds throughout (8.1), giving

� = Gal(" : Fix(�)), [" : Fix(�)] = |� |.

Now let ! ∈ F . We have

[" : Fix(Gal(" : !))] = |Gal(" : !) | = [" : !],

where the first equality follows from the previous paragraph by taking� = Gal(" :!), and the second follows from Theorem 7.2.19. But ! ⊆ Fix(Gal(" : !)) byLemma 8.1.2(iii), so ! = Fix(Gal(" : !)) by Workshop 3, q. 1. This completesthe proof of (i) and (ii).

We have already proved most of (iii) as Theorem 7.1.14(ii). It only remains toshow that whenever ! is an intermediate field such that Gal(" : !) is a normalsubgroup ofGal(" : ), then ! is a normal extension of . By Proposition 7.3.11,Fix(Gal(" : !)) : is normal. But by (i), Fix(Gal(" : !)) = !, so ! : isnormal, as required. �

The fundamental theorem of Galois theory is about field extensions that arefinite, normal and separable. Let’s take a moment to think about what thoseconditions mean.

An extension " : is finite and normal if and only if " is the splitting fieldof some polynomial over (Theorem 7.1.5). So, the theorem can be understoodas a result about splitting fields of polynomials.

Not every splitting field extension is separable (Example 7.2.15(iii)). However,we know of two settings where separability is guaranteed. The first is fields ofcharacteristic zero (Example 7.2.15(i)). The most important of these is Q, whichis our focus in this chapter: we’ll consider examples in which" : is the splittingfield extension of a polynomial over Q. The second is where the fields are finite(Example 7.2.15(ii)). We’ll come to finite fields in the final chapter.

105

Digression 8.2.2 Normality and separability are core requirements of Galoistheory, but there are extensions of the fundamental theorem (well beyond thiscourse) in which the finiteness condition on " : is relaxed.

The first level of relaxation replaces ‘finite’ by ‘algebraic’. Then Gal(" : )is no longer a finite group, but it does acquire an interesting topology. Oneexample is where " is the algebraic closure of , and Gal( : ) iscalled the absolute Galois group of . It contains all splitting fields ofpolynomials over , so to study it is to study all polynomials over at once.

Going further, we can even drop the condition that the extension is algebraic.In this realm, we need the notion of ‘transcendence degree’, which countshow many algebraically independent elements can be found in the extension.

You’ll want to see some examples! Section 8.3 is devoted to a single exampleof the fundamental theorem, showing every aspect of the theorem in all its glory.I’ll give a couple of simpler examples in a moment, but before that, it’s helpful toreview some of what we did earlier:

Remark 8.2.3 When working out the details of the Galois correspondence for apolynomial 5 ∈ [C], it’s not only the fundamental theorem that’s useful. Someof our earlier results also come in handy, such as the following.

i. Lemma 6.3.7 states that Gal ( 5 ) acts on the set of roots of 5 in SF ( 5 ), andthe injectivity part of Proposition 6.3.10 states that the action is faithful:if i, \ ∈ Gal ( 5 ) and i(U) = \ (U) for every root U of 5 in SF ( 5 ), theni = \. In words, an element of the Galois group is entirely determined bywhat it does to the roots.

ii. Corollary 6.3.15 states that |Gal ( 5 ) | divides :!, where : is the number ofdistinct roots of 5 in its splitting field.

iii. Let U and V be roots of 5 in SF ( 5 ). Then there is an element of the Galoisgroup mapping U to V if and only if U and V are conjugate over (havethe same minimal polynomial). This follows from Proposition 7.1.9. Inparticular, if 5 is irreducible then there is always an element of the Galoisgroup that maps U to V (Corollary 7.1.10).

iv. Let ! be an intermediate field of SF ( 5 ) : such that ! : is normal.Then every automorphism of ! over can be extended to an automorphismof SF ( 5 ) over . This was shown in the last paragraph of the proof ofTheorem 7.1.14.

Examples 8.2.4 i. Let " : be a normal separable extension of prime de-gree ?. By the fundamental theorem, |Gal(" : ) | = [" : ] = ?.

106

Every group of prime order is cyclic, so Gal(" : ) � �?. By the towerlaw, " : has no nontrivial intermediate fields, and by Lagrange’s the-orem, Gal(" : ) has no nontrivial subgroups. So F = {", } andG = {1,Gal(" : )}:

" 1

Gal(" : )

Both " and are normal extensions of , and both 1 and Gal(" : ) arenormal subgroups of Gal(" : ).

ii. Let 5 (C) = (C2 + 1) (C2 − 2) ∈ Q[C]. Put " = SFQ( 5 ) = Q(√

2, 8) and� = Gal(" : ) = GalQ( 5 ). Then " : is a finite normal separableextension, so the fundamental theorem applies. We already calculated � ina sketchy way in Example 6.3.6(ii). Let’s do it again in full, using what wenow know.First,

[" : ] =[Q

(√2, 8

): Q

(√2) ] [Q

(√2)

: Q]= 2 × 2 = 4

(much as in Example 5.2.2).Now consider how� acts on the set {±

√2,±8} of roots of 5 . The conjugacy

class of√

2 is {√

2,−√

2}, so for each i ∈ � we have i(√

2) = ±√

2.Similarly, i(8) = ±8 for each i ∈ �. The two choices of sign determine ientirely, so |� | ≤ 4. But by the fundamental theorem, |� | = [" : ] = 4, soeach of the four possibilities does in fact occur. So � = {id, i+−, i−+, i−−},where

i+−(√

2)=√

2, i−+(√

2)= −√

2, i−−(√

2)= −√

2,i+−(8) = −8, i−+(8) = 8, i−−(8) = −8.

The only two groups of order 4 are �4 and �2 × �2, and each element of �has order 1 or 2, so � � �2 × �2.The subgroups of � are

1

〈i+−〉 〈i−+〉 〈i−−〉

�

(8.2)

107

where lines indicate inclusions. Here 〈i+−〉 is the subgroup generated byi+−, which is {id, i+−}, and similarly for i−+ and i−−.What are the fixed fields of these subgroups? The fundamental theoremimplies that Fix(�) = Q. Also, i+−(

√2) =

√2, so Q(

√2) ⊆ Fix(〈i+−〉).

But [Q

(√2, 8

): Q

(√2) ]= 2 = |〈i+−〉| =

[Q

(√2, 8

): Fix(〈i+−〉)

](where the last step is by the fundamental theorem), soQ(

√2) = Fix(〈i+−〉).

Similar arguments apply to i−+ and i−−, so the fixed fields of the groups indiagram (8.2) are

Q(√

2, 8)

Q(√

2)

Q(8) Q(√

28)

Q

(8.3)

Equivalently, the groups in (8.2) are the Galois groups of Q(√

2, 8) over thefields in (8.3). For instance, Gal(Q(

√2, 8) : Q(8)) = 〈i−+〉.

Since the overall Galois group � � �2 × �2 is abelian, every subgroup isnormal. Hence all the extensions in diagram (8.3) are normal too.

Exercise 8.2.5 In this particular example, one can also see directlythat all the extensions in (8.3) are normal. How?

Like any big theorem, the fundamental theorem of Galois theory has someimportant corollaries. Here’s one.

Corollary 8.2.6 Let " : be a finite normal separable extension. Then for everyU ∈ " \ , there is some automorphism i of " over such that i(U) ≠ U.

Proof Theorem 8.2.1(i) implies that Fix(Gal(" : )) = . Now U ∉ , soU ∉ Fix(Gal(" : )), which is what had to be proved. �

Example 8.2.7 For any 5 ∈ Q[C] and irrational U ∈ SFQ( 5 ), there is somei ∈ GalQ( 5 ) that does not fix U. This is clear if U ∉ R, as we can take i to becomplex conjugation restricted to SFQ( 5 ). But it is not so obvious otherwise.

108

d

^b−b

b8

−b8

Figure 8.2: The roots of 5 , and the effects on them of d, ^ ∈ GalQ( 5 ).

8.3 A specific exampleChapter 13 of Stewart’s book opens with these words:

The extension that we discuss is a favourite with writers on Galoistheory, because of its archetypal quality. A simpler example wouldbe too small to illustrate the theory adequately, and anything morecomplicated would be unwieldy. The example is the Galois group ofthe splitting field of C4 − 2 over Q.

We go through the same example here. My presentation of it is different fromStewart’s, so you can consult his book if anything that follows is unclear.

Write 5 (C) = C4 − 2 ∈ Q[C], which is irreducible by Eisenstein’s criterion.Write � = GalQ( 5 ).

Splitting field Write b for the unique real positive root of 5 . Then the roots of5 are ±b and ±b8 (Figure 8.2). So SFQ( 5 ) = Q(b, b8) = Q(b, 8). We have

[Q(b, 8) : Q] = [Q(b, 8) : Q(b)] [Q(b) : Q] = 2 × 4 = 8,

where the first factor is 2 because Q(b) ⊆ R and the second factor is 4 because 5is the minimal polynomial of b over Q (being irreducible) and deg( 5 ) = 4. By thefundamental theorem, |� | = 8.

