Re: The algebraic closure of the rationals


On 6-Apr-2006, baez@xxxxxxxxxxxxxx (John Baez)
wrote in message <e13l06$sr9$1@xxxxxxxxxxxx>:

[...]

(Perhaps I should say Qbar is "an" algebraic closure, since
while they're all isomorphic, the full set of isomorphisms is
hard to understand, and this is part of the issue.)

[...]

2) On the other extreme: can we even prove the existence of Qbar
without the axiom of choice, or perhaps countable choice?

Pretty much every undergraduate algebra textbook has a proof that C
contains an algebraic closure of Q, which theorem is in fact almost
trivial given the Fundamental Theorem of Algebra.

I think you do need AC to show that all algebraic closures of a
given field are isomorphic, and perhaps also to show the existence
of an algebraic closure for fields other than subfields of C.

[...]

--
Jim Heckman

.

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