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Let $K/\mathbb{Q}$ be an imaginary quadratic extension with discriminant $-D$. Then there is an elliptic curve $E$ over $\overline{\mathbb{Q}}$ such that End$(E)^{0}: =$ End$(E) \otimes Q = K$.

Now let $p$ be a prime number. Is there any result (with some restrictions on $-D$ and $p$) which will make the above statement true (or false) when we replace $\overline{\mathbb{Q}}$ by $\overline{\mathbb{F}_p}$?

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There is such an elliptic curve in char $p$ if and only if $p$ splits in $K$. The elliptic curve in char. zero with $K$ as endomorphisms is CM so has potential good reduction everywhere and the reduction modulo a prime above $p$ is an elliptic curve in char $p$ with $K$ contained in the (field of fractions of) the endomorphism ring. If the reduction is ordinary, you get equality and you are done,if the reduction is supersingular you can't get equality (endomorphism ring is dim. 4). Finally, the reduction is ordinary if and only if $p$ splits in $K$.

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  • $\begingroup$ Just a note for the OP: the keyword for the final sentence is "Deuring's criterion". $\endgroup$ – Pete L. Clark Nov 30 '14 at 22:27
  • $\begingroup$ Thank you very much for the complete answer and the comment. $\endgroup$ – user114285 Dec 6 '14 at 5:30

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