(Well, it looks like I should stop making promises on this blog. There hasn’t been a single post about spectra yet. I hope that will change before next semester.)

So, today I am going to talk about the formal function theorem. This is more or less a statement that the properties of taking completions and taking cohomologies are isomorphic for proper schemes. As we will see, it is the basic ingredient in the proof of the baby form of Zariski’s main theorem. In fact, this is a very important point: the formal function theorem allows one to make a comparison with the cohomology of a given sheaf over the entire space and its cohomology over an “infinitesimal neighborhood” of a given closed subset. Now localization always commutes with cohomology on non-pathological schemes. However, taking such “infinitesimal neighborhoods” is generally too fine a job for localization. This is why the formal function theorem is such a big deal.

I will give the argument following EGA III here, which is more general than that of Hartshorne (who only handles the case of a projective scheme). The form that I will state today is actually rather plain and down-to-earth. In fact, one can jazz it up a little by introducing formal schemes; perhaps this is worth discussion next time.

1. Motivation

Let ${X}$ be a noetherian scheme and ${Z \subset X}$ a closed subset, defined by a coherent sheaf of ideals ${\mathcal{I}}$. (So ${Z = \mathrm{supp}(\mathcal{O}_X/\mathcal{I}}$.) Given a coherent sheaf ${\mathcal{F}}$ on ${X}$, we can consider the sheaves

$\displaystyle \mathcal{F}_k = \mathcal{F}/\mathcal{I}^k \mathcal{F},$

each of which is supported on the closed set ${Z}$.

We can thus consider their cohomologies ${H^n(X, \mathcal{F}_k) = H^n(Z, \mathcal{F}_k)}$; these form an inverse system over ${k}$ because the ${\mathcal{F}_k}$ do.

The question arises: what might this inverse system look like when we take the projective limit over ${k}$? In particular, can we get it directly from ${\mathcal{F}}$? Well, first of all, there are clearly natural maps

$\displaystyle H^n(X, \mathcal{F}) \rightarrow H^n(X, \mathcal{F}_k)$

for each ${k}$, which commute with the maps of the projective system. There is thus a map

$\displaystyle H^n(X, \mathcal{F}) \rightarrow \varprojlim_k H^n(X, \mathcal{F}_k) .$

Now, we can’t expect this map to be an isomorphism. The ${\mathcal{F}_k}$ are all supported on ${Z}$, while ${\mathcal{F}}$ probably isn’t.

So let ${I \subset \Gamma(X, \mathcal{I})}$ be an ideal. We know that ${I \subset \Gamma(X, \mathcal{O}_X)}$ and ${I}$ acts by multiplication on the cohomology groups ${H^n(X, \mathcal{F})}$. Furthermore, the ${k}$th element in the ${I}$-adic filtration, that is ${I^k H^n(X, \mathcal{F})}$, maps to zero in ${H^n(X, \mathcal{F}_k)}$ because it factors

$\displaystyle I^k H^n(X, \mathcal{F}) \rightarrow H^n( X, \mathcal{I}^k \mathcal{F}) \stackrel{0}{\rightarrow} H^n(X, \mathcal{F}_k) .$

In particular, for each ${k}$, we have maps

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

which induces a map on the inverse limits

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

where the first is the completion with respect to the ${I}$-adic topology.

This seems much more reasonable because, intuitively, completing at the ${I}$-adic topology is like taking a very small neighborhood of the subset ${Z}$. So we ask, when is this an isomorphism?

Well, first of all, completion is only really well-behaved for finitely generated modules. So we should have some condition that the cohomology groups are finitely generated. This we can do if there is a noetherian ring ${A}$ and a morphism ${X \rightarrow \mathrm{Spec} A}$ which is proper. In this case, it is a—nontrivial—theorem that the cohomology groups of any coherent sheaf on ${X}$ are finitely generated ${A}$-modules.

In addition, the ideal ${I}$ should somehow determine the ideal ${\mathcal{I}}$, not the other way around, since many different ${\mathcal{I}}$‘s could hypothetically have the same ${I}$. We could take ${\mathcal{I} = f^*(I)}$. It turns out that this is what we need.

2. The formal function theorem

So, motivated by the previous section, we state: Let ${X}$ be a proper scheme over ${\mathrm{Spec} A}$, for ${A}$ noetherian. Let ${I \subset A}$ be an ideal whose pull-back ${\mathcal{I}}$ is a sheaf of ideals on ${X}$. Fix a coherent sheaf ${\mathcal{F}}$ on ${X}$ and define as before the sheaves ${\mathcal{F}_k = \mathcal{F}/\mathcal{I}^k \mathcal{F}}$.

Theorem 2 Hypotheses as above, the natural morphism$\displaystyle \widehat{H^n(X, \mathcal{F})} \rightarrow \varprojlim H^n(X, \mathcal{F}_k)$

is an isomorphism of ${A}$-modules.

