GALOIS THEORY

1. Basic Ring Theory: definitions and examples

In this section we recall some basic definitions from rings and discuss some examples.

Recall: A group is a nonempty set G with a binary operation · : G×G→ G satisfying

(1) · is associative, i.e. a · (b · c) = (a · b) · c, for any a, b, c ∈ G.

(2) there exists 1 ∈ G such that 1 · a = a = a · 1 for all a ∈ G.

(3) for any a ∈ G, there exists a−1 such that a · a−1 = a−1 · a = 1.

Furthermore, (G, ·) is said to be abelian if a · b = b · a for any a and b.

Remark:

(1) In the definition of a group above, the element 1 is unique and is called the

identity of G. Given g ∈ G, the element g−1 is also unique and is called the

inverse of g.

(2) We often use + to denote the group operation for an abelian gorup and in this

case we denote the identity by 0 and the inverse of a by −a.

Definition 1.1. A ring is a set R with two binary operations + (addition) and ·

(multiplication) satisfying

(1) (R,+) is an abelian group.

(2) (R, ·) is a semigroup with an identity 1, i.e.,

– the operation is associative: a · (b · c) = (a · b) · c, for any a, b, c ∈ R;

– d · 1 = 1 · d = d, for any d ∈ R.

(3) (R,+, ·) satisfies the distributive laws, i.e. for any a, b, c ∈ R

– a · (b+ c) = a · b+ a · c;

– (b+ c) · a = b · a+ c · a.

A ring is said to be commutative if rs = sr, for any r, s ∈ R.

1

2 GALOIS THEORY

Remark:

(1) In Definition 1.1, the identity 1 is unique and is called the mutiplicative identity

of R, and 0 is called the additive identity (or the zero) of R.

(2) In this course, all rings, unless otherwise specified, are commutative and dif-

ferent from the trivial ring {0}.

(3) The symbol · for the multiplication operation is often omitted.

Definition 1.2. (1) A nonempty subset I ⊆ R is an ideal if −a, a+ b, ra ∈ I for

any r ∈ R, a, b ∈ I.

(2) An ideal I is called a principal ideal if I is generated by a single element a, i.e.

I = Ra = {ra|r ∈ R}.

Definition 1.3. A commutative ring R is

(1) an integral domain (ID), if R 6= {0}, and rs = 0 implies r = 0 or s = 0;

(2) a principal ideal domain (PID), if R is an ID, and any ideal of R is a principal

ideal.

(3) a field, if (R\{0}, ·) is a group.

Example 1.4. Z,Q,R,C,Z[x],C[x],C[x, y],Z2,Z4.

(1) All but Z4 are IDs.

(2) PIDs: Z,Q,R,C,C[x],Z2.

(3) Fields: Q,R,C,Z2.

Definition 1.5. (1) A subset S ⊆ R is a subring of R, if (S,+) is a subgroup of

(R,+), S is closed under the mulitplication · and S contains the identity 1.

(2) A nonzero subring S of a field F is a subfield of F if S is closed under taking

multiplicative inverses of nonzero elements.

Example 1.6. Among the rings in Example 1.4, we have

(1) a sequence of subrings: Z ⊆ Q ⊆ R ⊆ C ⊆ C[x] ⊆ C[x, y].

(2) a sequence of subfields: Q ⊆ R ⊆ C.

GALOIS THEORY 3

2. The polynomial ring F[x]

Let F be a field. In this section we recall some basic definitions on polynomials and

discuss the main properties of the polynomial ring F[x]. We write a polynomial f in

F[X] as f = anxn + an−1x

n−1 + · · ·+ a1x+ a0, where ai ∈ F.

2.1. Definitions and main properties. The degree of a polynomial f = anxn +

an−1xn−1 + · · ·+ a1x+ a0 is defined as

deg f =

m if f 6= 0 and m = max{i|ai 6= 0}

−∞ otherwise.

For any two polynomials f and g ,

deg fg = deg f + deg g.

