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I came across this point in a paper recently and I'm having difficulty seeing why it's true. Any explanations or hints would be appreciated.

For any prime $\mathfrak{p}$ of $\mathbb{F}_q [t]$ such that $\mathfrak{p}$ is not the pole of $t$, it is the case that for some $m_0 \in \mathbb{Z}_{>0}$, for all positive integer multiples $m$ of $m_0$ we have $\mbox{ord}_{\mathfrak{p}}(t^{p^m} - t) = 1$.

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Let $m_0$ be the degree of $\mathfrak p$, so the splitting field of $\mathfrak p$ is the extension of $\mathbb F_q$ of degree $m_0$. The set of roots of $t^{p^m}-t$ contains this field extension if and only if $m_0$ divides $m$, and this is equivalent to $\mathfrak p$ dividing $t^{p^m}-t$. The claim now follows since $t^{p^m}-t$ is separable, so $\mathfrak p$ divides this polynomial at most with multiplicity one.

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  • $\begingroup$ Thank you! Can I ask, what made you think of this solution? Was it incredibly obvious, or was there something that influenced your thinking? $\endgroup$
    – user221330
    Commented Jul 31, 2018 at 17:36
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    $\begingroup$ @user221330 It was straightforward, because by separability of $t^{p^m}-t$, the question simply was when $\mathfrak p$ divides this polynomial, which in turn is equivalent to each root of $\mathfrak p$ dividing it. $\endgroup$
    – Peter Mueller
    Commented Jul 31, 2018 at 17:56

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