The Torelli theorem states that the map $\mathcal{M}_g(\mathbb{C})\to \mathcal{A}_g(\mathbb{C})$ taking a curve to its Jacobian is injective. I've seen a couple of proofs, but all seem to rely on the ground field being $\mathbb{C}$ in some way. So:

Under what conditions does the Torelli Theorem hold?

Is algebraically closed necessary? Characteristic zero? It is known if there are any other base rings/schemes over which it is true?


1 Answer 1


The Torelli theorem holds for curves over an arbitrary ground field $k$ (in particular, $k$ need not be perfect). A very nice treatment of the "strong" Torelli theorem may be found in the appendix by J.-P. Serre to Kristin Lauter's 2001 Journal of Algebraic Geometry paper Geometric methods for improving the upper bounds on the number of rational points on algebraic curves over finite fields. It is available on the arxiv:


Here are the statements (translated into English):

Let $k$ be a field, and let $X_{/k}$ be a nice (= smooth, projective and geometrically integral) curve over $k$ of genus $g > 1$. Let $(\operatorname{Jac}(X),\theta_X)$ denote the Jacobian of $X$ together with its canonical principal polarization. Let $X'_{/k}$ be another nice curve.

Theorem 1: Suppose $X$ is hyperelliptic. Then for every isomorphism of polarized abelian varieties $(\operatorname{Jax}(X),\theta_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta_{X'})$, there exists a unique isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ such that $F = \operatorname{Jac} f$.

Theorem 2: Suppose $X$ is not hyperelliptic. Then, for every isomorphism $F: (\operatorname{Jax}(X),\theta_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta_{X'})$ there exists an isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ and $e \in \{ \pm 1\}$ such that $F = e \cdot \operatorname{Jac} f$. Moreover, the pair $(f,e)$ is uniquely determined by $F$.

  • $\begingroup$ Nice! I hadn't been expecting this...talked to one of the local arithmetic people and they weren't sure, but figured it'd be false. Thanks! $\endgroup$ May 7, 2010 at 22:38

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