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Let E be a generic elliptic curve over an algebraically closed field $k$ of characteristic $p>0$ (i.e. an elliptic curve corresponding to a geometric point over the generic point of $M_{1,1}$).

What is $End_{k}(E)$?

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The following will give you $\operatorname{End}(E)\otimes\mathbb{Q}$. One can probably then figure out which orders appear as $\operatorname{End}(E)$. (As noted by Will Sawin, the generic case that you've asked about is Case 1, so in your situation the endomorphism ring is $\operatorname{End}(E)=\mathbb Z$.)

Theorem (Deuring, see Mumford's Abelian Varieties Section 22, page 217) Let $E$ be an elliptic curve in characteristic $p>0$. We have the following equivalences:

  1. $E$ cannot be defined over a finite field if and only if $\operatorname{End}(E)\otimes\mathbb{Q}=\mathbb{Q}$.
  2. Suppose that $E$ is defined over a finite field. Then $\operatorname{End}(E)\otimes\mathbb{Q}$ is imaginary quadratic over $\mathbb{Q}$ if and only if $E[p]\cong\mathbb Z/p\mathbb Z$, i.e., if and only if $X$ is ordinary.
  3. If $E$ is supersingular, i.e., if $E[p]=0$, and thus necessarily defined over a finite field, then $\operatorname{End}(E)\otimes\mathbb{Q}$ is a quaterion algebra over $\mathbb Q$.
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    $\begingroup$ The question is about generic elliptic curves, which are never defined over finite fields, so only the first case appears. $\endgroup$
    – Will Sawin
    Apr 16 at 14:34
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    $\begingroup$ @WillSawin As you say, the posed question is answered by Case 1, but I thought that it was worth giving all three cases so that one can see the contrasting situations. $\endgroup$ Apr 16 at 15:02

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