MATH 314, Algebra II, Galois Theory

15 February 2009

Laurence Barker, Bilkent University.

Documentation for the course — past papers, course notes, and updates to this document —can be found on the internet by Googleing my name.

Classroom hours: Wednesday 09:40-10:30, Friday 10:40-12:30, Room SAZ19.

Office hours: Initially Wednesday 08:40 - 09:30, Fen A room 129, to be rescheduled if possible.

Assistant: Deniz Kutluay, with Office Hour on Friday, 14:40-15:30, Fen A room 144.

Source Texts: The course textbook is:

P. B. Bhattacharya et al, “Basic Abstract Algebra”, 2nd edition, CUP, 1994.

For a leisurely and readable account of Galois Theory is given in:

I. Stewart, “Galois Theory”, 3rd edition, Chapman & Hall, 2004.

A deeper and more concise account of Galois theory appears in:

I. M. Isaacs, “Algebra. A Graduate Course”, Brooks/Cole, 1994.

There are many other accounts of Galois Theory in textbooks and internet sources. I shallalso be providing some notes for a few selected parts of the course.

Prerequisites: The prerequisites for Galois theory are:• Linear algebra, already prepared for in the course MATH 223 Linear Algebra I.• Group theory, already prepared for in the course MATH 323 Algebra I.• Ring theory, which we shall have to cover as we go along.

Course Aim: To find out what algebra is.The above three components of preliminary material have always been pedagogical math-

ematics. Most of the definitions and results were cooked up by teachers who wished to slowlyand gently prepare students for various fields of research.

Galois theory — full name, Galois theory of polynomial equations — differs from those threepreliminaries in that it serves a definite purpose: It succeeds in describing something difficult,namely field extensions, in terms of something comparatively easy, namely groups. It alsoresolves some problems that were once foci of mathematical research. In fact, it revolutionizedalgebra, and it was an important part of the 19th century revolution in mathematics as awhole. Sure enough, it is nowadays just a pedagogical topic — after all, we are discussing it inthe context of an undergraduate course — but it is much more than a technical prerequisitefor various parts of present-day algebra and number theory. It is one of the major paradigmsof present-day mathematical research. That is to say, mathematicians see Galois theory assomething to try to imitate in their own work. Thus, those who have mastered Galois theorycan be said to be initiated into part of the culture of modern mathematics.

1

Assessment: The components of the assessment are:

• Homeworks, quizterms, quizzes: 25%. Selected questions from the homeworks willbe marked by the assistant. Some of the quizzes will be closely related to homeworks. Todiscourage copying, quiz marks will influence homework marks. For instance, if a student isabsent from a quiz or performs badly in a quiz, then the marks for a related homework questionmay be reset to zero or to the quiz mark.

• Midterm I, Midterm II, Final, 20%, 25%, 30%: Closed-book written exams. Theresults for Midterm II will announced shortly before the deadline for withdrawal from courses.

For the first time ever, I shall not be using the “curve” method for awarding grades.To obtain a pass, grade D or higher it will be necessary to understand the basic concepts,

such as the action of the Galois group on the roots to a polynomial, the notion of an algebraicfield extension. If you feel that you are unable to learn anything from me, then there aremany textbook and internet sources. Anyone unable to assimilate the basic material on GaloisTheory in Wikipedia ought not to graduate in mathematics at this university.

For a satisfactory assessment, grade C or higher, a rough minimal criterion would be anability to apply the Fundamental Theorem of Galois Theory to concrete examples. (Wikipediagives an excellent introduction to this.) To do that, one needs more than just a logical un-derstanding of the statement and the definitions of the terms. One needs some intuition ofand familiarity with the context of the theorem; and that can be gained only through theexperience of working through concrete examples and background theoretical results.

For a high grade, the student must satisfy herself rather than the teacher. For a start, oneshould completely master the the proof of the Fundamental Theorem of Galois Theory and theproof of the Unsolvability of the Quintic. There will almost certainly not be time to cover allof that in lectures, but full accounts can be found in many text-books. Another entertaininggame is to work through all the transitive subgroups of S4 and to realize them as Galois groupsof quartics over Q. (See the entry entitled “Inverse Galois Problem” in Wikipedia.)

Brief history: Algebra, of course, has its origin in Arabic and Persian mathematics. But,arguably, its first advance beyond the trivial was though the work of the 16th century Italianmathematicians; The formula for the solution to the cubic was initiated by del Ferro, developedby Tartaglia and, in 1545, essentially completed by Cardan. That also dealt with the quartic,which had been reduced to the cubic by Ferrari. Consolidation, including a clear notion of thenegative numbers and a clear notion of a formula, emerged a few decades later, though thework of Bombelli and Vieta.

The Fundamental Theorem of Algebra, recognized in the 17th century, provided a guaran-tee, in some sense, that all polynomial equations have solutions. But a formula for the quinticremained elusive.

Lagrange, I guess somewhere around the 1770s, used a rudimentary notion of symmetryto show how del Ferro’s reduction from the quartic to the cubic could be adapted to reducethe cubic to the quadratic. He noticed that the trick does not work for the quintic. Ruffini,making several offerings during the late 18th century, and Abel, some time around the 1810s or20s, developed those ideas to show that there does not exist a radical formula for the solutionto the quintic. (I am too lazy to look up more precise dates.)

Galois, inspired by the writings of Abel, conceived a new way of thinking about mathematicsin the late 1820s and early 30s. A striking piece of evidence for of the power of his approachis that, as an easy application of just some of the preliminary ideas behind his theory, he

2

essentially proved the impossibility of three constructions that had been sought in classicalgeometry: the duplication of the cube, the trisection of the angle, the squaring of the circle;except that, to complete the treatment of the last of those three, Lindemann’s Theorem isneeded, the transcendence of π, proved in 1882.

As a much deeper achievement, Galois improved the result of Abel and Ruffini by showingthat, not only is there no general radical formula for the quintic, but some particular quinticshave no radical solution.

Those applications — classical geometric constructions and formulas for solutions to poly-nomials — are no longer considered to important. The point is simply that those applicationsindicate the immense power of the theory. No-one ever makes use of the impossibility of squar-ing the circle or the unsolvability of the quintic. What is used very frequently, in present-dayresearch, is the theory itself, especially the Fundamental Theorem of Galois Theory.

Several decades were needed before this new kind of mathematics was taken up. Liouvillepresented the theory to the Paris Academy in the 1840s, but his promised write-up neverappeared; mathematicians tend to avoid a topic when a god has announced a forthcomingclean-up. Much of the stimulus for the new way of thinking about mathematics came fromelsewhere, Riemann and Cantor especially. Still, the influence of Galois was considerable.Klein and Lie, as students together in Berlin in about 1870, were partly motivated by the ideaof creating a kind of Galois theory for partial differential equations.

The following quote — I have abridged it severely without signaling the cuts — is from apreface Galois drafted, in 1831 or 1832, in preparation for an intended book on his theory ofpolynomial equations. He is not interested in methods for solving equations. His concern iswith the structures that lie behind the solutions. This was, indeed, to become the principaltheme of modern pure mathematics. From the courses you have taken already, you may havenoticed that pure mathematics does have a preoccupation with structure.

Incidentally, he also puts some perspective on the modern myth that all undergraduatelearning must derive from being taught.

“I do not say to anyone that I owe to their advice and their encouragement ev-erything that is good in my work. I do not say it because it would be a lie. If Ihad to address anything to the great in the world or the great in science, I swearit would not be in thanks.

Since Euler, calculations have become more and more necessary but more andmore difficult. Jump on calculations with both feet; group the operations, classifythem according to their difficulty and not according to their form; such, in myview, is the task of future geometers; such is the path I have embarked upon inthis work.

Here, one makes an analysis of analysis. There will be time to carry out thecalculations when the details of a question reclaim them.”

Syllabus Outline: The ordering of the syllabus is designed so as to bring out the flow ofconcepts rather than the flow of deductions. For that reason, results will sometimes be statedwithout proof, applied, and then proved later.

