[FOM] Algebraic closure of Q

Robert M. Solovay solovay at Math.Berkeley.EDU
Sun May 14 03:17:24 EDT 2006


 	I have several problems with your proposed proof in ZF that "if 
there is a countable sequence of finite sets whose union is not finite or 
countable" then " there is an algebraic closure of Q which is not 

 	I have three objections:

 	1) You claim that we can find a sequence, F_i, of splitting fields
of the equations

z^{p_i} - 2 = 0 whose p_i^th roots of 2 are precisely the set B_i.

 	This would be evident if we had the lemma:

The Galois group of the splitting field of the equation z^{p_i} - 2 = 0 is 
the symmetric group on p_i letters.

 	But this is clearly false for p_i > 3.

 	[The following property holds for some but not all triples of 
distinct roots: (x_2/x_1)^2 = x_3/x_1 ]

 	2) Even if you get this sequence of splitting fields, I don't see 
that there is a single field that contains them all. {In ZF. In ZFC, of 
course, this is evident.}

 	3) The proposition that every field has an algebraic closure seems 
to need the axiom of choice. {I believe it is indeed, known, that the use 
of AC here is essential.}

 	Of course, none of this shows that the result you are trying to 
prove isn't true. Indeed, perhaps there is a correct proof along the rough 
lines of your current attempt.

 	--Bob Solovay

On Fri, 12 May 2006, Harvey Friedman wrote:

> 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
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