Galois group We now look for the 8 elements of the Galois group. We’ll usethe principle that if i, \ ∈ � with i(b) = \ (b) and i(8) = \ (8) then i = \. To seethis, note that i(U) = \ (U) whenever U is a root of 5 , and since the action of �on the roots of 5 is faithful, i = \.

Complex conjugation on C restricts to an automorphism ^ of Q(b, 8) over Q,giving an element ^ ∈ � of order 2.

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I now claim that � has an element d satisfying d(b) = b8 and d(8) = 8. In thatcase, d will act on the roots of 5 as follows:

b ↦→ b8 ↦→ −b ↦→ −b8 ↦→ b

(Figure 8.2). This element d will have order 4.Proof of claim: since 5 is irreducible, � acts transitively on the roots of 5 in

SFQ( 5 ), so there is some i ∈ � such that i(b) = b8. The conjugacy class of 8over Q is {±8}, so i(8) = ±8. If i(8) = 8 then we can take d = i. If i(8) = −8 then

(i ◦ ^) (b) = i(b) = b8, (i ◦ ^) (8) = i(−8) = −i(8) = 8,

so we can take d = i ◦ ^.(From now on, I will usually omit the ◦ sign andwrite things like i^ instead. Of

course, juxtaposition is also used to mean multiplication, as in b8. But confusionshouldn’t arise: automorphisms are composed and numbers are multiplied.)

Figure 8.2 suggests that � is the dihedral group �4, the symmetry group ofthe square.

Warning 8.3.1 The symmetry group of a regular =-sided polygon has2= elements: = rotations and = reflections. Some authors call it �=

and others call it �2=. I will call it �=, as in the Group Theory course.

If this is right, we should have ^d = d−1^. (This is one of the defining equationsof the dihedral group; you saw it in Example 3.2.12 of Group Theory.) Let’s checkthis algebraically:

^d(b) = b8 = −b8, d−1^(b) = d−1(b) = −b8,^d(8) = ^(8) = −8, d−1^(8) = d−1(−8) = −8,

so ^d and d−1^ are equal on b and 8, so ^d = d−1^. It follows that ^dA = d−A^ forall A ∈ Z.

Figure 8.3 shows the effect of 8 elements of � on b, 8 and b8. Since no twoof them have the same effect on both b and 8, they are all distinct elements of �.Since |� | = 8, they are the only elements of �. So � � �4.

Warning 8.3.2 The ‘geometric description’ in Figure 8.3 applies onlyto the roots, not the whole of the splitting field Q(b, 8). For example,d2 is rotation by c on the set of roots, but it is not rotation by c on therest of Q(b, 8): it fixes each element of Q, for instance.

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i ∈ � i(b) i(8) i(b8) order geometric description(see Warning 8.3.2)

id b 8 b8 1 identityd b8 8 −b 4 rotation by c/2d2 −b 8 −b8 2 rotation by c

d3 = d−1 −b8 8 b 4 rotation by −c/2^ b −8 −b8 2 reflection in real axis

^d = d−1^ −b8 −8 −b 2 reflection in axis through 1 − 8^d2 = d2^ −b −8 b8 2 reflection in imaginary axis^d−1 = d^ b8 −8 b 2 reflection in axis through 1 + 8

Figure 8.3: The Galois group of C4 − 2 over Q.

Subgroups of the Galois group Since |� | = 8, any nontrivial proper subgroupof � has order 2 or 4. Let’s look in turn at subgroups of order 2 and 4, alsodetermining which ones are normal. This is pure group theory, with no mentionof fields.

• The subgroups of order 2 are of the form 〈i〉 = {id, i} where i ∈ � hasorder 2. So, they are

〈d2〉, 〈^〉, 〈^d〉, 〈^d2〉, 〈^d−1〉.

If you watched the video ‘What does it mean to be normal?’, you may beable to guess which of these subgroups are normal in�, the symmetry groupof the square. It should be those that can be specified without referring toparticular vertices or edges of the square. So, just the first should be normal.Let’s check.We know that ^d2 = d2^, so d2 commutes with both ^ and d, which generate�. Hence d2 is in the centre of � (commutes with everything in �). Itfollows that 〈d2〉 is a normal subgroup of �. On the other hand, for eachA ∈ Z, the subgroup 〈^dA〉 is not normal, since

d(^dA)d−1 = (d^)dA−1 = (^d−1)dA−1 = ^dA−2 ∉ 〈^dA〉.

• The subgroups of � of order 4 are isomorphic to either �4 or �2 ×�2, sincethese are the only groups of order 4.The only elements of � of order 4 are d±1, so the only subgroup of �isomorphic to �4 is 〈d〉 = {id, d, d2, d3 = d−1}.Now consider subgroups � of � isomorphic to �2 × �2.

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Exercise 8.3.3 Show that every such � must contain d2. (Hint:think geometrically.)

We have d2 ∈ �, and both other nonidentity elements of � have order 2, sothey are of the form ^dA for some A ∈ Z. The two such subgroups � are

〈^, d2〉 = {id, ^, d2, ^d2},〈^d, d2〉 = {id, ^d, d2, ^d−1}.

Finally, any subgroup of index 2 of any group is normal, so all the subgroupsof � of order 4 are normal.

Hence the subgroup structure of � � �4 is as follows, where a box around asubgroup means that it is normal in �.

1 order 1

〈^〉 〈^d2〉 〈d2〉 〈^d〉 〈^d−1〉 order 2

〈^, d2〉 � �2 × �2 〈d〉 � �4 〈^d, d2〉 � �2 × �2 order 4

� = 〈^, d〉 � �4 order 8

Fixed fields We now find Fix(�) for each � ∈ G , again considering the sub-groups of orders 2 and 4 in turn.

• Order 2: take Fix〈^〉 (officially Fix(〈^〉), but let’s drop the brackets). Wehave ^(b) = b, so b ∈ Fix〈^〉, so Q(b) ⊆ Fix〈^〉. But [Q(b, 8) : Q(b)] = 2,and by the fundamental theorem, [Q(b, 8) : Fix〈^〉] = |〈^〉| = 2, so Fix〈^〉 =Q(b).The same argument shows that for any i ∈ � of order 2, if we can spotsome U ∈ Q(b, 8) such that i(U) = U and [Q(b, 8) : Q(U)] ≤ 2, then

Finding fixed fields Fix〈i〉 = Q(U). For i = ^d2, we can take U = b8 (by Figure 8.3). We havedegQ(b8) = 4 since b8 is a root of 5 , so [Q(b8) : Q] = 4, or equivalently,[Q(b, 8) : Q(b8)] = 2. Hence Fix〈^d2〉 = Q(b8).

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Exercise 8.3.4 I took a small liberty in the sentence beginning‘The same argument’, because it included an inequality but theprevious argument didn’t. Prove the statement made in that sen-tence.

It is maybe not so easy to spot an U for ^d, but the geometric descriptionin Figure 8.3 suggests taking U = b (1 − 8). And indeed, one can check that^d fixes b (1 − 8). One can also check that b (1 − 8) is not the root of anynonzero quadratic over Q, so degQ(b (1 − 8)) is ≥ 4 (since it divides 8), so[Q(b, 8) : Q(b (1− 8))] ≤ 8/4 = 2. Hence Fix〈^d〉 = Q(b (1− 8)). Similarly,Fix〈^d−1〉 = Q(b (1 + 8)).Finally,

d2(b2) = (d2(b))2 = (−b)2 = b2, d2(8) = 8,so Q(b2, 8) ⊆ Fix〈d2〉. But [Q(b, 8) : Q(b2, 8)] = 2, so Fix〈d2〉 = Q(b2, 8).

• Order 4: for � = 〈^, d2〉, note that b2 is fixed by both ^ and d2, sob2 ∈ Fix(�), so Q(b2) ⊆ Fix(�). But b2 ∉ Q, so [Q(b2) : Q] ≥ 2, so[Q(b, 8) : Q(b2)] ≤ 4. The fundamental theorem guarantees that

[Q(b, 8) : Fix(�)] = |� | = 4,

so Fix(�) = Q(b2).The same argument applies to the other two subgroups � of order 4: ifwe can spot an element U ∈ Q(b, 8) \ Q fixed by the generators of �, thenFix(�) = Q(U). This gives Fix〈d〉 = Q(8) and Fix〈^d, d2〉 = Q(b28).

In summary, the fixed fields of the subgroups of � are as follows.