• Galois groups of quadratic polynomials

• Algebraic field extensions. Eisenstein’s Criterion. Ruler-and-compass constructions.

• Galois groups of irreducible cubic polynomials

3

• Polynomial rings and unique factorization domains.

• Normal extensions. The Fundamental Theorem of Galois Theory in characteristic zero, andapplications to concrete examples.

• Automorphisms of cyclic groups and cyclotomic fields.

• Statement of the criterion for solvability of a field extension. The Unsolvability of the Quintic.

• Separable extensions, Galois extensions, and statement of the Fundamental Theorem ofGalois Theory in general.

• Proof of the Fundamental Theorem of Galois Theory (if time permits).

• (Not on the examinable syllabus.) Proof of the criterion for solvability (if time permits,which is very unlikely).

The syllabus for the Final Exam will include proof of the Fundamental Theorem of Galoistheory, regardless of whether or not I present the material in class.

4

A Narrative of the Main Ideas in MATH 314, Algebra II,

or, How You could have invented Galois Theory.

These notes are intended as a guide, to lead the student though the main ideas. I shall omitsome of the proofs, and I shall also omit much of the abstract ring theory that lies in thebackground to the proofs.

These notes will not be useful for exam revision. It is not my intention, here, to providea key-facts revision-summary. The task of composing notes for exam revision is part of thelearning process, hence your task. If you first read these words on the night before Midterm I,then I suggest you abandon all hope and just go to bed. No-one, not even a reincarnation ofGalois, would be able to make much sense of the theoretical material without first acquiringan intuition by spending much time and effort on examples and exercises.

I should also mention that these notes are experiments with text, with a view to eventuallyproducing a book. Of course, since they are in a draft form, and somewhat experimental, theywill not have the quality or accuracy of a good textbook account.

Handout 1: Field extensions.

15 February 2009

Laurence Barker, Bilkent University.

Section 1.1: Symmetries of square roots

Consider the following little problem in school-level algebra.

Problem 1.1.1: Express (2 + i)10 in the form x + iy where x, y ∈ R.

Solution: The answer is (2 + i)10 = −237− 3116i. Indeed,

(2 + i)2 = 4− 1 + 4i = 3 + 4i , (2 + i)4 = (3 + 4i)2 = 9− 16 + 24i = −7 + 24i ,

(2 + i)8 = (−7 + 24i)2 = 49− 576− 336i = −527− 336i ,

(2 + i)10 = (−527− 336i)(3 + 4i) = −1581 + 1344− (2108 + 1008)i = −237− 3116i .

Since the numbers had a lot of digits, we might be in doubt about our result unless wecheck it in a different way. By the binomial theorem,

(2 + i)10 =(

210 −(

102

)28 +

(104

)26 −

(106

)24 +

(102

)22 − 1

)+

i

((101

)28 −

(103

)27 +

(105

)25 −

(107

)23 +

(109

)2)

= 1024− 45.256 + 210.64− 210.16 + 45.4− 1 + i(10.512− 120.128 + 252.32− 120.8 + 10.2)

= 1024− 11520 + 13440− 3360 + 180− 1 + i(5120− 15360 + 8064− 960 + 20) = −237− 3116i.

5

We had to do some fairly laborious work to find the answer and then to convince ourselvesthat the answer is correct. But no further calculation is needed to deal with the following verysimilar problem.

Problem 1.1.2: Express (2− i)10 in the form x + iy where x, y ∈ R.

Solution: Applying complex conjugation to both sides of the answer to the previous problem,(2− i)10 = −237 + 3116i.

The trick, here, is so habitual that one sometimes applies it without thinking about whyit works. Complex conjugation, recall, is the bijection C → C, written z 7→ z, such thatx + iy = x − iy for x, y ∈ R. In our solution to Problem 1.1.2, we made use of the fact thatcomplex conjugation preserves addition and multiplication, z1 + z2 = z1+z2 and z1.z2 = z1 . z2

for all z1, z2 ∈ C.

Problem 1.1.3: Express (1 +√

2)8 + (1−√2)4 in the form a + b√

2, where a, b ∈ Q.

Solution: The answer is (1 +√

2)8 + (1−√2)4 = 594 + 396√

2. To see this, first note that

(1 +√

2)2 = 1 + 2√

2 + 2 = 3 + 2√

2 ,

(1 +√

2)4 = (3 + 2√

2)2 = 9 + 8 + 12√

2 = 17 + 12√

2 ,

(1 +√

2)8 = (17 + 12√

2)2 = 289 + 288 + 408√

2 = 577 + 408√

2 .

By a similar argument, (1−√2)4 = 17− 12√

2. Therefore (1 +√

2)8 + (1−√2)4 = 577 + 17 +(408− 12)

√2 = 594 + 396

√2.

Again, we might be in doubt about our totting up of the digits unless we were to do atleast part of the calculation in a different way, for instance, by using the binomial theorem tocheck our evaluation of (1+

√2)8. But, as before, once that work has been done, the following

problem becomes very easy.

Problem 1.1.4: Express (1−√2)8 + (1 +√

2)4 in the form a + b√

2, where a, b ∈ Q.

Solution: Glancing at the previous problem, we see that, by a similar argument, (1−√2)8 +(1 +

√2)4 = 594− 396

√2.

Let me indulge in a personal anecdote. Long ago, a school-teacher loomed over my deskand frowned at what I was writing. I was replacing

√2 with −√2 in a complicated equation,

or something like that; I do not recall exactly. He asked me for a justification, and I replied“By symmetry.” He shook his head and said “What is symmetry? What do you mean?” Myexplanation, whatever it was, must have been unsatisfactory, because I remember him shakinghis head and informing me that I should write “By a similar argument.”

Twice, above, we used that phrase, “By a similar argument.” Is that really the only soundjustification, or is there some way of being more precise about the appeal to symmetry? Onemight try to argue that, just as i can be replaced by −i when all the other numbers are real,√

2 can be replaced by −√2 when all the other numbers are rational. But there is somethingdodgy here. One feels that the two square roots of −1 are entirely indistinguishable and hencealways interchangeable, but it is perhaps not so clear that the two square roots of 2 are quiteso indistinguishable. One property of

√2 is that it is the ratio of the length of a diagonal

to the length of a side of a square, whereas −√2 is not a ratio of any two lengths. Perhapsthat distinction between the two square roots of 2 could classified as “geometric” rather than

6

“algebraic”. We might propose that, in some “algebraic” sense, there is no distinction betweenthe two square roots of 2. But that raises the question: what do we mean by “algebraic”? Weshall return to that question in the next section.

Exercise 1.1.A: Let a, b be rational numbers, let n be a natural number, and let

x = (1 + ia√

2 + ib√

3)n + (1 + ia√

2− ib√

3)n + (1− ia√

2 + ib√

3)n + (1− ia√

2− ib√

3)n .

Without doing any calculations:

(1) Prove that x is real.

(2) Give a heuristic symmetry argument to indicate why x must be rational. (The term“heuristic” means not completely clear, not a proof, just reasonable and plausible. Despiteits supression from undergraduate mathematics, heuristic argument is extremely important inmathematical research. When work is in progress, mathematicians frequently employ it tomake good guesses as to potentially productive lines of study.)

Section 1.2: Rings and symmetries of rings

What is algebra? To answer this kind of question, one has to trace though the history. Aseveryone knows, the term algebra derives from part of the title of a book on quadratic equationsby the 9th century Arabic mathematician Al Khowarizmi. In the 16th century, the Italianmathematician Vieta proposed that the discipline should be given the more Christian-soundingname analysis. It came to be understood that the discipline was based on six operations:addition, subtraction, multiplication, division, the extraction of n-th roots, and the taking oflimits. During the 18th century, though, the terms algebra and analysis gradually acquired adistinction: ordinary algebra was understood to be concerned just with the first five operations,whereas infinite analysis was understood to include the much more subtle sixth operation. (Inthe above quote from Galois, he speaks of his topic as “analysis”, but — bearing in mind hisodd use of the word “geometer” — perhaps he has his tongue in his cheek.)

