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This question already has an answer here:

How do I prove that $X^{p^n}-X$ is the product of all monic irreducible polynomials in $\mathbb Z_p[X]$ of degree dividing $n$?

Let $\bar Z_p$ be an algebraic closure of $Z_p$.

Define $F=\{x\in \bar Z_p|x^{p^n}-x=0\}$.

Then $X^{p^n}-X=\prod_{\alpha\in F} X-\alpha$.

Now $S$ be the set of monic prime polynomials of $Z_p[X]$ of which degree divides $n$.

Then, I know that $F=\bigcup_{f\in S} \{x\in \bar Z_p| f(x)=0\}$.

With this information, how do I prove that $X^{p^n}-X$ is actually the product of all elements of $S$?

To assure the equality, I think it must be shown that $\{x\in \bar Z_p|f(x)=0\}$ are mutually disjoint, but I don't know how to show this

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marked as duplicate by Watson, Henrik - stop hurting Monica, Namaste abstract-algebra Dec 23 '16 at 14:24

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  • $\begingroup$ A polynomial is separable if and only if it is relatively prime to its derivative. What is the derivative of $X^{p^n} - X$ in characteristic $p$? $\endgroup$ – KCd Apr 19 '15 at 14:55
  • $\begingroup$ You need the facts that $F$ is the unique field of size $p^n$ inside $\overline{\Bbb{Z}_p}$ and that to each factor $d\mid n$ its has a subfield of size $p^d$. Then if $f(x)$ is an irreducible polynomial of degree $d$, its zeros lie inside the unique field of size $p^d$, $\subseteq F$. $\endgroup$ – Jyrki Lahtonen Apr 19 '15 at 14:57
  • $\begingroup$ @JyrkiLahtonen I have proven that. But still have a trouble concluding this exercise.. $\endgroup$ – Rubertos Apr 19 '15 at 14:59
  • $\begingroup$ @KCd I haven't learned "separable" yet $\endgroup$ – Rubertos Apr 19 '15 at 14:59
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    $\begingroup$ @JyrkiLahtonen, probably it's not a good idea in characteristic $p$ to be writing a polynomial as $p(x)$, especially when its leading term has an exponent that is a power of $p$. $\endgroup$ – KCd Apr 19 '15 at 15:39
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Some ideas:

Prove that $\;w\;$ is a multiple root of a non-zero polynomial $\;f\;$ iff $\;w\;$ is also a root of its derivative $\;f'\;$. Deduce that if $\;f\;$ is irreducible (over some field), then this happens iff $\;f'=0\;$, and thus over a field of characteristic $\;p>0\;$ this can happen iff all the non-zero coefficients of $\;f\;$ correspond to powers of $\;x\;$ which are multiples of $\;p\;$.

Thus, $\;T(x):=x^{p^n}-x\in\Bbb F_p[x]\;$ is separable, meaning: all its roots are simple.

Now, using your notation, show that $\;\Bbb F_{p^n}=\{w\in\overline{\Bbb F_p}\;:\;\;T(w)=0\}$ , and finally: show that $\;\Bbb F_{p^m}\le\Bbb F_{p^n}\iff m\mid n\;$ .

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  • $\begingroup$ I still don't get it even though I completely understand what you wrote here. What I don't get is to show that $f$ has only simple roots. How do I show this? Any root of $f$ is one of a root of $T$, but that "the roots of $T$ are simple" has nothing to do with showing the roots of $f$ are simple $\endgroup$ – Rubertos Apr 19 '15 at 19:40

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