## Serre’s affineness criterion

Since we’ve been discussing sheaf cohomology for the last few weeks of the algebraic geometry seminar, and I’m leaving Waterloo soon, I was thinking about possible topics for what will probably be the last seminar talk. I figured that having drudged through all this machinery, it would be nice to look at a cohomological characterisation of affine schemes: namely, the fact that a scheme $(X, \mathcal{O}_X)$ is affine if and only if all quasi-coherent sheaves $\mathscr{F}$ of $\mathcal{O}_X$-modules are acyclic, i.e. $H^i(X, \mathscr{F}) = 0$ for $i>0$. In this post I’ll go over the treatment in [Hartshorne III.3, “Cohomology of a Noetherian Affine Scheme”]. I will probably explain all this stuff more coherently in a video sometime down the road.

This is called Serre’s affineness criterion, and the key to the proof (or at least one direction of it) lies in the fact that if you start with an injective $A$-module $I$, and consider its associated sheaf of $\mathcal{O}_{\text{Spec } A}$-modules $\widetilde{I}$ (just defined by $X_f \mapsto I_f$), then in fact this is flasque. We saw before that flasque sheaves are acyclic for the global sections functor $\Gamma(X,-)$, so in particular we can use flasque resolutions to compute cohomology (this will be important later).

We also saw that injective sheaves are flasque, so one might be tempted to claim that the “key” we mentioned above is a mere triviality: indeed, why not just observe that (in view of the equivalence of categories) any injective $A$-module will give rise to an injective sheaf, and then finish? The problem with this argument is that the category of $A$-modules is equivalent to the category of quasicoherent sheaves, and not the full category of $\mathcal{O}_X$-modules. So yes, we will always have an injective of the former category, but we would need an injective of the latter category to conclude flasqueness — and in general this does not happen.

The starting point is a theorem of Krull from commutative algebra. The full statement concerns the $\mathfrak{a}$-adic topology on an $A$-module, and I don’t really know much (nor do I currently have time to read Atiyah-Macdonald) about completions. However, we only really need one containment:

Krull’s Theorem. Let $A$ be a Noetherian ring and $\mathfrak{a} \subseteq A$ be an ideal. If $M \subseteq N$ are finitely generated $A$-modules, then for any $n > 0$ there is $n' \geq n$ such that $\mathfrak{a}^n M \supseteq M \cap \mathfrak{a}^{n'} N$.

Now, define the following submodule of $M$:

$\Gamma_{\mathfrak{a}}(M) = \{ m \in M \mid \mathfrak{a}^n m = 0 \text{ for some } n > 0 \}.$

Before proceeding, let us mention a remark about injectives. We said an object $I$ of an abelian category $\mathfrak{A}$ was injective if the functor $\mathrm{Hom}(-,I)$ is exact. This (contravariant) functor is always left exact, so the important thing to take away is the following: “$I$ injective” means that if $M' \subseteq M$ is a submodule and $\varphi : M' \to I$ is a morphism, then $\varphi$ extends to a morphism $\widetilde{\varphi}: M \to I$.

Surprisingly, the above turns out to be equivalent to the following seemingly weaker condition (Baer’s criterion), namely: if $\mathfrak{b}$ is an ideal of $A$ and $\varphi : \mathfrak{b} \to I$ is a morphism, then $\varphi$ extends to a morphism $\widetilde{\varphi} : A \to I$. This equivalence is a basic result from commutative algebra.

This reminds me of a similar thing that came up when trying to formulate the universal property of the Stone-Cech compactification: in some sense the closed unit interval $[0,1]$ is a “good enough” representative of the class of all compact Hausdorff spaces (this is formalised in the fascinating notion of an injective cogenerator).

Lemma 1. Let $A$ be a Noetherian ring, let $\mathfrak{a} \subseteq A$ be an ideal. Then if $I$ is an injective $A$-module, then $J = \Gamma_{\mathfrak{a}}(I)$ is also an injective $A$-module.

