# endomorphisms of the jacobian of a general hyperelliptic curve

Let $C$ be a curve of genus $g$. If $C$ is very general, we know that the Jacobian $JC$ of $C$ is simple and thus $End(JC)=\mathbb{Z}$.

Do we know something about $End(JC)$ if $C$ is a very general hyperelliptic curve?

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I will assume you are working over the complex numbers (note that for instance, the statement of your first paragraph is false over the algebraic closure of a finite field).

It is sufficient to exhibit a single example in each genus. A theorem of Yuri Zarhin shows that if $P(x) \in \mathbb{Q}[x]$ is an irreducible polynomial of degree $d \geq 5$ with Galois group $S_d$ or $A_d$ (note that by Hilbert Irreducbility most polynomials over $\mathbb{Q}$ have this property!) then the endomorphism ring of the Jacobian of $y^2 = P(x)$ is $\mathbb{Z}$. The paper is:

MR1748293 (2001a:11097) Zarhin, Yuri G.(1-PAS) Hyperelliptic Jacobians without complex multiplication. Math. Res. Lett. 7 (2000), no. 1, 123–132.

Added: let me give a little more detail as to why the existence of one hyperelliptic curve (of any given genus $g \geq 2$) with this property implies that a very general hyperelliptic curve has this property. First, it is known that a very general principally polarized abelian variety has endomorphism ring $\mathbb{Z}$: for each strictly larger endomorphism ring $R$ there is a closed ("Shimura") subvariety of the Siegel space $\mathcal{A}_{g,1}$ which is the locus of all ppavs with endomorphism ring containing $R$. Every $R$ is a $\mathbb{Z}$-order in a finite-dimensional semisimple $\mathbb{Q}$-algebra, so there are at most (and in fact are) countably infinitely many possibilities. Therefore the locus of ppavs of dimension $g$ with endomorphism ring $\mathbb{Z}$ is what is called (by me, at least) an algebraic stratum: it is the complement in an irreducible variety $V$ of a countably union of proper, Zariski-closed subvarieties. In general, when one says "a very general point of $V$ has property $P$" one means that the set of points of $V$ which has property $P$ is an algebraic stratum. Notice that over $\mathbb{C}$ every algebraic stratum is nonempty, e.g. because its complement has "measure zero".

We are interested in hyperelliptic Jacobians, i.e., we are intersecting the algebraic stratum $A$ of $\mathcal{A}_{g,1}$ consisting of abelian varieties with $\operatorname{End} A \cong \mathbb{Z}$ with the closed subset $H$ of hyperelliptic Jacobians. In order to be sure this gives an algebraic stratum on $H$, it is necessary and sufficient to check that for each proper closed subvariety $W_R$ of $\mathcal{A}_{g,1}$ consisting of abelian varieties with endomorphism ring containing a strictly larger ring $R$ than $\mathbb{Z}$, then $H_R = H \cap W_R$ is still proper in $H$. But this is accomplished by finding even a single element of $H$ which does not lie on any $W_R$, and Zarhin's theorem does this (and more).

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Thank you very much for the reference and the explanation! I was indeed assuming that everything is defined over the complex numbers. Sorry for not mentioning that. – max Jul 7 '11 at 8:02

The following is the main result of Zarhin's paper Hyperelliptic Jacobian's without complex multiplication:

Let $K$ be a field of characteristic $0$, let $n \geq 5$ and let $f(x)$ be a polynomial of degree $n$ in $K[x]$ whose Galois group is $S_n$ or $A_n$. Then the endomorphism ring of the Jacobian of $$y^2 = f(x),$$ considered over the algebraic closure of $K$, is $\mathbb{Z}$.

Since a degree $n$ polynomial with coefficients in $\mathbb{Q}$ has Galois group $S_n$, except on a thin set, this shows that there are lots of hyperelliptic curves, defined over $\mathbb{Q}$, whose automorphism ring over $\overline{\mathbb{Q}}$ is $\mathbb{Z}$. I say "lots" rather than "very general", because I don't want to misuse a technical term, but I expect this is good enough for your purposes.

The first page of Zarhin's paper gives many references to earlier results, which may also be of use to you.

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well, that's kind of a coincidence. (My answer was written at the same time as yours, but my browser is being very slow at the moment..) – Pete L. Clark Jul 6 '11 at 18:49
Reading over your answer more carefully, it seems you do not claim that the property holds for a very general curve. But I claim that this follows from the existence of a single curve. I thought this was an obvious argument about each of the countably many closed subsets of $\mathcal{A}_{g,1}$ giving the locus of some particular larger endomorphism ring intersecting properly with the hyperelliptic Jacobian locus...but if it's not obvious to you, it's not obvious. Should I spell this out more explicitly? – Pete L. Clark Jul 6 '11 at 19:01
I would be glad to see this spelled out explicitly. It sounds right to me, but I didn't have enough confidence to state it. – David Speyer Jul 6 '11 at 19:07
@ David Speyer: Thank you as well for the reference! – max Jul 7 '11 at 8:04