Q(b, 8) degree 1

Q(b) Q(b8) Q(b2, 8) Q(b (1 − 8)) Q(b (1 + 8)) degree 2

Q(b2) Q(8) Q(b28) degree 4

Q degree 8

In the right-hand column, ‘degree’ means the degree of Q(b, 8) over the subfieldconcerned. The fundamental theorem implies that the Galois group of Q(b, 8)over each intermediate field is the corresponding subgroup of � in the earlierdiagram. For example, Gal(Q(b, 8) : Q(b2, 8)) = 〈d2〉. The fundamental theoremalso implies that the intermediate fields that are normal over Q are the boxed ones.

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Quotients Finally, the fundamental theorem tells us that

Gal(Q(b, 8) : Q)Gal(Q(b, 8) : !) � Gal(! : Q)

whenever ! is an intermediate field normal over Q.For ! = Q(b2, 8), this gives

�/〈d2〉 � Gal(Q(b2, 8) : Q). (8.4)

The left-hand side is the quotient of �4 by a subgroup isomorphic to �2. It hasorder 4, but it has no element of order 4: for the only elements of � of order 4are d±1, whose images in �/〈d2〉 have order 2. Hence �/〈d2〉 � �2 × �2. Onthe other hand, Q(b2, 8) is the splitting field over Q of (C2 − 2) (C2 + 1), which byExample 8.2.4(ii) has Galois group�2×�2. This confirms the isomorphism (8.4).

The other three intermediate fields normal over Q, I leave to you:Normal subgroupsand normalextensions Exercise 8.3.5 Choose one ofQ(b2),Q(8) orQ(b28), and do the same

for it as I just did for Q(b2, 8).

As you’ve now seen, it can take quite some time to work through a partic-ular example of the Galois correspondence. You’ll get practice at doing this inworkshop questions.

Beyond examples, there are at least two things we can do with the fundamentaltheorem of Galois theory. The first is to resolve the old question on solvabilityof polynomials by radicals, which we met back in Chapter 1. The second is towork out the structure of finite fields. We will carry out these two missions in theremaining two weeks.

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Chapter 9

Solvability by radicals

We began this course with a notorious old problem: can every polynomial besolved by radicals? Theorem 1.3.5 gave the answer and more: not only is itimpossible to find a general formula that does it, but we can tell which specificpolynomials can be solved by radicals.

Introduction toWeek 9

Theorem 1.3.5 states that a polynomial overQ is solvable by radicals if and onlyif it has the right kind of Galois group—a solvable one. In degree 5 and higher,there are polynomials that have the wrong kind of group. These polynomials arenot, therefore, solvable by radicals.

We’ll prove one half of this ‘if and only if’ statement: if 5 is solvable byradicals then GalQ( 5 ) is solvable. This is the half that’s needed to show that somepolynomials are not solvable by radicals. The proof of the other direction is inChapter 18 of Stewart’s book, but we won’t do it.

If you’re taking Algebraic Topology, you’ll already be familiar with the ideathat groups can be used to solve problems that seem to have nothing to do withgroups. You have a problem about some objects (such as topological spaces orfield extensions), you associate groupswith those objects (maybe their fundamentalgroups or their Galois groups), you translate your original problem into a problemabout groups, and you solve that instead. For example, the question of whether R2

and R3 are homeomorphic is quite difficult using only general topology; but usingalgebraic topology, we can answer ‘no’ by noticing that the fundamental group ofR2 with a point removed is not isomorphic to the fundamental group of R3 with apoint removed. In much the same way, we’ll answer a difficult question about fieldextensions by converting it into a question about groups.

For this chapter, you’ll need to remember something about solvable groups.At a minimum, you’ll need the definition, the fact that any quotient of a solvablegroup is solvable, and the fact that (5 is not solvable.

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9.1 RadicalsWe speak of square roots, cube roots, and so on, but we also speak about rootsof polynomials. To distinguish between these two related usages, we will use theword ‘radical’ for square roots etc. (Radical comes from the Latin word for root.A radish is a root, and a change or policy is radical if it gets right down to the rootsof the matter.)

Back in Chapter 1, I said that a complex number is called radical if ‘it can beobtained from the rationals using only the usual arithmetic operations [addition,subtraction, multiplication and division] and :th roots [for : ≥ 1]’. As an example,I said that

12 +

3√

7√2 − 2√7

4

√6 + 5

√23

(9.1)

is radical, whichever square root, cube root, etc., we choose (p. 12). Let’s nowmake this definition precise.

The first point is that the notation =√I or I1/= is highly dangerous:

Warning 9.1.1 Let I be a complex number and = ≥ 2. Then thereis no single number called =

√I or I1/=. There are = elements U of C

such that U= = I. So, the notation =√I or I1/= makes no sense if it is

intended to denote a single complex number. It is simply invalid.When I belongs to the set R+ of nonnegative reals, the convention isthat =√I or I1/= denotes the unique U ∈ R+ such that U= = I. There is

also a widespread convention that when I is a negative real and = isodd, =

√I or I1/= denotes the unique real U such that U= = I. In these

cases, there is a sensible and systematic way of choosing one of the=th roots of I. But for a general I and =, there is not.Complex analysis has a lot to say about different choices of =th roots.But we don’t need to go into that. We simply treat all the =th roots of Ion an equal footing, not attempting to pick out any of them as special.

With this warning in mind, we define the radical numbers without using nota-tion like =

√I or I1/=.

Definition 9.1.2 Let Qrad be the smallest subfield of C such that for U ∈ C,The definition ofradical number U= ∈ Qrad for some = ≥ 1 ⇒ U ∈ Qrad. (9.2)

A complex number is radical if it belongs to Qrad.

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So any rational number is radical; the sum, product, difference or quotient ofradical numbers is radical; any =th root of a radical number is radical; and thereare no more radical numbers than can be obtained by those rules.

For the definition ofQrad to make sense, we need there to be a smallest subfieldof C with the property (9.2). This will be true as long as the intersection of anyfamily of subfields of C satisfying (9.2) is again a subfield of C satisfying (9.2):for then Qrad is the intersection of all subfields of C satisfying (9.2).

Exercise 9.1.3 Check that the intersection of any family of subfieldsofC satisfying (9.2) is again a subfield ofC satisfying (9.2). (That anyintersection of subfields is a subfield is a fact we met back on p. 20;the new aspect is (9.2).)

Example 9.1.4 Consider again the expression (9.1). It’s not quite as random asit looks. I chose it so that the various radicals are covered by one of the twoconventions mentioned in Warning 9.1.1: they’re all =th roots of positive realsexcept for 3

√7√2 − 2√7, which is an odd root of a negative real. Let I be the

number (9.1), choosing the radicals according to those conventions.I claim that I is radical, or equivalently that I belongs to every subfield of C

satisfying (9.2).First, Q ⊆ since Q is the prime subfield of C. So 2/3 ∈ , and so 5

√2/3 ∈

by (9.2). Also, 6 ∈ and is a field, so 6 + 5√

2/3 ∈ . But then by (9.2) again,the denominator of (9.1) is in . A similar argument shows that the numerator isin . Hence I ∈ .

Definition 9.1.5 A polynomial overQ is solvable by radicals if all of its complexroots are radical.

The simplest nontrivial example of a polynomial solvable by radicals is some-thing of the form C= − 0, where 0 ∈ Q. The theorem we’re heading for is thatany polynomial solvable by radicals has solvable Galois group, and if that’s truethen the group GalQ(C= − 0) must be solvable. Let’s consider that group now. Theresults we prove about it will form part of the proof of the big theorem.

We begin with the case 0 = 1.

Lemma 9.1.6 For all = ≥ 1, the group GalQ(C= − 1) is abelian.

Proof Write l = 42c8/=. The complex roots of C= − 1 are 1, l, . . . , l=−1, soSFQ(C= − 1) = Q(l).

Let i, \ ∈ GalQ(C= − 1). Since i permutes the roots of C= − 1, we havei(l) = l8 for some 8 ∈ Z. Similarly, \ (l) = l 9 for some 9 ∈ Z. Hence

(i ◦ \) (l) = i(l 9 ) = i(l) 9 = l8 9 ,

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and similarly (\ ◦ i) (l) = l8 9 . So (i ◦ \) (l) = (\ ◦ i) (l). Since SFQ(C= − 1) =Q(l), it follows that \ ◦ i = i ◦ \. �

Exercise 9.1.7 In the last sentence of that proof, how exactly does it‘follow’?

Much more can be said about the Galois group of C= − 1, and you’ll see a bitmore at Workshop 5. But this is all we need for our purposes.

Now that we’ve considered C= − 1, let’s do C= − 0 for an arbitrary 0.Lemma 9.1.8 Let be a field and = ≥ 1. Suppose that C= − 1 splits in . ThenGal (C= − 0) is abelian for all 0 ∈ .

The hypothesis that C= − 1 splits in might seem so restrictive as to makethis lemma useless. For instance, it doesn’t hold in Q or even R (for = > 2).Nevertheless, this turns out to be the key lemma in the whole story of solvabilityby radicals.

