Timeline for Calculation of Frobenius on de Rham cohomology of elliptic curves with good reduction
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2023 at 8:18 | comment | added | Richard | Excuse me, I think $p$-power Frobenius is not an endomorphism of $\bar E$ since the residue field of $F$ (say, $\mathbb F_{p^r}$) is larger than $\mathbb F_p$. Hence $\varphi$ won't satisfy this equation, and instead we see $\varphi^{2r}-a_{p^r}\varphi^r+p^r=0$. Besides, supersingular is equivalent to $a_{p^r}\equiv 0 \pmod p$ (not for $a_p$). But with these I don't know how to deduce the statement then.. (for example I can't see if $x^2-a_{p^r}x+p^r$ is irreducible when $a_{p^r}\equiv 0\pmod p$) | |
| Oct 22, 2023 at 2:48 | comment | added | Richard | I have post it at here, thank you for pointing out. | |
| Oct 21, 2023 at 14:15 | comment | added | David Loeffler | If you have a follow-up question distinct from the original question, then ask it as a separate question. | |
| Oct 20, 2023 at 16:01 | comment | added | Richard | May I ask something further? By $p$-adic Hodge theory of admissible $\varphi$-modules I get: In good reduction case, $V$ is simple iff $E$ is supersingular ($V$ fails to be simple when $E$ has ordinary good reduction). I wonder am I right? | |
| Oct 20, 2023 at 14:56 | vote | accept | Richard | ||
| Oct 25, 2023 at 8:18 | |||||
| Oct 19, 2023 at 12:48 | history | answered | David Loeffler | CC BY-SA 4.0 |