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There is a well-known generalisation of Fermat's Little Theorem as for any $a\in\mathbb{Z}$ and $n\in\mathbb{N}$ we have $$\sum_{d\mid n} \mu(n/d) a^d\equiv 0\pmod n.$$ where $\mu$ is the Möbius function. I wonder if this formula still holds in function fields. Say $F$ be a finite field and $F[x]$ be the polynomial ring over $F$. Then $F[x]$ is an Unique Factorisation Domain. Then we can define $\mu$ ($F[x]$ is an Euclidean Domain so primes and irreducibles are same). What I wonder is the following formula holds and is it easy to see if it does? Let $f\in F[x]$ where $F$ is a finite field. Then $$\sum_{d\mid f} \mu(f/d)a^{|d|}\equiv 0 \pmod f$$ where $|d|=|F[x]/\langle d\rangle|$.

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  • $\begingroup$ Well, if $f$ is irreducible, then you have the usual proof of Fermat's Little Theorem apply to the image of $a$ in $F[x]/<f>$, so it should be OK. $\endgroup$ – Lev Borisov Feb 23 '16 at 0:17
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    $\begingroup$ Yes, such a formula holds. Take a proof of the classical identity in $\mathbf Z$ and try to carry it over. Did you attempt to do that? If not, you should. By the way, in the formula you ask about you should write that $d$ runs over monic factors of $f$. $\endgroup$ – KConrad Feb 23 '16 at 7:04

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