Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $X_{\mathbb{Q}}$ is a curve over $\mathbb{Q}$, we get a curve $X_{\mathbb{Q}_p}$ over $\mathbb{Q}_p$ for every prime $p$.

My question is about the reverse process. Say we are given curves $X_{\mathbb{Q} _ p}$ for every $p$ such that $X_ { \bar {\mathbb{Q}} _ p} \cong X_ {\bar{ \mathbb{Q}} _ q}$ for any two primes $p$ and $q$ (where the isomorphism is as schemes; alternatively, they are isomorphic when base-changed to an algebraically closed field that contains both $\mathbb{Q}_p$ and $\mathbb{Q} _ q$). Then is there a nice way to describe the obstruction for these models to come from a global curve $X_{\mathbb{Q}}$ over $\mathbb{Q}$?

share|improve this question
    
I don't have an answer, but maybe Dèbes and Emsalem (On fields of moduli of curves J. Algebra 211 (1999), no. 1, 42–56) is relevant to your question. –  inkspot Jan 26 '12 at 14:24
    
One obstruction in the case of elliptic curves is that you can choose transcendental $j$-invariants at every place, together with bizarre field isomorphisms. –  S. Carnahan Jan 27 '12 at 10:09

4 Answers 4

Interesting question. I guess the answer will be very different depending on genus $0,1,>1$. I think I can say something about genus zero.

The condition over the algebraic closures is vacuous for genus zero. For any field $K$ of characteristic zero, a curve of genus zero is isomorphic to a conic $ax^2+by^2=z^2$. When $K=\mathbb{Q}_p$ we can associate the Hilbert symbol $(a,b)_p$ which is $1$ if the conic has a point in $K$ and $-1$ if not. What's the condition that a collection $(a,b)_p$ (I'll include $p=\infty$ as well) comes from a global conic? You need that $(a,b)_p=1$ for all but finitely many $p$ and that $\prod (a,b)_p =1$ (Hilbert reciprocity). That these conditions are necessary and sufficient are proved, e.g., in Serre's Cours d'Arithmétique.

I am worried however that the question in general may not make sense, since the isomorphisms between the algebraic closures of the various $p$-adic fields are not unique. Say, in genus one, you need to give $p$-adic $j$-invariants for each $p$ all agreeing up to these isomorphisms. Does this mean that these come from a global $j$? Then is a question of local/global Weil-Chatelet groups. I don't know what happens there.

share|improve this answer

I agree with the previous opinions that this should fail for $g>0$. Here's an example. Let $C$ be a universal (hyper)elliptic curve $y^2=\prod_i (x-t_i)$. This is clearly defined over $K=\mathbb{Q}(t_1,\ldots)$, but it wouldn't be defined over $\mathbb{Q}$. (Suppose it were, then by specialization any (hyper)elliptic curve would be.) After choosing embeddings $K\subset \mathbb{ Q}_p$ for each $p$, we get a compatible family of $p$-adic curves which don't descend to the rationals.

share|improve this answer

You might be interested in Mazur's famous article On the passage from local to global in number theory.

Mazur conjectures that if a family of varieties $V_p$ over $\mathbf{Q}_p$ indexed by the places $p$ (including $p=\infty$) of $\mathbf{Q}$ comes from a variety $V$ over $\mathbf{Q}$, then there are only finitely many $V$ (up to $\mathbf{Q}$-isomorphism) with this property.

This is true for curves of genus $0$ (see Felipe's answer) and more generally for projective spaces or quadrics of any dimension. In these cases you have not only finiteness but uniqueness. For twisted forms of projective spaces, there is a criterion for existence : almost all the local invariants must vanish, and their sum must vanish.

But it is open for curves of genus $1$, although he relates this conjecture to the conjectural finiteness of the Shafarevich-Tate group of elliptic curves over $\mathbf{Q}$.

For curves of genus $>1$, it is relatively easy to prove Mazur's conjecture (because the automorphism group in question is finite).

Counterexamples (?) Here is one approach to constructing counterexamples (as in your question, not to Mazur's conjecture!) in every genus $g>0$.

