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For an algebraic number field $K$, let $\zeta_K(s)$ be the Dedekind zeta function associated to $K$, and let $\zeta_K'(s)$ be its derivative.

I believe that the following statement is true: $$\zeta_K\left(\dfrac12\right)\neq 0 \implies \zeta_K'\left(\dfrac12\right)\neq0.$$ I maganged to prove this for quadratic and cubic number fields, but I'm unsure how to prove the general case rigorously. Any thoughts or insight on how one could go about proving it would be really appreciated.

Thank you!

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    $\begingroup$ When you say "managed to deduce", it means: "managed to prove". So what is your question, really? $\endgroup$
    – GH from MO
    Commented Dec 2, 2022 at 7:49
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    $\begingroup$ @GHfromMO I could see it as meaning "managed to reduce [some research question] to", leaving this as the last step in a proof. $\endgroup$
    – user44191
    Commented Dec 2, 2022 at 18:30
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    $\begingroup$ From the symmetric functional equation of the $\zeta_{K}(s)$ you get that when $\zeta_{K}(1/2) \neq 0$ you can express the derivative at $s=1/2$ as some non-zero multiple of $\zeta_{K}(s)$…. I think it’s non-zero! $\endgroup$ Commented Dec 2, 2022 at 23:35
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    $\begingroup$ Are you the author of arXiv:2212.00324 ? See Theorem 2 there. $\endgroup$
    – GH from MO
    Commented Dec 3, 2022 at 16:40
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    $\begingroup$ What do we get when expressing $\zeta_K'(1/2)/\zeta_K(1/2)$ with the functional equation and simplifying the $\Gamma'/\Gamma$ values? $\endgroup$
    – reuns
    Commented Dec 3, 2022 at 17:49

1 Answer 1

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If $\zeta_K(1/2) \neq 0$ then $\zeta'_K(1/2) = 0$ if and only if $$ \log |D_K| = (\log(8\pi) + \gamma) n + \frac\pi2 r_1, $$ where $D_K$ is the discriminant of $K$, and $\gamma = 0.5772156649\ldots$ is Euler's constant.

We expect that this is impossible because such an equality would give a closed form for $\gamma$, which should not exist. Moreover, unless some $(r_1,r_2)$ yields $|D_K|$ surprisingly close to an integer, we expect to be able to prove that $[K:{\bf Q}] \leq N$ is impossible in time polynomial in $N$; for instance it takes only a few seconds in gp to do this for $N=200$ (the nearest is about $3.567 \cdot 10^{-5}$, for $(r_1,r_2) = (28,36)$). But I doubt that we can expect to prove this for all $r_1$ and $r_2$ in the foreseeable future.

One might hope instead to prove that $\zeta'_K(1/2) \neq 0$ by showing that $|D_K|$ would have to be too small for a number field of given $r_1,r_2$. Unfortunately this is not known either, even under the Generalized Riemann Hypothesis (GRH): the best we can prove is that if $\zeta_K$ satisfies GRH then $$ \log |D_K| > (\log(8\pi) + \gamma - o(1)) n + (\frac\pi2 - o(1)) r_1 $$ as $n \to \infty$. (The appearance of the same linear combination of $r_1$ and $r_2$ is not coincidental; see for instance these lecture notes on the discriminant bound: https://abel.math.harvard.edu/~elkies/M229.19/disc.pdf.)

To derive the condition on $D_K$, start from the functional equation $\xi_K(s) = \xi_K(1-s)$ where $$ \xi_K(s) = \Gamma(s/2)^{r_1} \Gamma(s)^{r_2} (4^{-r_2} \pi^{-n} |D_K|)^{s/2} \zeta_K(s). $$ (This was already suggested in the comments by edward cornfoot and reuns.) Thus $\xi'_K(1/2) = 0$. Hence if $\zeta_K(1/2) \neq 0$ then $\xi_K(1/2) \neq 0$, so the logarithmic derivative $\xi'_K / \xi_K$ also vanishes at $s = 1/2$. But this logarithmic derivative is $$ \frac{r_1}{2}( \psi(1/4) - \log\pi) + r_2 (\psi(1/2) - \log 2\pi) + \frac12 \log |D_K| + \frac{\zeta'_K(1/2)}{\zeta_K(1/2)} $$ where $\psi$ is the logarithmic derivative of the Gamma function. Therefore $\zeta'_K(1/2) = 0$ if and only if $$ \log|D_K| = r_1 (\log \pi - \psi(1/4) ) + 2r_2 (\log 2\pi - \psi(1/2)). $$ Our condition then follows from the known special values $$ \psi(1/2) = -\log 4 - \gamma, \quad \psi(1/4) = -\frac\pi2 - \log 8 - \gamma. $$

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