Picard group of genus one curve

Question

Is there a known example (or at least moral reason why such a thing should exist) of a genus $1$ curve $C/k$ over a field (assume perfect if you want) with no rational points such that $\mathrm{Pic}^0(C_{k^{sep}})^G\neq \mathrm{Pic}^0(C)$?

This condition is in a paper I'm reading, and I'm wondering how strong it is. Of course, if a rational point exists then the Leray spectral sequence splits canonically, so it always holds. I assume that in general there must be some obvious case where it doesn't hold.

Answer 1

Yes, there are examples of inequality. As it happens this is right up my alley, but I don't know how to tell you about the examples without introducing some other results that you may or may not be familiar with. Let me write $\overline{k}$ for a separable closure of $k$ and $G$ for $\operatorname{Aut}(\overline{k}/k)$.

For a genus one curve $C/k$ the Leray spectral sequence yields an exact sequence

$0 \rightarrow \operatorname{Pic(C)} \rightarrow \operatorname{Pic}(C_{\overline{k}})^G \stackrel{\delta}{\rightarrow} \operatorname{Br}(k) \rightarrow \operatorname{Br}(C),$

and similarly a map $\delta^0$ upon restriction to degree zero divisor classes. Let's put

$\operatorname{Br}(V/k) = \operatorname{Im}(\delta)$

$\operatorname{Br}^0(V/k) = \operatorname{Im}(\delta^0)$.

Your question can be rephrased as asking for an example where $\operatorname{Br}^0(V/k)$ is nonzero.

Now I refer you to (a special case of) Proposition 24 of this paper of mine:

There is a short exact sequence $0 \rightarrow \operatorname{Br}^0(C/k) \rightarrow \operatorname{Br}(C/k) \rightarrow C(I/P) \rightarrow 0$,

where (here) $C(I/P)$ is a finite cyclic of group equal to the index of $C$ -- i.e., the least positive degree of a divisor on $C$ -- divided by the period of $C$ -- i.e., the least positive degree of a divisor on $C$. Therefore if the period equals the index and $\operatorname{Br}(C/k)$ is nontrivial, then the group $\operatorname{Br}^0(C/k)$ is nontrivial. Now take any genus one curve $C$ over $\mathbb{Q}_p$ without a rational point. (By Tate Duality, the Weil-Chatelet group of an elliptic curve over $\mathbb{Q}_p$ is Pontjragin dual to the compact totally disconnected group $E(\mathbb{Q}_p)$, so we do not lack for examples.) By a theorem of [Lichtenbaum68], we have $I = P = n > 1$, say, so $C(I/P) = 1$. His paper also shows that $\operatorname{Br}(C/k)$ is cyclic of order $n$, so there you go.