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

*L*_{1} and

*L*_{2} are both splitting fields
of

*f* over

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

isomorphism *F:L*_{1}-->L_{2}
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__(a_{n}x^{n} + ... + a_{0}) =
F(a_{n})x^{n} + ... + F(a_{0})

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 *e*_{i} be
a basis for *M* as an *L*-vector space
and let
*f*_{j} be
a basis for *L* as a *K*-vector space.
It is easy to check that *e*_{i}f_{j} 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 *L*_{2} is a splitting field of *f* over *K*
we have the natural inclusion *i:K-->L*_{2}. By
the previous lemma since __i__(f)=f splits over
*L*_{2} we can extend *i* to a ring homomorphism
*H:L*_{1}-->L_{2}. Since *f* splits over
*L*_{1} then __i__(f)=f splits over
*H(L*_{1}). Since *L*_{2} 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.