Let $A = \mathbb{F}_5[T, \sqrt{T(T-1)(T+1)}]$. My question is, what is the easiest way to see that$$\zeta_A(s) = {{1 + 2 \cdot 5^{-s} + 5^{1 - 2s}}\over{1 - 5^{1 - s}}}?$$Much thanks in advance. Perhaps Gauss and Jacobi sums would be helpful? Elliptic curves over finite fields?

EDIT: We are defining $\zeta_A(s)$ as$$\zeta_A(s) = \prod_{m \in \text{max}(A)} {1\over{1 - \#(A/m)^{-s}}},$$for a finitely generated commutative ring $A$ over $\mathbb{Z}$. See here and here.

  • $\begingroup$ How do you define $\zeta_A$ for an $\mathbb F_5$-algebra $A$? $\endgroup$
    – Ferra
    Jul 14 '15 at 10:14
  • $\begingroup$ We are defining it as $\zeta_A(s) = \prod_{m \in \text{max}(A)} {1\over{1 - \#(A/m)^{-s}}}$. $\endgroup$
    – user254398
    Jul 14 '15 at 14:42

Note that $\mathrm{Spec}\, A$ is a curve given by equation $y^2=x(x-1)(x+1)$. It is the affine part of an elliptic curve $E$. The way to compute zeta function of elliptic curve is given by the Weil conjectures(actually, here we need a very special case of it which was proven much earlier).

Namely, $Z_E(t)=\frac{1-at+qt^2}{(1-t)(1-qt)}$ for an elliptic curve $E/\mathbb{F}_q$, where $a$ is some integer. To compute $a$ it is enough to calculate the linear term of $Z_E$. Simple search gives $\# E(\mathbb F_5)=8$ so $Z_E(t)=\exp(8t+o(t))=1+8t+o(t)$.

From the other hand, $Z_E(t)=(1-at+qt^2)(1+t+\dots)(1+5t+\dots)=(1-at)(1+6t+\dots)=1+(6-a)t+\dots$ so $a=-2$ and $Z_E(t)=\frac{1+2t+5t^2}{(1-t)(1-5t)}$. The desired zeta function of the affine curve $\mathrm{Spec}\,A$ therefore is equal to $\frac{1+2t+5t^2}{1-5t}=\frac{1+2\cdot5^{-s}+5^{1-2s}}{1-5^{1-s}}$.

UPD:A numerical error is corrected thanks to Jyrki Lahtonen


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