Recall: In an integral domain R, a nonzero element r is called a unit if there exists

s ∈ R such that rs = 1 and a nonzero nonunit element r is called an irreducible

element if r = st implies that s or t is a unit. If it is in the polynomial ring F[x] over

a field F, an irreducible polynomial can be understood as follows.

Definition 2.1. Let f = anxn + an−1x

n−1 + · · · + a1x + a0 ∈ F[x] be a polynomial

with an 6= 0.

(1) f is irreducible over F (or in F[X]) if degf ≥ 1 and f = gh implies that either

f or g has degree 0.

(2) f is monic if an = 1.

For any f, g, h ∈ F[x], if f = gh, then we say that g divides f or f is divisible by g,

denoted by g|f , and call g a factor of f .

Definition 2.2. A polynomial d is called the greatest common divisor (gcd) of two

given polynomials f and g if d is monic, d|f, d|g, and that c|f and c|g implies c|d.

Theorem 2.3. The polynomial ring F[X] has the following important properties.

(1) (Division with remainder) For f, 0 6= g ∈ F[X], there exist r, s ∈ F[X] such

that f = gs+ r and deg r < deg g.

4 GALOIS THEORY

(2) F[X] is a PID and thus a unique factorisation domain (UFD), i.e., any 0 6=

f ∈ K[X] can be uniquely factorised as f = uf1 · · · fn up to the reordering of

f1, · · · , fn, where u has degree 0, and f1, · · · , fn are monic irreducible over F.

(3) Suppose that d, f, g ∈ F[x] and d is the gcd of f and g, then there exist

polynomials h, l ∈ F[x] such that

d = fh+ gl.

Definition 2.4. An element a ∈ F is called a root of f ∈ F[X] if f(a) = 0.

Lemma 2.5. Let f ∈ F[x]. An element a ∈ F is a root of f ∈ F[X] if and only if

(X − a) divides f .

Proof. By Theorem 2.3, there exists g, r ∈ F[x] with degr ≤ 0, such that

f = (x− a)g + r.

Hence

a is a root of f , i.e. , 0 = f(a) ⇔ r = 0, as f(a) = r(a) and r is constant

⇔ x− a divides f , as f = g(x− a) + r.

�

2.2. A construction of fields. We first recall the construction of quotient rings.

Let R be a ring and I an ideal of R. Define a relation ∼ on R by

a ∼ b if a− b ∈ I.

Then ∼ is an equivalence relation (check!). Denote by a the equivalence class of a

and let

R/I = {a|a ∈ R}.

Define operations + : R/I ×R/I → R/I and · : R/I ×R/I → R/I by

a+ b = a+ b and a · b = ab

Theorem 2.6. (1) + and · are well-defined.

(2) (R/I,+, ·) is a ring with 0 the zero and 1 the multiplicative identity.

GALOIS THEORY 5

The ring (R/I,+, ·) is called a quotient ring of R. For instance, Zn = Z/nZ is a

quotient ring of Z.

Let f ∈ F[x] and (f) the ideal generated by f . Then

(f) = {gh | g ∈ F[x]}

and any 0 6= h ∈ (f) has degree bigger than or equal to deg f .

Theorem 2.7. Suppose that f is a nonzero polynomial in F[x]. Then the quotient

ring F[x]/(f) is a field if and only if f ∈ F[x] is irreducible over F.

Proof. (“⇒”). Assume for contradiction that f is not irreducible. Then f is either

(i) a nonzero constant or (ii) f = gh for some polynomials g and h of positive degree.

Case (i): (f) = F[x] and so F[X]/(f) = F[X]/F[X] = {0} is not a field.

Case (ii): as deg g > 0, deg h > 0 and deg f = deg g+ deg h, the polynomials g and

h are not zero and their degrees are smaller than deg f . So g and h are not in (f).

Thus

g 6= 0 and h 6= 0

in F[x]/(f). On the other hand,

0 = f = gh = g · h

So F[X]/(f) is not an ID and thus not a field.

Either case leads to a contradiction and so f is irreducible.

(“⇐”). As F[x]/(f) is a commutative ring, we need only to prove that any nonzero

element x ∈ F[x]/(f) has a multplicative inverse. AS

0 6= a,

f does not divide a and so

gcd(f, a) = 1.

