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Let $k$ be an infinite field and $A$ be an abelian variety over $k$. Can $A$ be embedded into a Jacobian variety $J$ over $k$?

In these notes by William Stein this is stated without proof in remark 1.5.8; it is attributed to personal conversation with Brian Conrad. Unfortunately I was unable to locate or come up with any proof.

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4 Answers 4

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You can find a detailed proof here (theorem 1.2) in the case of principally polarized abelian varieties. One reduces to this case using the Zarhin's trick. The assumption of $k$ being infinite should not be necessary (see remark 1.3 in the paper)

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    $\begingroup$ The main point of Theorem 1.2 in that Cadoret-Tamagawa paper was to bound $\dim J$ in terms of $\dim A$. The existence of $J$ is older, going back at least to Corollary 2.5 in this article of Gabber $\endgroup$
    – Bjorn Poonen
    Commented Jul 9, 2021 at 22:16
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Embed the dual abelian variety into projective space. Take a smooth hyperplane section and interate until it's one-dimensional, obtaining a smooth curve $C$. By Lefschetz $C$ is irreducible, and the natural map $H_1(C, \mathbb Z) \to H_1(A^\vee, \mathbb Z)$ is surjective. Because $H_1(C, \mathbb Z)= H_1(J(C), \mathbb Z)$, the natural map $H_1(J(C), \mathbb Z) \to H_1(A^\vee, \mathbb Z)$ is surjective as well. Now this does imply that the map of dual abelian varieties is surjective because abelian varieties over $\mathbb C$ are Pontryagin dual to the integral homology of the dual abelian varieties.

In characteristic $p$, a similar argument should work using the injectivitity on etale cohomology with torsion coefficients and crystalline cohomology with torsion coefficients / algebraic de Rham cohomology, but I didn't check the details.

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Let me give an answer for $k = \mathbb{C}$.

By a theorem of Matsusaka, every abelian variety $A$ over an algebraic closed field $k$ is a quotient of a Jacobian.

Now just apply Matsusaka's theorem to $A^{\vee}$, and dualize. Since we are over $\mathbb{C}$, dualization sends surjective morphisms of Abelian varieties into injective ones, so we are done.

I think that the slightly weaker version where the field is infinite is proved somewhere in Milne's lecture notes on Abelian Varieties.

Reference.

T. Matsusaka: On a generating curve of an Abelian variety, Nat. Sci. Rep. Ochanomizu Univ. 3 1-4, (1952).

Quoting from P. Samuel review on MathSciNet:

An abelian variety $A$ is said to be generated by a variety $V$ (and a mapping $f$ of $V$ into $A$) if $A$ is the group generated by $f(V)$. It is proved that every abelian variety $A$ may be generated by a curve defined over the algebraic closure of $\mathrm{def}(A)$.

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  • $\begingroup$ Thanks for the quick answer. Sorry for not adding this to the initial question, but it is unclear to me how the result directly follows: As far as I know, dualizing $J \to A^{\vee} \to 0 $ leaves you with $0\to K \to A \to J$ for a finite group scheme $K$. Am I mistaken? $\endgroup$
    – Johann Haas
    Commented Jul 5, 2018 at 12:36
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    $\begingroup$ Ok, I think it is safer to work over $\mathbb{C}$, in which case dualization sends surjective morphisms of Abelian varieties (or complex tori) into injective ones, and conversely. Maybe someone else can fix the argument for an arbitrary algebraically closed field $k$ (if this is possible). $\endgroup$
    – Francesco Polizzi
    Commented Jul 5, 2018 at 12:52
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    $\begingroup$ I don't think its true that duals of surjective morphisms are injective. What about twice the identity on an elliptic curve? There are no injective homomorphisms of elliptic curves apart from isomorphisms. $\endgroup$
    – Neil Strickland
    Commented Jul 5, 2018 at 13:34
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    $\begingroup$ You are right, there is a gap in the argument. We need to know that the kernel of the surjective morphism is connected (hence an abelian variety), and then we are done since the kernel of a morphism and the kernel of its dual (at least over $\mathbb{C}$) have the same number of connected components [Debarre, Tores et varietes abeliennes complexes, Proposition 4.6 p. 71]. $\endgroup$
    – Francesco Polizzi
    Commented Jul 5, 2018 at 14:13
  • $\begingroup$ Is there no reduction process to achieve this? Or maybe it comes for free from Matsusaka's theorem? $\endgroup$
    – Francesco Polizzi
    Commented Jul 5, 2018 at 14:19
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Let $A^*=Pic^0(A)$ and apply $Pic^0$ to a surjective homomorphism $J\to A^*$. Added: as noted, the existence of a surjective homomorphism goes back to a theorem of Matsusaka. For the proof, see Milne, Jacobian Varieties, 1986, 10.1. The argument in Kleiman, Algebraic cycles and the Weil conjectures, 1968, 2A7, shows that the kernel of the dual homomorphism is finite and not divisible by any $l$ prime to the characteristic. I expect that the argument can be made to work also for $p$, but I haven't checked this.

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    $\begingroup$ The existence of the surjection is not trivial, you need Matsusaka's theorem. $\endgroup$
    – Francesco Polizzi
    Commented Jul 5, 2018 at 12:21

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