For a given infinite set of primes, not too big, eg, satisfying Lang-Trotter conjecture, can we always find an E.C. with supersingular reduction (at least) at these primes? How about E.C. without CM?
1 Answer
No. In fact the set can be arbitrarily sparse, i.e. the $n$-th prime in the set can be chosen to exceed $a_n$ for any sequence $\{a_n\}$. This is because the rationals are countable. Fix an enumeration $j_1,j_2,j_3,\ldots$ of $\bf Q$. For each $n$ let $p_n$ be the smallest prime such that $p_n > a_n$ and $j_n$ is not supersingular mod $p_n$ (this is possible because the density of ordinary primes is positive, namely $1/2$ if $j_n$ is one of the thirteen CM $j$-invariants and $1$ otherwise). Then no rational number is the $j$-invariant of an elliptic curve with supersingular reduction at every $p_n$.
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$\begingroup$ What do you mean by "thirteen CM j-invariants"? I don't know this result. Could you explain it or give me some reference? $\endgroup$ Nov 24, 2013 at 7:11
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3$\begingroup$ @user42690: What Noam Elkies is refering to is that there exixts exactly $13$ values of $j$ which are in $\mathbb{Q}$ and whose corresponding elleptic curves is CM. These are completely explicit and correspond to the $13$ orders in imaginary quadratic fields which are principal domains. $\endgroup$ Nov 24, 2013 at 8:46
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$\begingroup$ @user42690: Look at Noam's answer there mathoverflow.net/questions/149702/… $\endgroup$– ACLNov 24, 2013 at 9:57
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$\begingroup$ This seems to answer a slightly different question than what is being asked, or am I mistaken? The questions asks whether for each sequence of primes $(p_n)$ there exists an elliptic curve that is supersingular at (at least) all $p_n$. This answer seems to show that for each sequence $(a_n)$ there exists a sequence $(p_n)$ of primes such that there is no elliptic curve that is supersingular at all primes $p_n$. In particular, the sequence $(p_n)$ does not appear to be arbitrary: it depends on both $(a_n)$ and the fact that we can choose $j_n$ to not be s.s. mod p_n. Am I missing something? $\endgroup$– ArbutusJan 23, 2023 at 15:17