By Theorem 2.3, there exist r, s ∈ F[X] such that

1 = ra+ sf.

6 GALOIS THEORY

Then

1 = ra+ sf = ra+ s0 = ra.

This shows that a has a multiplicative inverse r, as required. �

2.3. Criteria for irreducibility of polynomials.

Lemma 2.8 (Gauss). Let p be a prime number and let f, g ∈ Z[x]. Then p | fg

implies p | f or p | g.

Proof. Suppose that

f = b0 + b1x+ · · ·+ brxr, g = c0 + c1x+ · · ·+ csx

s, fg = a0 + a1x+ · · ·+ anxn.

Assume for contradiction that p divides neither f nor g. Then there exist coefficient

bi and cj with i, j minimal and not divisible by p. Note that

ai+j = b0ci+j + b1ci+j−1 + · · ·+ bi−1cj+1 + bicj + · · ·+ bi+jc0,

where we let ct and bl be 0, if t > r and l > s. Since p divides ai+j, b0, . . . , bi−1,

c0, . . . , cj−1, p divides bicj. As p is a prime number, p divides bi or cj, contradicting

that bi and cj are not divisible by p. So p divides f or g. �

Proposition 2.9. Let f ∈ Z[x] ⊆ Q[x] with degf ≥ 1. If f is irreducible over Z,

then it is irreducible over Q.

Proof. Assume for contradiction that f is not irreducible Q. As deg f ≥ 1, f is

not a unit. So f = gh for some polynomials g and h of positive degree in Q[x].

Let m = p1 . . . pr and n = pr+1 . . . ps with each pi a prime number such that mg

and nh are polynomials in Z[x]. We have p1 . . . psf = (p1 . . . prg)(pr+1 . . . psh) with

(p1 . . . prg) and (pr+1 . . . psh) in Z[x], denoted by g0 and h0, respectively. As p1 divides

p1 . . . psf = g0h0 and p1 is a prime number, p1 divides either g0 or h0. So

p2 . . . psf = g1h1

for some g1 and h1 in Z[x], of the same degree as g and f , respectively. Continue in

this manner for s times,

f = gshs

GALOIS THEORY 7

for some gs and hs in Z[x], of the same degree as g and f , respectively. This contradicts

that f is irreducible over Z. Therefore f is irreducible over Q. �

Note: the converse is not true, e.g. f = 2x is irreducible in Q[x], but it is reducible

in Z[x], as 2 is not a unit in Z.

Deducing from the proof of Proposition 2.9, we have the following.

Corollary 2.10. Let f ∈ Z[x] and f = g1g2 with g1, g2 ∈ Q[x]. Then there exist

polynomials h1, h2 ∈ Z[x], of the same degree as g1 and g2, respectively, such that

f = h1h2.

Theorem 2.11 (Eisenstein’s irreducibility criterion). Let f = a0 + a1x+ · · ·+ anxn

be a polynomial in Z[x] with 0 6= an and n ≥ 1. Suppose that there is a prime p such

that p 6 | an, p|a0, p|a1 · · · , p|an−1 and p2 6 | a0. Then f is irreducible over Q.

Proof. Assume for contradiction that f is not irreducible over Q. As deg f = n ≥ 1,

deg f is at least 2 and f has a factorization

a0 + a1x+ · · ·+ anxn = (b0 + · · ·+ brx

r)(c0 + · · ·+ csxs),

where brcs 6= 0 and 1 ≤ r, s < n. By Corollary 2.10, we may assume that all the

coefficients bi and cj are integers. As a0 = b0c0 is divisible by p but not by p2, exactly

one of b0 and c0 is divisible by p. WLOG, suppose

(*): p divides b0 and p does not divide c0.

As p does not divide an = brcs, p does not divide br. Let i be the smallest positive

integer such that p does not divide bi. Notice that 0 < i ≤ r < n and

ai = b0ci + b1ci−1 + · · ·+ bi−1c1 + bic0.