Above, we were discussing school mathematics, and school mathematics does consist, more-or-less, of the easiest bits of the mathematics of the 18th century. So let us review those sixoperations. Certainly, we can put aside the limit operation, which has nothing to do with thesymmetry trick that we have been considering. Sure enough, extraction of n-roots is relevantto our concerns but, already, in view of what we have seen above, that operation seems to besomewhat mysterious, so let us put that aside too. Division and subtraction are not reallyfundamental, because they can be described in terms of addition and multiplication. Whatremains, then, are the addition and multiplication operations. Now let us leave the 18thcentury behind us.

We define a ring to be a set R equipped with two operations R×R → R, one of them calledaddition, written as (a, b) 7→ a + b, the other called multiplication, written as (a, b) 7→ ab,the two operations being required to satisfy the following three axioms:

Additive Group Axiom: Under the addition operation, R is an abelian group.

Multiplicative Associativity Axiom: The multiplication operation is associative.

Distributivity Axiom: Given a, b, c ∈ R then a(b + c) = ab + ac and (a + b)c = ac + bc.

Recall that, when the operation on an abelian group is written as addition, the identityelement is called the zero element, written as 0. In particular, any ring has a zero element.

7

A ring R is said to be unital provided it satisfies the following further axiom.

Unity Axiom: There is an element 1 ∈ R such that 1 6= 0 and 1.a = a = a.1 for all a ∈ R.

It is easy to check that, for a unital ring, the element 1 is unique. We call it the unityelement of R.

We mention that there are variations in the literature. Most of the rings that appear inapplications are unital, and sometimes the Unity Axiom is included in the definition of a ring.The clause 0 6= 1 is sometimes omitted from the Unity Axiom.

Very often, mathematical objects of some kind or another are defined in this way, as setsequipped with some kind of structure. And then, very often, homomorphisms are defined to bethe functions that preserve the structure in some suitable sense. For example, the structuralequipment of a vector space consists of the addition and the scalar multiplication; homomor-phisms of vector spaces, called linear maps, are the functions which preserve the addition andscalar multiplication. As another example, the structural equipment of a topological space isthe set of open sets; homomorphisms of topological spaces, called continuous maps, preserveopen sets but in a slightly odd way, namely the preimage of an open set is required to be open,or equivalently, maps of topological spaces are required to send closed sets to closed sets. Asone more example, the structural equipment of a group is the group operation; homomorphismsof groups are required to preserve the group operation.

Let us do the same for rings. Given rings S and R, a homomorphism from S to R isdefined to be a function θ : S → R such that

θ(a + b) = θ(a) + θ(b) , θ(ab) = θ(a)θ(b)

for all a, b ∈ S. If θ is bijective, we call θ an isomorphism. In that case, the inverse functionθ : S → R is also an isomorphism. When there exists an isomorphism from S to R, we saythat S and R are isomorphic to each other, and we write S ∼= R.

A homomorphism R → R is called an endomorphism of R. An isomorphism R → R iscalled an automorphism of R. As a trivial example, for any ring R, the identity function onR is an automorphism of R. Another example of an automorphism is the complex conjugationoperation on C.

We can now go back to the question that appeared in the anecdote above. What is sym-metry? The mathematical device for expressing symmetry is the notion of a group. To bringthat notion into play, we make the following remark.

Remark 1.2.1: Given a ring R, then the set of automorphisms of R, denoted Aut(R), is agroup under the composition operation.

Proof: We have already noted that the identity function on R is an automorphism of R. Plainly,the automorphisms of R are closed under multiplication and inversion. The composition op-eration is associative. ut

Some further notation will be convenient. Given a ring R and a subset S ⊆ R, we call S asubring of R, writing S ≤ R, when S is a ring with the addition and multiplication operationscoming from R.

While we are discussing subrings, let us mention a device that is often very useful. Wecall it the embedding trick. Actually, it is used throughout mathematics but, to indicatethe idea, let us discuss it just in the context of rings. When S is a subring of R, the inclusionS → R is an injective ring homomorphism. Conversely, as we shall explain in a moment, the

8

following obvious remark will give us a way of regarding any injective ring homomorphism asan inclusion.

Remark 1.2.2: Let S and R be rings, and let θ : S → R be an injective homomorphism. Thenthe image Im(θ) = θ(s) : s ∈ S is a subring of R, and there is an isomorphism S → Im(θ)given by s 7→ θ(s).

In the situation of the remark, we can identify S with Im(S) by identifying each elements ∈ S with the element θ(s) ∈ Im(θ). In this way, S becomes a subring of R. The injectivehomomorphism θ now becomes the inclusion S → R. When we make use this device, wecall θ an embedding. Note that this use of the word embedding is really jargon rather thana properly defined term. Generally, when we use the word, we are referring to an injectivefunction that is being regarded as an inclusion. The same idea can be applied, of course, togroups, vector spaces, topological spaces and many other kinds of mathematical objects; wecan speak of embeddings of groups, embeddings of vector spaces and so on.

Sometimes, we write the unity element of a unital ring R as 1R rather than 1. Thisnotation can be used to avoid ambiguity when the unity element 1R of R is distinct from theunity element 1S of a subring S. For example, letting R be the ring of 2× 2 matrices over C,and letting S be the subring

S =(

λ 00 0

): λ ∈ C

then R and S are both unital rings, but their unity elements are different, indeed,

1R =(

1 00 1

), 1S =

(1 00 0

).

Of course there is no need for a similar notation to distinguish zero elements: for any subringS of a ring R, the zero element 0 of S coincides with the zero element 0 of R.

Consider a subring S ≤ R, and let x ∈ R. We write S[x] to denote the smallest subring ofR such that S ≤ S[x] 3 x. If R is commutative then the elements of S[x] are those elementsof R that can be written in the form ∞∑

i=0

aixi

where each ai ∈ S and only finitely many of the ai are non-zero. Since only finitely many ofthe ai are non-zero, every element of S[x] have the form

anxn + an−1xn−1 + ... + a2x

2 + a1x + a0

for some natural number n. Note that the rule for multiplying two such elements is

(anxn + ...a1x + a0)(bmxm + ... + b1x + b0) = cn+mxn+m + ... + c1x + c0

where ci = a0bi + a1bi−1 + ... + aib0. As an example, C = R[i] = x + iy : x, y ∈ R. In fact,every element of C can be written uniquely in the form x + iy where x, y ∈ R. That is to say,if x + iy = x′ + iy′, then x = x′ and y = y′.

More generally, given a finite subset X = x1, ..., xn ⊆ R, we write S[X] or S[x1, ..., xn]to denote the smallest subring of R such that S ≤ S[X] ⊇ X. Thus, if R is commutative, thenthe elements S[T ] are those elements of R that can be expressed in the form

∞∑

i1,...,in=0

ai1,...,in xi11 ...xin

n

9

where only finitely many of the ai1,...,in are non-zero.

Example 1.2.3: Every element of Q[√

2] can be written uniquely in the form a + b√

2, wherea, b ∈ Q.

Proof: Plainly, every element can be written in that form. The uniqueness follows from theirrationality of

√2. ut

Another anecdote: A certain Cambridge don used to claim that, when interviewing candi-dates for the undergraduate programme in mathematics, he would sometimes ask for a proofof the irrationality of

√2, and any candidate who failed to do so would be instantly rejected

on the grounds that he or she could not possibly be interested in mathematics.We have already noted that complex conjugation is an automorphism of C. Here is another

example of a ring automorphism.

Example 1.2.4: There is an automorphism σ of Q[√

2] given by σ(a + b√

2) = a − b√

2 fora, b ∈ Q.

Proof: Given a, b, a′, b′ ∈ Q, then

σ((a +√

2b)(a′ +√

2b′)) = σ(aa′ + 2bb′ +√

2(ab′ + a′b))

= aa′ + 2bb′ −√

2(ab′ + a′b) = (a−√

2b)(a′ −√

2b′) = σ(a +√

2b) σ(a′ +√

2b′) .

