TITLE: Divisorial contraction: when is the image an algebraic space or a stack?
QUESTION [5 upvotes]: Let $X$ be a smooth projective surface (in the category of varieties, or schemes), and let $C\subset X$ be a curve (a priori not irreducible, but the irreducible case in itself is already interesting).
There are classical notions that say when it is possible to have a morphism $X\to Y$, where $Y$ is a variety (or a scheme) which contracts $C$ (onto points) and which restricts to an isomorphism on $X\backslash C$. The matrix of intersection numbers of the components of $C$ need in particular to be negatively defined.
1) If we admit $Y$ to be an algebraic space, are the conditions weaker? (I think that the matrix has again to be negatively defined, reading Artin, "Algebraic spaces" Theorem 4.5, but are there other conditions that are weaker?)
2) If we admit $Y$ to be an algebraic stack, is the matrix again negatively defined or are there counterexamples?

REPLY [10 votes]: In order to have a contraction morphism $X\to Y$, the intersection matrix must be negative definite.
Conversely, if the intersection matrix is negative definite, the contraction morphism exists in the category of algebraic spaces. It may not exist in the category of schemes: blowing up a smooth cubic in the projective plane in $10$ appropriately chosen points, you can contract the resulting curve of genus $1$ and self-intersection number $-1$ on the blown-up surface only within the category of algebraic spaces. Quite generally, the contraction morphism exists in the category of schemes if the contraction gives rise to a rational singularity.
For details, proofs, and references, you might want to look up Badescu's book on algebraic surfaces, Chapter 3