Artin has a theorem (10.1 in Laumon, MoretBailly) that if $X$ is a stack which has separated, quasicompact, representable diagonal and an fppf cover by a scheme, then $X$ is algebraic. Is there a version of this theorem that holds for fpqc covers?

4$\begingroup$ Any algebraic space which is Zariskilocally quasiseparated, e.g. any scheme!, is automatically an fpqc sheaf, by Lemma of tag 023Q in de Jong's project; at least for those algebraic spaces it should be that the theorem holds with "fpqc" instead of "fppf". $\endgroup$– Thanos D. PapaïoannouCommented Feb 12, 2010 at 18:35
1 Answer
It is false. I'm not sure what the comment about algebraic spaces has to do with the question, since algebraic spaces do admit an fpqc (even \'etale) cover by a scheme. This is analogous to the fact that the failure of smoothness for automorphism schemes of geometric points is not an obstruction to being an Artin stack. For example, $B\mu_n$ is an Artin stack over $\mathbf{Z}$ even though $\mu_n$ is not smooth over $\mathbf{Z}$ when $n > 1$. Undeterred by this, we'll make a counterexample using $BG$ for an affine group scheme that is fpqc but not fppf.
First, we set up the framework for the counterexample in some generality before we make a specific counterexample. Let $S$ be a scheme and $G \rightarrow S$ a $S$group whose structural morphism is affine. (For example, if $S = {\rm{Spec}}(k)$ for a field $k$ then $G$ is just an affine $k$group scheme.) If $X$ is any $G$torsor for the fpqc topology over an $S$scheme $T$ then the structural map $X \rightarrow T$ is affine (since it becomes so over a cover of $T$ that splits the torsor). Hence, the fibered category $BG$ of $G$torsors for the fpqc topology (on the category of schemes over $S$) satisfies effective descent for the fpqc topology, due to the affineness requirement.
The diagonal $BG \rightarrow BG \times_S BG$ is represented by affine morphisms since for any pair of $G$torsors $X$ and $Y$ (for the fpqc topology) over an $S$scheme $T$, the functor ${\rm{Isom}}(X,Y)$ on $T$schemes is represented by a scheme affine over $T$. Indeed, this functor is visibly an fpqc sheaf, so to check the claim we can work locally and thereby reduced to the case $X = Y = G_T$ which is clear.
Now impose the assumption (not yet used above) that $G \rightarrow S$ is fpqc. In this case I claim that the map $S \rightarrow BG$ corresponding to the trivial torsor is an fpqc cover. For any $S$scheme $T$ and $G$torsor $X$ over $T$ for the fpqc topology, the functor $$S \times_{BG} T = {\rm{Isom}}(G_T,X)$$ on $T$schemes is not only represented by a scheme affine over $T$ (namely, $X$) but actually one that is an fpqc cover of $T$. Indeed, to check this we can work locally over $T$, so passing to a cover that splits the torsor reduces us to the case of the trivial $G$torsor over the base (still denoted $T$), for which the representing object is $G_T$.
So far so good: such examples satisfy all of the hypotheses, and we just have to prove in some example that it violates the conclusion, which is to say that it does not admit a smooth cover by a scheme. Take $S = {\rm{Spec}}(k)$ for a field $k$, and let $k_s/k$ be a separable closure and $\Gamma = {\rm{Gal}}(k_s/k)^{\rm{opp}}$. (The "opposite" is due to my implicit convention to use left torsors on the geometric side.) Let $G$ be the affine $k$group that "corresponds" to the profinite group $\Gamma$ (i.e., it is the inverse limit of the finite constant $k$groups $\Gamma/N$ for open normal $N$ in $\Gamma$). To get a handle on $G$torsors, the key point is to give a more concrete description of the ``points'' of $BG$.
