1
$\begingroup$

I have a question about the content of remark 2.6.6. (i) (p 18) from M. Brion's notes on structure of algebraic groups.

Let $G$ be a group scheme over certain fixed base field $k$ (as all other involved schemes) and $f: X \to Y$ be a $G$-torsor.

Remark 2.6.6 refers to the preceding result, Proposition 2.6.5, which states that $f$ is finite (resp. affine, proper, of finite presentation) if and only if so is the scheme $G$.
Then Remark 2.6.6(i) says:

As a consequence of the above proposition, every torsor $f: X \to Y$ under an algebraic group $G$ is of finite presentation. In particular, $f$ is surjective on $\overline{k}$-rational points, i.e., the induced map $X(\overline{k}) \to Y (\overline{k})$ is surjective. But $f$ is generally not surjective on $S$-points for an arbitrary scheme $S$ (already for $S = \operatorname{Spec}(k))$.

Question: This part I do not understand. Precisely, why $f$ being of finite presentation implies that the induced map $f(\overline{k}): X(\overline{k}) \to Y (\overline{k})$ is surjective on the level of $\overline{k}$-valued points, but in general not for other $S$-valued points, e.g., certain field extensions of base field $k$?

Could somebody unravel how being of finite presentation gives rise to such dichotomous behavior?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

If $f\colon X \to Y$ is a surjective locally finite type (not necessarily finite presentation) morphism of $\overline{k}$-schemes, then $f(\overline{k})$ is surjective. To see this, note that if $y \in Y(\overline{k})$ then $f^{-1}(y)$ is a nonempty locally finite type $\overline{k}$-scheme, hence has a $\overline{k}$-point by the Nullstellensatz.

This proof indicates why surjectivity can fail for $f(k)$: there are finite type $k$-schemes with no $k$-points. For an explicit example, let $\pi\colon \mathrm{SL}_n \to \mathrm{PGL}_n$ be the quotient map. There is an exact sequence \begin{equation} \mathrm{SL}_n(k) \to \mathrm{PGL}_n(k) \to k^\times/(k^\times)^n \to 1 \end{equation} where the map $\mathrm{PGL}_n(k) \to k^\times/(k^\times)^n$ is given by sending $\overline{M}$ to $\det(M)$, where $M$ is any lift of $\overline{M}$ to $\mathrm{GL}_n(k)$. So whenever $\overline{M}$ maps to a nontrivial element of $k^\times/(k^\times)^n$, the scheme-theoretic fiber $\pi^{-1}(\overline{M})$ has no $k$-points and $\overline{M}$ is not in the image of $\pi(k)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ just from point of curiosity, it also possible to capture the (non) surjectivity of $X(\overline{k}) \to Y (\overline{k})$ cohomologically in terms of appropriate $H^1(\text{bla})$ obstruction group (at least if $f$ is $G$-torsor)? In other words, does the latter fit in appropriate exact sequence? $\endgroup$
    – user267839
    Commented May 6 at 15:19
  • 2
    $\begingroup$ Yes: in general, if $f$ is a $G$-torsor, then one gets an exact sequence $X(k) \to Y(k) \to \mathrm{H}^1(k, G)$, where the right map is given by sending $y$ to the class of the $G$-torsor $f^{-1}(y)$. For example, the map $\pi$ in my answer is a $\mu_n$-torsor, and $\mathrm{H}^1(k, \mu_n) \cong k^\times/(k^\times)^n$. (It is a coincidence that the obstruction map is surjective in this case, coming from the fact that $\mathrm{H}^1(k, \mathrm{SL}_n) = 1$.) $\endgroup$
    – SeanC
    Commented May 6 at 15:24

Not the answer you're looking for? Browse other questions tagged or ask your own question.