So σ preserves multiplication. It is even easier to show that σ preserves addition. We haveestablished that σ is an endomorphism of Q[

√2]. But, plainly, σ is a bijective, so σ is an

automorphism. utAt last, we can avoid the “By a similar argument” clause that appeared in the solutions

to the third and fourth problems above. Instead, we can apply the automorphism σ. Theadvantage may seem slender, but see Exercise 1.2.C, where the “By a similar argument” clausecannot be applied.

Exercise 1.2.A: Show that Aut(Q) ∼= Aut(R) ∼= C1, where C1 denotes the trivial group.(Hint: Show that, given ρ ∈ Aut(R) and x ∈ R, then ρ(x) > 0 if and only if x > 0.)

Exercise 1.2.B: Show that Aut(Q[√

2]) ∼= C2. (Hint: First show that any automorphism ofQ[√

2] restricts to an automorphism of Q. Warning: Aut(C) is not isomorphic to C2, despitethe fact that C = R[i] and Aut(R) ∼= C1. Using techniques that are beyond the scope of thiscourse, it can be shown that Aut(C) is infinite.)

Exercise 1.2.C: Let f and g be polynomials over Q, and suppose that f(√

2) = g(√

2). Showthat f(−√2) = g(−√2). Generalize this conclusion to the scenario of Example 1.7. (Comment:This is close to the way Galois saw the symmetries, that is, as symmetries of polynomials ratherthan symmetries of field extensions.)

Exercise 1.2.D: Show that every element of Q[i,√

2] can be written uniquely in the forma + bi + c

√2 + di

√2 where a, b, c, d ∈ Q.

Exercise 1.2.E: Show that Aut(Q[i,√

2]) ∼= V4, the Klien 4-group.

Section 1.3: Fields

10

In the previous section, we characterized “algebra” in an 18th century way — and coinciden-tally in a school-mathematics kind of way — as something to do with addition, subtraction,multiplication, division and the extraction of n-th roots. Those operations do seem to be rel-evant to our concerns. Indeed, Problems 1.1.1 and 1.1.2 involved the two solutions ±i to thequadratic equation x2 + 1 = 0. To deal with those two solutions, we extended the ring R tothe ring C = R[i]. Similarly, Problems 1.1.3 and 1.1.4 involved the two solutions ±√2 to thequadratic equation x2 − 2 = 0, and we extended the ring Q to the ring Q[

√2]. Generally, a

quadratic equation has the formax2 + bx + c = 0

where a 6= 0. The mantra that we we all learned to chant in kindergarten is, of course,

x =−b±√b2 − 4ac

2a.

All five of the operations under discussion are involved in this formula.We have already introduced the notion of a ring, defined in terms of addition and multipli-

cation. Of course, rings admit subtraction too. For division, we need the existence of inverses.A ring R is said to be a division ring provided it satisfies the following axiom.

Inversion Axiom: For all non-zero elements x of R, there exists an element x−1 such thatx−1.x = 1 = x.x−1.

We call x−1 the inverse of x. Note that if R is a division ring then, for u, v ∈ R satisfyingux = 1 = xv, we have v = x−1xv = x−1 and similarly u = x−1. In particular, for any non-zerox, the inverse x−1 is unique.

One important example of a division ring — it arises naturally in many different contexts— is the ring of quaternions H = R ⊕ Ri ⊕ Rj ⊕ Rk, where the multiplication is determinedby the equation i2 = j2 = k2 = ijk = −1. The multiplication is not commutative, sinceij = k 6= −k = ji.

But, for all the rings considered in Sections 1 and 2, the multiplication was commuta-tive. Let us trumpet that property by giving it an axiomatic status. A ring R is said to becommutative provided it satisfies the following axiom.

Multiplicative Commutativity Axiom: The multiplication operation is commutative.

We define a field to be a commutative division ring. In introductory courses in linearalgebra, two fields that tend to be given especial attention are R and C, partly because of theirusefulness in real and complex analysis. Usually, in a first course on real analysis, R and C areconstructed from the field Q. The next remark gives another example of a field.

Example 1.3.1: The ring Q[√

2] is a field.

Proof: Plainly, Q[√

2 is commutative unital ring. It remains only to show that any non-zeroelement x of Q[

√2] has an inverse. Write x = a + b

√2 where a, b ∈ Q. Since x 6= 0, we have

a 6= 0 or b 6= 0, hence a2 − 2b2 6= 0. By direct calculation, x−1 = (a− b√

2)/(a2 − 2b2). utThe field in the next remark may be familiar from an introductory course in classical

number theory. Recall that, for an integer n ≥ 2, the ring of modulo n integers, denotedZ/n, is defined as follows. As a set, we define Z/n to be the set of equivalence classes underthe relation of modulo n congruence. We make Z/n become a ring with the addition and

11

multiplication operations inherited from Z. That is to say, writing [a]n and [b]n for modulo ncongruence classes of integers a and b, then

[a]n + [b]b = [a + b]n , [a]n [b]n = [ab]n .

Obviously, Z/n is a commutative unital ring.

Remark 1.3.2: Let n ∈ Z with n ≥ 2. Given a ∈ Z, then the element [a]n ∈ Z/n has aninverse if and only if a is coprime to n. In particular, Z/n is a field if and only if n is prime.

Proof: First suppose that a is coprime to n. Then there exist integers u and v such thatua + vn = 1. We have [u]n[a]n = [ua]n = [1− vn]n = [1]n. Thus [a]n has inverse [u]n.

Now suppose a is not coprime to n. Let b ≥ 2 be an integer dividing a and n. Then, forany integer u, the product ua is divisible by b. Perforce, ua is not congruent to 1 modulo n,and [u]n is not an inverse of [a]n. ut

Thus, for each prime p, there is a field Fp = Z/p with order p. One reason for using thenotation Fp rather than Z/p is that we wish to emphasize the fact that Fp is a field.

The smallest field is F2, which has only two elements, 0 and 1. In F2, we have 1+1 = 0. Thisraises a problem for us, because we might like to consider quadratic equations over arbitraryfields. However, the above formula for the solution to ax2 + bx + c = 0 does not make sensewhen a, b, c ∈ F2. Evidently, we shall have to be careful.

Nevertheless, it seems that fields might provide a good context in which to investigate thenotions of symmetry that we started to explore in Section 1.1. Now, if the notion of a field is tobe of much concern to us then, surely, homomorphisms between fields must also be of concernto us. But now we collide with a surprising result. Later, we shall find that the followingproposition can be proved very quickly and elegantly using the notion of a quotient ring. Fornow, though, just to familiarize ourselves with the notion of a ring homomorphism, let us givean ad hoc proof.

Proposition 1.3.3: Given a field F and a ring R, then any non-zero ring homomorphismF → R is injective.

Proof: For a contradiction, suppose that ν : F → R is a non-zero non-injective ring homo-morphism. Since ν is non-zero, there exists an element a ∈ F such that ν(a) 6= 0. Thenν(1)ν(a) = ν(1.a) = ν(a) 6= 0, hence ν(1) 6= 0.

On the other hand, since ν is non-injective, there exist elements b, c ∈ F such that b 6= cand ν(b) = ν(c). Putting d = b − c, then d 6= 0 but ν(d) = 0. But F is a field, so d has aninverse d−1 and ν(1) = ν(d−1.d) = ν(d−1)ν(d) = 0. This is a contradiction, as required. ut

In the situation of the proposition, we can apply the embedding trick mentioned in theprevious section. Via the embedding ν, each element x ∈ F is identified with the elementν(x) ∈ R. Thus F is identified with the subring ν(F ) of R.

We note another property of homomorphisms in the case where one of the rings is a field.

Lemma 1.3.4: Let F be a field and let R be a unital ring.

(1) Given a non-zero ring homomorphism µ : R → F , then µ(1R) = 1F .

(2) If R is a subring of F , then 1R = 1F .

