See the algebraically closed writeup to see the precise statement of the uniqueness and background. The guts of the proof is in the following technical lemma.

Lemma Let f:K-->L be a ring homomorphism of fields. Suppose in addition that L is algebraically closed field and that M is an algebraic field extension 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.

Proof: 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 and h extends g. Since T contains (K,f) it is not empty. We have to show that a totally ordered subset has an upper bound. So let S be such a subset. The union H of all F with (F,g) in S is clearly a field. Further we can define a map a:H-->L by making a agree with with g on F for all (F,g) in S. Obviously this is a ring homomorphism. Thus, Zorn's Lemma provides us with a maximal element (k,t) of S.

By construction 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 M-->L.

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 g:L1-->L2. Since g extends f we have g(a)=a for all a in K, as needed.

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