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When extensions are not normal, we can try to recover normality by making the extensions larger.

Definition 11.5. Let L be a finite extension of K. A normal closure of L : K is an extension N of L such that

Normal Closures 147 (1) N : K is normal;

(2) If L ⊆ M ⊆ N and M : K is normal, then M = N.

Thus N is the smallest extension of L that is normal over K.

The next theorem assures us of a sufficient supply of normal closures, and shows that (working inside C) they are unique.

Theorem 11.6. If L : K is a finite extension in C, then there exists a unique normal closure N⊆ C of L : K, which is a finite extension of K.

Proof. Let x₁, . . . , x_rbe a basis for L over K, and let m_jbe the minimal polynomial of x_jover K. Let N be the splitting field for f = m₁m₂. . . m_rover L. Then N is also the splitting field for f over K, so N : K is normal and finite by Theorem 9.9. Suppose that L ⊆ P ⊆ N where P : K is normal. Each polynomial m_jhas a zero x_j∈ P, so by normality f splits in P. Since N is the splitting field for f , we have P = N. Therefore Nis a normal closure.

Now suppose that M and N are both normal closures. The above polynomial f splits in M and in N, so each of M and N contain the splitting field for f over K. This splitting field contains L and is normal over K, so it must be equal to both M and N.

Example 11.7. Consider Q(α):Q where α is the real cube root of 2. This extension is not normal, as we have seen. If we let K be the splitting field for t³− 2 over Q, contained in C, then K = Q(α, αω, αω²) where ω = (−1 + i√

3)/2 is a complex cube root of unity. This is the same as Q(α, ω). Now K is the normal closure for Q(α ) : Q. So here we obtain the normal closure by adjoining all the ‘missing’ zeros.

Normal closures let us place restrictions on the image of a monomorphism.

Lemma 11.8. Suppose that K ⊆ L ⊆ N ⊆ M where L : K is finite and N is the normal closure of L: K. Let τ be any K-monomorphism L → M. Then τ(L) ⊆ N.

Proof. Let α ∈ L. Let m be the minimal polynomial of α over K. Then m(α) = 0 so τ (m(α )) = 0. But τ (m(α )) = m(τ (α )) since τ is a K-monomorphism, so m(τ (α )) = 0 and τ(α) is a zero of m. Therefore τ(α) lies in N since N : K is normal. Therefore τ (L) ⊆ N.

This result often lets us restrict attention to the normal closure of a given ex-tension when discussing monomorphisms. The next theorem provides a sort of con-verse.

Theorem 11.9. For a finite extension L : K the following are equivalent:

(1) L : K is normal.

(2) There exists a finite normal extension N of K containing L such that every K-monomorphism τ : L → N is a K-automorphism of L.

(3) For every finite extension M of K containing L, every K-monomorphism τ : L→ M is a K-automorphism of L.

Proof. We show that (1) ⇒ (3) ⇒ (2) ⇒ (1).

(1) ⇒ (3). If L : K is normal then L is the normal closure of L : K, so by Lemma 11.8, τ(L) ⊆ L. But τ is a K-linear map defined on the finite-dimensional vector space L over K, and is a monomorphism. Therefore τ(L) has the same dimen-sion as L, whence τ(L) = L and τ is a K-automorphism of L.

(3) ⇒ (2). Let N be the normal closure for L : K. Then N exists by Theorem 11.6, and has the requisite properties by (3).

(2) ⇒ (1). Suppose that f is any irreducible polynomial over K with a zero α ∈ L.

Then f splits over N by normality, and if β is any zero of f in N, then by Proposi-tion 11.4 there exists an automorphism σ of N such that σ (α) = β . By hypothesis, σ is a K-automorphism of L, so β = σ (α ) ∈ σ (L) = L. Therefore f splits over L and L: K is normal.

Our next result is of a more computational nature.

Theorem 11.10. Suppose that L : K is a finite extension of degree n. Then there are precisely n distinct K-monomorphisms of L into the normal closure N of L: K, and hence into any given normal extension M of K containing L.

Proof. Use induction on [L : K]. If [L : K] = 1, then the result is clear. Suppose that [L : K] = k > 1. Let α ∈ L\K with minimal polynomial m over K. Then

∂ m = [K(α ) : K] = r > 1

Now m is an irreducible polynomial over a subfield of C with one zero in the normal extension N, so m splits in N and its zeros α1, . . . , αrare distinct. By induction there are precisely s distinct K(α)-monomorphisms ρ1, . . . , ρs : L → N, where s = [L : K(α)] = k/r. By Proposition 11.4, there are r distinct K-automorphisms τ1, . . . , τrof Nsuch that τi(α) = αi. The maps

φi j= τiρj (1 ≤ i ≤ r, 1 ≤ j ≤ s) are K-monomorphisms L → N.