Proof If 0 = 0 then Gal (C= − 0) is trivial; suppose otherwise.Choose a root b of C= − 0 in SF (C= − 0). For any other root a, we have

(a/b)= = 0/0 = 1 (valid since 0 ≠ 0), and C= − 1 splits in , so a/b ∈ .It follows that SF (C= − 0) = (b). Moreover, given i, \ ∈ Gal (C= − 0), we

have i(b)/b ∈ (since i(b) is a root of C= − 0), so

(\ ◦ i) (b) = \(i(b)b· b

)=i(b)b· \ (b) = i(b)\ (b)

b.

Similarly, (i◦\) (b) = i(b)\ (b)/b, so (\◦i) (b) = (i◦\) (b). Since SF (C=−0) = (b), it follows that i ◦ \ = \ ◦ i. �

Warning 9.1.9 For 0 ∈ Q, the Galois group of C= − 0 over Q isnot usually abelian. For instance, you saw on Assignment 4 thatGalQ(C3 − 2) is the nonabelian group (3.

Exercise 9.1.10 What does the proof of Lemma 9.1.8 tell you aboutthe eigenvectors and eigenvalues of the elements of Gal (C= − 0)?

Exercise 9.1.11 Use Lemmas 9.1.6 and 9.1.8 to show thatGalQ(C=−0)is solvable for all 0 ∈ Q.This is harder than most of these exercises, but I recommend it as away of getting into the right frame of mind for the theory that’s comingin Section 9.2.

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Digression 9.1.12 We’re only going to do the theory of solvability by radicalsover Q. It can be done over any field, but Q has two special features. First, Qcan be embedded into an algebraically closed field that we know very well,namely, C. This makes some things easier. Second, charQ = 0. For fields ofcharacteristic ?, the proof that any polynomial with a solvable Galois groupis solvable by radicals has some extra complications.

9.2 Solvable polynomials have solvable groupsHere we’ll prove that every polynomial over Q that is solvable by radicals hassolvable Galois group.

You know by now that in Galois theory, we tend not to jump straight frompolynomials to groups. We go via the intermediate stage of field extensions, as inthe diagram

polynomial ↦−→ field extension ↦−→ group

that I first drew after the definition of Gal ( 5 ) (page 77). That is, we understandpolynomials through their splitting field extensions.

So it shouldn’t be a surprise that we do the same here, defining a notion of‘solvable extension’ and showing (roughly speaking) that

solvable polynomial ↦−→ solvable extension ↦−→ solvable group.

In other words, we’ll define ‘solvable extension’ in such a way that (i) if 5 ∈ Q[C]is a polynomial solvable by radicals then SFQ( 5 ) : Q is a solvable extension, and(ii) if " : is a solvable extension then Gal(" : ) is a solvable group. Henceif 5 is solvable by radicals then GalQ( 5 ) is solvable—the result we’re aiming for.Solvable

polynomials havesolvable groups:

a map

(I glossed over some details in that paragraph; we’ll get to those.)

Definition 9.2.1 Let " : be a finite normal separable extension. Then " : is solvable (or " is solvable over ) if there exist A ≥ 0 and intermediate fields

= !0 ⊆ !1 ⊆ · · · ⊆ !A = "

such that !8 : !8−1 is normal and Gal(!8 : !8−1) is abelian for each 8 ∈ {1, . . . , A}.

Exercise 9.2.2 Let # : " : be extensions, with # : " , " : and# : all finite, normal and separable. Show that if # : " and " : are solvable then so is # : .

We will focus on subfields of C, where separability is automatic (Exam-ple 7.2.15(i)).

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Example 9.2.3 Let 0 ∈ Q and = ≥ 1. Then SFQ(C= − 0) : Q is a finite normalseparable extension, being a splitting field extension over Q. I claim that it issolvable.

Proof: if 0 = 0 then SFQ(C= − 0) = Q, and Q : Q is solvable (taking A = 0 and!0 = Q in Definition 9.2.1). Now assume that 0 ≠ 0. Choose a complex root b ofC= − 0 and write l = 42c8/=. Then the complex roots of C= − 0 are

b, lb, . . . , l=−1b.

So SFQ(C= − 0) contains (l8b)/b = l8 for all 8, and so C= − 1 splits in SFQ(C= − 0).Hence

Q ⊆ SFQ(C= − 1) ⊆ SFQ(C= − 0).Now SFQ(C= − 1) : Q is normal (being a splitting field extension) and has abelianGalois group by Lemma 9.1.6. Also SFQ(C=−0) : SFQ(C=−1) is normal (being thesplitting field extension of C= − 0 over SFQ(C= − 1), by Lemma 6.2.13(ii)), and hasabelian Galois group by Lemma 9.1.8. So SFQ(C= − 0) : Q is a solvable extension,as claimed.

The definition of solvable extension bears a striking resemblance to the defini-tion of solvable group. Indeed:

Lemma 9.2.4 Let " : be a finite normal separable extension. Then

" : is solvable ⇐⇒ Gal(" : ) is solvable.

Proof We will only need the ⇒ direction, and that is all I prove here. For theconverse, see Workshop 5.

Suppose that " : is solvable. Take intermediate fields

= !0 ⊆ !1 ⊆ · · · ⊆ !A = "

as in Definition 9.2.1. For each 8 ∈ {1, . . . , A}, the extension " : !8−1 is finite,normal and separable (by Corollary 7.1.6 and Lemma 7.2.17), so we can apply thefundamental theorem of Galois theory to it. Since !8 : !8−1 is a normal extension,Gal(" : !8) is a normal subgroup of Gal(" : !8−1) and

Gal(" : !8−1)Gal(" : !8)

� Gal(!8 : !8−1).

By hypothesis, the right-hand side is abelian, so the left-hand side is too. So thesequence of subgroups

Gal(" : ) = Gal(" : !0) ⊇ Gal(" : !1) ⊇ · · · ⊇ Gal(" : !A) = 1

exhibits Gal(" : ) as a solvable group. �

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Exercise 9.2.5 Prove the ⇐ direction of Lemma 9.2.4. It’s a verysimilar argument to the proof of⇒.

According to the story I’m telling, solvability by radicals of a polynomialshould correspond to solvability of its splitting field extension. Thus, the subfieldsof C that are solvable over Q should be exactly the splitting fields SFQ( 5 ) ofpolynomials 5 that are solvable by radicals. (This is indeed true, though we won’tentirely prove it.) Now if 5 , 6 ∈ Q[C] are both solvable by radicals then so is 5 6,and SFQ( 5 6) is a solvable extension of Q containing both SFQ( 5 ) and SFQ(6).So it should be the case that for any two subfields of C solvable over Q, there issome larger subfield, also solvable over Q, containing both. We now prove this.

Lemma 9.2.6 Let ! and " be subfields of C such that the extensions ! : Q and" : Q are finite, normal and solvable. Then there is some subfield # of C suchthat ! ∪ " ⊆ # and # : Q is also finite, normal and solvable.

Proof Take subfields

Q = !0 ⊆ · · · ⊆ !A = !, Q = "0 ⊆ · · · ⊆ "B = "

such that !8 : !8−1 is normal with abelian Galois group for each 8, and similarlyfor " 9 . There is a chain of subfields

Q = !0 ⊆ · · · ⊆ !A = ! = "0(!) ⊆ · · · ⊆ "B (!) = " (!) (9.3)

of C, where " 9 (!) is the subfield of C generated by " 9 ∪ ! (an instance of thenotation (. ) of Definition 4.1.10). Put # = " (!). Certainly ! ∪ " ⊆ # . Weshow that # : Q is finite, normal and solvable.

Since ! : Q is finite and normal, ! = SFQ( 5 ) for some 5 ∈ Q[C], and similarly," = SFQ(6). Now # is the subfield of C generated by !∪" , so it is generated bythe roots of 5 and the roots of 6, or equivalently the roots of 5 6. So # = SFQ( 5 6),which is finite and normal over Q.

To see that # : Q is solvable, we show that each successive extension in (9.3)is normal with abelian Galois group. For those to the left of !, this is im-mediate. For those to the right, let 9 ∈ {1, . . . , B}. Since " 9 : " 9−1 is fi-nite and normal, " 9 = SF" 9−1 (ℎ) for some ℎ ∈ " 9−1 [C]. Then " 9 (!) =SF" 9−1 (!) (ℎ) by Lemma 6.2.13(i), so " 9 (!) : " 9−1(!) is normal. Its Ga-lois group is Gal" 9−1 (!) (ℎ), which by Corollary 6.3.13 is isomorphic to a sub-group of Gal" 9−1 (ℎ). But Gal" 9−1 (ℎ) is Gal(" 9 : " 9−1), which is abelian, soGal(" 9 (!) : " 9−1(!)) is abelian. �

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Digression 9.2.7 This proof, and some others in this section, can be unclut-tered slightly using the notion of compositum. By definition, the composi-tum of subfields !, " of a field � is the smallest subfield containing ! and" . For example, the # of the proof is the compositum of ! and " .

