First read the splitting field writeup. We are going to prove the uniqueness of splitting fields. What does this mean? Well suppose that f in K[x] is a polynomial over a field K. Suppose that L1 and L2 are both splitting fields of f over K. We will prove that that there is an isomorphism F:L1-->L2 such that F(k)=k for all k in K.

We need some notation and some basic remarks. Let F:L-->M be a ring homomorphism between fields L and M. First notice that F is injective. This is because the kernel of F is an ideal and a field is a simple ring. Second, we can extend F to a ring homomorphism F:L[x]-->M[x] if we define

F(anxn + ... + a0) = F(an)xn + ... + F(a0)
Next a trivial lemma about dimensions of field extensions.

Lemma If K<=L<=M are field extensions then [M:K]=[M:L][L:K]

There is only something to prove if the field extensions are all finite dimensional, so suppose that. Let ei be a basis for M as an L-vector space and let fj be a basis for L as a K-vector space. It is easy to check that eifj is a basis for M over K, hence the result.

The following lemma contains the guts of the uniqueness of splitting fields.

Lemma Let L be a splitting field over K of f in K[x] and let F:K-->M be a ring homomorphism of fields. Then F extends to a ring homomorphism L-->M iff F(f) splits over M.

Proof: (==>) This direction is easy. If F extends to a ring homomorphism L-->M then clearly F(L) is a splitting field for F(f) over F(K). In particular, F(f) splits over M.

(<==) We proceed by induction on [L:K]. If [L:K]=1 then there is nothing to prove, the base case. So we assume that [L:K]>1 Let g be an irreducible factor of f over K that is not linear. Clearly F(f) splits into linear factors over M. Let a be a root of g in L and let b be a root of F(f) in M. As explained in the field extension writeup, evaluation at a induces an isomorphism of rings T:K[x]/gK[x]-->K(a). Likewise, evaluation at b induces an isomorphism S:F(K)[x]/F(g)K[x]-->K(b). Let U:K(b)-->M be the natural inclusion map. Thus UST-1 is a ring homomorphism K(a)-->M. It is routine to check that it extends F, as required. Now consider [L:K(a)]. By the preceeding lemma and the fact (see field extension) that [K(a):K] is the degree of g which is > 1, we see that [L:K(a)]< [L:K] so by induction there is an extension of of UST-1 to a ring homomorphism K(a)-->M. Thus we have extended f as was required.

Proof of uniqueness of splitting fields: Since L2 is a splitting field of f over K we have the natural inclusion i:K-->L2. By the previous lemma since i(f)=f splits over L2 we can extend i to a ring homomorphism H:L1-->L2. Since f splits over L1 then i(f)=f splits over H(L1). Since L2 is a splitting field this means that H is surjective. Since it is injective it is therefore an isomorphism. Since H extends i it fixes the elements of K, as required.

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