Start with a curve $C$ over $\mathbf{Q}$ of genus $g$, and take $X_p$ to be $C_p=C\times_{\mathbf{Q}}\mathbf{Q}_p$ for every odd $p$ (and $p=\infty$, if you wish), but take $X_2$ to be a twist of $C_2$ (in the sense that they become isomorphic over an algebraic closure of $\mathbf{Q}_2$) but such that the jacobian $\mathrm{Jac}(X_2)$ of $X_2$ is not isogenous to $\mathrm{Jac}(C_2)$; presumably this can be arranged. If so, the family $(X_p)_p$ cannot come from a curve $X$ over $\mathbf{Q}$ : for any such $X$, $\mathrm{Jac}(X)$ will have to be isogenous to $\mathrm{Jac}(C)$, whereas they are not isogenous at the place $2$ by construction.

share|improve this answer
    
Mazur's conjecture is related to the OP's question but is a bit different. There he asks if the set of global varieties everywhere locally isomorphic to a fixed global one is finite. –  Felipe Voloch Jan 29 '12 at 16:48
    
That's what I wrote, Felipe. –  Chandan Singh Dalawat Jan 30 '12 at 3:51
    
I got confused by how you phrased it, sorry. –  Felipe Voloch Jan 30 '12 at 12:42

I think the following interpretation of your question is false for cardinality reasons. It occurred to me before I saw Donu's answer but seems to have a similar flavour.

I will say that two curves over a field are twists of each other if they become isomorphic over an algebraic closure.

Interpretation. Fix a curve $C$ over $\mathbf{Q}$. For every place $v$ of $\mathbf{Q}$, let $X_v$ be an arbitrary (random) twist of $C_v$. Does there exist a curve $D$ over $\mathbf{Q}$ such that $D_v$ is isomorphic to $X_v$ for every $v$ ?

In genus $0$, not every such family can come from a conic over $\mathbf{Q}$, as Felipe has remarked.

In genus $g>0$, if we start with a hyperelliptic curve $C$, then we get uncountably many families $(X_v)_v$ by taking random quadratic twists at each place $v$. All these families satisfy your hypotheses, and some cannot come from the countably many genus-$g$ curves over $\mathbf{Q}$.

Examples. Suppose we have an elliptic curve $E$ over $\mathbf{Q}$ which is the only curve in its isogeny class, and assume moreover that Ш$(E)$ is trivial. It follow that if a genus-$1$ curve $C$ is such that its jacobian $J$ is isomorphic to $E$ almost everywhere locally, then $J$ is $\mathbf{Q}$-isogenous to $E$, and hence $\mathbf{Q}$-isomorphic to $E$, and hence $C$ is isomorphic to $E$ (because Ш$(E)$ is trivial).

Construct the family $(X_v)_v$ by taking $X_v=E_v$ for every place $v\neq2$ of $\mathbf{Q}$, and perversely take $X_2$ to be a quadratic twist of $E_2$, so that $X_2$ and $E_2$ are not $\mathbf{Q}_2$-isomorphic.

If there were a genus-$1$ curve $C$ such that $C_v$ is $\mathbf{Q}_v$-isomorphic to $X_v$ at every $v$, then $C$ would have to be $\mathbf{Q}$-isomorphic to $E$ (by the choice of $E$, as explained above), which is impossible because they are not $\mathbf{Q}_2$-isomorphic.

It remains to find such an $E$. I'm sure this can be done by looking at the tables made by Cremona and Stein. Could someone please confirm this hunch ?

share|improve this answer
    
Here's a concrete example: take an elliptic curve $E/\mathbf{Q}$, and for every $p$, take "the" quadratic twist $F_p$ of $E$ at $p$ which is not $E$. Then the collection $(F_p)$ can not come from an elliptic curve over $\mathbf{Q}$. –  Denis Chaperon de Lauzières Jan 31 '12 at 5:52
    
Which one is "the" quadratic twist of $E_p$ ? Aren't there as many quadratic twists of $E_p$ as $\mathbf{Q}_p$ has quadratic extensions, namely seven when $p=2$, three when $p$ is an odd prime, and one only when $p=\infty$ ? Or do you mean "the unramified quadratic twist" ? –  Chandan Singh Dalawat Jan 31 '12 at 8:20
    
Sorry, I was thinking about the twist at the level of finite fields, not $Q_p$, so yes, the unramified twist... (The point being that the Hasse-Weil $L$-function of an $F$ with these completions will look like the inverse of the $L$-function of $E$, up to the critical line.) –  Denis Chaperon de Lauzières Feb 1 '12 at 6:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.