[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.
> Agree?

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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