The Stacks project

Lemma 56.7.4. Let $f : V \to X$ be a quasi-finite separated morphism of Noetherian schemes. If there exists a coherent $\mathcal{O}_ V$-module $\mathcal{K}$ whose support is $V$ such that $f_*\mathcal{K}$ is coherent and $R^ qf_*\mathcal{K} = 0$, then $f$ is finite.

Proof. By Zariski's main theorem we can find an open immersion $j : V \to Y$ over $X$ with $\pi : Y \to X$ finite, see More on Morphisms, Lemma 37.43.3. Since $\pi $ is affine the functor $\pi _*$ is exact and faithful on the category of coherent $\mathcal{O}_ X$-modules. Hence we see that $j_*\mathcal{K}$ is coherent and that $R^ qj_*\mathcal{K}$ is zero for $q > 0$. In other words, we reduce to the case discussed in the next paragraph.

Assume $f$ is an open immersion. We may replace $X$ by the scheme theoretic closure of $V$. Assume $X \setminus V$ is nonempty to get a contradiction. Choose a generic point $\xi \in X \setminus V$ of an irreducible component of $X \setminus V$. Looking at the situation after base change by $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }) \to X$ using flat base change and using Local Cohomology, Lemma 51.8.2 we reduce to the algebra problem discussed in the next paragraph.

Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $A$-module whose support is $\mathop{\mathrm{Spec}}(A)$. Then $H^ i_\mathfrak m(M) \not= 0$ for some $i$. This is true by Dualizing Complexes, Lemma 47.11.1 and the fact that $M$ is not zero hence has finite depth. $\square$

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