So last time, we introduced the first form of the formal function theorem. We said that if {X } was a proper scheme over {\mathrm{Spec} A} with structure morphism {f}, and {\mathcal{I} = f^*(I)} for some ideal {I \subset A}, then there were two constructions one could do on a coherent sheaf {\mathcal{F}} on {X} that were in fact the same. Namely, we could complete the cohomology {H^n(X,  \mathcal{F})} with respect to {I}, and we could take the inverse limit {  \varprojlim H^n(X,  \mathcal{F}/\mathcal{I}^k \mathcal{F})}. The claim was that the natural map

\displaystyle  \widehat{H^n(X, \mathcal{F})} \rightarrow \varprojlim H^n(X,  \mathcal{F}/\mathcal{I}^k \mathcal{F})

was in fact an isomorphism. This is a very nontrivial statement, but in fact we saw yesterday that a reasonably straightforward proof could be given via diagram-chasing if one appeals to a strong form of the proper mapping theorem.

1. Formal functions, jazzed up

Now, however, we want to jazz this up a little. I won’t do this as much as possible because I don’t want to talk too much about formal schemes yet. On the other hand, I want to replace cohomology groups with higher direct images.

First, however, we need to introduce a notion. Let {Z' \subset  Z} be a closed subscheme of the scheme {Z}, defined by a sheaf of ideals {\mathcal{J}}. Given a sheaf {\mathcal{F}} on {Z}, we can define the completion along {Z'}, denoted {\hat{\mathcal{F}}}, to be the inverse limit {\varprojlim  \mathcal{F}/\mathcal{J}^k \mathcal{F}}; this is a sheaf supported on the underlying set of {Z'}. If one is working with noetherian schemes, this depends only on the underlying set of {Z'}, since any two such ideals will have comparable powers. Now we can state the general formal function theorem:

Theorem 1 Let {f: X \rightarrow Y} be a proper morphism of noetherian schemes. Let {Y' \subset  Y} be a closed subscheme and {X' = X \times_Y Y'} the pull-back. If {\mathcal{F}} is a coherent sheaf on {X}, then there is a canonical isomorphism\displaystyle  \widehat{R^n f_*(\mathcal{F})} \simeq \varprojlim R^n  f_*(\mathcal{F}/\mathcal{I}^k \mathcal{F}).

Now one wants to say further that this is isomorphic to {R^n  f_*(\mathcal{F})}. This is also true, but requires a close analysis of exactly how cohomology behaves with respect to projective limits. I don’t want to go there. The claim is that we’ve already proved this jazzed-up version of the formal function theorem. Indeed, to compute the higher direct images, one just takes ordinary cohomology over inverse images and sheafifies. So let {U = \mathrm{Spec} A \subset Y} be an affine over which {Y'} is given by an ideal {I}. Then we can evaluate both sides of the above equation at {U}. On the left, we get the {I}-adic completion

\displaystyle  \widehat{ H^n(f^{-1}(U), \mathcal{F})};

(the hat should be over the whole expression) on the right, we get

\displaystyle  \varprojlim  \Gamma(U, R^n f_*(\mathcal{F}/\mathcal{I}^k \mathcal{F})) =   \varprojlim H^n(f^{-1}(U), \mathcal{F}/\mathcal{I}^k \mathcal{F}).

Note that taking sections commutes with taking inverse limits of sheaves. We know that these two are canonically isomorphic by the formal function theorem though, as {f^{-1}(U) \rightarrow U} is proper. Now we just sheafify.

2. An example

As an example, let {f: X \rightarrow Y} be a proper morphism of noetherian schemes, and let {\mathcal{F}} be a coherent sheaf on {X}. Suppose furthermore that {y  \in Y}. We will show that there is a canonical isomorphism between

\displaystyle \widehat{ (R^n f_*( \mathcal{F}))_{y}} \simeq \varprojlim H^n(f^{-1}(y),  \mathcal{F}/\mathfrak{m}_y^k \mathcal{F}).

This is to be interpreted as follows. The first term is the completion of the stalk {(R^n f_*(\mathcal{F})_y} at the maximal ideal {\mathfrak{m}_y \subset  \mathcal{O}_y}. The second is the inverse limit of the cohomologies of the pull-backs of {\mathcal{F}} to “infinitesimal neighborhoods” of the fiber. This is, incidentally, the version in which Hartshorne states the formal function theorem.

We can prove this as follows. First, let us make the flat base change {\mathrm{Spec}  \mathcal{O}_y \rightarrow Y}. The formation of higher direct images commutes with flat base change for reasonable schemes, and consequently we can reduce to the case where {Y = \mathrm{Spec} \mathcal{O}_y}. As a result, {y} is a closed point.

In this case, the result is a direct restatement of the formal function theorem. The only thing to note is that {H^n(f^{-1}(y), \mathcal{F}/  \mathfrak{m}_y^k  \mathcal{F}) = H^n(X, \mathcal{F}/\mathfrak{m}_y^k  \mathcal{F})} because {\left\{y\right\}} is closed.

Corollary 2 Let {f: X \rightarrow Y} be a proper morphism whose fibers are of dimension {\leq r}. Then {R^n f_*(\mathcal{F})=0} for any coherent {\mathcal{F}} if {n>r}.

Indeed, if {\mathcal{F}, n} are as above, we know that

\displaystyle  H^n(f^{-1}(y), \mathcal{F}/\mathfrak{m}_y^k \mathcal{F}) = 0

for all {k} because sheaf cohomology is trivial in dimension higher than the combinatorial dimension, by a theorem of Grothendieck. The above version of the formal function theorem implies that the coherent sheaf the {R^n f_*(\mathcal{F})} has trivial stalks at each {y} (since completion is faithful for finite modules), and consequently is trivial.

Example 1 (Hartshorne exercise) Let {f: X \rightarrow Y} be a proper morphism of noetherian schemes. Let {\mathcal{F}} be a coherent sheaf on {X}, flat over {Y}. Let us suppose that {H^n(X_y,  \mathcal{F} \otimes k(y)) = 0}; that is, the cohomology is trivial over the fiber at {y}. We will show more generally that {R^n f_*(\mathcal{F})} is trivial in a neighborhood of {y}. To do this, it is sufficient to show that the completion of the stalk {(R^n f_*(\mathcal{F}))_y} vanishes. By the formal function theorem, it is sufficient to see that

\displaystyle  H^n(X_y, \mathcal{F} \otimes \mathcal{O}_y/\mathfrak{m}_y^k) = 0, \quad,  \forall k.

It will suffice by the long exact sequence and induction to show that

\displaystyle  H^n(X_y, \mathcal{F} \otimes \mathfrak{m}_y^k/\mathfrak{m}_y^{k+1}) = 0, \quad,  \forall k,

since {\mathcal{F}} is flat. But {\mathfrak{m}_y/\mathfrak{m}_y^{k+1}} is isomorphic to a direct sum of copies of {\mathcal{O}_y/\mathfrak{m}_y}, so we are done.

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