# Divisorial contraction: when is the image an algebraic space or a stack?

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?

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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

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Thanks for the answer. When you say "the intersection matrix must be negative definite", what do you assume on $Y$ ? To be an algebraic space? What happens if we accept to have a stack? Is there some natural category where we accept that the intersection matrix is not negative definite? PS: Badescu's book is nice but only concerns contractions $X\to Y$ with both $X,Y$ projective algebraic surfaces, so not really the question I asked. –  Jérémy Blanc Aug 28 '12 at 13:39
The arguments for algebraic spaces and schemes are the same. For a precise statements, see Section 4 and in particular, Theorem 4.5, of M. Artin: "Algebraic Spaces", Yale University Press (1971). –  Christian Liedtke Aug 28 '12 at 14:03
P.S.: if you assume $X$ to be an algebraic space, and $Y$ to be an Artin stack with finite inertia, then, I think, you can argue via the coarse algebraic space of $Y$ (which exists by Keel-Mori), thereby reducing to the case that $Y$ was an algebraic space to start with. Right? –  Christian Liedtke Aug 28 '12 at 14:29
I am not expert with stacks, but it seems that your argument works. What about a stack without finite inertia? –  Jérémy Blanc Aug 29 '12 at 13:52