Since p divides ai, b0, . . . , bi−1 and p does not divide bi, p divides c0, contradicting (*).

So f is irreducible over Q. �

Example 2.12. Determine the irreducibility of f = 29x5 + 5

3x4 + x3 + 1

3over Q.

8 GALOIS THEORY

First, note that f is irreducible over Q if and only if 9f is irreducible over Q. Now,

9f = 2x5 + 15x4 + 9x3 + 3, and 3 6 | 2, 3 | 15, 3 | 9, 3 | 3 and 32 6 | 3, so by Eisenstein’s

criterion, f is irreducible over over Q.

Proposition 2.13. Let p be a prime number, f = a0 + a1x + · · ·+ anxn ∈ Z[x] and

f = a0 + a1x+ · · ·+ anxn ∈ Zp[x] with an 6= 0 and n ≥ 1. If f is irreducible over Zp,

then f is irreducible over Q.

Proof. As an 6= 0, an 6= 0 and deg f = deg f = n ≥ 1. Assume (for contradiction)

that f is not irreducible. As deg f ≥ 1, f is not a constant. So f can be written as

f = g1g2

with gi ∈ Q[x] of positive degree, ∀i. By Corollary 2.10, we may assume that gi ∈ Z[x],

∀i. Note that

f = g1g2, deg f = deg f and deg gi ≤ deg gi.

So

deg gi = deg gi ≥ 1.

This shows that f is reducible over Zp, a contradiction. So f is irreducible over Q. �

Example 2.14. Determine the irreducibility of f = x3 + 11x+ 100 over Q.

Consider f = x3 + 2x+ 1 ∈ Z3[x], which is of degree 3. So if f is reducible, then it

has a linear factor and so it has a root in Z3. We can check that in Z3

f(0) = 1, f(1) = 1, f(2) = 1.

Therefore f has no roots in Z3 and so is irreducible over Z3. By Proposition 2.13, f

is irreducible over Q.

GALOIS THEORY 9

3. Field extensions

3.1. Prime fields.

Definition 3.1. The characteristic of a field a ring R, denoted by charR, is the

smallest natural number p such that p · 1 = 0, if such a number exists, and it is zero

otherwise.

For example, the characteristics of Q, R, and C are 0, the characteristic of Zp is p.

Remark: the only ring (up to isomorphism) of characteristic 1 is the trivial ring {0}.

In this course, unless otherwise specified, a ring is different from the trivial ring.

Proposition 3.2. The characteristic of a field is always zero or a prime number.

Proof. Assume for a contradiction that the characteristic n is neither zero nor a prime

number. Then n > 1 and there exist 1 < r, s < n such that n = rs. Thus 0 = n · 1 =

(r · 1)(s · 1), and so r · 1 = 0 or s · 1 = 0, contradicting the minimality of n. �

Definition 3.3. The prime field K of a field F is the intersection of all subfields of F.

For instance the prime field of Q is itself and the prime field of R is Q.

Remark:

(1) a prime field is a field.

(2) the prime field of a field is unique.

(3) a field F has no proper subfield if and only if the prime field of F is itself, e.g.

Zp (p is a prime) and Q .

Recall: a ring homomorphism is a map φ : R→ S such that

φ(a+ b) = φ(a) + φ(b), φ(ab) = φ(a)φ(b), φ(1R) = φ(1S).

Furthermore, φ is called an isomorphism if φ is bijective.

Theorem 3.4. The prime field of a field F is isomorphic to Q if charF = 0, and to

Zp if charF = p > 0 for some prime p ∈ Z.

Proof. Let K be the prime field of F. Define

φ : Z→ K, n 7→ n · 1.

10 GALOIS THEORY

As K is a subfield, any n · 1 ∈ K and so φ is well-defined. Further, φ is a ring

homomorphism (check!).

First consider the case where charF = p > 0. Then p is a prime, Kerφ ∼= pZ and

Imφ ∼= Z/pZ

is a subfield of K. As K is the prime field of F,

K ∼= Zp.

