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Suppose I have an Artin stack $\mathfrak M$ locally of finite-type over $\mathbb C$ with presentation $M\rightarrow \mathfrak M$. Suppose further that $\mathfrak M$ "represents" (in the stack sense) some set of objects which are bounded, i.e. roughly there is a family of such objects $\mathcal F$ over a scheme $S$ of finite type over $\mathbb C$ such that every element of our set of objects occurs as $\mathcal F_s$ for some closed points $s\in S$. Why does it follow that $\mathfrak M$ is actually of finite-type over $\mathbb C$?

Specifically, I've seen it argued that $S$ induces a surjection $S\rightarrow M$ of schemes over $\mathbb C$, which thus makes $M$ of finite-type over $\mathbb C$. But I don't see why we get an induced map to $M$ instead of just to $\mathfrak M$. Indeed the family $\mathcal F$ should induce a surjective morphism $S\rightarrow \mathfrak M$, but I don't see why this should factor through $M$.

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  • $\begingroup$ There is generally no such factorization (think about the case when $\mathfrak{M}$ is a scheme and $S = \mathfrak{M}$!). The argument you've seen should have the source replaced by $S \times_{\mathfrak{M}} M$ (which is, strictly speaking, an algebraic space and not necessarily a scheme, but nonetheless is of finite type over $M$ and has the "expected" set of $\mathbf{C}$-points, so you're good to go). $\endgroup$
    – user30379
    Commented Apr 4, 2013 at 20:49
  • $\begingroup$ Why is the fiber product of finite type over $M$? Are you claiming that since $S$ is of finite-type over $\mathbb C$, it must then also be of finite-type over $\mathfrak M$, and then base-change? $\endgroup$
    – HNuer
    Commented Apr 4, 2013 at 21:03
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    $\begingroup$ Any map $f:X \rightarrow Y$ between quasi-separated algebraic spaces locally of finite type over an affine scheme $B$ with $X$ of finite type over $B$ is itself of finite type. Indeed, $Y$ is covered by q-c opens $U_i$, preimages of which are open in $X$, so finitely many preimages cover $X$. Hence, $f$ lands inside a quasi-compact open $U \subset Y$, and $U \rightarrow B$ is finite type (being q-c and lft), so $X \rightarrow U$ is finite type. Thus, we want $U \hookrightarrow Y$ to be finite type. For $V \rightarrow Y$ with $V$ affine, $U \times_Y V$ is q-c since $Y$ is quasi-separated. $\endgroup$
    – user30379
    Commented Apr 4, 2013 at 21:56

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Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times_{\frak M} M$ surjecting onto $S$. The image of $U$ in $M$ is quasi-compact; let $V$ be a quasi-compact open subscheme of $M$ containing it. Since $S$ surjects onto $\frak M$, so does $U$. So $V \to \frak M$ is smooth and surjective, and $V$ is of finite type, so $\frak M$ is of finite type.

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  • $\begingroup$ How do you see the existence of the open subset $U$ of $M$ with the properties you claim? $\endgroup$
    – HNuer
    Commented Apr 4, 2013 at 20:58
  • $\begingroup$ The map from the fiber product onto $S$ is smooth, hence open $\endgroup$
    – Angelo
    Commented Apr 4, 2013 at 21:04
  • $\begingroup$ So is your $U$ an open of $M$ or $S\times_{\mathfrak M} M$? $\endgroup$
    – HNuer
    Commented Apr 4, 2013 at 21:11
  • $\begingroup$ Yes, you are right, my explanation is incorrect. I edited the answer. $\endgroup$
    – Angelo
    Commented Apr 4, 2013 at 21:19
  • $\begingroup$ Got it, nice slick argument. Thanks $\endgroup$
    – HNuer
    Commented Apr 4, 2013 at 21:27

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