I will try to explain the proof of this fact. The first step, as usual, is to draw a diagram. Let ${k \in \mathbb{Z}_{\geq 0}}$. Then we have an exact sequence of sheaves

$\displaystyle 0 \rightarrow \mathcal{I}^k \mathcal{F} \rightarrow \mathcal{F} \rightarrow \mathcal{F}_k \rightarrow 0,$

and in fact, for ${k \leq k'}$, we can draw a commutative diagram of exact sequences

From this, we can draw a diagram of long exact sequences in cohomology:

So this is a good thing. Indeed, we already know one of the maps: the second vertical map is just the identity. By a careful analysis of this diagram, we will be able to deduce the formal function theorem. The point is that we are going to take an inverse limit of this system of exact sequences.

The first thing we need to do is to make the sequence short exact. To do that, let’s introduce some notation. Since ${n}$ is fixed throughout this discussion, let’s just write ${H = H^n(X, \mathcal{F})}$. Since ${k}$ varies, let’s also write ${H_k = H^n(X, \mathcal{F}_k)}$ and ${R_k}$ for the image of ${H^n(X, \mathcal{I}^k\mathcal{F}) \rightarrow H^n(X, \mathcal{F})}$. And, finally, let’s write ${Q_k}$ for the image ${H^n(X, \mathcal{F}_k) \rightarrow H^{n+1}(X, \mathcal{I}^{k} \mathcal{F})}$; this is equivalently a kernel:

$\displaystyle Q_k = \ker( H^{n+1}(X, \mathcal{I}^k \mathcal{F}) \rightarrow H^{n+1}(X, \mathcal{F})).$

OK. Given this notation, it is clear that we have a family of semi-short exact sequences, which forms an inverse system:

$\displaystyle 0 \rightarrow R_k \rightarrow H \rightarrow H_k \rightarrow Q_k \rightarrow 0.$

The goal is to take the inverse limit of this and somehow get the formal function theorem.

We are going to show two things. One is that the ${R_k \subset H}$ form basically the ${I}$-adic filtration on ${H}$. Not quite, but the point is that they induce the ${I}$-adic topology. This is reasonable, since the ${R_k}$‘s are defined via ${H^n(X, \mathcal{I}^k \mathcal{F})}$ and this kind of looks like multiplication by ${\mathcal{I}^k}$, although it is in the wrong spot. Second, we are going to show that the morphisms between the ${Q_k}$‘s are eventually zero. After this, when we take the inverse limit of

$\displaystyle 0 \rightarrow H/R_k \rightarrow H_k \rightarrow Q_k \rightarrow 0$

it will follow (since the ${Q_k}$‘s are basically trivial) that ${\varprojlim H/R_k = \varprojlim H_k}$. Since the ${R_k}$ form the ${I}$-adic topology on ${H}$, we will have simply derived the formal function theorem.

2.1. Step one: the ${R_k}$

Let us start off by analyzing the ${R_k}$. We are going to show that the ${R_k}$ form a filtration on ${H}$ that is equivalent to the ${I}$-adic one. On the one hand, it is clear that ${I^k H \subset R_k}$ because if ${x \in I^k}$, the map

$\displaystyle x: H \rightarrow H$

factors as

$\displaystyle H^k(X, \mathcal{F}) \stackrel{x}{\rightarrow} H^k(X, \mathcal{I}^k \mathcal{F}) \rightarrow H^k(X, \mathcal{F}).$

So the ${R_k}$ are at least as large as the ${I}$-adic filtration. For the other inclusion, namely that the ${R_k}$ are small, we shall need to invoke a fairly big result.

Theorem 3 Let ${f: X \rightarrow Y}$ be a morphism of proper noetherian schemes. Then if ${\mathcal{F}}$ is a coherent sheaf on ${X}$, the higher direct images ${R^i f_*(\mathcal{F})}$ are coherent on ${Y}$.

So this is basically a relative and scheme-theoretic version of the fact that on a complete (e.g. projective) variety over a field, the cohomology groups are always finite-dimensional. I don’t want to prove it here, and instead refer you to EGA III.3 for the (rather beautiful) argument. Actually, however, we need a stronger variant of the above direct image theorem:

Theorem 4 Let ${f: X \rightarrow Y}$ be a morphism of proper noetherian schemes and let ${\mathcal{S}}$ be a quasi-coherent, finitely generated algebra over ${\mathcal{O}_Y}$. Then if ${\mathcal{F}}$ is a quasi-coherent sheaf on ${X}$ which is a finitely generated ${f^*(\mathcal{S})}$-module, the higher direct images ${R^i f_*(\mathcal{F})}$ are finitely generated ${\mathcal{S}}$-modules.

This is a strengthening of the usual proper mapping theorem, but it is in fact not terribly difficult to deduce from it by simply forming the relative ${\mathrm{Spec}}$ of ${\mathcal{S}}$ and invoking the usual one. Nonetheless, I won’t repeat the details in EGA III.3.

