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I'd like to show the following:

$X^{p^n}-X$ is the product of all irreducible, normalized $f \in \mathbb F_p[X]$ with $deg(f) \mid n$ in $\mathbb F_p[X]$

To me, the claim above feels kinda similar to this theorem:

An irreducible polynomial $f \in \mathbb F_p[X]$ is a divisor of $X^{p^n}-X$ iff $deg(f(x)) = d$ divides $n$.

But I don't know how I can connect those two (or is this even the correct approach?). I'd appreciate any kind of help.

Yasuduck
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  • What exactly is the difference between the two, according to you? – Servaes Dec 23 '16 at 00:41
  • @Servaes: The theorem does not tell you the multiplicities of the irreducible factors of $X^{p^n}-X$. – Eric Wofsey Dec 23 '16 at 00:59
  • @EricWofsey I agree there are differences concerning the multiplicities as well as the leading coefficients of the factors. But the phrasing of the two statements obscures this, which is why I tried to suggest pinpointing what this difference is exactly (as it seems there was little effort put into this). Once this difference is clear, your very clean answer shows that it does not take much to verify that the claim indeed follows from the theorem. – Servaes Dec 23 '16 at 01:05
  • @Servaes: Ah, that's fair. I read your question as implying there wasn't any real difference and so you were confused about what OP was missing. – Eric Wofsey Dec 23 '16 at 01:10
  • Does this answer your question? Every irreducible polynomial f over perfect field F is separable – Watson Feb 23 '21 at 10:46
  • Sorry it is a duplicate of https://math.stackexchange.com/questions/106721/product-of-all-monic-irreducibles-with-degree-dividing-n-in-mathbbf-q – Watson Feb 23 '21 at 12:05

1 Answers1

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By the theorem and unique factorization in $\mathbb{F}_p[X]$, the factorization of $g(X)=X^{p^n}-X$ into irreducible polynomials must have the form $$g(X)=cf_1(X)^{e_1}f_2(X)^{e_2}\dots f_m(X)^{e_m}$$ where $c$ is a constant, $f_1,\dots f_m$ are all of the irreducible monic polynomials of degree dividing $n$, and $e_1,e_2,\dots,e_m$ are all at least $1$. Since $g(X)$ is monic, we must have $c=1$, so we just need to show $e_i=1$ for each $i$. That is, we want to show $g(X)$ has no repeated irreducible factors.

A common trick to understand whether a polynomial has repeated factors is to consider its derivative. Note that the derivative of $g(X)$ is $$g'(X)=p^nX^{p^n-1}+1=1.$$ If $e_i>1$ for some $i$, then $f_i(X)$ would divide $g'(X)$ (since when you differentiate the factorization of $g(X)$ using the product rule, every term will include a factor of $f_i(X)$). This is impossible, since $f_i(X)$ is not a unit. Thus we must have $e_i=1$ for all $i$, as desired.

Eric Wofsey
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