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Topic: The algebraic closure of the rationals
Replies: 10   Last Post: May 10, 2006 12:30 PM

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John Baez

Posts: 542
Registered: 12/6/04
The algebraic closure of the rationals
Posted: Apr 6, 2006 2:00 PM
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Vaughan Pratt and I were discussing the fundamental theorem of
algebra, and we were led to some puzzles neither of us knows
anything about. Let Qbar stand for the algebraic closure of
the field of rational numbers. To what extent can we get our
hands on Qbar in an explicit way?

(Perhaps I should say Qbar is "an" algebraic closure, since
while they're all isomorphic, the full set of isomorphisms is
hard to understand, and this is part of the issue.)

For example:

1) Is there a way to enumerate the elements of Qbar such that
relative to this enumeration, the field operations are computable?
If so, how efficiently can they be computed? If not, how far
up the hierarchy of impossible-to-actually-compute functions do
we have to go?

2) On the other extreme: can we even prove the existence of Qbar
without the axiom of choice, or perhaps countable choice?

3) I've heard that it's even hard to get an "explicit" description
of the algebraic closures of finite fields - are there any
theorems to this effect?

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