**Introduction:** The definition of "good categorical quotient" in geometric invariant theory (given below) seems fairly ad hoc to me, except that it looks very similar to the coequalizer of the action in the category of locally ringed spaces. I'm trying to understand this similarity.

**Some background:** Suppose we have a group action $G\times X \to X$ in the category $\bf Sch$ of schemes. The category $\bf LRS$ of locally ringed spaces is cocomplete (see propositiion 1.12 in A. Vezzani's thesis), so in particular, $G\times X \rightrightarrows X$ has a $\bf LRS$-coequalizer; call it $Q$. One can construct it as the coequalizer in the category of topological spaces equipped with the sheaf of $G$-invariant functions to make it a locally ringed space, as Vezzani's proposition explains more carefully.

If $G\times X \rightrightarrows X$ has a $\bf Sch$-coequalizer $Y$, then GIT calls $Y$ a "categorical quotient"; see, for example, R. Birkner's Introduction to GIT. Note that here we get a universal map $u: Q\to Y$ in $\bf LRS$.

$Y$ is called a "** good categorical quotient**" if it satisfies 3 properties, also found in Birkner's notes, which I prefer to re-express here in terms of the universal map $u$:

i) $u^\sharp : {\cal O}_Y \to u _* {\cal O}_Q$ is an isomorphism.

ii) $u: Q\to Y$ is a closed map (at the level of topological spaces).

iii) $u$ takes disjoint closed sets to disjoint closed sets.

(These together imply that $X\to Y$ is a categorical quotient, i.e. $\bf Sch$-coequalizer; see Mumford's GIT I.2 remark 6.)

All of these properties are clearly satisfied if $u$ is an isomorphism. That is, **if the $\bf LRS$-coequalizer is a scheme, that scheme is a good categorical quotient.**

Towards a converse, (i) implies that $u$ must be dominant, and hence by (ii) it is surjective. And (iii) implies it is injective on closed points, so it's looking a lot like an isomorphism. Can anything more be said?

1) When is a good categorical quotient an $\bf LRS$-coequalizer of the group action? I.e., are there nice and general conditions where $u$ must be an isomorphism?

2) If the answer is not always, what's a counterexample? (

Answered by Anton)

Thanks!

Groupes algebriques(Demazure, Gabriel), 1970. I don't know the answer to your problem, but usually the notions concerning colimits of schemes are motivated by explicit applications, and not by abstract nonsense. I doubt that there is any connection. $\endgroup$ – Martin Brandenburg Mar 28 '11 at 10:19