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Giuga's conjecture (1950), which is still open and has strong numerical support, reads :

Let $n$ be a positive integer. If $1+\sum_{k=1}^{n-1}k^{n-1} \equiv 0\pmod{n}$ then $n$ is prime.

What would an analog for function fields be?

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The integers 1 to $n-1$ are the non-zero elements of $\mathbb Z/n\mathbb Z$. So one could take an ideal $I$ in a Dedekind domain $R$ and ask whether $$ (*)\qquad 1+\sum_{\substack{a\in R/I\\ a\ne0\\}} a^{\#R/I - 1} \equiv 0 \pmod{I} $$ is equivalent to $I$ being a prime ideal. I have no idea whether this is a reasonable question, since I have not done any experiements (I'll let you try), but this or something similar seems like a natural generalization.

Addendum: I did some experiments with $R=\mathbb F_p[x]$ and $I=(f(x))$. For the cases $$ p=2~\text{and}~\deg(f)\le8,\qquad p=3~\text{and}~\deg(f)\le4,\qquad p=5~\text{and}~\deg(f)\le3, $$ it is true that $$ \text{$(*)$ is true}\quad\Longleftrightarrow\quad \text{$f(x)$ is irreducible in $\mathbb F_p[x]$.} $$

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  • $\begingroup$ Wouldn't you need $1+$ on the left side for it to be analogous? $\endgroup$ Sep 16 at 23:44
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    $\begingroup$ @GerryMyerson Oops, thanks for noticing that I left off the 1 in front. I've fixed it. $\endgroup$ Sep 17 at 0:42
  • $\begingroup$ Thank you very much! $\endgroup$ Sep 17 at 5:24
  • $\begingroup$ Interesting experiments indeed! $\endgroup$ Sep 17 at 17:23
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Here is a proof of the experimental fact Joe Silverman discovered.

Let $f$ be a polynomial of degree $d$ in $R:=\mathbb{F}_q[x]$. For $I=(f)$, the sum $$1+\sum_{a \in R/I} a^{\# R/I - 1}$$ is equal, modulo $f$, to $$1+\sum_{g \in R,\, \deg g < d} g^{q^d-1}.$$ Let $$F_d(x) := 1+\sum_{g \in R,\, \deg g < d} g^{q^d-1} \in R.$$ We shall show that if $f$ is irreducible than $f \mid F_d$ and if $f$ is reducible than $f \nmid F_d$.

First, suppose that $f$ is reducible. Let $\alpha$ be a root of $f$, with minimal polynomial $m_{\alpha} \mid f$ and $\deg m_{\alpha} = e < d$. We can write any $g \in R$ of degree $<d$ uniquely as $m_{\alpha} g_0 + g_1$ where $\deg g_1 < e$ and $\deg g_0 < d-e$. Hence $$F_d(\alpha) = 1+\sum_{g_0,g_1 \in R,\, \deg g_0 < d-e, \, \deg g_1 < e} (m_{\alpha}(\alpha) g_0(\alpha)+g_1(\alpha))^{q^d-1} = 1+q^{d-e}\sum_{g_1 \in R,\, \deg g_1 < e} g_1(\alpha)^{q^d-1}=1$$ since we are in characteristic $p \mid q$. Hence $F_d$ cannot be divisible by $f$.

Next suppose that $f$ is irreducible and let $\alpha$ be a root of $f$. The map $g \mapsto g(\alpha)$ from $\mathbb{F}_q[x]/I$ to $\mathbb{F}_q[\alpha] = \mathbb{F}_{q^d}$ is an isomorphism, and so $$F_d(\alpha) = 1+\sum_{\beta \in \mathbb{F}_{q^{d}}} \beta^{q^d-1}=q^d =0$$ by Lagrange's Theorem applied to the multiplicative group of the field. Hence $f$ divides $F_d$, as needed.

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    $\begingroup$ Very nice. Would the same proof work for, say, the affine coordinate ring of any smooth algebraic curve over $\mathbb F_q$? (Seems as if it should, albeit not quite so explicitly.) $\endgroup$ Sep 17 at 19:52
  • $\begingroup$ Thank you! Could you please explain what $p$ is, since, unless mistaken, it wasn't introduced before? $\endgroup$ Sep 17 at 21:05
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    $\begingroup$ It's the characteristic of the base field. $\endgroup$ Sep 17 at 22:06
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    $\begingroup$ @JoeSilverman Good question, I don't have an answer from the top of my head. I think this is worthy of a separate MO question. $\endgroup$ Sep 17 at 23:05
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    $\begingroup$ @ThomasSauvaget $q$, the size of the finite field, is a power of a prime; that prime is $p$. It must also equal the characteristic of the field $\mathbb{F}_q$. $\endgroup$ Sep 17 at 23:07

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