In our usual notation, the compositum of ! and " is " (!) or ! ("). Thestandard notation for the compositum is !" , which has the advantage ofbeing symmetric and the disadvantage of being misleading: it is not the set{UV : U ∈ !, V ∈ "} (although it is the smallest subfield of � containingthat set).

The various fields used in the proof of Lemma 9.2.6, together with the fields!8 (") = !8" , can be drawn like this:

!"

!"B !A"

. .. . . .

!"0 !0"

! "

!A "B

. . . . ..

!0 "0

Q

I have chosen not to use the compositum explicitly in this course, but onceyou know about it, you’ll notice how often it appears implicitly in Galoistheory proofs.

The heart of the proof that solvable polynomials have solvable Galois groupsis the following lemma (which in turn depends on Lemmas 9.1.6 and 9.1.8 on theGalois groups of C=−1 and C=−0). Loosely, it says that the set of complex numbersthat can be reached fromQ by solvable extensions is closed under taking =th roots.Write

Qsol = {U ∈ C : U ∈ ! for some subfield ! ⊆ C that is finite, normal and solvableover Q}.

In fact, Qsol = Qrad, but we don’t know that yet.

Lemma 9.2.8 Let U ∈ C and = ≥ 1. If U= ∈ Qsol then U ∈ Qsol.

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The proof (below) is slightly subtle. Here’s why.Let ! be a subfield of C that’s finite, normal and solvable over Q, and take

U ∈ C and = ≥ 1 such that U= ∈ !. To find some larger " that contains Uitself and is also solvable over Q, we could try putting " = SF! (C= − U=). Now" : Q is indeed solvable (as can be shown using Exercises 9.1.11 and 9.2.2), butthe trouble is that " : Q is not in general normal. And normality is part of thedefinition of Qsol, ultimately because it’s an essential requirement if we want touse the fundamental theorem of Galois theory.

An example should clarify.

Example 9.2.9 Put U = 4√2 and take = = 2. We have U2 =√

2 ∈ Q(√

2), andQ(√

2) : Q is finite, normal and solvable (since its Galois group is the abeliangroup �2), so U2 ∈ Qsol. Hence, according to Lemma 9.2.8, U = 4√2 should becontained in some finite normal solvable extension " of Q.

How can we find such an "? We can’t take " = SFQ(√2) (C2 −√

2), since thisis Q( 4√2), which is not normal over Q. (You may have already contemplated theextensions Q( 4√2) : Q(

√2) : Q in Workshop 4, q. 4.)

To find a bigger " , still finite and solvable over Q but also normal, we haveto adjoin a square root not just of

√2 but also of its conjugate, −

√2. This is the

crucial point: the whole idea of normality is that conjugates are treated equally.(Normal behaviour means that anything you do for one element, you do for allits conjugates.) The result is Q( 4√2, 8) = SFQ(C4 − 2), which is indeed a finite,solvable and normal extension of Q containing 4√2.

Proof of Lemma 9.2.8 Write 0 = U= ∈ Qsol. Choose a subfield of C such that0 ∈ and : Q is finite, normal and solvable.

Step 1: enlarge to a field in which C= − 1 splits. Put ! = SF (C= − 1) ⊆ C.Since : Q is finite and normal, = SFQ( 5 ) for some 5 ∈ Q[C], and then

! = SFQ((C= − 1) 5 (C)

). Hence ! : Q is finite and normal. It follows from

Corollary 7.1.6 that ! : is normal. Its Galois group is Gal (C= − 1), which isisomorphic to a subgroup of GalQ(C= − 1) (by Corollary 6.3.13), which is abelian(by Lemma 9.1.6). Hence Gal(! : ) is abelian. Also : Q is solvable, so ! : Qis solvable.

In summary, ! is a subfield of C such that 0 ∈ ! and ! : Q is finite, normaland solvable, and, moreover, C= − 1 splits in !. We now forget about .

Step 2: adjoin the =th roots of the conjugates of 0. Write < ∈ Q[C] for theminimal polynomial of 0 over Q, and put " = SF! (<(C=)) ⊆ C. Then U ∈ " , as<(U=) = <(0) = 0. We show that " : Q is finite, normal and solvable.

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Since ! : Q is finite and normal, ! = SFQ(6) for some 6 ∈ Q[C]. Then" = SFQ(6(C)<(C=)), so " : Q is finite and normal. It follows that " : ! is finiteand normal too (by Corollary 7.1.6).

To show that " : Q is solvable, it is enough to show that " : ! is solvable (byExercise 9.2.2). Since ! : Q is normal and < ∈ Q[C] is the minimal polynomialof 0 ∈ !, it follows by definition of normality that < splits in !, say

<(C) =B∏8=1(C − 08)

(08 ∈ !). Define subfields !0 ⊆ · · · ⊆ !B of C by

!0 = !

!1 = SF!0 (C= − 01)!2 = SF!1 (C= − 02)...

!B = SF!B−1 (C= − 0B).Then

!8 = !({V ∈ " : V= ∈ {01, . . . , 08}

}).

In particular, !B = " . For each 8 ∈ {1, . . . , B}, the extension !8 : !8−1 is finiteand normal (being a splitting field extension), and its Galois group is abelian (byLemma 9.1.8 and the fact that C= − 1 splits in ! ⊆ !8−1). So " : ! is solvable. �

Now we can relate the set Qrad of radical numbers, defined in terms of basicarithmetic operations, to the set Qsol, defined in terms of field extensions.Proposition 9.2.10 Qrad ⊆ Qsol. That is, every radical number is contained insome subfield of C that is a finite, normal, solvable extension of Q.

As I mentioned, Qrad and Qsol are actually equal, although we won’t prove this.Proof By definition of Qrad, it is enough to show that Qsol is a subfield of Cwith the property that U= ∈ Qsol ⇒ U ∈ Qsol. We have just proved that property(Lemma 9.2.8), so it only remains to show that Qsol is a subfield of C.

The argument is similar to the proof that the algebraic numbers form a field(Proposition 5.3.7). Let U, V ∈ Qsol. Then U ∈ ! and V ∈ " for some !, " thatare finite, normal and solvable overQ. By Lemma 9.2.6, U, V ∈ # for some # thatis finite, normal and solvable overQ. Then U+V ∈ # , so U+V ∈ Qsol, and similarlyU · V ∈ Qsol. This shows that Qsol is closed under addition and multiplication. Theother parts of the proof (negatives, reciprocals, 0 and 1) are straightforward. �

This brings us to the main result of this chapter. Notice that it doesn’t mentionfield extensions: it goes straight from polynomials to groups.

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Theorem 9.2.11 Let 5 ∈ Q[C]. If the polynomial 5 is solvable by radicalsthen the group GalQ( 5 ) is solvable.

Proof Suppose that 5 is solvable by radicals. Write U1, . . . , U= ∈ C for its roots.For each 8, we have U8 ∈ Qrad (by definition of solvability by radicals), henceU8 ∈ Qsol (by Proposition 9.2.10). So each of U1, . . . , U= is contained in somesubfield of C that is finite, normal and solvable over Q. By Lemma 9.2.6, there issome subfield " of C that is finite, normal and solvable over Q and contains all ofU1, . . . , U=. Thus,

SFQ( 5 ) = Q(U1, . . . , U=) ⊆ ".By Lemma 9.2.4, Gal(" : Q) is solvable. Now SFQ( 5 ) : Q is normal, so

by the fundamental theorem of Galois theory, Gal(SFQ( 5 ) : Q) is a quotient ofGal(" : Q). But Gal(SFQ( 5 ) : Q) = GalQ( 5 ), and a quotient of a solvable groupis solvable, so GalQ( 5 ) is solvable. �

Examples 9.2.12 i. For 0 ∈ Q and = ≥ 1, the polynomial C= − 0 is solvable byradicals, so the group GalQ(C=−0) is solvable. You may already have provedthis in Exercise 9.1.11. It also follows fromExample 9.2.3 and Lemma 9.2.4.

ii. Let 01, . . . , 0: ∈ Q and =1, . . . , =: ≥ 1. Each of the polynomials C=8 − 08 issolvable by radicals, so their product is too. Hence GalQ

(∏8 (C=8 − 08)

)is a

solvable group.

Theorem 9.2.11 is most sensational in its contrapositive form: ifGalQ( 5 ) is notsolvable then 5 is not solvable by radicals. That’s the subject of the next section.

Digression 9.2.13 The converse of Theorem 9.2.11 is also true: if GalQ( 5 )is solvable then 5 is solvable by radicals. You can even unwind the proofto obtain an explicit formula for the solving the quartic by radicals (Stewart,Chapter 18).

To prove this converse statement, we have to deduce properties of a fieldextension from assumptions about its Galois group. A solvable group isbuilt up from abelian groups, and every finite abelian group is a direct sumof cyclic groups. The key step in the proof of the converse has come to beknown as ‘Hilbert’s Theorem 90’ (Stewart’s Theorem 18.18), which givesinformation about any field extension whose Galois group is cyclic.

