# What is a function field analog of Giuga's conjecture?

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?

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 $$\text{(*) is true}\quad\Longleftrightarrow\quad \text{f(x) is irreducible in \mathbb F_p[x].}$$

• Wouldn't you need $1+$ on the left side for it to be analogous? Sep 16 at 23:44
• @GerryMyerson Oops, thanks for noticing that I left off the 1 in front. I've fixed it. Sep 17 at 0:42
• Thank you very much! Sep 17 at 5:24
• Interesting experiments indeed! Sep 17 at 17:23

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 $$ uniquely as $$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.

• 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.) Sep 17 at 19:52
• Thank you! Could you please explain what $p$ is, since, unless mistaken, it wasn't introduced before? Sep 17 at 21:05
• It's the characteristic of the base field. Sep 17 at 22:06
• @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. Sep 17 at 23:05
• @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 $\mathbb{F}_q$. Sep 17 at 23:07