Re: algebra with finite field and isomorphic.
- From: "mina_world" <mina_world@xxxxxxxxxxx>
- Date: Sun, 3 Sep 2006 16:41:01 +0900
"mina_world" <mina_world@xxxxxxxxxxx> writes:
hello sir~
show that two finite fields of the same order p^n
are isomorphic.
[hint : let p(x) in Z_p[x] be irreducible of degree n.
show every field of p^n elements is isomorphic
to Z_p[x]/<p(x)>.]
---------------------------------------
yes, i try it.
i had three questions.
proof 1)
lemma 1) The multiplicative group <F^*,.> of
nonzero elements of a finite field F is cyclic.
Let F and F' be two finite fields of the order p^n.
These fields must have characteristic p, for a prime p,
so, they contain Z_p as a subfield.
by lemma 1, unit group of F is cyclic.
so, F^* = <a>. namely "a" is a generator.
so, F = Z_p(a) is a finite extension of Z_p.
so, F = Z_p(a) is a algebraic extension and simple extension.
since "a" is algebraic over Z_p,
let f(x) be the minimal(irreducible) polynomial of "a" over Z_p.
so, F = Z_p(a) ~ Z_p[x]/<f(x)>.
the elements of F and F' are exactly the roots of
the polynomial h(x) = x^p^n -x.
since a in F, h(a) = 0.
so, f(x) is one of the irreducible factors of h(x).
so, there exists "b" in F' such that f(b) = 0.
since "b" is algebraic over Z_p
so, Z_p(b) ~ Z_p[x]/<f(x)>.
but i mush show that F' = Z_p(b).
i can't this. how do you show it ?
b is a root of the irreducible polynomial f of degree n over Z_p,
why ?
Which part of the assertion are you querying?
You have already said that f is irreducible.
You have already said that f(b) = 0, so b is a root of f.
And f has degree n because F = F_p(a) with |F| = p^n and "a" a root of f.
yes, so, |Z_p(b)| = p^n = |F'|.
and Z_p(b) in F'
how do you show that F' in Z_p(b) ?
i try again with hint.
let F be finite field of the order p^n.
F contain Z_p as a subfield.
let p(x) in Z_p[x] be irreducible of degree n.
so, there exists "a" in algebraic closure of Z_p such that
p(a) = 0.
since "a" is algebraic over Z_p,
Z_p(a) ~ Z_p[x]/<p(x)>.
since the basis of Z_p[x]/<p(x)> is {1, a, a^2,...,a^(n-1)},
|Z_p[x]/<p(x)>| = p^n.
but i can't show that F= Z_p(a).
how do you show it ?
and this question still holds, too.
.
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