Timeline for surjectivity of rational points induced by surjective map from affine space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2012 at 15:08 | comment | added | Jason Starr | For a generically 'etale morphism, and assuming your local field is non-Archimedean, you can just use Hensel's lemma to prove that the image contains a sufficiently small adic neighborhood of the image point wherever it is 'etale. So the subgroup generated by the image will contain a sufficiently small adic neighborhood of the identity. | |
| Jun 29, 2012 at 1:39 | comment | added | ronggang | @ Will: It is not surjective even in this case since in general $V^+$ or $V(k)^+$ is not equal to $V(k)$. e.g. PGL_2. | |
| Jun 28, 2012 at 16:02 | comment | added | Will Sawin | If you're thinking of those specific maps, then it should be surjective. The reason is that the map from $\mathbb A^n$ to a connected unipotent group is not just surjective, it is an isomorphism. The algebraic inverse comes from the formal power series for $\log$. Isomorphisms are surjective on rational points. You might run into additional troubles if the decomposition into unipotents is not canonical, though. | |
| Jun 28, 2012 at 14:33 | comment | added | ronggang | I mean the group generated by the set $f(k^n)$ has finite index in the group $V(k)$. Here is the main example in my mind. Suppose $V$ is generated by connected unipotent groups defined over $k$. Then we get such a map $f$ according to I Porp. 2.2 of Borel's algebraic groups. Then the question is whether the group generated by $k$ points of these unipotent groups has finite index in $V(k)$. If $V$ is semisimple then the question is whether the group they generate contains $V^+$ in the sense of 1973 paper of Borel and Tits( Homomorphismes "Abstraits" de Groupes Algebriques Simples). | |
| Jun 28, 2012 at 12:35 | comment | added | Jason Starr | @ronggang -- What precisely do you mean by "index"? Why should the image of f be a subgroup of the group of $k$-points? For instance, the image on $k$-points of the map $f(x) = x^2 + 1$ is not a subgroup of the additive group. | |
| Jun 28, 2012 at 8:24 | comment | added | ronggang | Sorry, finite index in $V(k)$. | |
| Jun 28, 2012 at 8:20 | comment | added | ronggang | Thanks. In fact what I want to know is the following: Assume in addition that $V$ is a linear algebraic group. Is it true that the group generated by $f(k^n)$ has finite index in $V$? | |
| Jun 28, 2012 at 8:03 | vote | accept | ronggang | ||
| Jun 28, 2012 at 8:22 | |||||
| Jun 28, 2012 at 7:16 | history | answered | Olivier Benoist | CC BY-SA 3.0 |