Next suppose that charK = 0. Then Kerφ = 0, i.e. φ is injective, and φ induces a

morphism

φ : Q→ K,n

m7→ φ(n)(φ(m)−1).

Again, φ is injective, and so Q ∼= Imφ is a subfield of K. As K is prime,

K ∼= Q.

This finishes the proof. �

3.2. Field extensions and extension degrees. Recall that the definition of a vec-

tor space over a field K is a set V with an addition

+ : V × V → V, (u, v) 7→ u+ v

and a scalar multiplication

· : K× V → V, (k, v) 7→ k · v

such that for all a, b ∈ K, v, w ∈ V , the following hold.

(i) (V,+) is an abelian group.

(ii) 1 · v = v.

(iii) a · (b · v) = (ab) · v.

(iv) a · (v + w) = a · v + a · w and (a+ b) · v = a · v + b · v.

Suppose that K is a subfield of F. Then we can view F as a vector space over K.

The addition in F is the usual field addition in F and the scalar multiplication is the

GALOIS THEORY 11

multiplication of elements in K with elements in F. For example: R ⊂ C and C is a

vector space over R and R is a vector space over Q.

Definition 3.5. Let F be a field and let K be a subfield of F. Then

(1) F is called an extension field of K.

(2) The pair K ⊆ F is called a field extension.

(3) The dimension of F as a vector space over K, denoted by [F : K], is called the

degree of the field extension K ⊆ F.

For example, R ⊆ C = R ⊕ iR and [C : R] = 2. Every field is an extension field

of its prime field, and so it is a vector space over its prime field. Thus any field is a

vector space over Q or Zp for some prime p.

Theorem 3.6. Let f be an irreducible polynomial of degree d in F[x].

(1) The map φ : F→ F[x]/(f), a 7→ a, is an injective ring homomorphism. Thus

F can be viewed as a subfield of the field F[x]/(f).

(2) The set {1, x, · · · , xd−1} is a basis of F[x]/(f) over F and so the extension

degree [F[x]/(f) : F] = d.

(3) The element x is a root of f in F[x]/(f).

Proof. (1) As the addition and multiplication in F[x]/(f) is induced from those in F[x]

and F is a subring of F[x], φ is a ring homomorphism. We show that it is injective.

Suppose that φ(a) = a = 0. Then a = gf for some g ∈ F[x]. Comparing the degrees

of a and gh gives a = g = 0. So φ is injective and we can view F as a subfield of

F ⊆ F[x]/(f), which by Theorem 2.7 is a field.

(2) Note for any g ∈ F[x], there exist h, r ∈ F[x] with deg r < deg f such that

g = fh+ r.

So g = r ∈ span{1, x, · · · , xd−1}.

Further, 1, x, · · · , xd−1 are linearly independent. Indeed,∑d−1

i=0 aixi = 0 implies that

h =∑d−1

i=0 aixi = gf for some g ∈ F[x]. Comparing the degrees of h and gf gives

h = g = 0. Thus ai = 0 for all i. Therefore {1, x, · · · , xd−1} is a basis of F[x]/(f) over

F and [F[x]/(f) : F] = d.

12 GALOIS THEORY

(3) f(x) = f = 0 and so x is a root of f in F[x]/(f). �

For any irreducible polynomial f ∈ F[x] of degree ≥ 2, it has no roots in F, but

Theorem 3.6 tells us that there is a natural extension field F ⊆ F[x]/(f) such that f

has a root in the extension field F[x]/(f).

Example 3.7. (1) K[x]/(x− a) ∼= K as fields, for any a ∈ K.

(2) x2 + 1 is irreducible in R[x]. R[x]/(x2 + 1) = R ⊕ Rx as vector spaces and

R[x]/(x2 + 1) ∼= C as fields.

Lemma 3.8 (The tower law). Let K ⊆ L ⊆ F be a tower of fields. Suppose that

{xi}i∈I is a basis of L over K, {yj}j∈J is a basis of F over L. Then {xiyj}i∈I,j∈J is a

basis of F over K. In particular, [F : K] = [F : L][L : K].