So now we are going to consider the case of interest in the formal function theorem. Namely, we have ${f: X \rightarrow \mathrm{Spec} A}$ a proper morphism, ${I \subset A}$ an ideal, ${f^*(I) = \mathcal{I}}$ an ideal of ${\mathcal{O}_X}$, and ${\mathcal{F}}$ a coherent sheaf on ${X}$. Now we can consider the blowup algebra ${A \oplus I \oplus I^2 \oplus \dots}$, which is a finitely generated ${A}$-algebra; it can be given a sheafish version ${\mathcal{S}}$ on ${\mathrm{Spec} A}$ and pulled back to ${X}$, when it is the ${\mathcal{O}_X}$-algebra

$\displaystyle f^*(\mathcal{S})= \mathcal{O}_X \oplus \mathcal{I} \oplus \mathcal{I}^2 \oplus \dots.$

Now the quasi-coherent sheaf

$\displaystyle \mathcal{F} \oplus \mathcal{I}\mathcal{F} \oplus \dots$

is a finitely generated (sheaf of) module(s) over ${f^*(\mathcal{S})}$. It follows that the cohomology

$\displaystyle H^n(X,\mathcal{F}) \oplus H^n(X, \mathcal{I}\mathcal{F}) \oplus \dots$

is a finitely generated module over the blowup algebra ${A \oplus I \oplus I^2 \oplus \dots}$, by the second coherence theorem. By finite generation, it follows that

$\displaystyle I H^n(X, \mathcal{I}^k \mathcal{F}) = H^n(X, \mathcal{I}^{k+1}\mathcal{F})$

for large ${k}$. The images of these are the ${R_k}$, so it is now clear that ${I R_k = R_{k+1}}$ for ${k}$ large. This implies that the ${\left\{R_k\right\}}$ form a filtration equivalent to the ${I}$-adic one. This is precisely what we wanted to see.

2.2. Step two: The ${Q_k}$

So one step is done. We have analyzed the ${R_k}$. Now, we need to analyze the ${Q_k}$. Namely, we need to show that the maps ${Q_{k'} \rightarrow Q_k}$ between them are basically zero, for ${k' \gg k}$. Recall the definition

$\displaystyle Q_k = \ker( H^{n+1}(X, \mathcal{I}^{k}\mathcal{F})) \rightarrow H^{n+1}(X, \mathcal{F}))$

or

$\displaystyle Q_k = \mathrm{Im}( H^n(X, \mathcal{F}_k) \rightarrow H^{n+1}(X, \mathcal{I}^k \mathcal{F})).$

We will play both off against each other together with the proper coherence theorem. So the first thing to do is to note that since everything here is an ${A}$-module, and by the first definition there are canonical maps ${Q_{k} \times I \rightarrow Q_{k+1}}$ given by multiplication, we find that

$\displaystyle Q = \bigoplus Q_k$

is a module over the blowup algebra ${A \oplus I \oplus \dots}$, as before. In fact, we find that ${Q}$ is a submodule of ${\bigoplus H^n(X, \mathcal{I}^{k+1} \mathcal{F})}$.

Since we know that the module ${\bigoplus H^n(X, \mathcal{I}^{k+1} \mathcal{F})}$ is finitely generated by the (second) coherence theorem, we find that ${Q}$ is finitely generated over the blowup algebra. However, it is also true, from the second definition we may see this, that ${Q_k}$ is as an ${A}$-module annihilated by ${I^k}$, because ${\mathcal{F}_k}$ is. It follows that there is a high power ${N}$ such that the ideal ${I^N \oplus I^{N+1} \oplus \dots}$ in the blowup algebra annihilates the entire module ${Q}$.

Now the map ${Q_{k+r} \rightarrow Q_k}$ is induced by the usual map ${H^n(X, \mathcal{I}^{k+r} \mathcal{F}) \rightarrow H^n(X, \mathcal{I}^k \mathcal{F})}$, We have a sequence

$\displaystyle Q_{k} \otimes I^r \rightarrow Q_{k+r} \rightarrow Q_k$

where the first comes from the blowup multiplication and the second is the canonical map. The first map is surjective for ${r}$ by finite generation over the blowup algebra, and the composite is zero for ${r \geq N}$ by the above reasoning. It follows that

$\displaystyle Q_{k+r} \rightarrow Q_k$

is zero if ${r \gg 0}$.

2.3. Completion of the proof

So we have established two things. One, that the ${R_k}$ form basically an ${I}$-adic filtration on ${H = H^n(X, \mathcal{F})}$. Second, that the maps ${Q_{k'} \rightarrow Q_k}$ are zero for ${k' \gg k}$, and in particular, the inverse limit ${\varprojlim Q_k}$ is zero. Now we have an exact sequence for each ${k}$,

$\displaystyle 0 \rightarrow H/R_k \rightarrow H_k \rightarrow Q_k \rightarrow 0,$

and we can take the inverse limit, which is a left-exact functor. We find an exact sequence

$\displaystyle 0 \rightarrow \hat{H} \rightarrow \varprojlim H_k \rightarrow 0,$

which is precisely what the formal function theorem states.