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I have a question somewhat related to my last question. Suppose $C$ and $C'$ are two genus $1$ curves (smooth, projective, geom conn.) over a perfect field $k$ with no $k$-rational points and that $C$ and $C'$ have the same Jacobian (maybe a subtlety will arise with what that technically means when I bring up cohomology, so "canonically identified" might be the right term?).

Let $S(C)$ be the set of field extensions of $k$ such that $C$ obtains a rational point. Is there an example where $S(C)=S(C')$ but $C$ is not isomorphic to $C'$ as a $k$-curve?

It seems like it should be easy to produce examples in the following sense. Suppose $k=\mathbf{Q}$ and that the Jacobian is an elliptic curve $E$ with $j$-invariant not $0$ or $1728$. I can realize $C$ and $C'$ as classes in $H^1(G, E)$ (for the pedantic out there $G=Gal(\overline{\mathbf{Q}}/\mathbf{Q})$), since they are torsors under $E$. By $j$-invariant assumptions I know that the only way $C$ and $C'$ are isomorphic as $k$-curves is if their classes are "negative" of eachother (compose the torsor structure with the only automorphism $[-1]:E\to E$).

Since there are only two inequivalent torsor structures on $C$ in this situation we just need to know there are more than $2$ classes in $H^1(G, E)$ that become the trivial class over the same set of field extensions. Note: they clearly have the same index (and maybe period? do the classes necessarily generate the same cyclic subgroup?) but the assumption is stronger. Is this well-known? Is there a rephrasing of this that turns it into something well-known?

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up vote 3 down vote accepted


Hint: for any elliptic curve $E$ and any field extension $l/k$, $H^1(l/k,E)$ is subgroup of $H^1(k,E)$. So if $[C,\iota] \in H^1(k,E)$ and $l$ is a splitting field for $[C,\iota]$, it is also a splitting field for any element of the finite subgroup generated by $[C,\iota]$.

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Oh. I see. So take some large cyclic subgroup generated by $C$, then necessarily there must be some $C'$ not isomorphic to $C$ that generates the same cyclic subgroup, and then they will have this property. Thanks! I feel silly now. – Matt Feb 27 '13 at 18:53

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