Take the 2-minute tour ×  In this MO question it is claimed that
a catchphrase for "algebraic spaces" could be that they are "the
result of looking at the orbit space of the action of a finite
group on a scheme".  Hence my question:  Given an algebraic
space $X$, is it true that there exists a scheme $S$ and an
action of a finite group $G$ on $S$ such that $X=S/G$ ?  If I
remember correctly, every algebraic space is the quotient of an
affine scheme by an étale equivalence relation, so I tend to
think that there could exist such equivalence relations that are
not "implemented" by a group action...  share|improve this
question Do you mean a group scheme which is of finite type over
$S$? –  Martin Brandenburg Jun 29 '11 at 20:10 I don't know :)
Probably, I mean that everything is -say- over $\mathbb{C}$ and
$G$, as a $\mathbb{C}$-scheme, is a finite discrete collection
of closed points. Anyway, I mean any situation that resembles
the setting in the question quoted above. –  Qfwfq Jun 29 '11 at
22:50 If X is a ge