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Suppose $E_{/\mathbb{Q}}$ is an elliptic curve over $\mathbb{Q}$ without CM. By Elkies' theorem, there exist infinitely many primes $p$ for which $E$ has supersingular reduction at $p$.

Question. Is it possible to choose an imaginary quadratic field $K$ such that every supersingular prime of $E$ is inert in $K$? Less ambitiously, is it possible to choose $K$ so that infinitely many supersingular primes are inert in $K$?

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  • $\begingroup$ I think this question might be equivalent to: "is there a modulus $m$ such that infinitely many of the supersingular primes are congruent to $1$ mod $m$?" $\endgroup$
    – Lior Silberman
    Commented Feb 25, 2021 at 4:17
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    $\begingroup$ @LiorSilberman I don't see why your statement should be equivalent to the original statement, although there is a plausible stronger statement which implies both (that the supersingular primes should be equidistributed among residue classes mod $m$ for any $m$). $\endgroup$
    – David Loeffler
    Commented Feb 25, 2021 at 8:03
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    $\begingroup$ Side question of a lazy person: is there a summary of Elkies' proof available anywhere, possibly one that would be more accessible to an algebraic geometer? I looked at Elkies' paper two or three times in my life (admittedly, with no good reason to learn the proof other than curiosity) and I always ran out of energy after having to consult Deuring's very long and very old paper. $\endgroup$
    – Piotr Achinger
    Commented Feb 25, 2021 at 8:17
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    $\begingroup$ If $E$ has a point of order $n$ then $a_p \equiv 1 + p \mod n$ so $E$ is supersingular only if $p \equiv -1 \mod n$ which forces $p$ to be inert in $\mathbb Q( \sqrt{ \pm \ell})$ for any odd prime $\ell \mid n$ (with the sign based on the congruence class of $\ell$ mod $4$) or $\mathbb Q(i)$ if $4\mid n$. So at least for some $E$ the answer is "yes" (and the answers to Lior's and David's questions are "no"). For $E$ with maximal Galois image, I think there are no obstructions of this type, so I would guess the answers are "no" and "yes, every" but am not very sure. $\endgroup$
    – Will Sawin
    Commented Feb 25, 2021 at 20:23

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