# [FOM] Algebraic closure of Q

Harvey Friedman friedman at math.ohio-state.edu
Fri May 12 19:34:27 EDT 2006

On 5/12/06 1:44 PM, "Timothy Y. Chow" <tchow at alum.mit.edu> wrote:

>> 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.

First of all, what is an algebraic closure? I assume that we are taking it
to be an algebraically closed extension in which every element is algebraic
in some finite number of ground elements.

In this case, this means that every element is algebraic. I.e., is a
solution to a nontrivial polynomial with rational coefficients, or is 0.
Since nontrivial polynomials have only finitely many solutions, isn't the
field a countable union of finite sets? If we assume only that a countable
union of finite sets is countable, then the field is countable. Of course
then any two are isomorphic.

For the other direction, suppose any two algebraic closures of Q are
isomorphic. Let A_1,A_2,... be finite sets of cardinalities n_1,n_2,... .
Wlog, assume the A's are disjoint and disjoint from Q. Let p_1 < p_2 < ...
be prime numbers such that each n_i < p_i. Then expand the sets to
B_1,B_2,..., still keeping them disjoint and disjoint from Q, but now where
Bi has exactly p_i elements.

Can't we consider the equations

z^p = 2

for p = p_1,p_2,..., and then use the sets B_i for the solutions where p =
p_i? Then make this into a field. And then take an algebraic closure. The
result will be also an algebraic closure of Q.

Now compare it to the usual algebraic closure of Q that is countable. Then
we get that the union of the B's is countable, and hence the union of the
A's is countable. Therefore a countable union of finite sets is countable.

Believe this?

Harvey