# Jacobians defined over smaller fields

Let $L/K$ be an extension of number fields.

Let $X$ be a curve over $L$ which can not be defined over $K$. Let $J(X)$ be the Jacobian of $X$ over $L$.

In general, the Jacobian $J(X)$ probably doesn't admit a model over $K$. But one could imagine that this does happen sometimes.

Question. Does there exist an example where $J(X)$ admits a model over $K$?

In other words, can "the" field of definition of $J(X)$ be smaller than "the" field of definition of $X$?

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A not-very-interesting example would be a curve $X$ of genus zero over $L$ which does not lift to $K$. –  Tyler Lawson Oct 4 '12 at 22:44
The genus-1 curve $ax^3 + by^3 + cz^3 = 0$ has Jacobian (away from characteristics 2 and 3) depending only on $abc$, so you can probably make some explicit examples based on that. –  grp Oct 4 '12 at 23:28
My guess is that this is impossible if the polarization of $J(X)$ is also defined over $K$, due to Torelli. So I would look for an abelian surface over $K$ with a principal polarization defined over $L$ that does not descend to $K$ (I think that any p.p. abelian surface is a Jacobian of a genus 2 curve). –  Piotr Achinger Oct 4 '12 at 23:31
@Piotr: It sounds like you ask just that the p.p. does not descend to $K$ respecting the given $K$-structure on the abelian surface. In principle, it might happen that the p.p. abelian surface over $L$ admits another $K$-descent as a p.p. abelian surface (i.e., with a $K$-structure different from the given one on the abelian surface), so the associated curve over $L$ would then be defined over $K$. So to make an example in this way one needs a stronger "does not descend to $K$" property. Perhaps you were already aware of this, in which case all I'm doing is clarifying your comment. –  grp Oct 5 '12 at 0:06
Here is an approach to $g=1$ that similar to grp's. It's not so explicit as grp's, but it is well-organized by some machinery; and thus motivates the machinery. Given an elliptic curve $E$ over a local field $K$, the group of genus $1$ curves with $J(C)=E$ is approximately $K/O_K$. If $K$ is degree $d$ over $\mathbb Q_p$, then $K/O_K=(\mathbb Q_p/\mathbb Z_p)^d$ .... (nb $\mathbb Q_p/\mathbb Z_p=$"$\mathbb Z/p^\infty$" the Prüfer group). Thus there are lots of torsors defined over $L$ that aren't defined over $K$. –  Ben Wieland Oct 5 '12 at 5:17

We can make example in genus 1 using any nontrivial extension of number fields $L/K$ with $[L:K]$ odd, and any non-CM elliptic curve $E$ over $K$. (By "non-CM" I mean no CM over $\overline{K}$.) More concretely, taking $K = \mathbf{Q}$, if $E$ is any non-CM elliptic curve over $\mathbf{Q}$ and $L$ is any number field of odd degree $> 1$ then there exist infinitely many a genus-1 curves $X$ over $L$ with Jacobian $E_L$ such that $X$ cannot descend to $\mathbf{Q}$ as an abstract curve.

To see this, suppose for such data $L/K$ and $E$ we knew that the natural restriction map $$r:H^1(K,E) \rightarrow H^1(L,E)$$ is not surjective. A class in $H^1(L,E)$ corresponds to a genus-1 curve $X$ over $L$ such that its Jacobian is equipped with an identification with $E_L$. Suppose it happens that abstractly $X \simeq Y_L$ for a genus-1 curve $Y$ over $K$. Then $J(Y)$ is a $K$-descent of $J(X) = E_L$, which is to say that $J(Y)$ is a $K$-form of $E$ that becomes isomorphic to it over $L$. Since $E$ is not CM, so its geometric automorphic group is of order 2, the $K$-isomorphism class of $J(Y)$ corresponds to an element in $H^1(K,\mathbf{Z}/(2))$ that is split by the odd-degree $L/K$ and hence is trivial. That is, $J(Y)$ is $K$-isomorphic to $E$ and (with a moment's thought) $Y$ defines a class in $H^1(K,E)$ mapping to the class of $X$ in $H^1(L,E)$. Thus, if we could find a class in $H^1(L,E)$ not hit by $r$ then the corresponding $X$ would meet the desired conditions.

Pick an arbitrary nontrivial extension of number fields $L/K$ and any abelian variety $E \ne 0$ over $K$. I claim that the map $r$ as above is never surjective. To prove this, let $A$ be the abelian variety ${\rm{R}}_{L/K}(E_L)$ of dimension $[L:K]\dim(E) > \dim(E)$ given by Weil restriction, so there is a natural inclusion of $E$ as an abelian subvariety of $A$ with non-zero cokernel $B = A/E$. For a prime $\ell$ not dividing $[L:K]$, the map $r$ is on $\ell$-primary parts is injective with cokernel given by the $\ell$-primary part of $H^1(K,B)$ since the "norm" map $A \rightarrow E$ restricts to multiplication on $E$ by $[L:K]$ which is coprime to $\ell$.

