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Possible Duplicates:
Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
algebraic closure of commuting pairs of matrices

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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    $\begingroup$ Take a look at mathoverflow.net/questions/46566/… and the links given there. $\endgroup$ Sep 18, 2012 at 16:16
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    $\begingroup$ 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. $\endgroup$ Sep 18, 2012 at 16:19
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    $\begingroup$ Ah. Thenk you very much for your notifications. I am sorry for duplication of this Question. I didn't found it in MO. $\endgroup$
    – Ali Reza
    Sep 18, 2012 at 16:43
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    $\begingroup$ Incidentally, the algebraic closure is not unique. $\endgroup$
    – anon
    Sep 18, 2012 at 20:47

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