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.

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    $\begingroup$ Please define $X$ and $G$. $\endgroup$ – user18119 Feb 20 '13 at 19:43
  • $\begingroup$ I think it's pretty clear that $X=C$ was a typo, and of course $G=Gal(k^{sep}/k)$, the only natural group action possible here (also the one that comes out of the Leray s.s.). $\endgroup$ – Matt Feb 20 '13 at 21:41

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.

  • $\begingroup$ Thank you so much. This is fantastic. Do you have a feel for whether or not "most" genus one curves without a rational point don't have equality there? In other words, is equality somehow really close to having a rational point? $\endgroup$ – Matt Feb 21 '13 at 2:22
  • $\begingroup$ In/equality is not "close to having a rational point" in any sense known to me. For one thing, the examples constructed above can have arbitrarily large index. Unless I am forgetting, I really only understand the quantity you're asking about insofar as it fits into the above exact sequence: each of the other two quantities I have a better feel for. Another extreme is an element of the Shafarevich-Tate group of a global field: then even $\operatorname{Br}(C/k)= 0$. Honestly, I recommend you read my WCI and WCII papers and my "period, index and potential shah" paper: that's what I know! $\endgroup$ – Pete L. Clark Feb 21 '13 at 2:54
  • $\begingroup$ Also, if actually $C(k) \neq \varnothing$, then $\operatorname{Br}(C/k) = 0$: you probably knew this already. $\endgroup$ – Pete L. Clark Feb 21 '13 at 2:55

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