algebra with finite field and isomorphic.

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 ?
---------------------------------------
proof 2)

the elements of F and F' are exactly the roots of
the polynomial h(x) = x^p^n -x.

so, F = F'
thus, isomorphic.

is this a foolish thinking ?

i think that it means that
F=F' is unique splitting field of [x^p^n -x] over Z_p.
no ?
----------------------------------------
proof 3)

anyway, i don't use the hint.
[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)>.]

i want to prove with hint.
so, i need your advice.
i also need the advice of Arturo Magidin.


.

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