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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

user26857
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Rubertos
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  • 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 $p$? – KCd Apr 19 '15 at 14:55
  • 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 $p^d$, $\subseteq F$. – Jyrki Lahtonen Apr 19 '15 at 14:57
  • @JyrkiLahtonen I have proven that. But still have a trouble concluding this exercise.. – Rubertos Apr 19 '15 at 14:59
  • @KCd I haven't learned "separable" yet – Rubertos Apr 19 '15 at 14:59
  • @JyrkiLahtonen Hence for $f\in S$ and a root $\alpha$ of $f$, $Z_p(\alpha)$ is the unique subfield of size $p^{deg(f)}$. How do I apply this to have a conclusion? – Rubertos Apr 19 '15 at 15:02
  • So you know that zeros of such an $f(x)$ are also zeros of $x^{p^n}-x$, right (and neither have zeros of multiplicity higher than 1 by KCd's argument)? – Jyrki Lahtonen Apr 19 '15 at 15:02
  • @JyrkiLahtonen Yes I know that. I wrote that in my post.., but no to the second one. I don't understand KCd's argument. I haven't learned "separable" yet. I know that there are $p^n$ distinct roots for $X^{p^n}-X$ in $F$, but for arbitrary $f\in S$, I don't know why it has $\deg(f)$ distinct roots – Rubertos Apr 19 '15 at 15:04
  • I think it is needed in this exercise. Separability simply means that all the zeros of all the polynomials in $S$ are simple. Have you seen that? It actually follows if you know that $g(x)=x^{p^n}-x$ has $p^n$ distinct zeros. Otherwise the gcd of $f(x)$ and $g(x)$ would be a proper factor of $f$. – Jyrki Lahtonen Apr 19 '15 at 15:09
  • @JyrkiLahtonen I know that $X^{p^n}-X$ has $p^n$ distinct roots. The thing I have a trouble with is to assert that $f=\prod X-\alpha$ where $\alpha$ ranges over ${x\in \bar Z_p | f(x)=0}$ and ${x\in \bar Z_p|f(x)=0}$ are mutually disjoint for $f's$.. – Rubertos Apr 19 '15 at 15:17
  • If you know that $x^{p^n}-x$ has $p^n$ distinct roots, then they are all simple, right? So all the elements of $F$ have minimal polynomials with simple roots. OTOH all the prime polynomials of degree $d \mid n$ are such minimal polynomials. Ergo? – Jyrki Lahtonen Apr 19 '15 at 15:17
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    @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 $p$. – KCd Apr 19 '15 at 15:39
  • A good point, @KCd. I edited the comment. – Jyrki Lahtonen Apr 19 '15 at 15:42
  • @KCd Let $\bar F$ be an algebraic closure of a field $F$ and $\alpha\in \bar F$. Is it true the the multiplicity of $\alpha$ of the minimal polynomial of $\alpha$ must be $1$? – Rubertos Apr 19 '15 at 15:43
  • For fields in general no, but for finite fields and all fields of characteristic zero yes. Concrete counterexample: if $F = {\mathbf F}_p(u)$, where $u$ is an indeterminate, then $X^p - u$ is irreducible in $F[X]$ but it has only one root: if $r$ is a root then $X^p - u = X^p - r^p = (X-r)^p$ because the $p$th power function is additive in characteristic $p$. Hence $X^p - u$ has $r$ has its only root. – KCd Apr 19 '15 at 15:49
  • You say you don't know what separable polynomials are. So please learn about them. They are covered in lots of algebra books. – KCd Apr 19 '15 at 15:51
  • How do I prove that for finite fields? That's the one and only one I didn't know and it is exactly what I kept asking for.. – Rubertos Apr 19 '15 at 15:51

1 Answers1

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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 $\;\Bbb F_{p^m}\le\Bbb F_{p^n}\iff m\mid n\;$ .

Timbuc
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  • 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 – Rubertos Apr 19 '15 at 19:40