See the algebraically closed
to see the precise statement
of the uniqueness and background.
The guts of the proof is
in the following technical lemma.
be a ring homomorphism of fields.
Suppose in addition that L
is algebraically closed field and that
M is an algebraic
of K. Then we can extend
f to a ring homorphism
M-->L. If further,
M is algebraically closed
and L is algebraic over
f(K) then this extension
is an isomorphism.
Before we start recall that any ring homomorphism
of fields is forced to be injective (see proof of
the uniqueness of splitting fields for details).
The proof is an application of
Zorn's Lemma. Consider the set
T which consists of all pairs
(F,g) where F is a
subfield of M and a field
extension of K, and g:F-->L
is a ring homomorphism. Define a partial order
on T by setting
(F,g) <= (H,h)
if and only if F<=H
. Since T
it is not empty.
We have to show that a totally ordered subset has an
. So let S
be such a subset.
The union H
of all F
is clearly a field. Further we can define a map
by making a
for all (F,g)
. Obviously this is a ring homomorphism.
Thus, Zorn's Lemma provides us with a maximal
t:k-->L extends f. Now I claim
that we have k=M. If not, choose
a that lies in M but not k.
Now the idea in the lemma in proof of the uniqueness of splitting fields
applies here and shows that t can be extended to a
ring homomorphism k(a)-->L. This contradicts
the maximality of of (k,t). So we have established
that there is an extension of f to a ring homomorphism
Now suppose that M is algebraically closed and L
is algebraic over f(K). So f(M) is algebraically
closed and L is algebraic over f(M).
It follows that L=f(M), as needed.
Proof of the uniqueness of algebraic closure:
This is easy now. For suppose that L1
and L2 are both algebraic closures of
K. The inclusion of K in L2
gives an injective ring homomorphism f:K-->L2.
So by the lemma it extends to an isomorphism
Since g extends f we have g(a)=a for all
a in K, as needed.