Contracting divisors to a point This is quite possibly a stupid question, but it is pretty far from what I normally do, so I wouldn't even know where to look it up.
If $X$ is a projective variety over an algebraically closed field of arbitrary characteristic and $Y\subset X$ a smooth divisor. Under which conditions can I contract $Y$ to a point, i.e. under which conditions is there a projective (smooth!?) variety $V$, and a morphism $f:X\rightarrow V$, such that $f$ is an isomorphism away from $Y$, and $Y$ is mapped to a point.
What can one say if $Y$ is a strict normal crossings divisor? 
Hints and references are very appreciated!
 A: If you ask V to be smooth and if the divisor Y is irredicible then Y should be isomorphic to the projective space and an intersection on X of the divisor Y with a line L on Y have to be equal to $-1.$ 
A: For a smooth $Y$, a necessary condition for contractibility is that the conormal line bundle $N_{Y,X}^\*$ is ample. It is also sufficient for contracting to an algebraic space. The reference is Algebraization of formal moduli. II. Existence of modifications. by M. Artin. 
$Y$ can be contracted to a point on an algebraic (projective) variety if in addition $Y=\mathbb P^{n-1}$, $n=\dim X$. You can prove this easily by hands. Start with an ample divisor $H$ and then prove that an appropriate linear combination $|aH+bY|$ is base point free and is zero exactly on $Y$. You will find the argument in Matsuki's book on Mori's program for example.
So if $X$ is a surface and $Y=\mathbb P^1$ with $Y^2<0$ then it is contractible to a projective surface. For a reducible divisor $Y=\sum Y_i$ a necessary condition (which is also sufficient in the category of algebraic spaces) is that the matrix $(Y_i.Y_j)$ is negative definite. The strongest elementary sufficient condition for contractibility to a variety is that $\sum Y_i$ is a rational configuration of curves. This is contained in On isolated rational singularities of surfaces by M. Artin. 
This paper also contains an example of an elliptic curve $Y$ with $Y^2=-1$ which is not contractible to an algebraic surface. The surface $X$ is the blowup of $\mathbb P^2$ at 10 sufficiently general points lying on a smooth cubic, $Y$ is the strict preimage of that cubic.
Finally, for an irreducible divisor $Y$ the resulting space $V$ is smooth iff $Y=\mathbb P^{n-1}$ and $N_{Y,X}=\mathcal O(-1)$. Indeed, $X\to V$ has to factor through the blowup of $V$ at a point by the universal property of the blowup. But then $X$ has to coincide with this blowup by Zariski main theorem. And on the blowup at a point the exceptional divisor is $\mathbb P^{n-1}$ with the normal bundle $\mathcal O(-1)$.
A: In the case of smooth surfaces, this is already a tough question : 
If E is a rational curve, then E can be contracted to a point in a smooth variety if  the self-intersection $E\cdotp E=-1$ (thm of Castelnuovo).
If you don't impose the smoothness of the variety, then a sufficient condition is $E\cdotp E<0$.
Moreover, there exists a theorem, due to Grauert, which states that if $\sum E_i$ is a divisor on a smooth surface such that the matrix of intersections $(E_i\cdotp E_j)$ is negative, then there exists $f:S \to S'$ birational such that the exceptional locus of $f$ is exactly $\cup E_i$. But this theorem is valid only in the analytic setting, which means that $S'$ is a priori not an algebraic variety.
Concerning the general question of contraction of curves in an algebraic variety, the main results come from Mori's theory (cf the book of Kollar and Mori, Birational geometry of algebraic varieties)
