The algebraic closure of the rationals consists of the algebraic numbers, and is a countable field. The complex numbers are an uncountable field. In between the two lie the algebraic closures of the various transcendental extensions of the rationals. -- Gene On Friday, July 21, 2017, 5:46:55 PM PDT, Henry Baker <hbaker1@pipeline.com> wrote: Thanks, Dan. The algebraic closure of the infinite sequence GF(p),GF(p^2),GF(p^3),... as described in the Wikipedia article is pretty ugly -- at least as described. I was hoping for something as elegant as the complex numbers with their usual topology. Perhaps I need to check back in about 300 years? At 10:08 AM 7/21/2017, Dan Asimov wrote:
Every field K — if it does not already contain all the roots of polynomials in K[x] — is a proper subfield of a larger field that is algebraically closed.
E.g., see <https://en.wikipedia.org/wiki/Finite_field#Algebraic_closure>.
I don't know how such an algebraic closure is usually visualized.
But I believe that, for any prime p, the Galois group of [the algebraic closure of the finite field F(p)] over F(p) is isomorphic to the same thing for any other prime.
—Dan
From: Henry Baker <hbaker1@pipelline.com> Jul 21, 2017 9:06 AM: -----
OK, if I extend the rationals with the root alpha of an irreducible polynomial p[x], I can plot alpha on the complex plane; indeed, I can plot *all* of the roots of p[x] on the complex plane. So all of these "extension roots" live in the complex plane.
Is there an analogous (single) place/field where all extension roots of GF(p) live -- i.e., a larger field which includes all of the extension fields of GF(p) ?
There seems to be a problem, since there are many (isomorphic) ways to extend GF(p); perhaps these are all different in this larger field?
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