# Existence of algebraic closure and Axiom of choice [duplicate]

we need zorn's lemma for proving that every field $F$ has a unique algebraic closure. but I haven't seen a converse for this important Theorem.

From the above illustration my question is:

Is it true that the existence of The unique algebraic closure is equevalent to *axiom of choice*$(AC)$?

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Take a look at mathoverflow.net/questions/46566/… and the links given there. –  Andres Caicedo Sep 18 '12 at 16:16
In particular, the answers over there show that the answer to this question is negative, since the existence of ultrafilters is known to be strictly weaker than AC. –  Joel David Hamkins Sep 18 '12 at 16:19
Ah. Thenk you very much for your notifications. I am sorry for duplication of this Question. I didn't found it in MO. –  Ali Reza Sep 18 '12 at 16:43
Incidentally, the algebraic closure is not unique. –  anon Sep 18 '12 at 20:47