Digression 9.2.14 The proof of Theorem 9.2.11 might not have ended quitehow you expected. Given my explanations earlier in the chapter, you mightjustifiably have imagined we were going to show that when the polynomial5 is solvable by radicals, the extension SFQ( 5 ) : Q is solvable. That’s not

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what we did. We showed that SFQ( 5 ) is contained in some larger subfield" such that " : Q is solvable, then used that to prove the solvability of thegroup GalQ( 5 ).But all is right with the world: SFQ( 5 ) : Q is a solvable extension. Indeed, itsGalois group GalQ( 5 ) is solvable, so Lemma 9.2.4 implies that SFQ( 5 ) : Qis solvable too.

9.3 An unsolvable polynomialHere we give a specific example of a polynomial over Q that is not solvable byradicals. By Theorem 9.2.11, our task is to construct a polynomial whose Galoisgroup is not solvable. The smallest non-solvable group is �5 (of order 60). Ourpolynomial has Galois group (5 (of order 120), which is also non-solvable.

Finding Galois groups is hard, and we will use a whole box of tools and tricks,from Cauchy’s theorem on groups to Rolle’s theorem on differentiable functions.

First we establish a useful fact on the order of Galois groups.

Lemma 9.3.1 Let 5 be an irreducible polynomial over a field , with SF ( 5 ) : separable. Then deg( 5 ) divides |Gal ( 5 ) |.

Proof Let U be a root of 5 in SF ( 5 ). By irreducibility, deg( 5 ) = [ (U) : ],which divides [SF ( 5 ) : ] by the tower law, which is equal to |Gal ( 5 ) | byTheorem 7.2.19 (using separability). �

Next, we need some results about the symmetric group (=. I assume you knowthat (= is generated by the ‘adjacent transpositions’ (12), (23), . . . , (=−1 =). Thismay have been proved in Fundamentals of Pure Mathematics, and as the GroupTheory notes say (p. 58):

This is intuitively clear: suppose you have = people lined up and youwant them to switch into a different order. To put them in the orderyou want them, it’s clearly enough to have people move up and downthe line; and each time a person moves one place, they switch placeswith the person next to them.

Here’s a different way of generating (=.

Lemma 9.3.2 For = ≥ 2, the symmetric group (= is generated by (12) and(12 · · · =).

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Proof We have(12 · · · =) (12) (12 · · · =)−1 = (23),

either by direct calculation or the general fact that f(01 · · · 0: )f−1 =

(f(01) · · ·f(0: )) for any f ∈ (= and cycle (01 · · · 0: ). So any subgroup �of (= containing (12) and (12 · · · =) also contains (23). By the same argument, �also contains (34), . . . , (= − 1 =). But the adjacent transpositions generate (=, so� = (=. �

Lemma 9.3.3 Let ? be a prime number, and let 5 ∈ Q[C] be an irreduciblepolynomial of degree ? with exactly ? − 2 real roots. Then GalQ( 5 ) � (?.

Proof Since charQ = 0 and 5 is irreducible, 5 is separable and therefore has ?distinct roots in C. By Proposition 6.3.10, the action of GalQ( 5 ) on the roots of5 in C defines an isomorphism between GalQ( 5 ) and a subgroup � of (?. Since5 is irreducible, ? divides |GalQ( 5 ) | = |� | (by Lemma 9.3.1). So by Cauchy’stheorem, � has an element f of order ?. Then f is a ?-cycle, since these are theonly elements of (? of order ?.

The complex conjugate of any root of 5 is also a root of 5 , so complexconjugation restricts to an automorphism of SFQ( 5 ) over Q. Exactly two of theroots of 5 are non-real; complex conjugation transposes them and fixes the rest.So � contains a transposition g.

Without loss of generality, g = (12). Since f is a ?-cycle, fA (1) = 2 forsome A ∈ {1, . . . , ? − 1}. Since ? is prime, fA also has order ?, so it is a ?-cycle.Now without loss of generality, fA = (123 · · · ?). So (12), (12 · · · ?) ∈ �, forcing� = (? by Lemma 9.3.2. Hence GalQ( 5 ) � (?. �

Exercise 9.3.4 Explain why, in the last paragraph, fA has order ?.

Theorem 9.3.5 Not every polynomial over Q of degree 5 is solvable by radi-cals.

Proof We show that 5 (C) = C5 − 6C + 3 satisfies the conditions of Lemma 9.3.3.Then GalQ( 5 ) is (5, which is not solvable, so by Theorem 9.2.11, 5 is not solvableby radicals.

Evidently deg( 5 ) is the prime number 5, and 5 is irreducible by Eisenstein’scriterion with prime 3. It remains to prove that 5 has exactly 3 real roots. This iswhere we use some analysis, considering 5 as a function R→ R (Figure 9.1).

We have

limG→−∞

5 (G) = −∞, 5 (0) > 0, 5 (1) < 0, limG→∞

5 (G) = ∞,

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Figure 9.1: The function G ↦→ G5 − 6G + 3.

and 5 is continuous on R, so by the intermediate value theorem, 5 has at least 3real roots. On the other hand, 5 ′(G) = 5G4 − 6 has only 2 real roots (± 4

√6/5), so

by Rolle’s theorem, 5 has at most 3 real roots. Hence 5 has exactly 3 real roots,as required. �

Exercise 9.3.6 Prove that for every = ≥ 5, there is some polynomialof degree = that is not solvable by radicals.

Digression 9.3.7 We now know that some polynomials 5 over Q are notsolvable by radicals, which means that not all their complex roots are radical.

Could it be that some of the roots are radical and others are not? Yes: simplytake a polynomial 6 that is not solvable by radicals and put 5 (C) = C6(C).Then the roots of 5 are 0 (which is radical) together with the roots of 6(which are not all radical).

But what if 5 is irreducible? In that case, either all the roots of 5 are radicalor none of them are. This follows from the fact that the extension Qrad : Q isnormal, which we will not prove.

Digression 9.3.8 There are many similarities between the theory of con-structibility of points by ruler and compass and the theory of solvability ofpolynomials by radicals. In both cases, the challenge is to construct somethings (points in the plane or roots of polynomials) using only certain tools(ruler and compass or a machine for taking =th roots). In both cases, therewere difficult questions of constructibility that remained open for a very longtime, and in both cases, they were solved by Galois theory.

The solutions have something in common too. For the geometry problem,we used iterated quadratic extensions, and for the polynomial problem, weused solvable extensions, which could reasonably be called iterated abelianextensions. For the geometry problem, we showed that the coordinates of any

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point constructible by ruler and compass satisfy a certain condition on theirdegree overQ (Corollary 5.4.3); for the polynomial problem, we showed thatany polynomial solvable by radicals satisfies a certain condition on its Galoisgroup. There are other similarities: compare Lemmas 5.4.4 and 9.2.6, forexample, and maybe you can find more similarities still.

We have now used the fundamental theorem of Galois theory to solve a majorproblem about Q. What else can we do with it?

The fundamental theorem is about separable extensions. Our twomain sourcesof separable extensions are:

• fields of characteristic 0 such asQ (Example 7.2.15(i)), whichwe’ve exploredextensively already;

• finite fields (Example 7.2.15(ii)), which we’ve barely touched.

In the next and final chapter, we’ll use the fundamental theorem and other resultswe’ve proved to explore the world of finite fields. In contrast to the complicatedworld of finite groups, finite fields are almost shockingly simple.

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Chapter 10

Finite fields

This chapter is dessert. Through this semester, we’ve developed a lot of sophisti-cated theory for general fields. All of it works for finite fields, but becomes muchsimpler there. It’s a miniature world in which life is sweet. For example:

Introduction toWeek 10 • If we want to apply the fundamental theorem of Galois theory to a field

extension " : , we first have to ask whether it is finite, and whether itis normal, and whether it is separable. When " and are finite, all threeconditions are automatic (Lemma 10.4.2).

• There are many fields of different kinds, and to classify them all would be anear-impossible task. But for finite fields, the classificiation is very simple,as we’ll see. We know exactly what finite fields there are and how manyelements they have.

• The Galois correspondence for arbitrary field extensions can also be com-plicated. But again, it’s simple for finite fields. Their Galois groups are veryeasy (they’re all cyclic), we know what their subgroups are, and it’s easy todescribe all the subfields of any finite field.

So although the world of finite fields is not trivial, there’s a lot about it that’ssurprisingly straightforward.

Two aspects of finite fields may seem counterintuitive. First, they always havepositive characteristic, which means they satisfy some equation like 1+ · · · +1 = 0.Second (and relatedly), ?th powers and ?th roots behave strangely in fields ofcharacteristic ?—at least, ‘strange’ if, likemost of us, characteristic 0 is what you’remost familiar with. But the behaviour of ?th roots and powers is fundamental toall the nice properties of finite fields.