We claim they are distinct. Suppose φi j= φkl. Then τ_k⁻¹τi= ρlρ⁻¹_j . The ρj fix K(α), so they map α to itself. But ρ_j is defined by its action on α, so ρlρ⁻¹_j is the identity. That is, ρ_l= ρj. So τ_k⁻¹τiis the identity, and τ_k= τi. Therefore i = k, j = l, so the φi jare distinct. They therefore provide rs = k distinct K-monomorphisms L → N.

Finally, we show that these are all of the K-monomorphisms L → N. Let τ : L → N be a K-monomorphism. Then τ(α) is a zero of m in N, so τ(α) = αifor some i. The map φ = τ_i⁻¹τ is a K(α )-monomorphism L → N, so by induction φ = ρjfor some

j. Hence τ = τiρj= φi jand the theorem is proved.

We can now calculate the order of the Galois group of a finite normal extension, a result of fundamental importance.

Corollary 11.11. If L : K is a finite normal extension inside C, then there are pre-cisely[L : K] distinct K-automorphisms of L. That is,

|Γ(L : K)| = [L : K]

Exercises 149 Proof. Use Theorem 11.10.

From this we easily deduce the important:

Theorem 11.12. Let L : K be a finite extension with Galois group G. If L : K is normal, then K is the fixed field of G.

Proof. Let K₀be the fixed field of G, and let [L : K] = n. Corollary 11.11 implies that

|G| = n. By Theorem 10.5, [L : K₀] = n. Since K ⊆ K0we must have K = K₀. An alternative and in some ways simpler approach to Corollary 11.11 and Theo-rem 11.12 can be found in Geck (2014).

There is a converse to Theorem 11.12, which shows why we must consider nor-mal extensions in order to make the Galois correspondence a bijection. Before we can prove the converse, we need a theorem whose statement and proof closely resemble those of Theorem 11.10.

Theorem 11.13. Suppose that K ⊆ L ⊆ M and M : K is finite. Then the number of distinct K-monomorphisms L→ M is at most [L : K].

Proof. Let N be a normal closure of M : K. Then the set of K-monomorphisms L → Mis contained in the set of K-monomorphisms L → N, and by Theorem 11.10 there are precisely [L : K] of those.

Theorem 11.14. If L is any field, G any finite group of automorphisms of L, and K is its fixed field, then L: K is finite and normal, with Galois group G.

Proof. By Theorem 10.5, [L : K] = |G| = n, say. There are exactly n distinct K-monomorphisms L → L, namely, the elements of the Galois group.

We prove normality using Theorem 11.9. Thus let N be an extension of K con-taining L, and let τ be a K-monomorphism L → N. Since every element of the Ga-lois group of L : K defines a K-monomorphism L → N, the GaGa-lois group provides n distinct K-monomorphisms L → N, and these are automorphisms of L. But by The-orem 11.13 there are at most n distinct K-monomorphisms L → N, so τ must be one of these monomorphisms. Hence τ is an automorphism of L. Finally, L : K is normal by Theorem 11.9.

If the Galois correspondence is a bijection, then K must be the fixed field of the Galois group of L : K, so by the above L : K must be normal. That these hypotheses are also sufficient to make the Galois correspondence bijective (for subfields of C) will be proved in Chapter 12. For general fields we need the additional concept of

‘separability’, see Chapter 17.

EXERCISES

11.1 Suppose that L : K is finite. Show that every K-monomorphism L → L is an automorphism. Does this result hold if the extension is not finite?

11.2 Construct the normal closure N for the following extensions:

(a) Q(α):Q where α is the real fifth root of 3 (b) Q(β ):Q where β is the real seventh root of 2 (c) Q(√

2,√ 3):Q (d) Q(α,√

2): Q where α is the real cube root of 2 (e) Q(γ):Q where γ is a zero of t³− 3t²+ 3

11.3 Find the Galois groups of the extensions (a), (b), (c), (d) in Exercise 11.2.

11.4 Find the Galois groups of the extensions N : Q for their normal closures N.

11.5 Show that Lemma 11.8 fails if we do not assume that N : K is normal, but is true for any extension N of L such that N : K is normal, rather than just for a normal closure.

11.6 Use Corollary 11.11 to find the order of the Galois group of the extension Q(

√ 3,√

5,√

7): Q. (Hint: Argue as in Example 6.8.) 11.7 Mark the following true or false.

(a) Every K-monomorphism is a K-automorphism.

(b) Every finite extension has a normal closure.

(c) If K ⊆ L ⊆ M and σ is a K-automorphism of M, then the restriction σ |L

is a K-automorphism of L.

(d) An extension having Galois group of order 1 is normal.

(e) A finite normal extension has finite Galois group.

(f) Every Galois group is abelian (commutative).

(g) The Galois correspondence fails to be bijective for non-normal exten-sions.

(h) A finite normal extension inside C, of degree n, has Galois group of order n.

(i) The Galois group of a normal extension is cyclic.

Chapter 12

The Galois Correspondence

We are at last in a position to establish the fundamental properties of the Galois correspondence between a field extension and its Galois group. Most of the work has already been done, and all that remains is to put the pieces together.