Let the extention $GF(p^m) \supset GF(p)$ that contains roots of $p(x)=x^{p^{m}}-1$. Show that those roots are distinct and that forms a field

Question

I want to rewrite a question not so well written on this site and clarified by Mr. Lahtonen (thank you again).

So here the question:

Let the extention $GF(p^m) \supset GF(p)$ that contains roots of $p(x)=x^{p^{m}}-1$. Show that those roots are distinct and that forms a field

I know that the roots of $p(x)=x^{p^{}}-1$ are contained in $p(x)=x^{p^{m}}-1$, but then?

edit: probably the correct exercise was $p(x)=x^{p^{m}-1}$

Answer 1

I am not sure if you understood the proof completely or approximately, so I will include a full proof.

$P(X)=X^{p^m-1}-1$ vanishes at every nonzero point of the field, and there are $p^m-1$ of them. So the product of all $X-a$, where $a$ runs through all nonzero elements of the field, whoch we denote as $Q$, has degree $p^m-1$ and divides $P$. Since $P$ and $Q$ are monic and have the same degree, they are equal. Thus the root of $XP$ are pairwise distinct and are exactly the elements of the field.

Answer 2

Consider the splitting field $E=GF(p^m)$ of $f(x)=x^{p^m}-x$ over $\mathbb{Z}_p$

We will show that $|E|=p^m.$ Since $f(x)$ splits in $E$, we know that $f(x)$ has exactly $p^m$ zeros in $E$, counting multiplicity. Moreover, by the Theorem

A polynomial $f(x)$ over a field $F$ has a multiple zero in some extension $E$ if and only if $f(x)$ and $f^{'}(x)$ have a common factor of positive degree in $F[x]$.(Refer Gallian Theorem $20.5$ for the proof)

Every zero of $f(x)$ has multiplicity $1$. Because $f^{'}(x)=p^mx^{p^m-1}-1=0.x^{p^m-1}-1=-1$ Thus $f(x)$ and $f^{'}(x)$ does not have any common factor of positive degree. Thus, $f(x)$ has $p^m$ distinct zeros in $E=GF(p^m)$

On the other hand, the set of zeros of $f(x)$ in $E$ is closed under addition, subtraction, multiplication, and division by nonzero elements so that the set of zeros of $f(x)$ is itself an extension field of $Z_p$ in which $f(x)$ splits. Thus, the set of zeros of $f(x)$ is $E$ and, therefore, $|E|=p^m$.