Timeline for Zeta function of $X = \mathbb{F}_p \mathbb{P}^1$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2022 at 18:38 | comment | added | user30211 | I see. I guess that gives an alternative approach given that $\text{Pic}(Y) = \mathbb{Z}$. | |
| Aug 20, 2022 at 17:59 | comment | added | Will Sawin | @BobZinckel The exact sequence $\mu_{\ell^n} \to \mathbb G_m \to \mathbb G_m$ gives a long exact sequence $H^1(Y, \mu_{\ell^n} ) \to H^1( Y, \mathbb G_m) \to H^1(Y, \mathbb G_m)$ with the map given by multiplication by $\ell^n$. Since the kernel of multiplication by $\ell^n$ on $\mathbb Z$ is zero, this is consistent with $H^1(Y, \mu_{\ell^n})=0$, which implies $H^1(Y,\mathbb Q_\ell)=0$. | |
| Aug 20, 2022 at 17:28 | comment | added | user30211 | @WillSawin One more thing - could you explain why $H^1(Y, \mathbb{G}) \cong \mathbb{Z}$ is consistent with $H^1(Y, \mathbb{Q}_l) \cong 0$? | |
| Aug 19, 2022 at 2:22 | comment | added | user30211 | Thanks for your time Will Sawin. I'm happy to have met you! | |
| Aug 19, 2022 at 2:21 | comment | added | Will Sawin | @BobZinckel The arithmetic Frobenius acts on the cyclotomic character by multiplication by $q$, since it raises roots of unity to the $q$th power, so the geometric Frobenius acts on the cyclotomic character by $q^{-1}$, so the geometric Frobenius acts on the inverse $\mathbb Q_\ell(-1)$ of the cyclotomic character by $q$. | |
| Aug 19, 2022 at 2:19 | comment | added | user30211 | "$H^2(Y, \mathbb{Q}_l) \cong \mathbb{Q}_l(-1)$, and thus the unique Frobenius eigenvalue on it is..." could you spell this out a bit more for me? | |
| Aug 19, 2022 at 2:18 | comment | added | Will Sawin | @BobZinckel Yes. | |
| Aug 19, 2022 at 2:17 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Aug 19, 2022 at 2:16 | comment | added | user30211 | You mean $q^{2/2} = q$? | |
| Aug 19, 2022 at 2:13 | vote | accept | CommunityBot | moved from User.Id=30211 by developer User.Id=481663 | |
| Aug 19, 2022 at 2:01 | comment | added | paul garrett | :) I was contemplating writing a more amateurish version of this, but then thought, "wait, maybe Will S. will see this..." :) Best wishes. | |
| Aug 19, 2022 at 1:15 | history | answered | Will Sawin | CC BY-SA 4.0 |