Re: Algebra with root and Z_2[x].

On Nov 25, 5:17 am, Bill Dubuque <w...@xxxxxxxxxxxxxxxxxxxx> wrote:
Muhammad <mzafrul...@xxxxxxx> wrote:

Have you not solved the following problem also?
Let p be a prime and let E = GF(p^4). Then every polynomial of degree
1, 2, 4 in Z_p [X] has a root in E.
I think the reason is: Given an irreducible polynomial f of degree 2
in Z_p [X] we have Z_p [X]/(f) =GF(p^2). So there is a root of f that
satisfies x^{p^2}-x = 0, but x^{p^2}-x divides x^{p^4}-x.

Similar ideas also lead to an easy irreducibility test, namely the
polynomial analog of the Pocklington-Lehmer test, see my prior posthttp://google.com/groups?selm=56965i$gho%40joseph.cs.berkeley.edu

--Bill Dubuque

Thanks. I looked up the link. Indeed there are lot of good results out
there. I will try to look up the paper you mention in that other post.
Muhammad
.