[FOM] Algebraic closure of Q
Timothy Y. Chow
tchow at alum.mit.edu
Fri May 12 13:44:46 EDT 2006
> I think the uniqueness should be equivalent, over a weak fragment of ZF,
> to any countable union of finite sets is countable.
Hmmm...so, given any algebraically closed field containing Q, I guess one
can enumerate the polynomials (without choice) and let F_n be the subfield
generated by the first n polynomials. Then I guess one checks that the
proof that the union of the F_n is algebraically closed doesn't require
choice. So it should be provable in ZF that every algebraic closure of Q
is a countable union of countable sets.
On sci.math.research, David Madore sketched the (choice-free) argument
that any two countable algebraic closures of Q are isomorphic.
So I think I see how it suffices to assume that a countable union of
*countable* sets is countable. How do you get this down to a countable
union of *finite* sets?
And for the converse, I guess the idea is that given a countable union of
finite sets S_n, one can simply build a tower of extensions where the nth
extension has degree |S_n| over the last one?
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