The Stacks project

Lemma 70.10.9. Let $k$ be a field. Let $f : X \to Y$ be a monomorphism of algebraic spaces over $k$. If $Y$ is locally quasi-finite over $k$ so is $X$.

Proof. Assume $Y$ is locally quasi-finite over $k$. By Lemma 70.10.8 we see that $Y = \coprod \mathop{\mathrm{Spec}}(A_ i)$ where each $A_ i$ is an Artinian local ring finite over $k$. By Decent Spaces, Lemma 66.19.1 we see that $X$ is a scheme. Consider $X_ i = f^{-1}(\mathop{\mathrm{Spec}}(A_ i))$. Then $X_ i$ has either one or zero points. If $X_ i$ has zero points there is nothing to prove. If $X_ i$ has one point, then $X_ i = \mathop{\mathrm{Spec}}(B_ i)$ with $B_ i$ a zero dimensional local ring and $A_ i \to B_ i$ is an epimorphism of rings. In particular $A_ i/\mathfrak m_{A_ i} = B_ i/\mathfrak m_{A_ i}B_ i$ and we see that $A_ i \to B_ i$ is surjective by Nakayama's lemma, Algebra, Lemma 10.19.1 (because $\mathfrak m_{A_ i}$ is a nilpotent ideal!). Thus $B_ i$ is a finite local $k$-algebra, and we conclude by Lemma 70.10.8 that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06S1. Beware of the difference between the letter 'O' and the digit '0'.