[FOM] Algebraic closure of Q

Stephen G Simpson simpson at math.psu.edu
Fri Jun 2 16:12:40 EDT 2006

I wrote:

 > > Therefore, it appears that Hodges must be talking about some
 > > other context and/or some other sense of algebraic closure.  I am
 > > not familiar with Hodges' result, so could you please explain the
 > > context.
 > > 
 > > Define an algebraic closure of a countable field K to be a
 > > countable algebraically closed field L together with a
 > > monomorphism h: K --> L such that every element of L is a root of
 > > a nonzero polynomial whose coefficients are images under h of
 > > elements of K.

Timothy Chow wrote on May 13, 2006:

 > I believe that the difference between your definition and the
 > definition used by Hodges is that you require the algebraic closure
 > of a countable field to be countable *by definition*, whereas
 > Hodges takes an algebraic closure of K to be any extension of K
 > that is (1) algebraically closed and (2) contains no proper
 > subfield that is algebraically closed.  If a countable union of
 > finite sets is not necessarily countable, then an algebraic closure
 > (in the sense of Hodges) of a countable field K need not be
 > countable.

I have a further question about this.  

Hodges' result is that, if you use Hodges' definition of algebraic
closure, then ZF without the Axiom of Choice does not suffice to prove
the existence of the algebraic closure of Q, the rationals.

On the other hand, as I pointed out, there is a perfectly reasonable
alternative definition of algebraic closure, such that under this
alternative definition, it is provable in ZF without the Axiom of
Choice (and indeed in much weaker systems, e.g., relatively weak
subsystems of second-order arithmetic) that for every countable field
(such as Q) there exists a unique algebraic closure.

My question is, given this state of affairs, what is the interest of
Hodges' result?

After all, it is well known that, when proving a counterpart of a
classical mathematical theorem in an alternative foundational
framework, it is often necessary to modify or restrict the definitions
appropriately.  And, as pointed out above, Hodges' result provides a
nice example or illustration of this general principle.  Here I am
taking ZFC to be the classical foundational framework, and ZF without
the Axiom of Choice to be an alternative framework.  I think this is a
standard point of view.

My question is, does Hodges' result have any mathematical or
foundational significance, beyond the above?  For example, is Hodges'
model of ZF (used to prove his result) of additional interest in
algebra, or perhaps in combinatorics, or in some other branch of

Of course I admire Hodges' ingenuity in constructing an appropriate
model of ZF to prove his result.  My question concerns only the wider
significance of the result.


Stephen G. Simpson

Professor of Mathematics, Pennsylvania State University

Research interests: mathematical logic, foundations of mathematics

Web page: http://www.math.psu.edu/simpson/

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