TITLE: Quotients of Schemes by Free Group Actions
QUESTION [26 upvotes]: I've often seen people in seminars justify the existence of a quotient of a scheme by an algebraic group by remarking that the group action is free. However, I'm pretty sure they are also invoking something else. So my question is: when you can you quotient a scheme by a free action and get a scheme? In particular, when do the coset spaces of a subgroup of an algebraic group exist as a scheme? And in these cases, how do you construct the quotient?

REPLY [17 votes]: In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, proposition 1.8.
In particular, if $X$ is a closed subscheme of projective space, then for any finite set of  points of $X$ (in particular, any orbit of a single point) we can find a hyperplane not containing any of those points.  The complement will be an affine open of $X$ containing the orbit.  This covers the existence of the quotient in the cases that $X$ is projective (and by a similar argument, quasi-projective).   
Here is a sketch of the argument in SGA I:
If the condition is satisfied, one first shows that this implies that $X$ is covered by $G$-invariant open affines, then by taking invariants, constructs the quotient of each affine by $G$, and then glues together to get the global quotient $X/G$. 
Conversely, if $X/G$ is a scheme then for any open affine $V \subseteq X/G$ the inverse image of $V$ under the morphism $X\longrightarrow X/G$ is an open affine of $X$ stable under $G$.  Since we can cover $X$ by such opens, the orbit of any point of $X$ is contained in some affine open set.
As in George McNinch's answer there is no need for the $G$-action to be free.  (This discussion was limited to quotients by finite groups though!)