Given a moduli problem, it appears that nonexistence of automorphisms is a necessary condition for existence of a fine moduli space(is this strictly true?).
In any case, assuming the above, what additional condition on a moduli problem in algebraic geometry will make sure that a coarse moduli space is in fact a fine moduli space?
In the n-lab page on Deligne-Mumford, the following appears.
Deligne-Mumford stacks correspond to moduli problems in which the objects being parametrized have finite automorphism groups.
Also, a few problematic examples I have heard of, had infinite automorphism groups.
Therefore, is it true that for a moduli problem in which the stack is Deligne-Mumford, and where there are no automorphisms, existence of a coarse moduli space would imply the existence of a fine moduli space?
Best Answer
I think this is an instructive question. Here are some partial answers.
A category fibered in groupoids whose fibers are sets (e.g. no automorphisms) is a presheaf. Strictly speaking, I mean equivalent to the fibered category associated to a presheaf. This truly follows from the definitions, and is a good exercise to do when one is learning the basic machinery behind stacks. Similarly, a stack which is equivalent to a presheaf is a sheaf (i.e. the descent condition collapses to the sheaf condition; this is a little harder but still tautological). And a pre-stack with no automorphisms is a separated pre-sheaf.
The other claim is more interesting and not tautological, and reflects the fact that the diagonal of a morphism of stacks is way more interesting than in the case of schemes. For instance, an algebraic stack is DM iff its diagonal is unramified. (I believe this is in Champs Algebriques, but there's a nice discussion in Anton's notes from Martin Olsson's stacks course.
The point is that the diagonal of a stack carries information about automorphisms of the objects it parameterizes. So for instance the condition (in the definition of a stack) that the diagonal is representable is equivalent to the statement that Isom(X,Y) (and thus Aut(X)) is representable by an algebraic space. Also tautological is the statement that the diagonal being unramified is equivalent to the statement that there are no infintesmal automorphisms (e.g. non-trivial automorphisms of an object over $k[\epsilon]/\epsilon^2$ which reduce to the identity map over $k$). So here it is now clear where one uses finiteness: the $k[\epsilon]/\epsilon^2$ points of a finite groups scheme are the same as the $k$ points; on the other hand this fails if the automorphism scheme is, say, $\mathbf{G}_m$.
Finally, while the tautological answer above answers your question, it is instructive to see how automorphisms cause $M_g$ (moduli of genus g curves) to not be representable. Let H be a hyperelliptic curve given by $y^2 = f(x)$ defined over a field $k$. Then the curve $H_d$ given by $dy^2 = f(x)$ is not isomorphic to $H$ over $k$ if d is not a square in k. Call this a `twist' of H; in general twists of a variety X are given by the Galois cohomology group $H^1(G_k,Aut X)$ which is non-zero in non-trivial situations (alternatively you can use torsors and etale cohomlogy), and a generic hyperelliptic curve has automorphism group $\{\pm 1\}$. Thus H and $H_d$ give two different $k$ points of $M_g$ which become the same point over a finite extension; thus $M_g$ fails the sheaf condition in the etale topology, (and so in general existence of automorphisms causes, for cohomological reasons, failure of your moduli problem to even be a sheaf).
Last comment (to clarify others' comments): fine moduli space should certainly allow algebraic spaces for a correct answer to your question.