must be X ’ v, hence v ∈ K(w) and so u = w ’ tv ∈ K(w). This shows that K(u, v) K(w)

and therefore K(w) = K(u, v).

Corollary 3.76. Let L/K be a ¬nite separable extension of a ¬eld of characteristic 0.

Then L has a primitive element.

Proof. Since Q K, K is in¬nite and by Example 3.67 L/K is separable.

To ¬nd a primitive element we can always use the menu provided in the proof of Theo-

rem 3.75, however a ˜try it and see™ approach will usually su¬ce.

√

Example 3.77. Find a primitive element for the extension Q( 3, i)/Q.

√ √ √

Solution. Consider 3 + i. Then working over the sub¬eld Q( 3) Q( 3, i) we ¬nd that

√

i ∈ Q( 3) R and

/

√ √ √ √

(X ’ ( 3 + i))(X ’ ( 3 ’ i)) = X 2 ’ 2 3X + 4 ∈ Q( 3)[X],

hence √

X 2 ’ 2 3X + 4 = minpolyQ(√3),√3+i (X).

Now taking √ √

(X 2 ’ 2 3X + 4)(X 2 + 2 3X + 4) = X 4 ’ 4X 2 + 16 ∈ Q[X],

we see that minpolyQ,√3+i (X) | (X 4 ’ 4X 2 + 16) in Q[X]. Notice that

√ √

√ 1√ √

( 3 ’ i) ( 3 ’ i)

( 3 + i)’1 = √ √ = = ( 3 ’ i) ∈ Q( 3 + i),

3+1 4

( 3 + i)( 3 ’ i)

√ √

since ( 3 + i)’1 ∈ Q( 3 + i). Hence

√ 1√ √ 1√ √

3 = (( 3 + i) + ( 3 ’ i)), i = (( 3 + i) ’ ( 3 ’ i)),

2 2

√ √ √ √ √

are both in Q( 3 + i), showing that Q( 3, i) Q( 3 + i) and so Q( 3, i) = Q( 3 + i). Thus

we must have deg minpolyQ,√3+i (X) = 4, and so minpolyQ,√3+i (X) = X 4 ’ 4X 2 + 16.

There is a general phenomenon illustrated by Example 3.77.

Proposition 3.78. Let u ∈ K be separable over K. Then

minpolyK,u (X) = (X ’ ±1 (u)) · · · (X ’ ±d (u)),

where ±1 , . . . , ±d are the elements of MonoK (K(u), K). In particular, the polynomial

(X ’ ±1 (u)) · · · (X ’ ±d (u)) ∈ K[X]

is in K[X] and is irreducible therein.

Proof. Since K(u) is separable then by Lemma 3.52,

d = deg minpolyK,u (X) = [K(u) : K] = (K(u) : K).

In Example 3.77 we have

√ √ √ √

[Q( 3, i) : Q] = [Q( 3, i) : Q( 3)][Q( 3) : Q] = 2 · 2 = 4.

√ √

There are four monomorphisms ±k : Q( 3, i) ’’ Q( 3, i) given by

√ √ √ √ √ √

3 ’’ 3 3 ’’ ’ 3 3 ’’ ’ 3

±1 = id, ±2 = , ±3 = , ±4 = .

i ’’ ’i i ’’ i i ’’ ’i

Then

√ √ √ √ √ √

±2 ( 3 + i) = ( 3 ’ i), ±3 ( 3 + i) = (’ 3 + i), ±4 ( 3 + i) = (’ 3 ’ i),

so √ √ √ √

3 ’ i)(X ’ 3 + i)(X + 3 ’ i)(X + 3 + i) = X 4 ’ 4X 2 + 16 ∈ Q[X].

(X ’

√ √ √

Hence this polynomial is irreducible. So we have [Q( 3 + i) : Q] = 4 and Q( 3 + i) = Q( 3, i).

45

3.7. Normal extensions and splitting ¬elds

Definition 3.79. A ¬nite extension E/K is normal if •E = E for every • ∈ MonoK (E, K).

