I am trying to prove that a functor $\mathcal{F}:(Sch/S)^{op}\rightarrow (Sets)$ is representable by an $S$-scheme $F$. My intuition says that it is indeed representable. I have been reading Fulton's "Hurwitz schemes and irreducibility of moduli of algebraic curves", where he uses Corollary 2 of "Representation of unramified functors" by Murre to show that a functor $\mathcal{H}:(Sch/S)^{op}\rightarrow (Sets)$ is representable by a scheme $H$ which is étale over $P$.

However, I know that if $F$ exist it cannot be étale over $S$ ($S$ and the hypothetical $F$ would be the same dimension, but the fibers of the structure morphism would vary in dimension). My question is:

Is there some general criteria to show that such an $\mathcal{F}$ is representable by a scheme $F$ which is not necessarily étale over $S$? I found some criteria for $\mathcal{F}$ to be representable by an algebraic space, but I am looking for this stronger representability result.


  • $\begingroup$ There are some necessary conditions: for instance, it should be a Zariski (or, even better, fpqc) sheaf. $\endgroup$ – Zhen Lin Jan 22 '14 at 20:11
  • $\begingroup$ Right. That's encoded in some of Murre's criteria. He lists some necessary conditions, says that they are not sufficient, but then derives additional conditions (which end up being sufficient) when assuming $\mathcal{F}$ is representable by an unramified, and later étale, scheme over $S$. $\endgroup$ – user16544 Jan 22 '14 at 22:53
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    $\begingroup$ Perhaps you already know this, but one common technique is to start with a functor which you know is representable such that the functor you're interested in is a subfunctor of. Then one needs to impose extra conditions to "get down" to the subfunctor, and many times the conditions are open or closed (giving you open/closed subschemes of the parent functor). Often this ends up involving GIT and can get very technical. $\endgroup$ – RghtHndSd Jan 23 '14 at 13:19
  • $\begingroup$ That is of course a good technique, but unfortunately as far as I can tell there isn't any functor my $\mathcal{F}$ should be a subfunctor or quotient of. $\endgroup$ – user16544 Jan 23 '14 at 20:19
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    $\begingroup$ I'm afraid since the question is too general nothing more can be said except for the "trivial" criterion which says that $F$ is a sheaf which is covered by open representable subfunctors. $\endgroup$ – Martin Brandenburg May 26 '14 at 22:02

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