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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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marked as duplicate by Benjamin Steinberg, Joel David Hamkins, Qiaochu Yuan, George Lowther, Asaf Karagila Sep 18 '12 at 21:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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
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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
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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
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Incidentally, the algebraic closure is not unique. –  anon Sep 18 '12 at 20:47