Re: Finite extension of a finite field

In article <3a05a386-fe15-4f17-8d60-9b0369e5c023@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<sanchopancho80@xxxxxx> wrote:
On 15 Aug., 17:49, Angus Rodgers <twir...@xxxxxxxxxxx> wrote:
On Fri, 15 Aug 2008 08:44:00 -0700 (PDT), sanchopanch...@xxxxxx
wrote:

I've seen somebody writing F_p[a^(1/p)] with a in the finite field
F_p. What does this mean? More precisely: What does a^(1/p) mean?

Just guessing: it means F_p[b], where b is a root of x^p - a in an
extension field.

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril

I have thought that too at first, but why are the results isomorphic
for different roots? The polynomial x^p - a doesn't have to be
irreducible.

Let K be an algebraic closure of F_p; let r be a root of x^p-a in
K. Then r^p = a. And, in K, we have (x-r)^p = x^p - r^p = x^p - a
because K is of characteristic p.

That is: the polynomial has only one root, repeated p times, in any
algebraic closure of F_p. So it does not matter whether the polynomial
is irreducible or not: the only extension you can get is the one where
you adjoin (the unique) p-th root of a. The degree of the extension
will depend on the irreducible factors of x^p-a, of course, but not
the extension.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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