Proof. Denote by X the linear space span{xiyj|i ∈ I, j ∈ J}, i.e. the space of finite

linear combinations of xiyj over K. Then F ⊇ X . Next for any λ ∈ F,

λ =∑j∈J

ajyj and aj =∑i∈I

aijxi ∈ L,

where aij ∈ K. Then

λ =∑

i∈I,j∈J

aijxiyj ∈ X and so F = X.

It remains to prove the linear independence of xiyj (i ∈ I, j ∈ J). As (yj)j∈J is a

basis of F over L and (xi)i∈I is a basis of L over K,

0 =∑

i∈I,j∈J

aijxiyj =∑j∈J

(∑i∈I

aijxi)yj

implies ∑i∈I

aijxi = 0 and so aij = 0

for all j ∈ J, i ∈ I. So xiyj (i ∈ I, j ∈ J) are linearly independent and thus they form

a basis of F over K. Consequently,

[F : K] = |I||J | = [F : L][L : K].

�

GALOIS THEORY 13

4. Finitely generated field extensions

Let K ⊆ F be a field extension and let {s1, · · · , sn} ⊆ F. Denote by K[x1, · · · , xn]

the polynomial ring over K in n variables. Define

φ : K[x1, · · · , xn]→ F, by f 7→ f(s1, · · · , sn).

Then φ is a ring homomorphism (check). Define

K[s1, · · · , sn] = Imφ

and

K(s1, · · · , sn) = {ab−1 | a, b ∈ K[s1, · · · , sn], b 6= 0}.

By construction, K[s1, . . . , sn] is a subring of F and K(s1, . . . , sn) is a subfield of F.

Moreover, K[s1, · · · , sn] is the smallest subring of F containing K and s1, · · · , sn and

K(s1, · · · , sn) is the smallest subfield of F containing K and s1, · · · , sn.

Example 4.1. Q(√

2) = Q[√

2] = Q⊕Q√

2. Indeed, first note that Q[√

2] is a field,

which we can check directly or show that it it is isomorphic to the field Q[x]/(x2− 2)

and it contains Q and√

2. By definition Q[√

2] ⊆ Q(√

2) and so by the minimality

of Q(√

2),

Q(√

2) = Q[√

2].

The second identity follows from the fact that any power of√

2 is either a power of 2

or an integral multiple of√

2.

By definition, K[s1, · · · , sn] ⊆ K(s1, · · · , sn). Example 4.1 shows that K[s1, · · · , sn]

and K(s1, · · · , sn) can be the same. We will discuss when they are the same in next

section.

Definition 4.2. A field extension K ⊆ F is said to be

(1) finite if the extension degree [L : F] <∞.

(2) finitely generated if there exist finitely many elements s1, · · · , sn ∈ F such that

F = K(s1, · · · , sn).

(3) a simple extension if F = K(s) for some s ∈ F.

14 GALOIS THEORY

Remark:

(1) By definition, a simple field extension is finitely generated.

(2) A finite field extension K ⊆ F is also finitely generated.

Indeed, if F = span{s1, . . . , sn} with s1, . . . , sn a basis of F over K, then

F ⊆ K[s1, . . . , sn] ⊆ K(s1, . . . , sn) ⊆ F,

thus F = K(s1, . . . , sn) and so K ⊆ F is finitely generated.

(3) Both converses of the statements in (1) and (2) are not necessarily true.

Example 4.3. Consider the tower of fields: Q ⊆ Q(√

2) ⊆ Q(√

2,√

3) ⊆ R. What is

the extension degree [Q(√

2,√

3) : Q]?

By Example 4.1,

Q(√

2) = Q[√

2] = Q + Q√

2

and so [Q(√

2) : Q] = 2 and similarly

[Q(√

2)(√

3) : Q(√

2)] = 2.

Note that Q(√

2,√

3) = Q(√

2)(√

3). So by the Tower Law

[Q(√

2,√

3) : Q] = [Q(√

2)(√

3) : Q(√

2)][Q(√

2) : Q] = 4.

Question: Can Q ⊆ Q(√

2,√

3) be a simple extension?