Claim: If $A$ is a $k$algebra and $B$ is an $A$algebra, then to give a $G$torsor structure to ${\rm{Spec}}(B)$ over ${\rm{Spec}}(A)$ is the same as to give a right $\Gamma$action on the $A$algebra $B$ that is continuous for the discrete topology such that for each open normal subgroup $N \subseteq \Gamma$ the $A$subalgebra $B^N$ is a right $\Gamma/N$torsor (for the fpqc topology, and then equivalently the \'etale topology).
Proof: Descent theory and a calculation for the trivial torsor. QED Claim
Example: $A = k$, $B = k_s$, and the usual (right) action by $\Gamma$.
Corollary: If $A$ is a strictly henselian local ring then every $G$torsor over $A$ for the fpqc topology is trivial.
Proof: Let ${\rm{Spec}}(B)$ be such a torsor. By the Claim, for each open normal subgroup $N$ in $\Gamma$, $B^N$ is a $\Gamma/N$torsor over $A$. Since $A$ is strictly henselian, this latter torsor is trivial for each $N$. That is, there is a $\Gamma/N$invariant section $B^N \rightarrow A$. The nonempty set of these is finite for each $N$, so by set theory nonsense with inverse limits of finite sets (ultimately not so fancy if we take $k$ for which there are only countably many open subgroups of $\Gamma$) we get a $\Gamma$invariant section $B \rightarrow A$. QED Corollary
Now suppose there is a smooth cover $Y \rightarrow BG$ by a scheme. In particular, $Y$ is nonempty, so we may choose an open affine $U$ in $Y$. I claim that $U \rightarrow BG$ is also surjective. To see this, pick any $y \in U$ and consider the resulting composite map $${\rm{Spec}} \mathcal{O}_{Y,y}^{\rm{sh}} \rightarrow BG$$ over $k$. By the Corollary, this corresponds to a trivial $G$torsor, so it factors through the canonical map ${\rm{Spec}}(k) \rightarrow BG$ corresponding to the trivial $G$torsor. This latter map is surjective, so the assertion follows. Hence, we may replace $Y$ with $U$ to arrange that $Y$ is affine. (All we just showed is that $BG$ is a quasicompact Artin stack, if it is an Artin stack at all, hardly a surprise in view of the (fpqc!) cover by ${\rm{Spec}}(k)$.)
OK, so with a smooth cover $Y \rightarrow BG$ by an affine scheme, the fiber product $$Y' = {\rm{Spec}}(k) \times_{BG} Y$$ (using the canonical covering map for the first factor) is an affine scheme since we saw that $BG$ has affine diagonal. Let $A$ and $B$ be the respective coordinate rings of $Y$ and $Y'$, so by the Claim there is a natural $\Gamma$action on $B$ over $A$ such that the $A$subalgebras $B^N$ for open normal subgroups $N \subseteq \Gamma$ exhaust $B$ and each $B^N$ is a $\Gamma/N$torsor over $A$. But $Y' \rightarrow {\rm{Spec}}(k)$ is smooth, and in particular locally of finite type, so $B$ is finitely generated as a $k$algebra. Since the $B^N$'s are $k$subalgebras of $B$ which exhaust it, we conclude that $B = B^N$ for sufficiently small $N$. This forces such $N$ to equal $\Gamma$, which is to say that $\Gamma$ is finite.
Thus, any $k$ with infinite Galois group does the job. (In other words, if $k$ is neither separably closed nor real closed, then $BG$ is a counterexample.)

$\begingroup$ Is there an example if X is assumed to be locally of finite presentation (commutes with filtered colimits of rings)? Also, should the second sentence of paragraph 4 should read S > BG instead of G > BG? $\endgroup$– BhargavCommented Feb 14, 2010 at 18:19

3$\begingroup$ My comment clarified an earlier comment by user fpqc—also known as Harry Gindi—about the kind of descent that algebraic spaces satisfied automatically; but that comment is now deleted, so mine stands incongruous. $\endgroup$ Commented Feb 14, 2010 at 19:53

2$\begingroup$ Is at least this variant of Artin's theorem true? Let $X$ be a quasiseparated étale sheaf which admits a surjective fpqc morphism from a scheme; then $X$ is an algebraic space, i.e. $X$ admits a surjective étale cover by a scheme. $\endgroup$ Commented Feb 14, 2010 at 20:10