Proof: For any non-zero element x ∈ F such that x2 = x, we have x = x−1x2 = x−1x = 1F .

12

We have 12R = 1R. So, under the hypothesis of part (2), 1R = 1F . Under the hypothesis

of part (1), µ(1R)2 = µ(1R). By the first paragraph of the previous proof, µ(1R) 6= 0. Henceµ(1R) = 1F . ut

Note that parts (1) and (2) are, in some sense, equivalent to each other. Part (2) followsfrom part (1) by considering the inclusion R → F . Conversely, part (1) follows from part (2)by the embedding trick.

Combining the latest two results, we obtain the following corollary.

Corollary 1.3.5: Let F and E be fields, and let ν : F → E be a ring homomorphism. Then νis injective. Furthermore, ν(1F ) = 1E and, in particular, if F is a subfield of E, then 1F = 1E.

We define a field extension to be a pair (E, F ) where E is a field and F is a subfield of E.In a standard notation, field extensions are usually written more briefly in the form E/F ratherthan (E,F ). The importance of this definition comes from the fact that field extensions areessentially the same things as non-zero homomorphisms between fields. Given a field extensionE/F then, of course, the inclusion F → E is a non-zero homomorphism. Conversely, given anon-zero homomorphism ν : F → E, then we can view ν as an embedding and, via ν, we canregard F as a subfield of E. Thus, we obtain a field extension E/F .

Evidently, the study of field extensions amounts to the same thing as the study of ho-momorphisms between fields. The reason for speaking of field extensions E/F rather thanhomomorphisms between fields F → E is just that it makes things easier if we regard F as asubfield of E.

The next result tells us that every field is an extension of Q or Fp for some prime p.

Proposition 1.3.6: Any field F has a unique minimal subfield S, and either S ∼= Q or elseS ∼= Fp, where p is a prime.

Proof: Let θ : Z → F be the ring homomorphism such that θ(1) = 1. First suppose that θ isinjective. Applying the embedding trick, we can regard Z as a subring of F via θ. By Lemma1.3.3, 1Z = 1F , so we can write 1 unambiguously. Given n,m ∈ Z with m 6= 0, then m has aninverse m−1 in F . Letting S be the subset of F consisting of the elements, n/m = nm−1, thenS is a field, indeed, S is an isomorphic copy of the field Q. By Corollary 1.3.4, any subfieldL of E has the same unity element as E, so L contains S. In other words, S is the minimalsubfield of E.

Now suppose that θ is non-injective, say, θ(p) = 0 for some non-zero integer p. Thenθ(−p) = 0, so we may assume that p is positive. Let us now also also assume that p is as smallas possible. If p = nm as a product of positive integers, then θ(n)θ(m) = θ(p) = 0, so θ(n) = 0or θ(m) = 0. By the minimality of p, we deduce that n = p or m = p. We have shown that pis prime. Therefore, the image Im(θ) = 0, θ(1), ..., θ(p − 1) is isomorphic to Fp. Arguing asbefore, we see that S is the unique minimal subfield of E. ut

When S ∼= Q, we define char(F ) = 0, and when S ∼= Fp, we define char(F ) = p. Thenatural number char(F ) is called the characteristic of F . Note that, for any field extensionE/F , the unique minimal subfield of F must also be the minimal subfield of E. In particular,char(E) = char(F ), so we can speak unambiguously of the characteristic char(E/F ) of thefield extension E/F .

As we noted earlier, the above formula x = (−b ±√b2 − 4ac)/2a does not make sense forthe field F2. In fact, the formula does not make sense for any field with characteristic 2. In

13

our study of quadratic equations in the next section, it will be convenient to avoid fields withcharacteristic 2.

Exercise 1.3.A: Show that there exists a field F4 with order 4. (Hint: The multiplicativegroup F − 0 must have order 3. Using that observation, work out what the addition andmultiplication tables have to be, and then just observe that this yields a ring, it is unital, it iscommutative, and it is a division ring.)

Exercise 1.3.C: Show that, for all a, b, c ∈ F2 with a 6= 0, there exists an element α ∈ F4 suchthat aα2 + bα + c = 0. (Hint: there are only four case to consider.)

Exercise 1.3.D: Let t be the real cube root of 2.

(1) Show that t and t2 are irrational.

(2) Show that Aut(Q[t]) ∼= C1. (Hint: Use Exercise 1.B. Given σ ∈ Aut(Q[t]), consider theelement σ(t) ∈ Q[t].)

(3) Show that any element of Q[t] can be written uniquely in the form a + bt + ct2 wherea, b, c ∈ Q. (Hint: Suppose, for a contradiction, that there exist rational numbers a, b, c, notall of them zero, such that a + bt + ct2 = 0. Use parts (1) and (2), and also use Example 1.7.)

(4) Rewrite the equation (a+ bt+ ct2)(x+ yt+ zt2) = x′+ y′t+ z′t2 in the form M(x, y, z)T =(x′, y′, z′)T , where M is a 3 × 3 matrix. (The T here denotes transposition, just a littledevice of notation for expressing column vectors in a line of text.) Show that det(M) =a3 + 2b3 + 4c3 − 6abc.

(5) Show that there do not exist non-zero rational numbers a, b, c such that a3 + 2b3 + 4c3 −6abc = 0. (Hint: the trick, using parities, is similar to the well-known ancient Greek proof ofpart (1).)

(6) Deduce that there exist rational numbers x, y, z such that M(x, y, z)T = (1, 0, 0).

(7) Hence deduce that Q[t] is a field.

Section 1.4: Quadratic polynomials

We began to consider quadratic equations in the previous section. One important example isthe quadratic equation x2 + 1 = 0, which has solutions i and −i in C. The terminology thatwe have been using is potentially confusing, because there seems to be a distinction betweenthe equation x2 + 1 = 0 and the equality x2 + 1 = 0. The equation is something that hastwo solutions in C, namely x = i and x = −i. On the other hand, the equality is a statementwhich holds for x ∈ C if and only if x = i or x = −i. Usually, a statement is understood to besomething that is either true or else false; a statement does not have “solutions”. Instead ofdebating this philosophical subtlety, let us simply avoid it by introducing a cunning device ofnotation.

Let R be a commutative unital ring. We define a polynomial over R to be an infinitesequence (a0, a1, a2, ...) where each ai ∈ R and only finitely many of the ai are non-zero. Wemake the set of polynomials over R become a ring with addition and multiplication given by

(a0, a1, ...) + (b0, b1, ...) = (a0 + b0, a1 + b1, ...) ,

(a0, a1, ...)(b0, b1, ...) = (c0, c1, ...)

14

where ci = a0bi + a1bi−1 + ... + aib0. In the notation introduced just prior to Example 1.2, thering of polynomials over R is precisely the ring R[X], where X = (0, 1, 0, 0, 0, ...) (with onlyone non-zero term). Indeed,

(a0, a1, a2, ...) = a0 + a1X + a2X2 + ... .

Furthermore, since only finitely many of the ai are non-zero, we can write

(a0, a1, a2, ...) = anXn + an−1Xn−1 + ... + a2X

2 + a1X + a0

where n is such that ai = 0 for i > n. We now introduce a confusing but standard device ofnotation. Elements of R[X] are sometimes denoted by expressions such as f(X) or g(X) orh(X). The reason for adding the (X) at the end will become apparent in a moment. Consideran element f(X) ∈ R[X], say,

f(X) = anXn + an−1Xn−1 + ... + a2X

2 + a1X + a0 .

We define the polynomial function associated with f(X) to be the function R → R suchthat

x 7→ anxn + an−1xn−1 + ... + a2x

2 + a1x + a0

for x ∈ R. The polynomial function associated with f(X) is denoted by the symbol f . Thus,

f(x) = anxn + an−1xn−1 + ... + a2x

2 + a1x + a0 .

