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@pipeline.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? -----