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.



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> 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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Understanding a valuation property of function fields
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.