In some contexts, the distinction between a polynomial and a polynomial function donot matter very much. For instance, two polynomials f(X) and g(X) over C are equal toeach other if and only if the polynomial functions f : C → C and g : C → C are equal toeach other. This is why, in the differential calculus, polynomials are treated as functions.Nevertheless, a polynomial f(X) ∈ R[X] is not the same thing as the associated polynomialfunction f : R → R. To illustrate this point in a striking way, let us note that, if R is finite,then there are only finitely many functions R → R and, perforce, there are only finitely manypolynomial functions R → R, but there are infinitely many polynomials over R.

Now let f(X) be a non-zero polynomial over R, and write

f(X) = anXn + an−1Xn−1 + ... + a2X

2 + a1X + a0

where an 6= 0. We call n the degree of f(X). We call a0, a1, ..., an the coefficients of f(X),and we call an the leading coefficient. We mention that polynomials of degrees 1, 2, 3, 4, 5are called linear, quadratic, cubic, quartic, quintic, respectively. An element r ∈ R suchthat f(r) = 0 is called a root to the polynomial f(X).

One obvious motive for defining the addition and multiplication operations on R[X] is sim-ply that those operations are naturally suggested by our experience of manipulating formulas;the defining formula for the multiplication operation on R[X] is very similar to a formula thatappeared in Section 1.1. Given elements x, α, β ∈ R, then (x−α)(x−β) = x2− (α+β)x+αβ.The ring structure that we have imposed on R[X] allows us to write (X − α)(X − β) =X2 + (α + β)X + αβ, which is a quadratic polynomial with roots α and β. Later, we shall bemaking a study of R[X] as a ring, and we shall find that some theory emerges which can beapplied to the study of field extensions. For the time-being, however, we shall just be usingthe notion of a polynomial as a convenience of terminology that avoids any possible confusionbetween equations and equalities.

15

Proposition 1.4.1: Let F be a field with char(F ) 6= 2, and let f(X) = aX2 + bX + c be aquadratic polynomial over F . Write ∆ = b2 − 4ac. Then:

(1) If ∆ = 0, then f(X) has exactly one root, namely −b/2a.

(2) If ∆ 6= 0 and ∆ does not have a square root in F , then f(X) does not have a root in F .

(3) If ∆ 6= 0 and ∆ has a square root δ in F , then there are exactly two distinct square rootsof ∆ in F , namely δ and −δ, furthermore, there are exactly two distinct roots to F (X) in F ,namely (−b + δ)/2a and (−b− δ)/2a.

Proof: Suppose that f(X) has a root γ ∈ F . Putting ε = 2aγ + b, then γ = (−b + ε)/2a andε2 = 4a2α2 + 4abα + b2 = ∆. So part (2) holds.

Now suppose that ∆ has a square root δ ∈ F . By direct calculation, it is easy to confirmthat the elements α = (−b + δ)/2a and β = (−b − δ)/2a are roots to f(X). Letting γ and εbe as in the previous paragraph, then (ε− δ)(ε + δ) = ε2 − δ2 = 0, hence ε = ±δ and we haveε = α or ε = β. Therefore, parts (1) and (3) hold. ut

But our concern is with the situation where one field is extended to a larger field byadjoining a root to a polynomial. Given a field extension E/F and a polynomial f(X) over F ,then f(X) is also a polynomial over E, and f(X) may have roots that belong to the set E−F(the set of elements of E that do not belong to F ).

Proposition 1.4.2: Let E/F be a field extension with char(E/F ) 6= 2, let f(X) = aX2+bX+cbe a quadratic polynomial over F , and suppose that f(X) has a root in E − F . Write ∆ =b2 − 4ac. Then:

(1) ∆ has no square roots in F but exactly two square roots δ and −δ in E,

(2) f(X) has no roots in F but exactly two roots in E, namely α = (−b + δ)/2a and β =(−b− δ)/2a.

(3) We have F [α] = F [β] = F [δ]. Any element of F [α] can be written uniquely in the forms + tα where s, t ∈ F . The ring F [α] is a field.

Proof: Parts (1) and (2) are immediate from the previous proposition. We have

F [α] = s + tα : s, t ∈ F = u + vδ : u, v ∈ F = F [δ]

and similarly F [β] = F [δ]. We claim that every element of the ring F [α] = F [β] = F [γ] canbe expressed uniquely in the form u + vδ. Suppose that u + vδ = u′ + v′δ with u, v, u′, v′ inF . We must show that u = u′ and v = v′. If v = v′ then plainly u = u′. For a contradiction,suppose that v 6= v′. Then δ = (u−u′)/(v−v′) which belongs to F , hence the two roots α andβ of f(X) belong to F . But this contradicts the hypothesis that f(X) has a root in E − F .The claim is established. It follows that every element of F [α] can be expressed uniquely inthe form s + tα.

Plainly, F [α] is a commutative unital ring. It remains only to show that any non-zeroelement u + vδ has an inverse. We have u 6= 0 or v 6= 0. Neither of the two square roots ±δ of∆ belong to F , so u2 − v2∆ 6= 0. We have (u + vδ)−1 = (u− vδ)/(u2 − v2∆). ut

For positive integer n and a field F , an element x ∈ F satisfying xn = 1 is called an n-throot of unity. If n is the smallest integer such that xn = 1, then we call x a primitive n-throot of unity. For instance, the primitive 4-th roots of unity are i and −i. Instead of speaking

16

of 3-rd roots and 2-nd roots of unity, we speak of cube roots and square roots of unity. Theprimitive n-th roots of unity in C are

ω = e2πi/3 =−1 + i

√3

2, ω2 = e4πi/3 =

−1 + i√

32

.

Example 1.4.3: With the notation above, ω and ω2 are the roots to the quadratic polynomialX2 + X + 1 = 0. The ring F [ω] = F [ω2] = F [i

√3] a field.

Proof: The first sentence of the assertion is obvious. The second sentence is a special case ofthe Proposition 1.4.2. ut

Let us mention that only part of the conclusion to Proposition 1.4.2 still holds for char-acteristic 2. Consider a field extension E/F with char(E/F ) = 2. As before, let f(X) be aquadratic polynomial over F , and suppose that f(X) has a root α in E − F . Later, we shallgive an example where α is the unique root to f(X) in E. However, it can also be shown thatthe most important part of the conclusion to the proposition still holds: the ring F [α] is still afield. In fact, we shall later be proving that, given any field extension E/F and any root α ∈ Eto a polynomial over F , then F [α] is a field. To achieve that result, though, we shall needsome more sophisticated technology that will enable us to study roots of polynomials withoutactually solving the polynomial equations. In the words of Galois quoted above, we shall needto “jump on calculations with both feet”.

Exercise 1.4.A: Let ω and t be as in Example 1.4.3 and Exercise 1.3.C. Using Proposition1.4.2, show that:

(a) Any element of Q[t, ω] can be written uniquely in the form a + bt + ct2 + dω + eωt + fωt2

where a, b, c, d, e, f ∈ Q. (Hint: consider the field extension Q[t, ω]/Q[t].)

(b) Show that Q[t, ω] is a field.

(c) By direct calculation, multiplying out the brackets of the right-hand side, show that

X3 − 2 = (X − t)(X − ωt)(X − ω2t)

(Warning: you may feel that the equation is obvious, since t and ωt and ω2t are the cube rootsof 2 in C. There is some good intuition behind that observation but we cannot make use of ityet, because we have not yet proved any results about polynomial rings.)

Section 1.5: Galois groups

We are now ready to use the notion of a group to clarify the idea of symmetry thatwas introduce in Section 1.1. Let us work with the situation in Proposition 1.4, where wehave a formula for the solution to the polynomial equation under consideration. Thus, weare considering a field extension E/F with char(E/F ) 6= 2, a quadratic polynomial f(X) =aX2 + bX + c over F , we are supposing that f(X) has a root in E−F . As we have seen, f(X)has exactly two distinct roots in E, namely α = (−b + δ)/2a and β = (−b − δ)/2a in E − F ,where ±δ are the two square roots of b2 − 4ac.

