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Let $X$ be an abelian variety over a finite field of characteristic $p$ such that the $X[p]=0$. In other words, none of the Newton slopes are $0,1$.

QUESTIONS.

(a) Is it possible for the endomorphism algebra $End(X)$ to be commutative?

(b) If so, are there examples that are fairly easy to construct?

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    $\begingroup$ If you count only endomorphisms defined over the base field, it happens already for supersingular curves over $\mathbb F_p$. $\endgroup$
    – Will Sawin
    Oct 14, 2016 at 21:26
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    $\begingroup$ Otherwise you have to go to genus $3$ surfaces. Examples are relatively easy to construct using Honda-Tate theory by fixing the characteristic polynomial of Frobenius - e.g. for most values of $a,b$, an abelian variety with characteristic polynomial $x^6 + a p x^4 - b p x^3 + a p^2 x^2 + p^3$ will have abelian endomorphism ring. $\endgroup$
    – Will Sawin
    Oct 14, 2016 at 21:29
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    $\begingroup$ Apologies. I meant endomorphisms over the algebraic closure of the finite field. $\endgroup$
    – Student88
    Oct 14, 2016 at 21:29

1 Answer 1

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(a) Oh, yes. This is proven in a paper of Hendrik Lenstra and Frans Oort http://www.sciencedirect.com/science/article/pii/0022404974900292 .

(b) A construction of the corresponding CM-field of endomorphisms is described in the paper mentioned above.

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