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10.1 ?th roots in characteristic ?Recall from Lemma 2.2.11 that every finite field has positive characteristic, which,by Lemma 2.2.5, must be a prime number ?.

Square roots usually come in pairs: how many times in your life have youwritten a ± sign before a

√? But in characteristic 2, plus and minus are the same,

so the two square roots become one. We’ll see that this pattern persists.

Proposition 10.1.1 Let ? be a prime number and ' a ring of characteristic ?.

i. The function\ : ' → '

A ↦→ A ?

is a homomorphism.

ii. If ' is a field then \ is injective.

iii. If ' is a finite field then \ is an automorphism of '.

Proof For (i), certainly \ preserves multiplication and 1. To show that \ preservesaddition, let A, B ∈ ': then by Lemma 3.3.15 and the hypothesis that char ' = ?,

\ (A + B) = (A + B)? =?∑8=0

(?

8

)A8B?−8 = A ? + B? = \ (A) + \ (B).

Now (ii) follows since every homomorphism between fields is injective, and (iii)since every injection from a finite set to itself is bijective. �

The homomorphism \ : A ↦→ A ? is called the Frobenius map, or, in the caseof finite fields, the Frobenius automorphism.

That \ is a homomorphism is a shocker. Writing ‘(G + H)= = G= + H=’ is aclassic mistake in school-level algebra. But here, it’s true!

Examples 10.1.2 i. The Frobenius automorphism of F? = Z/〈?〉 is not veryinteresting. When� is a finite group of order =, Lagrange’s theorem impliesthat 6= = 1 for all 6 ∈ �. Applying this to the multiplicative groupF×? = F? \ {0} gives 0?−1 = 1 whenever 0 ≠ 0 ∈ F?. It follows that 0? = 0for all 0 ∈ F?. That is, \ is the identity. Everything is its own ?th root!

ii. The proof that the Frobenius map preserves addition may seem familiar.We essentially did it in Example 7.2.4, in the case of the ring (F? (D)) [C] ofpolynomials over the field F? (D) of rational expressions over F?.

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iii. We don’t have many examples of finite fields yet (but we will soon). Apartfrom those of the form F?, the simplest is F2(U), where U is a root of the irre-ducible polynomial 1+ C + C2 over F2 (Example 4.3.8(ii)). By Theorem 5.1.5,

F2(U) = {0 + 1U : 0, 1 ∈ F2} = {0, 1, U, 1 + U}.

Since 1 + U + U2 = 0 and F2(U) has characteristic 2,

U2 = 1 + U, (1 + U)2 = U.

So the Frobenius automorphism of F2(U) interchanges U and 1+U. Like allautomorphisms, it fixes 0 and 1.

Exercise 10.1.3 Write out the addition and multiplication tables ofF2(U).

Corollary 10.1.4 Let ? be a prime number.

i. In a field of characteristic ?, every element has at most one ?th root.

ii. In a finite field of characteristic ?, every element has exactly one ?th root.

Proof Part (i) says that the Frobenius map is injective, and part (ii) says that it isbijective, as Proposition 10.1.1 states. �

Examples 10.1.5 i. In a field of characteristic 2, every element has at mostone square root.

ii. In C, there are ? different ?th roots of unity. But in a field of characteristic?, there is only one: 1 itself.

iii. Let be a field of characteristic ? and 0 ∈ . Corollary 10.1.4(i) says that0 has at most one ?th root. It may have none. For instance, Exercise 7.2.5asked you to show that in F? (D), the element D has no ?th root.

iv. In the 4-element field F2(U) of Example 10.1.2(iii), the only square root ofU is 1 + U, and the only square root of 1 + U is U. Each is the square root ofthe other.

Exercise 10.1.6 Work out the values of the Frobenius automorphismon the field F3(

√2), which you first met in Exercise 4.3.9.

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10.2 Classification of finite fieldsIf you try to write down a formula for the number of groups or rings with a givennumber of elements, you’ll find that it’s hard and the results are quite strange. Forinstance, more than 99% of the first 50 billion groups have order 1024.

But fields turn out to be much, much easier. We’ll obtain a complete classifi-cation of finite fields in the next two pages.

The order of a finite field " is its cardinality, or number of elements, |" |.

Warning 10.2.1 Order and degree mean different things. For in-stance, if the order of a field is 9, then its degree over its primesubfield F3 is 2.

Lemma 10.2.2 Let " be a finite field. Then char" is a prime number ?, and|" | = ?= where = = [" : F?] ≥ 1.

In particular, the order of a finite field is a prime power.

Proof By Lemmas 2.2.5 and 2.2.11, char" is a prime number ?. ByLemma 2.2.10, " has prime subfield F?. Since " is finite, 1 ≤ [" : F?] < ∞;write = = [" : F?]. As a vector space over F?, then, " is =-dimensional and soisomorphic to F=?. But |F=? | = |F? |= = ?=, so |" | = ?=. �

Example 10.2.3 There is no field of order 6, since 6 is not a prime power.

Lemma 10.2.2 prompts two questions: given a prime power ?=, is there somefield of order ?=? And if so, how many are there?

The answer to the first question is yes:

Lemma 10.2.4 Let ? be a prime number and = ≥ 1. Then the splitting field ofC ?= − C over F? has order ?=.

Proof Put 5 (C) = C ?= − C ∈ F? [C] and " = SFF? ( 5 ). Then � 5 = −1 (since

= ≥ 1), so by (i)⇒(ii) of Lemma 7.2.10, 5 has no repeated roots in " . Hence "has at least ?= elements.

Write \ for the Frobenius map of " . The set ! of roots of 5 in " is the set offixed points of \= = \ ◦ · · · ◦ \, which is Eq{\=, id"}. Since \ is a homomorphism,! is a subfield of " (by Lemma 7.3.2), and contains all the roots of 5 in " . Henceby definition of splitting field, ! = "; that is, every element of " is a root of 5 .Since deg( 5 ) = ?=, this implies that " has at most ?= elements. �

As for the second question, there is exactly one field of each prime power order.To show this, we need a lemma.

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Lemma 10.2.5 Let " be a finite field of order @. Then U@ = U for all U ∈ " .

The proof uses the same argument as in Example 10.1.2(i).

Proof Themultiplicative group"× has order @−1, so Lagrange’s theorem impliesthat U@−1 = 1 for all U ∈ "× = " \ {0}. Hence U@ = U whenever 0 ≠ U ∈ " , andclearly the equation holds for U = 0 too. �

Exercise 10.2.6 Verify that V4 = V for all V in the 4-element fieldF2(U) of Examples 10.1.2(iii) and 10.1.5(iv),

Lemma 10.2.7 Every finite field of order @ is a splitting field of C@ − C over F?.

Proof Let " be a field of order @. By Lemma 10.2.2, @ = ?= for some prime ?and = ≥ 1, and char" = ?. Hence " has prime subfield F?. By Lemma 10.2.5,every element of " is a root of 5 (C) = C ?= − C. So 5 has |" | = ?= = deg( 5 )distinct roots in " , and therefore splits in " . The set of roots of 5 in " generates" , since it is equal to " . Hence " is a splitting field of 5 . �

Together, these results completely classify the finite fields.

Theorem 10.2.8 (Classification of finite fields) i. Every finite field hasorder ?= for some prime ? and integer = ≥ 1.

ii. For each prime ? and integer = ≥ 1, there is exactly one field of order?=, up to isomorphism. It has characteristic ? and is a splitting field forC ?= − C over F?.

Proof This is immediate from the results above together with the uniqueness ofsplitting fields (Theorem 6.2.12(ii)). �

When @ > 1 is a prime power, we write F@ for the one and only field of order @.

Warning 10.2.9 F@ is not Z/〈@〉 unless @ is a prime. It can’t be,becauseZ/〈@〉 is not a field (Example 2.2.15). Tomy knowledge, thereis no description of F@ simpler than the splitting field description.

We now know exactly how many finite fields there are of each order. But inalgebra, it’s important to think not just about the objects (such as vector spaces,groups, modules, rings, fields, . . . ), but also the maps (homomorphisms) betweenobjects. So now that we’ve counted the finite fields, it’s natural to try to count thehomomorphisms between finite fields. Field homomorphisms are injective, so thisboils down to counting subfields and automorphisms. Galois theory is very wellequipped to do that! We’ll come to this in the final section. But first, we consideranother way in which finite fields are very simple.

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10.3 Multiplicative structureThe multiplicative group × of a finite field is as easy as can be:

Proposition 10.3.1 For an arbitrary field , every finite subgroup of × is cyclic.In particular, if is finite then × is cyclic.

The multiplicativegroup of a finitefield is cyclic

Proof This was Theorem 5.1.13 and Corollary 5.1.14 of Group Theory. �

Example 10.3.2 In examples earlier in the course, we frequently used the =th rootof unity l = 42c8/= ∈ C, which has the property that every other =th root of unityis a power of l.