In some sense, the full group of symmetries of the field F [α] is the automorphism groupAut(F [α]). Given an automorphism σ ∈ Aut(F [α]), then σ(α) = (−σ(b) + σ(δ))/2σ(a).However, the idea behind our quick solutions to Problems 1.1.2 and 1.1.4 was to interchangei and −i or to interchange

√2 and −√2 while leaving all the other numbers unchanged.

17

To capture that idea a little more generally, our concern must be with the automorphismσ ∈ Aut(F [α]) which interchanges δ and −δ while leaving a, b, c unchanged. That is to say,our desired automorphism σ ought to satisfy

σ(δ) = −δ , σ(−δ) = δ , σ(a) = a , σ(b) = b , σ(c) = c .

Such an automorphism σ, if it exists, must also interchange α and β. That is to say, σ(α) = βand σ(β) = α.

In a moment, we shall prove that such an automorphism σ does indeed exist. But first,let us introduce a definition which expresses the idea of, so to speak, keeping all the othernumbers unchanged. Consider an arbitrary field extension E/F . Let α be an automorphismof E. If α(x) = x for all x ∈ F , we say that α fixes F and we call α an F -automorphismof E. The F -automorphisms of E comprise a subgroup of Aut(E) called the Galois group ofE/F , denoted Gal(E/F ).

The next theorem collects together some material on quadratic polynomials that we alreadyestablished in the previous section, it establishes that our desired automorphism σ does indeedexist, and it also describes the Galois group associated with a quadratic polynomial. To provethe theorem, we shall just have to generalize the arguments we gave in Example 1.2.4 andExercise 1.2.B. Since we lack a formula for the solutions to a quadratic equation in characteristic2, we must exclude that case.

Theorem 1.5.1: Let E/F be a field extension with char(E/F ) 6= 2. Let f(X) = aX2 + bX +cbe a quadratic polynomial with a root in E − F , and let ∆ = b2 − 4ac. Then:

(1) The element ∆ ∈ F has precisely two square roots ±δ in E. Furthermore, δ ∈ E − F .

(2) The polynomial f(X) has precisely two roots in E, namely α = (−b + δ)/2a and β =(−b− δ)/2a. Furthermore, α, β ∈ E − F .

(3) We have F [α] = F [β] = F [δ] and, moreover, this ring is a field.

(4) There is an F -automorphism σ of F [α] such that σ(δ) = −δ, in other words, σ(α) = βand σ(β) = α.

(5) We have Gal(F [α]/F ) = 1, σ and Gal(F [α]/F ) ∼= C2.

Proof: Parts (1) and (2) were established in Propositions 1.4.1 and 1.4.2. In the proof ofProposition 1.4.2, we also established part (3), and we showed that every element of the fieldF [α] = F [β] = F [δ] can be expressed uniquely in the form u+vδ with u, v ∈ F . So there existsa well-defined function σ : F [α] → F [α] such that σ(u + vδ) = u− vδ. For the multiplication,

σ((u + vδ)(u′ + v′δ)) = σ(uu′ + vv′∆ + (uv′ + u′v)δ)

= uu′ + vv′∆− (ab′ + a′b)δ = σ(u + vδ) σ(u′ + v′δ) .

Preservation of the addition is established similarly. So σ is a homomorphism. Plainly, σ isbijective and σ acts as the identity function on F . Therefore σ is an F -automorphism. Part(4)is established.

Finally, given any F -automorphism ρ of F [α], then ρ is determined by its value at δ,indeed, ρ(u+ vδ) = ρ(u)+ ρ(v)ρ(δ) = u+ vρ(δ). We have ρ(δ)2 = ρ(δ2) = ρ(∆) = ∆, so eitherρ(δ) = δ or else ρ(δ) = −δ. So either ρ = 1 (the identity automorphism of F [α]) or else ρ = σ.Evidently, Gal(F [α]/F ) ∼= C2. ut

18

In a moment, we shall give a little application of the theorem. First, let us go back to someschool-level algebra.

Problem 1.5.2: Express (√

2 + i)8 in the form a + b√

2 + ci + di√

2 where a, b, c, d ∈ Q.

Solution: We have (√

2 + i)8 = 17− 56i√

2. Indeed,

(√

2 + i)2 = 2− 1 + 2√

2i = 1 + 2√

2i ,

(√

2 + i)4 = (1 + 2i√

2)2 = 1− 8 + 4√

2i = −7 + 4i√

2 ,

(√

2 + i)8 = (−7 + 4i√

2)2 = 49− 32− 56i√

2 = 17− 56i√

2 .

Problem 1.5.3: Express (√

2− i)8 in the form a + b√

2 + ci + di√

2 where a, b, c, d ∈ Q.

Solution: Applying complex conjugation to the solution to the previous problem, (√

2− i)8 =17 + 56i

√2.

In neither of those two problems did we make use of any of the theory that we have beendeveloping. We could have produced both of those answers when we were school-children. Butthe next problem is a little different.

Problem 1.5.4: Express (−√2 + i)8 in the form a + b√

2 + ci + di√

2 where a, b, c, d ∈ Q.

Solution: The quadratic polynomial X2 − 1 has roots ±√2, which do not belong to Q[i].Invoking Theorem 1.5.1, let σ be the non-trivial element of Gal(Q[i,

√2]/Q[i]. Then σ(i) = i

and σ(√

2) = −√2. Applying σ to the solution to Problem 1.5.2, we obtain (−√2 + i)8 =17− 56i

√2.

Here, as in Problem 1.2.4, we made use of an automorphism which interchanges the twosquare roots of 2. But we also needed to ensure that there exists an automorphism whichinterchanges the two square roots of 2 yet without interchanging the two square roots of −1.Of course, this application of Theorem 1.5.1 is rather trivial, and anyway, we could still havesolved Problem 1.5.4 when we were at school just by writing down the answer and appendingthe justification, “By a similar argument”.

The material that we have been developing will become more useful when we apply it topolynomials with higher degree. Instead of trying to find formulas for the solutions, we shallstudy polynomials and field extensions using more sophisticated techniques. We shall look atthe structures that arise, and we shall look at the symmetries of those structures.

The following two exercises are accessible using only the material that we have introducethus far, together with some group theory from Algebra I. But they are hard. So hard, in fact,that I would be amazed if anyone were able to do them at an early stage of the course. Thewill become gradually easier, though, as we acquire more theory in the subsequent parts of thecourse.

For each of these two exercises, if one person succeeds in correctly completing it beforeanyone else does, then that person will receive a significant bonus to their total credit for thecourse. However, if the first correct solution is produced by a syndicate, the members of thesyndicate will be congratulated but will not receive a bonus credit.

Exercise 1.5.A: Let ω and t be as in Exercise 1.4.A.

(1) Show that Gal(Q[t]/Q) ∼= C1.

19

(2) Show that there exists a Q-automorphism σ of Q[t, ω] such that σ(ω) = ω2 and σ(t) = t.

(3) Show that any element of Q[t, ω] can be written uniquely in the form x0 +x1 t+x2 t2 whereeach xj ∈ Q[ω].

(4) Let ρ be the function Q[t, ω] → Q[t, ω] such that

ρ(x0 + x1 t + x2 t2) = x0 + x1 ωt + x2 ω2t2 .

Show that ρ is a Q[ω]-automorphism of Q[t, ω].

(5) Show that, given Q-automorphisms σ1 and σ2 of Q[t, ω] such that σ1(t) = σ2(t) andσ1(ωt) = σ2(ωt) and σ1(ω2t) = σ2(ω2t), then σ1 = σ2.

(6) Hence deduce that Gal(Q[t, ω]/Q) ∼= S3.

Exercise 1.5.B: Let ε = e2πi/7. For this question, you may assume the following result whichwill be proved later: every element of Q[ε] can be written uniquely in the form a0 +a1ε+a2ε

2 +... + a6ε6 where each aj ∈ Q.

(1) Show that Gal(Q[ε]/Q) ∼= C6.

(2) Show thatX3 −X2 − 2X − 1 = (X − c1)(X − c2)(X − c3)

where cj = 2 cos(2πij/7) = εj + ε−j .

(3) Let E = R ∩Q[ε]. Show that E = Q[c1, c2, c3].

(4) Show that there exists a Q-automorphism σ of E such that ρ(c1) = c2 and ρ(c2) = c3 andρ(c3) = c1.

(5) Show that c2 = c21 − 2 and c3 = c2

2 − 2 and c1 = c23 − 2. Hence show that there does not

exist a Q-automorphism σ of E such that σ(c1) = c2 and σ(c2) = c1.

(6) Hence deduce that Gal(E/Q) ∼= C3.

Section 1.6: The programme

Warning: in the current draft, this section has been written in haste, and it probably hasan even higher density of slips and typos than the previous sections.

To indicate the programme, let us state two theorems which will be proved later.

Theorem 1.6.1: Let F be a field and let f(X) = anXn + ... + a1X + a0 be a polynomial overF with degree n. Then f(X) has an most n distinct roots in F . Furthermore, if f(X) has ndistinct roots α1, ..., αn in F , then f(X) = an(X − α1)...(X − αn).

When f(X) has exactly n distinct roots in F , we say that f(X) splits completely in F .

Theorem 1.6.2: Let E/F be a field extension, let α ∈ E and suppose that α is a root to apolynomial over F . Then F [α] is a field.

The theorem tells us that F [α]/F is a field extension. We shall sometimes be studyingGalois groups having the form Gal(F [α]/F ). Sometimes, though, it will be convenient toconsider a slightly more general situation. When we have proved the above theorem, thefollowing corollary will follow easily.

20

Corollary 1.6.3: Let E/F be a field extension, let Ω be a finite subset of E, and suppose thatevery element of Ω is a root to a polynomial over F . Then F [Ω] is a field.

Proof: Write Ω = α1, ..., α)n and F0 = F and Fj = F [α1, ..., αj ]. Then Fn = F [Ω]. We shallprove that each Fj is a field. Assume, inductively, that Fj−1 is a field. We have Fj = Fj−1[αj ].But α is a root to a polynomial over F and, perforce, α is a root to a polynomial over Fj−1.By the previous theorem, Fj is a field. ut

So F [Ω]/F is a field extension. We shall often be studying Galois groups having the formGal(F [Ω]/F ). But we shall find that the theory tends to work best in the case where theelements α1, ..., αn of Ω are mutually distinct roots to a polynomial with degree n which splitscompletely in E.

Much of the essential idea of Galois Theory is expressed in the following two remarks. Thetwo remarks involve some concepts that were introduced in Algebra I. Recall that, given agroup G and a G-set X, we define a G-subset of X to be a subset Y of X such that the actionof G on X restricts to an action of G on Y . That is to say, for all g ∈ G and y ∈ Y , we havegy ∈ Y .

Two elements x1, x2 ∈ X are said to be G-conjugate to each other provided gx1 = x2 forsome g ∈ G. We say that X is transitive provided any two elements of X are G-conjugate toeach other. The G-conjugacy relation on X an equivalence relation. The equivalence classes,called the G-orbits in X, are precisely the transitive G-subsets of X. Of course, X is transitiveif and only if X consists of a single G-orbit.

Remark 1.6.4: Let E/F be a field extension, let f(X) be a polynomial over F , and let Ωbe the set of roots to f(X) in E. Then Ω is a Gal(E/F )-subset of E/F . That is to say, theaction of Gal(E/F ) on E/F restricts to an action of Gal(E/F ) on Ω.

Proof: Write f(X) = amXm + ... + a1X + a0. We are to show that, given σ ∈ Gal(E/F ) andα ∈ Ω, then σ(α) ∈ Ω. The hypothesis f(α) = 0 yields the conclusion f(σ(α)) = 0 because

f(σ(α)) = am σ(α)m + ... + a1 σ(α) + a0 = σ(amαm + ... + a1α + a0) = σ(f(α)) = 0 . 2

Let us make a further recollection concerning the background group theory. As before,consider a group G and a G-set X. The equality gx = ρ(g)x expresses a group homomorphismρ : G → Sym(X). We say that the G-set X is faithful provided every non-identity element ofG acts non-trivially on X. That is to say, for all g ∈ G − 1, there exists an element x ∈ Xsatisfying gx 6= x. This is equivalent to the condition that ρ is injective. In that case, we canapply the embedding trick: we can regard G as a subgroup of of Sym(X) by identifying eachg with ρ(g).

Suppose now that X is finite with order n, say. Writing X = x1, ..., xn and identifyingeach xi with i, we obtain an identification of Sym(X) with Sn. Thus, if X is a faithful G-set,then G becomes a subgroup of Sn and, in particular, G is finite.

Remark 1.6.5: Let E/F be a field extension, let f(X) be a polynomial over F , and let Ω bethe set of roots to f(X) in E. Suppose that E = F [Ω]. Then Gal(E/F ) acts faithfully on Ω.Furthermore, Ω is finite and Gal(E/F ) is finite.

Proof: Given a non-identity element σ ∈ Gal(E/F ) then the action of σ on Gal(E/F ) isdetermined by the action of σ on Ω. But σ acts non-trivially on Gal(E/F ), so σ acts non-trivially on Ω. We have shown that Ω is faithful. Theorem 1.6.1 tells us that Ω is finite. Bythe faithfulness, Gal(E/F ) is finite. ut

21

Example 1.6.6: We put E/F = C/R, and we let f(X) = X2 + 1. The set of roots tof(X) is Ω = i,−i. The Galois group is Gal(E/F ) = 1, σ where the automorphism σ iscomplex conjugation. In particular, Gal(E/F ) ∼= C2. The action of Gal(E/F ) on Ω is suchthat σ(i) = −i and σ(−i) = i. Thus Gal(E/F ) acts transitively on Ω.

Example 1.6.7: Let E/F be a field extension with char(E/F ) 6= 2. Let f(X) = aX2 +bX +cbe a quadratic polynomial with a root in E − F . Then, as we saw in Theorem 1.5.1, f(X)has exactly two roots α and β. We have Ω = α, β. The Galois group is Gal(E/F ) = 1, σwhere σ interchanges α and β. Again, Gal(E/F ) ∼= C2 and Gal(E/F ) acts transitively on Ω.

Example 1.6.8: Let E/F = Q[ω]/Q, where ω = e2πi/3. Let f(X) = X3 − 1 = (X − 1)(X2 +X + 1). Then Ω = 1, ω, ω2. Since ω and ω2 are the roots to the quadratic polynomialX2 + X + 1, the generator σ of the group Gal(E/F ) ∼= C2 interchanges ω and ω2. But, ofcourse, σ(1) = 1. So there are two orbits of Gal(E/F ) on Ω, namely 1 and ω, ω2. Inparticular, Gal(E/F ) does not act transitively on Ω.

Example 1.6.8: Let t be the real cube root of 2 In Exercise 1.3 D we proved that Q[t] is afield. Of course, this also follows from Theorem 1.6.2, because t is a root to the polynomialf(X) = Xs− 2. The set of roots to f(X) in Q[t] is Ω = t. In Exercise 1.5.A, we proved thatGal(Q[t]/Q) ∼= C1. Thus, the action of Gal(Q[t]/Q) on Ω is just the action of the trivial groupon the trivial permutation set. Obviously, Gal(Q[t]/Q) acts transitively on Ω.

Example 1.6.9: Let E/Q = Q[t, ω], and let f(X) = X3 − 2. Then Ω = t, ωt, ω2t. ByExercise 1.5.A, Gal(E/F ) ∼= S3 and Gal(E/F ) acts transitively on Ω.

Example 1.6.10: Let E = Q[c1, c2, c3]/Q in the notation of Exercise 1.5.B, and let f(X) =X3 −X2 − 2X − 1. Then Ω = c1, c2, c3. By Exercise 1.5.B Gal(E/F ) ∼= C3 and Gal(E/F )acts transitively on Ω.

22