0
$\begingroup$

Hi,

I have the following two questions about automorphism groups of schemes. First of all, let $S$ be a scheme, and $S^c$ its set of closed points. What is the connection between $Aut(S)$ and $Aut(S^c)$ ? Secondly, let $S$ be a $\mathbb{Z}$-scheme, and $S_k$ the base-extension to some field k. What is the precise relation between their automorphism groups ? Thanks,

THC
$\endgroup$
3
  • 2
    $\begingroup$ What do you mean by an automorphism of $S^c$? $\endgroup$ Commented Jan 16, 2012 at 11:36
  • $\begingroup$ Laurent, I interpreted the question asking whether two automorphisms that agree on closed points would agree, so accordingly one could define $Aut(S^c)$ as the image of $Aut(S)$ in the set of all self-maps of $S^c$... $\endgroup$ Commented Jan 16, 2012 at 11:59
  • $\begingroup$ I decided to interpret the question in terms of set-theoretic permutations, just in case Sándor's answer wasn't what THC had in mind. $\endgroup$
    – S. Carnahan
    Commented Jan 16, 2012 at 12:16

2 Answers 2

3
$\begingroup$

It depends on what kind of scheme $S$ is. If $S$ is (reduced) of finite type over a field, then the set of closed points is dense in $S$, so all morphisms are determined by their behaviour on the closed points. On the other hand if $S$ is a local scheme, then it has a single closed point, so you can't say much about automorphisms just from the closed points.

$\endgroup$
0
$\begingroup$

The functor from schemes to sets, given by taking the set of closed points, is neither full nor faithful. For example, the spectrum of a field with nontrivial automorphisms has a single closed point, but its automorphism group is nontrivial. On the other hand, the complex projective line has automorphism group $PGL_2(\mathbb{C})$, but the set of closed points is an abstract set of cardinality $2^{\aleph_0}$, and has an automorphism group of strictly larger cardinality. What you can say is that the functor induces a homomorphism (that may be neither injective nor surjective).

If you want to consider automorphism groups of base extensions, you might as well assemble them into the automorphism group sheaf $\underline{\operatorname{Aut}} (S)$, which eats a scheme $T$, and returns the automorphism group of $S \times T$ (as a $T$-scheme). When $S$ is projective, this sheaf is represented by a scheme, whose $k$-points are precisely the automorphisms of $S_k$. For example, $\mathbb{P}^1$ has automorphism group scheme $PGL_2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .