Timeline for Is the set of points on an abelian surface which project to rational points on the Kummer surface a subgroup?
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| Nov 3, 2023 at 2:45 | history | became hot network question | |||
| Nov 2, 2023 at 20:20 | comment | added | R.P. | See Lemma 5 in arxiv.org/abs/1404.3641. But it's really a very easy (and general) result, see the reference to Skorobogatov's book Torsors and rational points. | |
| Nov 2, 2023 at 20:12 | comment | added | R.P. | You should understand the set of $k$-rational points on $X$ as the disjoint union of the set of $k$-rational points on all quadratic twists of $J$ (where $k$ is any field extension of the base field). That is why multiples of a point on $J$ which projects to a rational point on $X$ also project to rational points on $X$, but sums of two such points in general do not, except when they come from the same twist of $J$. | |
| Nov 2, 2023 at 19:55 | vote | accept | Vik78 | ||
| Nov 2, 2023 at 19:48 | answer | added | R. van Dobben de Bruyn | timeline score: 4 | |
| Nov 2, 2023 at 19:37 | comment | added | R. van Dobben de Bruyn | The preimage of $X(\mathbf Q)$ is the set of points whose Galois orbit is contained in the fibre. This means that any Galois conjugate of $x$ is either $x$ or $-x$. Since $[n] \colon J \to J$ is defined over $\mathbf Q$, it is $\operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q)$-equivariant, and since $-([n]x) = [n](-x)$, we see that all multiples satisfy the same property. However, for sums the situation is more complicated: if $\{x,-x\}$ is defined over a quadratic field $K$ and $\{y,-y\}$ over a quadratic field $L$, then the orbit of $x+y$ in general contains $\pm x \pm y$, defined over $KL$. | |
| Nov 2, 2023 at 19:36 | answer | added | Joe Silverman | timeline score: 4 | |
| Nov 2, 2023 at 19:33 | history | edited | Vik78 | CC BY-SA 4.0 |
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| Nov 2, 2023 at 18:43 | history | asked | Vik78 | CC BY-SA 4.0 |