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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