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Recall that the analytic rank $r^{\rm an}(E)$ of a (modular) elliptic curve $E$ is defined to be the order of vanishing of its Hasse-Weil $L$-function $L(E,s)$ at $s=1$. A conjecture due to Ralph Greenberg in [MR1260957 (95a:11059)] implies in particular the following:

Basic fact: Let $\Sigma$ be a finite set of prime integers. Then, as $E$ ranges over all rational elliptic curves of conductor divisible only by primes in $\Sigma$, we have that $r^{\rm an}(E)$ is $0$ or $1$ except for finitely many $E$'s.

Immediate as it is, this fact is a special case of the (rather deep, as formulated in the paper above) Greenberg's conjecture, namely that of normalized newforms of weight $2$ with rational Fourier coefficients, and it shows that (rational) elliptic curves with larger and larger Mordell-Weil rank will tend to have larger and larger conductor; more precisely, such elliptic curves will tend to have conductor divisible by a larger and larger number of primes.

So I am wondering if we can give an easy answer to the following:

-Can we show that there exist rational elliptic curves with conductor divisible by an arbitrarily large number of distinct primes?

-Can we do better and given any finite set $\Sigma$ of distinct primes, show the existence of a rational elliptic curve with conductor divisible by each of the primes in $\Sigma$?

-Can we show that in fact there exist a number of non-isogenous such elliptic curves for each of the previous questions that grows (e.g. linearly!) with the cardinality of $\Sigma$?

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    $\begingroup$ I don't think I've understood your question. For each p, the power of p dividing the conductor of an elliptic curve is bounded above -- by 2, if p > 3. Right? So isn't your conjecture about a finite list of elliptic curves? $\endgroup$
    – JSE
    Apr 15, 2011 at 1:42
  • $\begingroup$ @JSE, I took the questions to be, first, is it true that for any $n$ there is an elliptic curve whose conductor is divisible by $n$ different primes? and, second, is it true that given any squarefree number $q$ there is an elliptic curve with conductor divisible by $q$? I don't see any way for a finite list of curves to answer these questions. $\endgroup$
    – Gerry Myerson
    Apr 15, 2011 at 1:57
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    $\begingroup$ Like JSE I find the beginning of your question confusing. You say that your "Basic Fact" is a special case of a deep conjecture of Greenberg. But actually your basic fact follows immediately from the fact that the set of $\mathbb{Q}$-isomorphism classes of elliptic curves in question is finite, by a 1965 theorem of Shafarevich. It doesn't have anything to do with analytic ranks. It is also obviously true that elliptic curves with "larger and larger Mordell-Weil rank" will have "larger and larger conductor" because in any infinite set of curves the conductors have to approach infinity. $\endgroup$
    – Pete L. Clark
    Apr 15, 2011 at 3:37
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    $\begingroup$ If the question is whether, given a set $\Sigma$, there is an elliptic curve over Q with exactly bad primes in $\Sigma$, then the answer is no. Take $\Sigma$ empty for example. :) Or $\Sigma=\{5\}$, or $\Sigma=\{13\}$, and such. $$ $$ If the question is as reads, then the answer is yes, via twisting. To get an elliptic curve divisible by all primes in $\Sigma$, consider the $2^\#\Sigma$ curves obtained by quadratic twists of subsets of $\Sigma$, and (at least) one will have the conductor divisible by all $p\in\Sigma$. There are $\infty$ many such curves, seen by twisting $p\not\in\Sigma$. $\endgroup$
    – Junkie
    Apr 15, 2011 at 4:34
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    $\begingroup$ "and it shows that (rational) elliptic curves with larger and larger Mordell-Weil rank will tend to have larger and larger conductor; more precisely, such elliptic curves will tend to have conductor divisible by a larger and larger number of primes." The latter is true in twist families, but in general depends on class group of the cubic also. Elkies/Watkins found a rank 11 curve with prime conductor (2004), namely 18031737725935636520843. Conductor in size must go up with rank $r$, see Mestre's paper, where $r=O(\log N)$ is shown. archive.numdam.org/article/CM_1986__58_2_209_0.pdf $\endgroup$
    – Junkie
    Apr 15, 2011 at 4:39

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The conductor is divisible by the same primes as the discriminant. The discriminant of $y^2=x^3+b$ is $-432b^2$. So you can certainly make the conductor divisible by an arbitrarily large number of distinct primes, and by whichever primes you want (as long as you want 2 and 3).

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    $\begingroup$ In general, the conductor is only divisible by SOME of the primes dividing the discriminant, because you might have the "wrong model". $\endgroup$
    – David Hansen
    Apr 14, 2011 at 17:17
  • $\begingroup$ I take your point. So, it is easy to work out the conductor of the curves $y^2=x^3+b$? or the curves $y^2=x^3+ax$? and do they suffice to answer the original questions? $\endgroup$
    – Gerry Myerson
    Apr 15, 2011 at 1:59
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    $\begingroup$ The only one I know off the top of my head is $y^2=x^3+a^2x$: if $a$ is odd and squarefree, the conductor is $32A^2$, where $A$ is the (absolute) discrimiant of $\mathbf{Q}(\sqrt{A})$. $\endgroup$
    – David Hansen
    Apr 15, 2011 at 3:30
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    $\begingroup$ Gah, that's wrong. The conductor of $y^2=x^3+a^2x$ is some power of $2$ times $a^2$ for odd squarefree $a$. So the answers to the OP's first two questions are both "yes". $\endgroup$
    – David Hansen
    Apr 15, 2011 at 3:54
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Can we do better and given any finite set Σ of distinct primes, show the existence of a rational elliptic curve with conductor divisible by each of the primes in Σ?

I doubt it. Conjecturally there are only supposed to be about N^{5/6 + eps} elliptic curves of conductor < N. So already there are many squarefree conductors that are missed, and somehow I don't think it very likely that throwing in the guys with some bad reduction changes very much.

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    $\begingroup$ @JSE, are you interpreting "divisible by each of the primes in $\Sigma$" to mean "divisible by each of the primes in $\Sigma$ and $\it only$ by the primes in $\Sigma$"? @monodromy, is that the intended interpretation? $\endgroup$
    – Gerry Myerson
    Apr 15, 2011 at 5:44
  • $\begingroup$ Yes, and now that I read more carefully, my interpretatino is probably wrong. Junkie's commment on the post is good. $\endgroup$
    – JSE
    Apr 15, 2011 at 12:47

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