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.
$\square$

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