Remark 3.80. If E/K is a normal extension then whenever an irreducible polynomial

p(X) ∈ K[X] has a root in E then it splits in E since by Lemma 3.52 each pair of roots of p(X)

is conjugate over K and one can be mapped to the other by a monomorphism K ’’ K which

must send E into itself.

Proposition 3.81. A ¬nite extension E/K is normal if and only if it is a splitting ¬eld

over K for some polynomial f (X) ∈ K[X].

Proof. Suppose that E/K is normal. Then there is a sequence of extensions

K K(u1 ) K(u1 , u2 ) ··· K(u1 , . . . , un ) = E

Construct a polynomial by taking

f (X) = minpolyK,u1 (X) minpolyK,u2 (X) · · · minpolyK,un (X).

Then by Remark 3.80, f (X) splits in E. Also E is generated by some of the roots of f (X).

Hence E is a splitting ¬eld for f (X) over K.

Now suppose that E is a splitting ¬eld for g(X) ∈ K[X], so that E = K(v1 , . . . , vk ), where

v1 , . . . , vk are the distinct roots of g(X) in E. Now any monomorphism θ ∈ MonoK (E, K) must

map these roots to θ(v1 ), . . . , θ(vk ) which are also roots of g(X) and therefore lie in E (see

Proposition 3.34). Since θ permutes the roots vj , we have

θE = θK(v1 , . . . , vk ) = K(θ(v1 ), . . . , θ(vk )) = K(v1 , . . . , vk ) = E.

This result makes it easy to recognize a normal extension since it is su¬cient to describe it

as a splitting ¬eld for some polynomial over K. In Chapter 4 we will see that separable normal

extensions play a central rˆle in Galois Theory, indeed these are known as Galois extensions.

o

Exercises on Chapter 3

3.1. Prove Proposition 3.2.

3.2. Finding splitting sub¬elds E C over Q and determine [E : Q] for each of the following

polynomials.

p1 (X) = X 4 ’X 2 +1, p2 (X) = X 6 ’2, p3 (X) = X 4 +2, p4 (X) = X 4 +5X 3 +10X 2 +10X+5.

[Hint: for p4 (X), consider p4 (Y ’ 1) ∈ Q[Y ].]

√

3.3. Prove that AutQ (Q( 3 2, ζ3 )) ∼ S3 , the symmetric group on 3 elements, as claimed in the

=

solution of Example 3.38. [Hint: work out the e¬ect of each automorphism on the three roots of

the polynomial X 3 ’ 2.]

3.4. Let k be a ¬eld of characteristic char k = p > 0 and k(T ) be the ¬eld of rational functions

in T over k. Show that the polynomial g(X) = X p ’ T ∈ k(T )[X] is irreducible and has a

multiple root in k(T ). How does g(X) factor in k(T )[X]?

√√ √ √

3.5. Find primitive elements for the extensions Q( 5, 10)/Q, Q( 2, i)/Q, Q( 3, i)/Q,

√

Q( 4 3, i)/Q, in each case ¬nding it minimal polynomial over Q. [Hint: look for elements of

high degree over Q, or use the method of proof of Theorem 3.75.]

3.6. Prove the following converse of Proposition 3.20:

Let L/K be a ¬nite extension. If there are only ¬nitely many subextensions F/K L/K, then

L/K is simple, i.e., L = K(w) for some w ∈ L.

[Hint: First deal with the case where L = K(u, v), then use induction on n to prove the general

case L = K(u1 , . . . , un ).]

3.7. Let K be a ¬eld. Show that every quadratic (i.e., of degree 2) extension E/K is normal.

Is such an extension always separable?

46 3. ALGEBRAIC EXTENSIONS OF FIELDS

3.8. Let f (X) ∈ Q[X] be an irreducible polynomial of odd degree greater than 1 and having

only one real root u ∈ R. Show that Q(u)/Q is not a normal extension.

CHAPTER 4

Galois extensions and the Galois Correspondence

In this Chapter we will study the structure of Galois extensions and their associated Galois

groups, in particular we will explain how these are related through the Galois Correspondence.

Throughout the chapter, let K be a ¬eld.

