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:



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> Under what conditions does the Torelli Theorem hold?
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Is algebraically closed necessary? Characteristic zero? It is known if there are any other base rings/schemes over which it is true?



When does the Torelli Theorem hold?
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:


<http://arxiv.org/abs/math/0104247>


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 $F$. 


