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Apr 19, 2021 at 19:13 comment added Tim Campion One obstruction to having a functorial construction of algebraic closures is the fact that the classifying space of the category of field extensions of $k$ is not typically homotopy equivalent to the category of algebraically closed field extensions of $k$ -- the former is contractible while the latter is $BGal(\bar k / k)$. To the extent that constructivism goes hand in hand with functoriality, and to the extent that this obstruction is encoded in the Galois group, I wonder if the answer to this question might be dependent on the inverse Galois problem...
Nov 20, 2010 at 12:01 answer added Joel David Hamkins timeline score: 32
Nov 19, 2010 at 22:39 answer added Eivind Dahl timeline score: 4
Nov 19, 2010 at 9:27 vote accept Qiaochu Yuan
Nov 19, 2010 at 2:18 comment added Andrés E. Caicedo Hi Timothy. Thanks for finding these posts, I thought they were more recent and couldn't find them.
Nov 19, 2010 at 2:10 comment added Timothy Chow Not exactly what you asked, but there was some discussion on the Foundations of Mathematics mailing list about whether the existence of an algebraic closure of Q is equivalent to the assumption that a countable union of finite sets is countable. See this post by Harvey Friedman cs.nyu.edu/pipermail/fom/2006-May/010541.html and this post by Andreas Blass cs.nyu.edu/pipermail/fom/2006-May/010551.html pointing out some difficulties.
Nov 19, 2010 at 1:31 history edited Qiaochu Yuan CC BY-SA 2.5
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Nov 19, 2010 at 1:29 comment added Willie Wong (BTW: credit where credit's due: I learned of that website from Andres mathoverflow.net/questions/45928/… )
Nov 19, 2010 at 1:29 answer added Andrés E. Caicedo timeline score: 25
Nov 19, 2010 at 1:28 comment added Willie Wong Every field has an algebraic closure is "Form 69" in consequences.emich.edu/conseq.htm , and it doesn't list any equivalent forms to it.
Nov 19, 2010 at 1:10 history asked Qiaochu Yuan CC BY-SA 2.5