Lemma 30.20.9. Let $f : X \to Y$ be a morphism of schemes. Let $y \in Y$. Assume

1. $Y$ locally Noetherian,

2. $f$ is proper, and

3. $\dim (X_ y) = d$.

Then for any coherent sheaf $\mathcal{F}$ on $X$ we have $(R^ pf_*\mathcal{F})_ y = 0$ for all $p > d$.

Proof. The fibre $X_ y$ is of finite type over $\mathop{\mathrm{Spec}}(\kappa (y))$. Hence $X_ y$ is a Noetherian scheme by Morphisms, Lemma 29.15.6. Hence the underlying topological space of $X_ y$ is Noetherian, see Properties, Lemma 28.5.5. Moreover, the underlying topological space of each infinitesimal neighbourhood $X_ n$ is the same as that of $X_ y$. Hence $H^ p(X_ n, \mathcal{F}_ n) = 0$ for all $p > d$ by Cohomology, Proposition 20.20.7. Hence we see that $(R^ pf_*\mathcal{F})_ y^\wedge = 0$ by Lemma 30.20.7 for $p > d$. Note that $R^ pf_*\mathcal{F}$ is coherent by Proposition 30.19.1 and hence $R^ pf_*\mathcal{F}_ y$ is a finite $\mathcal{O}_{Y, y}$-module. By Nakayama's lemma (Algebra, Lemma 10.20.1) if the completion of a finite module over a local ring is zero, then the module is zero. Whence $(R^ pf_*\mathcal{F})_ y = 0$. $\square$

Comment #7107 by Laurent Moret-Bailly on

Is there a reason for stating the result in terms of stalks, rather than concluding that $R^pf_*\mathcal{F}$ is zero in a neighbourhood of $y$? (The neighbourhood can be taken independently of $\mathcal{F}$ but I realize this has to wait for the semicontinuity property). The same comment applies to the algebraic space analogue 0A4T, except that (maybe) we need some "decent" assumption on the base.

Comment #7270 by on

Yes, this is one of those inumerable things that one could change. In this case however, the output of the argument is really that the stalk is zero! So I could claim that this is sort of the "correct" statement.

Comment #7273 by Laurent Moret-Bailly on

Dear Johan, you certainly have a point here. On the other hand, along with the "correct" statement there is the "most usfeul" one; my position would be to state both. (The situation is a bit similar to my comment #5007).

Comment #7333 by on

Yes, stating both would probably be optimal, but I will leave that for another day. Thanks for the comments!

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