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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?

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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? $\endgroup$ – user42690 Nov 24 '13 at 7:11
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    $\begingroup$ @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. $\endgroup$ – Filippo Alberto Edoardo Nov 24 '13 at 8:46
  • $\begingroup$ @user42690: Look at Noam's answer there mathoverflow.net/questions/149702/… $\endgroup$ – ACL Nov 24 '13 at 9:57

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