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?


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! $\endgroup$ – Charles Siegel May 7 '10 at 22:38

This is Corollary 12.2, from JS Milne's section on Jacobian varieties in Cornell-Silverman: Arithmetic Geometry:

Let C and C' be curves of genus at least two over a perfect ground field. If the canonically polarized Jacobians of C and C' are isomorphic over k, then so are C and C'.

See also the remark 12.5 immediately after for why $g \geq 2$ is in the assumptions -- the only obstructions in lower genus is that the curve may not have any rational points, while the Jacobian always does.

  • 3
    $\begingroup$ Perfect fields implies general case: an isom $f$ between $C$ and $C'$ over purely insep. ext'n $k'/k$ (such as perfect closure) is defined over $k$. To prove it, since $k'/k$ is directed union of finite subext'ns, WLOG $k'/k$ is finite. Let $k'' = k' \otimes_k k'$. By faith. flat descent, suffices to prove pullbacks $f_1$ and $f_2$ along the inclusions $k' \rightarrow k''$ coincide. The "difference" is $k''$-aut. of $C_{k''}$ whose pullback over diagonal $k'$-point is id. (as each $f_i$ is $f$ over that point). But Aut-scheme of $C$ is etale ($g > 1$!), so no non-triv deformations of id. $\endgroup$ – BCnrd May 8 '10 at 2:02
  • $\begingroup$ Thanks for the corrections, Pete / BCnrd. @BCnrd: Does your comment imply that there are other obstructions for the Torelli theorem to hold over non-perfect fields than lack of rational points? $\endgroup$ – Dan Petersen May 8 '10 at 3:15
  • $\begingroup$ @dan: no obstruction (with genus > 1!). Once one has proved a uniqueness result over alg. closed fields then by Galois descent (descent data due to the uniqueness) one gets existence & uniqueness over any perfect field. Then the argument I gave (which used that $k''$ is infinitesimal in the final step) pushes the result down to an arbitrary field from its perfect closure. So to handle general field, suffices to prove existence/uniqueness result over alg. closed fields. For genera 0 & 1 nothing to say over a general field when no rat'l pt, and result is trivially true when have rat'l pt. $\endgroup$ – BCnrd May 8 '10 at 3:43
  • $\begingroup$ Thanks! And oops, I was indeed asking something very trivial. $\endgroup$ – Dan Petersen May 8 '10 at 23:30

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