4.1. Galois extensions

Definition 4.1. A ¬nite extension E/K is a (¬nite) Galois extension if it is normal and

separable.

From Section 3.5 we know that for such a Galois extension E/K, [E : K] = (E : K) and also

every monomorphism • ∈ MonoK (E, K) maps E into itself, hence restricts to an automorphism

of E which will be denoted •|E .

K

˜>

• ˜˜

˜

˜˜

˜∼ =

E • •• •/ E

|E

=

/K

K

Also, by the Monomorphism Extension Theorem 3.49, every automorphism ± ∈ AutK (E) ex-

tends to a monomorphism E ’’ K ¬xing elements of K. So there is a bijection

MonoK (E, K) ←’ AutK (E)

and we have

(4.1) | AutK (E)| = (E : K) = [E : K].

Definition 4.2. For a ¬nite Galois extension E/K, the group Gal(E/K) = AutK (E) is

called the Galois group of the extension or the Galois group of E over K.

Notice that Equation (4.1) implies

(4.2) | Gal(E/K)| = (E : K) = [E : K].

We can also reformulate the notion of conjugacy introduced in De¬nition 3.51.

Definition 4.3. Let E/K a ¬nite Galois extension and u, v ∈ E. Then v is conjugate to u

if there is a • ∈ Gal(E/K) for which v = •(u); we also say that v is a conjugate of u.

It is easy to see that for u, v ∈ K, there is a ¬nite Galois extension E/K in which v is a

conjugate of u if and only v is a conjugate of u over K in the old sense.

4.2. Working with Galois groups

Let E/K be a ¬nite Galois extension. Then we know that E is a splitting ¬eld for some

polynomial over K. We also know that E is a simple extension of K and so it is a splitting

¬eld for the minimal polynomial of a generating element which has degree equal to [E : K]. It

is often convenient to use these facts to interpret elements of the Galois group as permutations

of the roots of some polynomial which splits over E.

√√

Example 4.4. Describe the Galois group Gal(Q( 2, 3)/Q) as a subgroup of the group of

permutations of the roots of (X 2 ’ 2)(X 2 ’ 3) ∈ Q[X].

47

48 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

Solution. We have

√√ √√ √ √

[Q( 2, 3) : Q] = [Q( 2, 3) : Q( 2)] [Q( 2) : Q] = 4,

and the following non-trivial elements of the Galois group together with the element identity

±1 = id:

«√ √ «√ √ «√ √

2 ’’ ’ 2 2 ’’ 2 2 ’’ ’ 2

√ √ √ √ √ √

¬’ 2 ’’ 2· ¬’ 2 ’’ ’ 2· ¬’ 2 ’’ 2·

±2 = ¬ √ √ · , ±3 = ¬ √ √ · , ±4 = ¬ √ √ ·.

3 ’’ 3 3 ’’ ’ 3 3 ’’ ’ 3

√ √ √ √ √ √

’ 3 ’’ ’ 3 ’ 3 ’’ 3 ’ 3 ’’ 3

√ √√ √

Writing the roots in the list 2, ’ 2, 3, ’ 3 and numbering them 1 to 4 accordingly, we see

that these automorphisms correspond to the following permutations in S4 expressed in cycle

notation:

±2 ←’ (1 2), ±3 ←’ (3 4), ±4 ←’ (1 2)(3 4).

Example 4.5. Using a primitive element u for the extension, describe the Galois group

√√

Gal(Q( 2, 3)/Q) as a subgroup of the group of permutations of the roots of minpolyQ,u (X) ∈

Q[X].

√√ √ √ √ √

Solution. We have Q( 2, 3) = Q( 2 + 3) and the conjugates of u = 2 + 3 are

√ √

± 2 ± 3. Listing these as

√ √√ √ √ √ √ √

2 + 3, 2 ’ 3, ’ 2 + 3, ’ 2 ’ 3,

and after numbering them accordingly, we ¬nd the correspondences

±2 ←’ (1 3)(2 4), ±3 ←’ (1 2)(3 4), ±4 ←’ (1 4)(2 3).

