# Etale endomorphisms of abelian varieties in positive characteristic

Let $K$ be the function field of a smooth curve over the algebraic closure $k$ of the finite field ${\bf F}_p$ (where $p>0$ is a prime number).

My question is : does there exist an abelian variety $A$ over $K$, with the following properties :

(a) the $K|k$-image of $A$ is trivial;

(b) there exists an étale $K$-endomorphism of $A$, whose degree is a power of $p$ ?

The condition on the $K|k$-image can be rephrased as : there are no non-zero $K$-homomorphisms
$A\to C_K$, where $C$ is an abelian variety over $k$.

For examples of abelian varieties satisfying condition (b) only, look at abelian varieties $C_K$, where $C$ is an ordinary abelian variety over ${\bf F}_p$. The abelian variety $C$ is endowed with the étale endomorphism given by the Verschiebung morphism.

Also, notice that if there is an abelian variety over $K$ satisfying (a) and (b), then the dimension of $A$ is larger than one (ie it is not an elliptic curve). Indeed, if an elliptic curve $E$ satisfies the above conditions, then $E$ has an endomorphism, which is not a multiplication by a scalar and thus it has complex multiplications; this implies that it is isogenous to an elliptic curve defined over $k$, by a theorem of Grothendieck (or by more direct arguments).

Finally, I would like to point out that if $A$ is an ordinary abelian variety over $K$, which has maximal Kodaira-Spencer rank, then $A[p]$$(K^{\rm sep})=0$ by a theorem of J-F Voloch (see p. 1093 in "Diophantine Approximation on Abelian Varieties in Characteristic $p$", Amer. J. Math., Vol. 117, No. 4., pp. 1089-1095); this shows that such an abelian variety cannot have an endomorphism as in (b).

• Unfortunately, yes, there are such examples. I was shown an explicit example by David Helm using the universal abelian variety over a suitable Shimura curve. I have to see if I can find the details. David lurks here on MO, maybe he will see this and write an answer. – Felipe Voloch Aug 22 '11 at 12:42
• @Voloch Brilliant ! I would be very interested to see the details. One reason why I am interested in this matter is the following fact : if $\#A[p](K^{\rm sep})=\infty$ then $A$ is isogenous to an abelian variety with an étale endomorphism as above (see arXiv:1103.2625, remark on p. 14). – Damian Rössler Aug 23 '11 at 10:40
• @Voloch ... sorry in my last comment, replace $\#A[p](K^{\rm sep})=\infty$ by $\#A[p^\infty](K^{\rm sep})=\infty$. – Damian Rössler Aug 23 '11 at 10:42
• Great! I've tried unsuccessfully to prove the result in your comment. – Felipe Voloch Aug 23 '11 at 11:44

If $p-1$ is divisible by $24$ then there is an explicit example of an ordinary $7$-dimensional abelian variety $X$, whose endomorphism algebra is the imaginary quadratic field $Q(\sqrt{-3})$; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. Namely, $K$ is the field of rational functions $k(t)$ and $X$ is the jacobian of the $K$-curve $y^3=x^9-x-t$. See Example 4.3 of arXiv:math/0606422 [math.NT] [MR2289628 (2007j:11077)] for details.
If $p>2$ and $g>1$ is an odd integer then there exists a $g$-dimensional ordinary abelian variety $X$ over a suitable $K$, whose endomorphism algebra is an imaginary quadratic field; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. See Theorem 1.5(i,ii) of the same paper that is based on a construction of Oort and van der Put. (Actually, the condition $p>2$ could be dropped.)
• @Yuri Zarhin : thank you very much for taking the time to write this answer. I do not understand one point though; how to you get the étale endomorphism of order a power of $p$ exactly ? In your first example, suppose that $p$ splits completely at $p$ so that $p=r\cdot t$ in ${\bf Q}(\sqrt{3})$; why is $r$ or $t$ étale ? Tell me if I am on the wrong track here. – Damian Rössler Aug 28 '11 at 7:46
• @ Damian Rössle: You are welcome. Actually, it seems that I've answered a different question inspired by your remark about ordinary variety. Indeed, if $X$ is not defined over a finite field then the Frobenius and Verschiebung morphisms (or their powers) are not endomorphisms of $X$ over any field. My apologies! – Yuri Zarhin Aug 28 '11 at 16:32