Over which fields does every smooth projective geometrically connected curve of genus one have a rational point?
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1$\begingroup$ Algebraically closed and finite fields are examples. You ask for which $K$ is $H^1(K,E)=0$ for all elliptic curves $E/K$. $\endgroup$– Chris WuthrichCommented Sep 30, 2021 at 13:29
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2$\begingroup$ Pseudo-algebraically closed fields are also examples. But I don't think there's a specific term. One construction which might interest you is this: start from any field $k_0$, let $E$ be a genus $1$ curve and $k_1 = k_0(E)$ its function field: then $E$ has a point over $k_1$; repeat transfinitely, taking inductive limits at limit steps: you will eventually get a field with the property you asked, in which $k_0$ is algebraically closed. $\endgroup$– Gro-TsenCommented Sep 30, 2021 at 13:43
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$\begingroup$ @Gro-Tsen I don't understand your construction. Each time you make an extension don't get new non-isotrivial curves? Why would they have a rational point? $\endgroup$– QasimCommented Sep 30, 2021 at 13:57
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$\begingroup$ Yes, each extension brings new curves, which is why the process must be iterated transfinitely (repeat with these new curves, and so on): but you run out of curves before you run out of ordinals. See arxiv.org/abs/math/0303168 theorem 2.3 and lemma 2.4 for a related situation (where we add points to Severi-Brauer varieties rather than curves of genus 1). $\endgroup$– Gro-TsenCommented Sep 30, 2021 at 18:11
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$\begingroup$ (Disclaimer: I didn't check too carefully that the argument which works for Severi-Brauer varieties also works in your case, so don't take my word for it. This is why I'm writing this as a comment and not as an answer.) $\endgroup$– Gro-TsenCommented Sep 30, 2021 at 18:14
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