2
$\begingroup$

Let \begin{equation*} \zeta(s):=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}} \end{equation*} be the Riemann zeta function. Van Gorder has shown that $\zeta$ satisfies a differential equation \begin{equation*} T(\zeta(s)-1)=\frac{1}{s-1}\text{,} \end{equation*} where $T$ is an infinite differential operator.

Robert A. Van Gorder, MR 3276353 Does the Riemann zeta function satisfy a differential equation?, J. Number Theory 147 (2015), 778--788.

Let $k=\mathbb{F}_{p}(x)[y]/\pi$ be a global field, where $\pi\in\operatorname{Spec}(\mathbb{F}_{p}(x)[y])$ is a maximal ideal, and let \begin{equation*} \zeta_{k}(s):=\prod_{\mathfrak{m}\in\operatorname{Spec}(\mathcal{O}_{k})\text{ maximal}}\frac{1}{1-N_{\mathfrak{m}}^{-s}} \end{equation*} be the associated zeta function, where $N_{m}:=\#\mathcal{O}_{k}/\mathfrak{m}$. I wonder if results similar to Van Gorder's can be obtained for $\zeta_{k}$.

Does the zeta function of a global field satisfy a differential equation?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Zeta functions of function fields are rational functions in $p^s$. $\endgroup$
    – Wojowu
    Commented May 7, 2022 at 10:06
  • $\begingroup$ I know. But is there a general way to derive a functional equation they satisfy? $\endgroup$
    – The Thin Whistler
    Commented May 7, 2022 at 10:08
  • 2
    $\begingroup$ Probably. An easy way to see there exists one is to note that for a function $F(s)=f(p^s)$, with $f(x)\in\mathbb C(x)$, both $F$ and $F'$ belong to $\mathbb C(p^s)$, which means they are algebraically dependent over $\mathbb C$. $\endgroup$
    – Wojowu
    Commented May 7, 2022 at 10:10

0

Reset to default

You must log in to answer this question.