Isomorphism of Splitting Fields of a Polynomial, Study notes of Mathematics

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Lecture Notes Math 371: Algebra (Fall 2006)

by Nathanael Leedom Ackerman November 14, 2006

TALK SLOWLY AND WRITE NEATLY!!

0.1 Proof of the Main Theorem

Let f (x) be a monic polynomial of degree n with coef- ficients in a field F. We recall that a splitting field of f (x) ∈ F [x] is a field of the form K = F (α 1 ,... , αn) such that f (x) = (x − α 1 ) · · · (x − αn) in K[x]. We now want to show that any two splitting fields of a given poly- nomial f (x) are isomorphic. This follows from the fact that a field extension of the form F (α) is determined by the irreducible polynomial for α over F and from some “bookkeeping”. The bookkeeping is notationally a little confusing but not hard.

Extending Isomorphism

Definition 0.1.0.1. Let ϕ : F → F be an isomor- phism. Then ϕ extends to an isomorphism F [x] → F [x]

The rings F [x] and F [x] are isomorphic as we saw, and since f and f correspond under this isomorphism, so do the ideals (f ) and (f ). Hence the rings F [x]/(f ) and F [x]/(f ) are isomorphic, and combining these isomor- phisms yields the isomorphism ϕ 1.

Isomorphism of Splitting Fields

Theorem 0.1.0.3. Let ϕ : F → F be an isomorphism of fields. Let f (x) ∈ F [x] be nonconstant and let f (x) be the corresponding polynomial in F [x]. Let K and K be the splitting fields for f (x) and f (x). Then there is an isomorphism ψ : K → K which restricts to ϕ on F.

Proof. If f (x) factors into linear factors over F , then f (x) also factors into linear factors. In the case K = F and K = F , so ϕ = ψ.

Now assume that f does not split completely. Choose an irreducible factor g(x) of f (x) of degree > 1. The corre- sponding polynomial g(x) will make an irreducible factor of f (x). Let α be a root of g in K and write F 1 = F (α). Make a similar choice for α and let F 1 = F (α) in K. Then by the previous lemma we can extend ϕ to an iso- morphism ϕ 1 : F 1 → F 1 which sends α √ α. Being a splitting field for f over F , K is also a splitting field of f over the larger field F 1 and similarly K is a splitting field for f over F 1. Therefore we may replace F, F , ϕ by F 1 , F 1 , ϕi and proceed by induction.

Corollary 0.1.0.4. Any two splitting fields of f (x) ∈ F [x] are isomorphic.

Proof. Set F = F and ϕ = id in the previous theorem.

Splitting Fields are Galois

ϕ 1 : F 1 → F 1 = F (α) by setting ϕ 1 (α) = α.

We use induction on [K : F ]. Since [K : F 1 ] < [K : F ] the induction hypothesis tells us that for this particular choice of ϕ 1 there are [K : F 1 ] extensions of ϕ 1 to an iso- morphism ψ : K → K. On the other hand, g has distinct roots in K because g, g are irreducible. So the number of choices for α is the degree of g which is [F 1 : F ]. So there are [F 1 : F ] choices for the isomorphism ϕ 1. This gives us a total of [K : F 1 ][F 1 : F ] = [K : F ] extensions of ϕ to ψ : K → K.

Galois Group of Polynomial

Definition 0.1.0.7. Since any two splitting fields K of f (x) ∈ F [x] are isomorphic, the Galois group G(K/F ) depends, up to isomorphism, only on f. It is often re- ferred to as the Galois group of the polynomial over F.

Collection of Equivalences

Corollary 0.1.0.8. Let K/F be a finite field exten- sion. the following are equivalent. (i) K is a Galois extension of F. (ii) K is the splitting field of an irreducible polynomial f (x) ∈ F [x]. (ii’) K is the splitting field of a polynomial f (x) ∈ F [x]. (iii) F is the fixed field for the action of the Galois group G(K/F ) on K.

(iii’) F is the fixed field for an action of a finite group of automorphisms of K.

Main Theorem of Galois Theory

Theorem 0.1.0.9 (Main Theorem). Let K be a Galois extension of a field F and let G = G(K/F ) be its

Let L be an intermediate field. The corresponding sub- group of G is H = G(K/L). By definition, H acts triv- ially on L, so L ⊆ KH^. On the other hand, K is a Galois extension of L by previous results. Hence [K : L] = |H|. However also by previous results we have |H| = [K : KH^ ] and so L = KH^.