To prove this, we only need to establish Baer’s criterion for $J$, and this is done by observing one can apply Krull’s theorem to the inclusion $\mathfrak{b} \subseteq A$, pulling back from $\mathfrak{b}/(\mathfrak{b} \cap \mathfrak{a}^{n'})$ to $A/\mathfrak{a}^{n'}$, and finally using the natural map $A \to A/\mathfrak{a}^{n'}$ to pull back to $A$ as required.

Lemma 2. Let $A$ be a Noetherian ring, and $I$ an injective $A$-module. Then for any $f \in A$, the natural map $I \to I_f$ to the localisation is surjective.

This lemma isn’t very difficult either. If $\mathfrak{b}_i$ is defined as the annihilator of $f^i$, then you get some ascending chain of ideals in $A$, but $A$ is Noetherian, so $\mathfrak{b}_r = \mathfrak{b}_{r+1} = \ldots$, yada yada. Then, letting $\theta : I \to I_f$ be the natural map, you take some $x \in I_f$, write $= \theta(y)/f^n$ for some $y \in I$ and $n \geq 0$ (you can do this by definition of localisation), and define a map $\varphi : (f^{n+r}) \to I$ by sending $f^{n+r} \mapsto f^r y$ (this turns out to be fine since $(f^{n+r}) \cong A/{\text{Ann } (f^{n+r})}$ as $A$-modules, and $\text{Ann } f^{n+r} = \mathfrak{b}_{n+r} = \mathfrak{b}_r$). Lift $\varphi$ to a map $\psi : A \to I$ by injectivity of $I$, and then let $z = \psi(1)$. Then $\theta(z) = x$. Magic.

Proposition. If $I$ is an injective $A$-module, then $\widetilde{I}$ is a flasque sheaf of $\mathcal{O}_X$-modules, where $X = \text{Spec } A$.

To establish this, we use Noetherian induction on the support of the sheaf $\widetilde{I}$ (call it $Y$). The basic idea is, for some open $U \subseteq X$, to choose some $f$ and consider some open of the form $X_f \subseteq U$. Noting that $\Gamma(X_f,\widetilde{I}) = I_f$, we can invoke the lemma above, and then the problem reduces to showing $\Gamma_Z(X,\widetilde{I}) \to \Gamma_Z(U,\widetilde{I})$ is surjective, where $Z = X \setminus X_f$. But this follows by induction (put $J = \Gamma_{(f)}(I) = \Gamma_Z(X,\widetilde{I})$ and note this is an injective $A$-module by the lemma, hence $\widetilde{J}$, whose support is strictly contained in $Y$, is flasque; at this point we win since $\Gamma_Z(U,\widetilde{J}) = \Gamma(U,\widetilde{I})$ for all opens $U$).

Theorem. Let $X = \text{Spec } A$ for some Noetherian ring $A$. Then if $i>0$, $H^i(X,\mathscr{F}) = 0$ for all quasicoherent sheaves $\mathscr{F}$ on $X$.

To see this, let $M = \Gamma(X,\mathscr{F})$. Take an injective resolution in the category of $A$-modules, apply the Serre functor $M \mapsto \widetilde{M}$ to get a flasque resolution of $\mathscr{F}$. Applying the global sections functor, we just get back the original resolution, so we’re done.

Theorem (Serre). Let $X$ be a Noetherian scheme. Then TFAE:

1. $X$ is affine.
2. $H^i(X,\mathscr{F}) = 0$ for any quasicoherent sheaf $\mathscr{F}$ on $X$ and $i>0$.
3. $H^1(X,\mathscr{I}) = 0$ for any coherent sheaf of ideals $\mathscr{I}$ on $X$.

We’ve already shown that (1) => (2), and (2) => (3) is easy. (3) => (1) can be proved using the following characterisation of affineness: $X$ is affine if and only if there are $f_1, \ldots, f_r \in A := \Gamma(X,\mathcal{O}_X)$ such that $\langle f_1, \ldots, f_r \rangle = A$, each set $X_f$ is affine, and $X$ is covered by $X_{f_1}, \ldots, X_{f_r}$.