Re: algebra with finite field and isomorphic.

In article <edbsr0$2h9$1@xxxxxxxxxxxxxxxx>,
"mina_world" <mina_world@xxxxxxxxxxx> writes:

<mareg@xxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:edbq1f$ba6$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
In article <edb4av$lra$1@xxxxxxxxxxxxxxxx>,
"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.

Derek Holt.

i only know the fact that
f(x) be the minimal(irreducible) polynomial of "a" or "b" over Z_p.


so the extension Z_p(b) has degree n over Z_p, so |Z_p(b)| = p^n = |F'|.


---------------------------------------
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 ?

It is not necessarily true that F = F', but this argument is basically
correct.
F and F' consist of the p^n roots of x^p^n - x, so they must both be
splitting
fields of x^p^n - x over Z_p. Now there is a theorem which says that any
two
splitting fields of an irreducible polynomial over a field are isomorphic,
so
it follows from that theorem that F and F' are isomorphic.

----------------------------------------
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 think the hint is very unhelpful! In order to use the hint, you first
have
to prove that there exists an irreducible polynomial of degree n over Z_p,
which is by no means trivial.


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 ?




.

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