In the other direction, suppose that we start with a sub- group H ⊂ G and let L = KH^. Then H ⊂ G(K/L). But |H| = [K : KH^ ] = [K : L] = |G(K/L)|. Therefore H = G(K/L).

We therefore have that these two maps are inverses. Fur- ther since K is a Galois extension of L = KH^ , [K : L] = |H| and [L : F ] = [G : H]

There are a few properties of the Main Theorem which

are worth discussing. First notice that the correspon- dence between fields and groups is order reversing. That is if L, L′^ are two fields and H = G(K/L) and H′^ =

G(K/L′) then L ⊂ L if and only if H ⊂ H′^ Images of Interm

Theorem 0.1.0.10. Let K/F be a Galois extension and let L be an intermediate field. Let H = G(K/L) be the corresponding subgroup of G = G(K/F ). Then

(a) Let σ be an element of G. The subgroup of G which corresponds to the conjugate subfield σL is the conjugate subgroup σHσ−^1. In other words G(K/σL) = σHσ−^1. (b) L is a Galois extension of F if and only if H is a normal subgroup of G. When this is so, then G(L/F ) is isomorphic to the quotient group G/H

*****SEE DIAGRAM ON PAGE 559 *****

Its kernel is the set of σ ∈ G which induce the identity on L, which is H. Therefore G/H is isomorphic to G(L/F ). Counting degrees and orders we have

[L : F ] = |G/H| ≤ |G(L/F )|

Hence L is a Galois extension and that G/H ≈ G(L/F )

Conversely suppose that L/F is Galois. Then L is a splitting field of some polynomial g(x) ∈ F [x]. So L = F (β 1 ,... , βk) where the βi are the roots of g(x) in K. An F -automorphism σ of K permutes these roots and there- fore carries L to itself. L = σl By Part (a), H = σHσ−^1 and thus H is normal.

0.2 Kummer Extensions

Now we will consider extensions which contain pth roots of unity where p is a prime. For now we will assume that all fields are subfields of C.

Definition 0.2.0.11. Let F ⊆ C be a subfield of C which contains a primitive pth root of unity ζp = e^2 πi/p.

Extensions Generated by Single Root of xp^ − a

Lemma 0.2.0.12. If α is a root of f (x) = xp^ − a then α, ζpα, ζ p^2 α,... , ζ pp −^1 α are the roots of f (x). So the splitting field of xp^ − a is generated by a single root K = F [α]

Proof. This is easy to see as (αζpm )p^ = a for all m. And further as ζpm 6 = ζ pm ′for all m′^6 = m these are all the roots. So it suffices to show that ζp ∈ F (α, αζp,... , αζ pp −^1 ). But this is true because ζp = (αζp)/α.

Splitting Field of xp^ − a

Theorem 0.2.0.13. Let F ⊆ C and let F contain a pth root of unity. Further let a ∈ F be an element which is not a pth power in F. Then the splitting field

irreducible over F , and that G(K/F ) is cyclic of order p.

Galois Extensions of Degree p

Theorem 0.2.0.14. Let F be a subfield of C which contains a pth root of unity ζp and let K/F be a Ga- lois extension of degree p. Then K is obtained by adjoining a pth root to F.

Proof. The Galois group G has prime order p = [K : F ] so it is a cyclic group. Any element σ, not the identity, will generate it. Let us view K as an F -vector space. Then σ is a linear operator on K. For, since σ is an F -automorphism,

σ(α + β) = σ(α) + σ(β) and σ(cα) = σ(c)σ(α) = cσ(α)

for all c ∈ F and α, β ∈ K. Since G is a cyclic group of order p, σp^ = 1. An eigenvalue λ for this operator must satisfy the relation λp^ = 1 which means that λ is a power

of ζ. By hypothesis, these eigenvalues are in the field F. Moreover, there is at least one eigenvalue different from

1. This is a fact about any linear operator T such that some power of T is the identity, because such a linear operator can be diagonalized. Its eigenvalues are the en- tries of the diagonal matrix A which represents it. If T is not the identity, as is the case here, then A 6 = I, so some diagonal entry is different from 1.

We choose an eigenvector α with an eigenvalue ζi^6 = 1. Then σ(α) = ζiα and hence σ(αp) = σ(α)p^ = (ζiα)p^ = ζipαp^ = αp. So σ fixes αp. Since σ generates G the ele- ment αp^ is in the fixed field KG^ which is F.

We have therefore found an element α ∈ K whose pth power is in F. Since σ(α) 6 = α the element α is not in F. But, since [K : F ] is prime, α generates K.