# [FOM] Algebraic closure of Q

Andreas Blass ablass at umich.edu
Mon May 15 10:25:50 EDT 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.

I had been discussing this question (how much choice follows from
uniqueness of the algebraic closure of Q) with Reinhard Boerger, and
I had proposed an idea somewhat similar to Harvey's.  Unfortunately,
Reinhard detected a problem, which seems to affect Harvey's proof as
well.  The problem is in the next-to-last sentence, where one needs
to take the algebraic closure of a field that isn't known to be
countable (because it already has the B_i embedded in it).  I think
this problem can be evaded by interleaving the adjunction of the
"desired" roots (the p-th roots of 2 in Harvey's proposal) with the
additional roots (of other polynomials) needed for algebraic
closure.  The idea is that each of these additional roots would be
added to a countable field, in which only finitely many of the
desired roots have been adjoined so far.
There is, however, another problem with Harvey's proposal: The set
of roots of the equation z^p = 2 has (for p at least 5 or so) a non-
trivial structure, specifically, a four place relation "xy = uv" that
holds of some but not all 4-tuples of distinct roots.  So if we are
to use the sets B_i for the solutions when p = p_i, the sets B_i
would have to support such structures.  For each B_i, the necessary
structure exists, but we'd need to choose one for each i, and that
may be impossible.
As far as I can see, avoiding this problem requires replacing the
sequence of polynomials z^{p_i} - 2 with some sequence such that the
Galois group of each one is the full symmetric group of permutations
of all its roots; furthermore, the Galois group must be this big not
only over Q but over the extension obtained by adjoining roots of
previous polynomials in our list.  (Recall that evading the first
problem meant that the list of "desired" polynomials is interleaved
with a list of all polynomials over Q; this interleaved list is what
"previous polynomials in our list" refers to in the previous
sentence.)  I believe that one can find polynomials with the required
Galois groups, but I have not checked this carefully.

Andreas Blass

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