TITLE: Is a sub-stack of a scheme a scheme?
QUESTION [8 upvotes]: 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 $Y$?

REPLY [2 votes]: 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.