# 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 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 $$p(x)=x^{p^{m}-1}$$

• I am sorry, but as is, $p(x)=(x-1)^{p^m}$. – Mindlack Dec 17 '18 at 12:00
• Mmm that because $(x-1)^{p^{m}} = x^{p^{m}}-1 \pmod p$, right? – Alessar Dec 17 '18 at 12:03
• Yes indeed... did you mean $x^{p^m-1}-1$? – Mindlack Dec 17 '18 at 12:04
• You mean in the question text? – Alessar Dec 17 '18 at 12:06
• Yes: since your polynomial $p$ is a power of an affine polynomial, it cannot have distinct roots. However, if $p$ instead means $x^{p^m-1}-1$ then your statement holds. – Mindlack Dec 17 '18 at 12:10

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.

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, $$|E|=p^m$$.