Can we find an analogue of l in an arbitrary field ? It’s not obvious howto generalize the formula 42c8/=, since the exponential is a concept from complexanalysis. But Proposition 10.3.1 solves our problem. For = ≥ 1, put

*= ( ) = {U ∈ : U= = 1}.

Then*= ( ) is a subgroup of ×, and is finite since its elements are roots of C=−1.So by Proposition 10.3.1, *= ( ) is cyclic. Let l be a generator of *= ( ). Thenevery =th root of unity in is a power of l, which is what we were aiming for.

Note, however, that *= ( ) may have fewer than = elements, or equivalently,the order of l may be less than =. For instance, if char = ? then*? ( ) is trivialand l = 1, by Example 10.1.5(ii).

Exercise 10.3.3 Let be a field and let � be a finite subgroup of ×of order =. Prove that � ⊆ *= ( ).

Example 10.3.4 The group F×? is cyclic, for any prime ?. This means that thereis some l ∈ {1, . . . , ? − 1} such that l, l2, . . . runs through all elements of{1, . . . , ? − 1} when taken mod ?. In number theory, such an l is called aprimitive root mod ? (another usage of the word ‘primitive’). For instance, youcan check that 3 is a primitive root mod 7, but 2 is not, since 23 ≡ 1 (mod 7).

Corollary 10.3.5 Every extension of one finite field over another is simple.

Proof Let " : be an extension with " finite. By Proposition 10.3.1, the group"× is generated by some element U ∈ "×. Then " = (U). �

This is yet another pleasant aspect of finite fields.

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Exercise 10.3.6 In the proof of Corollary 10.3.5, once we know thatthe group "× is generated by U, how does it follow that " = (U)?

Digression 10.3.7 In Digression 7.2.22, I mentioned the theorem of theprimitive element: every finite separable extension " : is simple. One ofthe standard proofs involves splitting into two cases, according to whether" is finite or infinite. We’ve just done the finite case.

Corollary 10.3.8 For every prime number ? and integer = ≥ 1, there exists anirreducible polynomial over F? of degree =.

Proof The field F?= has prime subfield F?. By Corollary 10.3.5, the extensionF?= : F? is simple, say F?= = F? (U). The minimal polynomial of U over F? isirreducible of degree [F? (U) : F?] = [F?= : F?] = =. �

This is not obvious. For example, can you find an irreducible polynomial ofdegree 100 over F31?

10.4 Galois groups for finite fieldsWe now work out the Galois correspondence for any extension of one finite fieldover another.

Warning 10.4.1 The term ‘finite field extension’ means an exten-sion " : that’s finite in the sense defined on p. 52: " is finite-dimensional as a vector space over . It doesn’t mean that " and are finite fields. But the safest policy is to avoid this term entirely.

The three hypotheses of the fundamental theorem of Galois theory are alwayssatisfied when both fields in the extension are finite:

Lemma 10.4.2 Let " : be a field extension.

i. If is finite then " : is separable.

ii. If " is also finite then " : is finite and normal.

Proof For (i), we show that every irreducible polynomial 5 over is separable.Write char = ? > 0, and suppose for a contradiction that 5 is inseparable. ByCorollary 7.2.12,

5 (C) = 10 + 11C? + · · · + 1A CA ?

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for some 10, . . . , 1A ∈ . For each 8, there is a (unique) ?th root 28 of 18 in , byCorollary 10.1.4(ii). Then

5 (C) = 2?0 + 2?

1 C? + · · · + 2?A CA ? .

But by Proposition 10.1.1(i), the function 6 ↦→ 6? is a homomorphism [C] → [C], so

5 (C) = (20 + 21C + · · · + 2A CA)? .This contradicts 5 being irreducible.

For (ii), suppose that " is finite. Write char" = ? > 0. By Theorem 10.2.8," is a splitting field over F?, so by Lemma 6.2.13(ii), it is also a splitting fieldover . Hence " : is finite and normal, by Theorem 7.1.5. �

Part (i) fulfils the promise made in Remark 7.2.13 and Example 7.2.15(ii), andthe lemma as a whole lets us use the fundamental theorem freely in the worldof finite fields. We now work out the Galois correspondence for the extensionF?= : F? of an arbitrary finite field over its prime subfield.

Proposition 10.4.3 Let ? be a prime and = ≥ 1. Then Gal(F?= : F?) is cyclic oforder =, generated by the Frobenius automorphism of F?= .

By Workshop 4, q. 7, Gal(F?= : F?) is the group of all automorphisms of F?= .

Proof Write \ for the Frobenius automorphism of F?= ; then \ ∈ Gal(F?= : F?).First we calculate the order of \. By Lemma 10.2.5, U?= = U for all U ∈ F?= , orequivalently, \= = id. If < is a positive integer such that \< = id then U?< = U forall U ∈ F?= , so the polynomial C ?< − C has ?= roots in F?= , so ?= ≤ ?<, so = ≤ <.Hence \ has order =.

On the other hand, [F?= : F?] = =, so by the fundamental theorem of Galoistheory, |Gal(F?= : F?) | = =. The result follows. �

Exercise 10.4.4 What is the fixed field of 〈\〉 ⊆ Gal(F?= : F?)?

In Fundamentals of Pure Mathematics or Group Theory, you presumably sawthat the cyclic group of order = has exactly one subgroup of order : for each divisor: of =. (And by Lagrange’s theorem, there are no subgroups of other orders.)

Exercise 10.4.5 Refresh your memory by proving this fact aboutsubgroups of cyclic groups.

In the case at hand, Gal(F?= : F?) = 〈\〉 � �=, and when : | =, the uniquesubgroup of order : is 〈\=/:〉.

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Proposition 10.4.6 Let ? be a prime and = ≥ 1. Then F?= has exactly one subfieldof order ?< for each divisor < of =, and no others. It is{

U ∈ F?= : U?<

= U}.

Proof The subfields of F?= are the intermediate fields of F?= : F?, which by thefundamental theorem of Galois theory are precisely the fixed fields Fix(�) ofsubgroups � of Gal(F?= : F?). Any such � is of the form 〈\=/:〉 with : | =, and

Fix〈\=/:〉 ={U ∈ F?= : U?

=/:= U

}.

The tower law and the fundamental theorem give

[Fix〈\=/:〉 : F?] =[F?= : F?]

[F?= : Fix〈\=/:〉]=

=

|〈\=/:〉|==

:,

so | Fix〈\=/:〉| = ?=/: . As : runs through the divisors of =, the quotient =/: alsoruns through the divisors of =, so putting < = =/: gives the result. �

Warning 10.4.7 The subfields of F?= are of the form F?< where <divides =, not < ≤ =. For instance, F8 has no subfield isomorphic toF4 (that is, no 4-element subfield), since 8 = 23, 4 = 22, and 2 - 3.

Let < be a divisor of =. By Proposition 10.4.6, F?= has exactly one subfieldisomorphic to F?< . We can therefore speak of the extension F?= : F?< withoutambiguity. SinceF?< = Fix〈\<〉 and 〈\<〉 � �=/<, it follows from the fundamentaltheorem that

Gal(F?= : F?<) � �=/< . (10.1)

So in working out the Galois correspondence for F?= : F?, we have accidentallyderived the Galois group of a completely arbitrary extension of finite fields.

In the Galois correspondence for F?= : F?, all the extensions and subgroupsinvolved are normal, either by Lemma 10.4.2 or because cyclic groups are abelian.For < | =, the isomorphism

Gal(F?= : F?)Gal(F?= : F?<)

� Gal(F?< : F?)

supplied by the fundamental theorem amounts to

�=

�=/<� �< .

Alternatively, substituting : = =/<, this is �=/�: � �=/: .

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Example 10.4.8 Consider the Galois correspondence for F?12 : F?, where ? isany prime. Writing \ for the Frobenius automorphism of F?12 , the subgroups of� = Gal(F?12 : F?) are

〈\12〉 � �1 � 1 order 1

〈\6〉 � �2 order 2〈\4〉 � �3 order 3

〈\3〉 � �4 order 4〈\2〉 � �6 order 6

� = 〈\〉 � �12 order 12

Their fixed fields are

F?12 degree 1

F?6 degree 2F?4 degree 3

F?3 degree 4F?2 degree 6

F? degree 12

Here, ‘degree’ means the degree of F?12 over the subfield, and (for instance) thesubfield of F?12 called F?4 is{

U ∈ F?12 : U?4= U

}� F?4 .

TheGalois groupGal(F?12 : F?4) is 〈\4〉 � �3, and similarly for the other subfields.

Exercise 10.4.9 What do the diagrams of Example 10.4.8 look likefor ?8 in place of ?12? What about ?432? (Be systematic!)

Ordered sets In Workshop 5, you’ll be asked to work through the Galois correspondence foran arbitrary extension F?= : F?< of finite fields, but there’s not much more to do:almost all the work is contained in the case < = 1 that we have just done.

∗ ∗ ∗

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