Timeline for Naive point count underestimates the number of mod $p$ points of an elliptic curve for infinitely many primes
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2020 at 18:48 | comment | added | Will Sawin | Yes, that works (and the proof that there are infinitely many non-supersingular primes uses Chebotarev). | |
| Jan 14, 2020 at 18:44 | comment | added | Asvin | In the CM by $\mathbb Q(\sqrt d')$ case, $a_p = 0 \iff d'$ is not a square mod $p$ so I guess that's easy if we choose $d$ carefully enough by quadratic reciprocity/Dirichlet. In the ordinary case, $a_p = 0 \iff p$ is a supersingular prime but this density is $0$ so ignoring the supersingular primes, we still have infinitely many primes. Does this not work? | |
| Jan 14, 2020 at 18:37 | comment | added | Will Sawin | The question is how to make the proof work, i.e. to prove that if $E$ does not have CM by $\mathbb Q(\sqrt{d})$, then $a_p \neq 0$ for infinitely many such $p$. I guess one can probably do this with Chebotarev. | |
| Jan 14, 2020 at 18:27 | comment | added | Asvin | That is a good point but we do have a lot of freedom in choosing $d$ so maybe there isn't such a large difference between $\geq$ and $>$ as long as we avoid $d$ so that $E$ has CM by $\mathbb Q(\sqrt d)$. After all, $E(\mathbb F_p) = p+1$ exactly when $p$ is supersingular and the density of such primes is $0$ in the ordinary case and in the supersingular case is given by congruence conditions that can be avoided by choosing a different $d$. | |
| Jan 14, 2020 at 18:19 | comment | added | Will Sawin | This works for $\geq$ but not for $>$. Note that if $E$ has CM by $\mathbb Q(\sqrt{d})$, then $E$ has $p+1$ points over every $\mathbb F_p$ where $d$ is not a square mod $p$. So the difference is really significant! | |
| Jan 14, 2020 at 15:40 | history | edited | Asvin | CC BY-SA 4.0 |
deleted 2 characters in body
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| Jan 13, 2020 at 19:01 | history | answered | Asvin | CC BY-SA 4.0 |