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Conjecture: If I have an elliptic curve with j-invariant 0 of the form $y^2 = x^3 + b$ over some prime-order field $\mathbb{F}_p$ (where $p$ is not 2 or 3), and the group of rational curvepoints has prime order $q$ (which is not 2 or 3), then there is some curve of the form $y^2 = x^3 + b'$ over $\mathbb{F}_q$ with prime order $p$.

I'm not sure about the characteristic 2-or-3 case, I guess I could check that exhaustively, but it's less interesting to me and I haven't taken the time to do it yet. For larger primes I've found a few dozen cases with small primes, a couple with large primes, and no counterexamples.

I have 3 questions:

  1. Is this conjecture true?
  2. What does it mean? This seems to give an equivalence relation between ground fields of different characteristic, which seems really weird to me... what am I looking at here? (Sorry this is so vague.)
  3. What is the natural way to generalize this to curves whose groups of rational points has composite order, and/or fields with non-prime order?
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    $\begingroup$ By the Hasse-Weil bound $p$ and $q$ have to be within $\pm 2 \sqrt{\text{min}(p, q)}$ of each other, and by the Sato-Tate conjecture this is more or less the only constraint asymptotically. So given Legendre's conjecture your conjecture is just about not implausible (although I wouldn't be surprised if it were false), but I don't think it uncovers any deep relationships between primes or anything. In any case, you need transitivity to have an equivalence relation and that's ruled out by the Hasse-Weil bound. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 15, 2016 at 19:50
  • $\begingroup$ @QiaochuYuan then perhaps when such large prime gaps appear the resulting curves cannot have prime order? $\endgroup$
    – Andrew Poelstra
    Commented Sep 15, 2016 at 19:52
  • $\begingroup$ I lied, the conjecture's probably false. The problem is that $p$ and $q$ could be far enough apart that the Hasse-Weil bound is satisfied in exactly one direction and not the other. So I expect there are large counterexamples. Or maybe small ones: take $p = 11, q = 5$. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 15, 2016 at 21:30
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    $\begingroup$ Not quite the same question, but Kate Stange and I wrote a paper a few years back about elliptic curves $E/\mathbb Q$ and primes $p$ and $q$ such that $\#E(\mathbb F_p)=q$ and $\#E(\mathbb F_q)=p$. It turned out that the case you're considering ($j=0$) was far and away the most complicated case. The paper is "Amicable pairs and aliquot cycles for elliptic curves", Exper. Math. 20(3) (2011), 329-357. $\endgroup$
    – Joe Silverman
    Commented Sep 15, 2016 at 21:49
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    $\begingroup$ There are no counterexamples when $p$ and $q$ both less than or equal to $10000$ (apart from the case of $p = 7$ and $q = 3$ - I'm not sure I know to formulate the question properly in characteristic 3). $\endgroup$
    – Jeremy Rouse
    Commented Sep 15, 2016 at 22:22

1 Answer 1

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If $\# E(\mathbb{F}_p) = q$ and $j=0$, then the endomorphism ring is an order in the field of third roots of unity so $(p+1-q)^2 - 4p = -3u^2$ for some integer $u$. Now note that $(p+1-q)^2 - 4p$ is symmetric in $p$ and $q$. Hence, if there is an elliptic curve at all over $\mathbb{F}_q$ with $p$ points, then it automatically has endomorphism ring by an order in the field of third roots of unity and thus has $j=0$ and is of the required form. So the remaining issue is whether $p$ lands in the Hasse interval for $q$. But, let's say $p < q$, then $q < (\sqrt{p}+1)^2$ so $\sqrt{p} > \sqrt{q}-1$ and $p$ is in the Hasse interval for $q$. The case $q<p$ is similar. This answers 1. I don't know about 2. I didn't really use the primality of $p,q$ in the proof except to ensure there is a finite field of that order. So, I guess 3. should go the same way. The only thing to watch out is that I am assuming $p,q$ are $1$ modulo $3$ for the curves to be ordinary. As supersingular curves over prime fields of cardinality $p>3$ have order $p+1$, hence not prime, this is not an issue but can become an issue when considering non-prime fields.

Edit: To get $j=0$ the endomorphism ring has to be the maximal order. I think my proof is incomplete but hopefully is OK.

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  • $\begingroup$ Surely being in the Hasse interval is necessary but not sufficient, right? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 15, 2016 at 23:03
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    $\begingroup$ @QiaochuYuan In the prime field case every value in the Hasse interval corresponds to a curve, by a result of Deuring. $\endgroup$
    – Felipe Voloch
    Commented Sep 15, 2016 at 23:07
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    $\begingroup$ Here's an idea. The trace of Frobenius on a $j=0$ elliptic curve (for $p \equiv 1 \pmod{3}$) is given by the representation of $p$ by the form $x^{2} + xy + y^{2}$, which is related to the representation of $4p$ in the form $x^{2} + 3y^{2}$. Since $(p+1-q)^2 + 3u^{2} = 4p$, then $(q+1-p)^{2} + 3u^{2} = 4q$. $\endgroup$
    – Jeremy Rouse
    Commented Sep 15, 2016 at 23:07
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    $\begingroup$ @JeremyRouse I was just thinking along the same lines and I think that does it. $\endgroup$
    – Felipe Voloch
    Commented Sep 15, 2016 at 23:08

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