Next we summarize the properties of Galois groups that can be deduced from what we have

established so far.

Recollection 4.6. Recall that an action of a group G on a set X is transitive if for every

pair of elements x, y ∈ X, there is an element g ∈ G such that y = gx (so there is only one

orbit); the action is faithful or e¬ective if for every non-identity element h ∈ G, there is a

element z ∈ X such that hz = z.

For an extension F/K and a polynomial f (X) ∈ K[X], recall that Roots(f, F ) denotes the

set of roots of f (X) in F .

Theorem 4.7. Let E/K be a ¬nite Galois extension. Suppose that E is the splitting ¬eld

of a separable irreducible polynomial f (X) ∈ K[X] of degree n. Then the following are true.

(i) Gal(E/K) acts transitively and faithfully on Roots(f, E).

(ii) Gal(E/K) can be identi¬ed with a subgroup of the group of permutations of Roots(f, E).

If we order the roots u1 , . . . , un then Gal(E/K) can be identi¬ed with a subgroup of Sn .

(iii) | Gal(E/K)| divides n! and is divisible by n.

As we have seen in Examples 4.4 and 4.5, in practise it is often easier to use a not necessarily

irreducible polynomial to determine and work with a Galois group.

Example 4.8. The Galois extension Q(ζ8 )/Q has degree [Q(ζ8 ) : Q] = 4 and it has the

following automorphisms apart from the identity:

3 5 7

± : ζ8 ’’ ζ8 , β : ζ8 ’’ ζ8 , γ : ζ8 ’’ ζ8 .

If we list the roots of the minimal polynomial

minpolyQ,ζ (X) = ¦8 (X) = X 4 + 1

357

in the order ζ8 , ζ8 , ζ8 , ζ8 , we ¬nd that these automorphisms correspond to the following permu-

tations in S4 :

± ←’ (1 2)(3 4), β ←’ (1 3)(2 4), γ ←’ (1 4)(2 3).

4.3. SUBGROUPS OF GALOIS GROUPS AND THEIR FIXED FIELDS 49

So the Galois group Gal(Q(ζ8 )/Q) corresponds to

{id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} S4 .

Noticing that

1 1

ζ8 = √ + √ i,

2 2

√ √ √

we√easily ¬nd that 2, i ∈ Q(ζ8 );√ hence Q( 2, i) Q(ζ8 ). Since [Q( 2, i) : Q] = 4, we have

Q( 2, i) = Q(ζ8 ). Notice that Q( 2, i) is the splitting ¬eld of f (X) = (X 2 ’ 2)(X 2 + 1) over

√ √

Q. Now list the roots of f (X) in the order 2, ’ 2, i, ’i, and observe that

«√ √ «√ √

2 ’’ ’ 2 2 ’’ ’ 2

√ √ √ √

¬’ 2 ’’ 2· ¬’ 2 ’’ 2·

±: ¬ · ←’ (1 2)(3 4), β: ¬ · ←’ (1 2),

i ’’ ’i i ’’ i

’i ’’ i ’i ’’ ’i

«√ √

√2 ’’ 2

√

¬’ 2 ’’ ’ 2·

γ: ¬ ·

i ’’ ’i ←’ (3 4).

’i ’’ i

√

In this description, the Galois group Gal(Q(ζ8 )/Q) = Gal(Q( 2, i)/Q) corresponds to the

subgroup

{id, (1 2), (3 4), (1 2)(3 4)} S4 .

While it can be hard to determine Galois groups in general, special arguments can sometimes

be exploited.

Example 4.9. Suppose that f (X) = X 3 + aX 2 + bX + c ∈ Q[X] is an irreducible cubic and

that f (X) has only one real root. Then Gal(Q(f (X))/Q) ∼ S3 .

=

Proof. Let u1 ∈ R be the real root of f (X) and let u2 , u3 be the remaining complex

roots. Then Q(f (X)) = Q(u1 , u2 , u3 ) and in fact [Q(f (X)) : Q] = 6 since [Q(f (X)) : Q] | 6

and u2 ∈ Q(u1 )