[FOM] Algebraic closure of Q
Timothy Y. Chow
tchow at alum.mit.edu
Sat May 13 12:49:08 EDT 2006
Stephen G Simpson <simpson at math.psu.edu> 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.
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.
> Then, the existence of an algebraic closure of any countable field K is
> provable in RCA_0. The uniqueness, up to isomorphism over K, is
> equivalent over RCA_0 to WKL_0.
>
> Define a strong algebraic closure of a countable field K to be an
> algebraic closure h: K --> L as above, such that the image of K under
> h exists as a subfield of L. Then, the existence of a strong
> algebraic closure of any countable field is equivalent over RCA_0 to
> ACA_0. The uniqueness is provable in WKL_0, hence in ACA_0.
Thanks for the information!
I wrote:
> 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?
Joe Shipman explained this to me. One really only needs to adjoin the
roots of each polynomial at each step, not the whole field generated by
them. One still gets a field anyway, as one can prove separately.
Tim
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