Thus, it suffices to show that if $B$ is any nonzero abelian variety over $K$ (say with dimension $g > 0$) and if $\ell$ is any prime whatsoever then $H^1(K,B)[\ell]$ is infinite. (Such infinitude is no surprise since we impose no "local conditions" on this cohomology group such as unramifiedness away from a fixed finite set of places. The baby version is that $H^1(K,\mu_m) = K^{\times}/(K^{\times})^m$ is obviously infinite for any $m > 1$.) No doubt this infinitude can be proved using results known at the time of Tate's work on global Galois cohomology, but out of "laziness" let's instead argue using the refinement given by the product formula of Wiles.

Let $B^{\vee}$ be the dual abelian variety, and let $S$ be a finite set of non-archimedean places of $K$ consisting of all $\ell$-adic places and some places $v\nmid \ell$ where $B^{\vee}[\ell]$ is locally split (so $B^{\vee}[\ell](K_v)$ has order $\ell^{2g}$ for such $v$). Let $\delta$ be the $\mathbf{F}_{\ell}$-dimension of $B(K)/(\ell)$. Let $S^{(\ell)}$ denote the set of places in $S$ away from $\ell$. Note that by Chebotarev applied to the splitting field of $B^{\vee}[\ell]$ over $K$ we can make $S^{(\ell)}$ as large as we wish. A straightforward application of the Kummer sequence and the Wiles product formula gives that the image in $H^1(K,B)[\ell]$ of the subgroup of $H^1(K,B[\ell])$ satisfying the Selmer condition away from $S$ has $\mathbf{F}_{\ell}$-dimension at least $$g[K:\mathbf{Q}]-\delta+2g|S^{(\ell)}|.$$ Now by making $S^{(\ell)}$ as large as we wish, we can make this lower bound arbitrarily large. Thus, $H^1(K,B)[\ell]$ is infinite.

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$\S 5$ of math.uga.edu/~pete/wcnotes.pdf contains an arguably more elementary proof that $H^1(K,B)$ contains infinitely many elements of order $n$ for any $n > 1$ (in fact over a more general class of fields). –  Pete L. Clark Nov 5 '12 at 8:00
@Pete Clark: The definition of PHS given in those notes (in the 2nd sentence of the definition of WC-groups there) is incorrect over every field of positive characteristic. For example, it doesn't detect the effect of making $A$ act on a PHS for an abelian variety $A'$ through composition with a purely inseparable isogeny $A \rightarrow A'$ that isn't an isomorphism. Also, I disagree with the first sentence of section 2 of those notes (if one reflects on how the degree-1 cohomology of $A$ comes up in practice), though the two questions you raise there are reasonable nonetheless. –  user27056 Nov 5 '12 at 11:37

As Piotr has suggested, Torelli essentially answers the question if the canonical principal polarization is included in the data. More accurately, this is a version of Torelli due to Serre: if $(A,\lambda)$ is a ppav of dimension at least $2$ over a field $K$ and if over some extension $L$ of $K$ there is a curve $C$ such that $(A,\lambda)_L$ is $L$-isomorphic to the Jacobian of $C$, then there is a curve $C_0$ over $K$ such that $C$ is isomorphic to $(C_0)_L$ and some quadratic twist of $(A,\lambda)$ is $K$-isomorphic to the Jacobian of $C_0$. Moreover, if $C$ is hyperelliptic then no twist is necessary.

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@xbnv: my edit crossed with your comment. I hope things are clearer now (I forgot the quadratic twist). –  inkspot Oct 7 '12 at 9:06
@inkspot: I deleted my previous comment. In your new final sentence, I think you meant "not hyperelliptic" (since the "ppav" functor on non-hyperelliptic curves is fully faithful and so Galois descent data transfers both ways). –  user27056 Oct 7 '12 at 9:22
@xbnv: I meant hyperelliptic. In this case the curve and its Jacobian have the same automorphism group and Galois descent data transfer both ways. In the non-hyperelliptic case the Jacobian has more automorphisms (the $-1$); you can then twist the Jacobian by this $-1$ and, as it were, "lose the curve". The reference to Serre is to arXiv 0104247, his paper with Lauter. –  inkspot Oct 7 '12 at 9:50
@inkspot: Thanks for correcting my mistaken memory. So the genus-1 case (for which compatibility with the p.p. is automatic, due to its uniqueness in such cases) is more special for this question than I had been expecting. –  user27056 Oct 7 '12 at 16:11