Let $E$ be an elliptic curve over $\mathbb{Z}[1/N]$ where $N$ is some non-zero integer. Can one show that that the integer $n_p-p-1$ (where $n_p$ is the number of points of $E$ mod $p$) is positive for infinitely many primes $p$ (not dividing $N$) without invoking the modularity theorem or the Sato-Tate conjecture?
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$\begingroup$ "not dividing $N$" seems to be an unnecessary condition, as there are only finitely many primes dividing $N$. $\endgroup$– Gerry MyersonCommented Jan 13, 2020 at 14:46
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1$\begingroup$ I will venture a guess: without modularity, no, one cannot show it. The question is closely related to the continuation of the L-series to the point 1, and I think there was no progress on this until modularity (where for CM forms we count the earlier modularity result) $\endgroup$– Dror SpeiserCommented Jan 13, 2020 at 20:21
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$\begingroup$ @GerryMyerson strictly speaking $n_p$ is not defined for primes dividing $N$. One could take Neron models or something like that but I would prefer not to invoke anything non-trivial unless necessary (and as you point out it is not necessary here). $\endgroup$– user145520Commented Jan 13, 2020 at 23:48
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2$\begingroup$ I just want to point out that one only needs the modularity theorem here, not the (much more difficult) Sato-Tate conjecture. $\endgroup$– Will SawinCommented Jan 14, 2020 at 18:20
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$\begingroup$ @WillSawin but I suppose you would agree that it follows more immediately from Sato-Tate than from modularity. $\endgroup$– user145520Commented Jan 15, 2020 at 4:39
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2 Answers
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Here's a weak variant that should really be a comment: For any twist $E^d: dy^2 = f(x)$ of the elliptic curve $y^2 = f(x)$, for either $E$ or $E^d$, there exist infinitely many primes such that your required condition is met.
Note that if $d$ is not a square mod $p$, that is, $E^d$ is a non trivial twist mod $p$, then $$E^d(\mathbb F_p)+E(\mathbb F_p) = 2(p+1).$$ This is because for every value of $x \in \mathbb F_p$, $f(x)$ is either a square in which case this contributes two points to $E(\mathbb F_p)$ or it is not a square, in which case it contributes two points to $E^d(\mathbb F_p)$. This shows that one of the two sets $E(\mathbb F_p)$ and $E^d(\mathbb F_p)$ has to be $\geq p+1$ and the other one has to be $\leq p+1$. Since there are infinitely many primes for which $d$ is not a square mod $p$, this proves the claim.
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$\begingroup$ 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! $\endgroup$ Commented Jan 14, 2020 at 18:19
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$\begingroup$ 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 $d$. $\endgroup$– AsvinCommented Jan 14, 2020 at 18:27
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$\begingroup$ 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. $\endgroup$ Commented Jan 14, 2020 at 18:37
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$\begingroup$ 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? $\endgroup$– AsvinCommented Jan 14, 2020 at 18:44
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$\begingroup$ Yes, that works (and the proof that there are infinitely many non-supersingular primes uses Chebotarev). $\endgroup$ Commented Jan 14, 2020 at 18:48
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I don't think that anyone has any idea how to prove such a theorem without also being able to prove that $E$ is (potentially) modular. For example you can ask the same question for curves of higher genus. For curves of genus $2$, this problem appears to have been open before 2018, and for curves of higher genus it is still open, see the introduction to http://www.math.uchicago.edu/~fcale/papers/CDM.pdf
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$\begingroup$ While comparing with curves of higher genus is not a bad point in itself, of course there are many things which are much easier for elliptic curves than for others (like RH for their zeta functions). I would not be surprised if this were also the case for this problem. $\endgroup$– WojowuCommented Jan 14, 2020 at 18:55
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$\begingroup$ @Wojowu The linked discussion also discusses the problem for elliptic curves as well, and even a special CM elliptic curve. My reading of it is that there are no known approaches to the current problem other than those establishing analytic statements related to modularity, and so presumably the original problem it is still open for elliptic curves over general number fields as well. $\endgroup$– Frob pCommented Jan 14, 2020 at 19:15
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$\begingroup$ Thanks, I see. I admit I didn't give the linked paper a good look before replying :P $\endgroup$– WojowuCommented Jan 14, 2020 at 19:38
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$\begingroup$ The paragraph after Exercise 1.1.2 in the linked Calegari paper is essentially my observation. I enjoyed the paper, thanks for linking it. $\endgroup$– AsvinCommented Jan 15, 2020 at 4:30