*f*in

*K[x]*is a polynomial over a field

*K*. Suppose that

*L*and

_{1}*L*are both splitting fields of

_{2}*f*over

*K*. We will prove that that there is an isomorphism

*F:L*such that

_{1}-->L_{2}*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

Next a trivial lemma about dimensions of field extensions.F(a_{n}x^{n}+ ... + a_{0}) = F(a_{n})x^{n}+ ... + F(a_{0})

**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 *e _{i}* be
a basis for

*M*as an

*L*-vector space and let

*f*be a basis for

_{j}*L*as a

*K*-vector space. It is easy to check that

*e*is a basis for

_{i}f_{j}*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,

*splits over*

__F__(f)*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

*in*

__F__(f)*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]/*. Let

__F__(g)K[x]-->K(b)*U:K(b)-->M*be the natural inclusion map. Thus

*UST*is a ring homomorphism

^{-1}*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*to a ring homomorphism

^{-1}*K(a)-->M*. Thus we have extended

*f*as was required.

**Proof of uniqueness of splitting fields:**
Since *L _{2}* is a splitting field of

*f*over

*K*we have the natural inclusion

*i:K-->L*. By the previous lemma since

_{2}*splits over*

__i__(f)=f*L*we can extend

_{2}*i*to a ring homomorphism

*H:L*. Since

_{1}-->L_{2}*f*splits over

*L*then

_{1}*splits over*

__i__(f)=f*H(L*. Since

_{1})*L*is a splitting field this means that

_{2}*H*is surjective. Since it is injective it is therefore an isomorphism. Since

*H*extends

*i*it fixes the elements of

*K*, as required.