Timeline for Finite-type Artin Stack over $\mathbb C$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2014 at 4:39 | history | edited | Emerton |
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| Apr 4, 2013 at 21:56 | comment | added | user30379 | 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. | |
| Apr 4, 2013 at 21:27 | vote | accept | HNuer | ||
| Apr 4, 2013 at 21:03 | comment | added | HNuer | 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? | |
| Apr 4, 2013 at 20:51 | answer | added | Angelo | timeline score: 5 | |
| Apr 4, 2013 at 20:49 | comment | added | user30379 | 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). | |
| Apr 4, 2013 at 20:35 | history | asked | HNuer | CC BY-SA 3.0 |