Timeline for Points of abelian varieties over purely transcendental extensions
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2018 at 15:08 | vote | accept | cll | ||
| Mar 12, 2018 at 13:38 | comment | added | Zhiyu | Could you explain why the restrictions of f are both group homomorphisms in Proof 2.? The rigidity lemma requires the source to be proper. | |
| Feb 21, 2018 at 18:54 | comment | added | R. van Dobben de Bruyn | @BenLim: any $k(x_1,\ldots)$-point of a finite type $k$-scheme factors through the finitely generated extension $k(x_1,\ldots,x_n)$ for some $n$. | |
| Feb 21, 2018 at 18:52 | comment | added | R. van Dobben de Bruyn | @AnnaAbasheva: if $L'$ is the function field of $\mathbb A^n$ (or $\mathbb P^n$), then any morphism $\operatorname{Spec} L' \to X$ to any finite type $k$-scheme $X$ spreads out to some open neighbourhood of the generic point in $\mathbb A^n$. If you don't know this, you should do this as an exercise. (Hint: first think about the case that $X$ is affine. Use that $X$ is of finite type.) | |
| Feb 21, 2018 at 17:34 | comment | added | David Benjamin Lim | Hi Remy, it seems that your argument uses the fact that the transcendence degree of L'/k is finite. However the OP claims this is true for any purely transcendental extension? | |
| Feb 21, 2018 at 8:34 | comment | added | cll | There is one thing I don't understand in your proof: why the morphism $\mathrm{Spec} L'\to A$ spreads out to a rational variety? | |
| Feb 21, 2018 at 5:19 | comment | added | R. van Dobben de Bruyn | @FelipeVoloch: Ah, I thought I had seen that question (and answer) here before, but I couldn't find it when I was writing. Thanks! | |
| Feb 20, 2018 at 21:41 | comment | added | Felipe Voloch | mathoverflow.net/questions/9066/… | |
| Feb 20, 2018 at 19:39 | comment | added | R. van Dobben de Bruyn | @JasonStarr: ah, that's pretty neat. I'm a little agnostic as to which proof is simpler, but the one you gave is certainly pretty nice. | |
| Feb 20, 2018 at 17:44 | comment | added | Jason Starr | I think the simplest proof of the lemma uses that there are no nontrivial finite, etale covers of $\mathbb{P}^1$. Thus, for every odd prime $\ell$ different from the characteristic, every morphism from $\mathbb{P}^1$ to $A$ factors through the "multiplication-by-$\ell$" morphism. For any ample invertible sheaf $L$ on $A$, this implies that the degree of the pullback of $L$ on $\mathbb{P}^1$ is divisible by $\ell$ (by the Theorem of the Cube). Thus, the degree is zero, and the morphism is constant. | |
| Feb 20, 2018 at 16:11 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |