Let $S$ be a scheme. Let $\mathcal X$ be a algebraic $S$-stack and be $Y$ a $S$-scheme. Let $f:\mathcal X\longrightarrow Y$ be a $S$-morphism of algebraic stacks which is an open embedding (resp. a closed embedding). Is $\mathcal X$ automatically a open(resp. closed) subscheme of $Y$?
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5$\begingroup$ Note that $S$ has nothing to do with anything here; may as well take $S = {\rm{Spec}}(\mathbf{Z})$. (An algebraic $S$-stack is nothing more or less than an algebraic $\mathbf{Z}$-stack equipped with a morphism to $S$.) Please see my comment to Thanos for a sketch of what to do, and for the omitted details I offer you the same advice as elsewhere (as befits your pseudonym). $\endgroup$– BCnrdCommented Aug 31, 2010 at 23:46
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Let $f:\mathcal{X}\rightarrow Y$ be a morphism from an Artin stack to a scheme such that $f$ is an immersion. Then $\mathcal{X}$ is automatically an algebraic space, so we're done by Knutson, Algebraic spaces, II.6.16.
Additions prompted by Brian's comment
Assume that $f:\mathcal{X}\rightarrow Y$ is a schematic map, and that $Y$ is a scheme; then $f$ is the pullback of $f$ over the map of schemes $\mathrm{id}_Y$, so $\mathcal{X}$ must be a scheme. Knutson needs lemma II.6.16 because he doesn't use the now-standard definition of schematic, but atlases instead.
When using immersion, I always mean $j\circ i$, where $i$ is a closed immersion and $j$ an open one, following EGA I. But I understand that this is not a better choice than the other way round, and that they are only equivalent when the morphism is quasicompact.
answered Aug 31, 2010 at 22:56
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2$\begingroup$ Thanos, don't use the word "immersion" in general without quasi-compactness qualifications if not assuming it to be an open or closed immersion (unclear if general non-qc subschemes satisfy fpqc or even etale descent; do you know a proof?). Let's stick to open or closed or q-c immersions. Then your method is killing a fly with a sledgehammer. Lean back in the chair and ask yourself: what is a (good) definition of that concept for a morphism of stacks? Upon jogging the memory cells, you'll see that the question posed becomes a tautology via descent theory for such morphisms of schemes. QED $\endgroup$– BCnrdCommented Aug 31, 2010 at 23:34
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1$\begingroup$ Dear Thanos: I wasn't quibbling about def'n of immersion, but rather noting that whatever def'n you wish to use for schemes without q-c assumptions on the morphism (when not open immersion) doesn't seem to even make sense for stacks in a useful way. Can you prove the concept is fppf-local or etale-local on the base in the case of schemes? If not then you cannot make good sense of such a concept for morphisms of stacks; the concept of "schematic" morphism of stacks is unworkable when not applied to a class of morphisms satisfying effective descent for schemes (for suitable topologies). $\endgroup$– BCnrdCommented Sep 1, 2010 at 0:42
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2$\begingroup$ Dear Thanos: For (q-sep'td) alg. spaces there's effective descent for fppf topology (Artin). That's why for stacks it's practical to ask if morphism is "rel. rep'tble in alg. spaces" (rather than in schemes). Concept of (non-qc) immersion can certainly be defined for stacks (anyone can make up def'ns...), but is impractical even for alg. spaces since not etale-local on the base for schemes. The defn is Zar-local on target always, but "good" concepts for morphisms of alg. spaces (resp. stacks) should be etale-local (resp. fppf-local) on the base, not just Zar-local: otherwise hard to check. $\endgroup$– BCnrdCommented Sep 1, 2010 at 1:00