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For a smooth proper curve $C$ over a finite field $\mathbb{F}_q$, local zeta function can be shown that $$ Z(C,t)=\frac{P(t)}{(1-t)(1-qt)} $$ with a polynomial of degree $2g$, where $g$ is the genus of $C$. We know that $P(t)$ decomposes as $\prod (1-\omega_i t)$, where $\omega_i$ is a complex number satisfying $\lvert\omega_i\rvert=q^{1/2}$.

My question: Is there a smooth proper curve $C$ such that $\omega_i= q^{1/2}$ for some $i$? I can prove that there is no such an elliptic curve.

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    $\begingroup$ I think there is such an elliptic curve (maybe you are implicitly assuming $q$ is a prime in your proof?): Take a supersingular elliptic curve over $\mathbb F_p$ for $p>3$ and base change to $\mathbb F_{p^4}$. $\endgroup$
    – Will Sawin
    Commented Jun 24, 2023 at 23:17
  • $\begingroup$ Oh I see. The "proof" in my mind only workes when q=p.... $\endgroup$
    – user145752
    Commented Jun 24, 2023 at 23:21
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    $\begingroup$ @WillSawin Thank you very much. Your example completely works. $\endgroup$
    – user145752
    Commented Jun 24, 2023 at 23:22

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There is such a curve on $\mathbb F_5$: $y^2=x^5+4x$. The L-function of its Jacobian is on LMFDB.

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