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