Last time, we proved an important theorem. Namely, for a proper morphism of noetherian schemes $X \to \mathrm{Spec} A$, we showed that the cohomology of a coherent sheaf $\mathcal{F}$ on $X$, flat over the base, could be described as the cohomology of a finite complex $K$ of flat, finitely generated modules, and moreover that if we base-changed to some other scheme $\mathrm{Spec B}$, we just had to compute the cohomology of $K \otimes_A B$ to get the cohomology of the base extension of the initial sheaf.

With this, it is not too hard to believe that as we vary over the fibers $X_y$, for $y$ in the base, the cohomologies will have somewhat comparable dimensions. Or, at least, their dimensions will vary somewhat reasonably. The precise statement is provided by the semicontinuity theorem.

Theorem 6 (The Semicontinuity theorem) Let ${f: X \rightarrow Y}$ be a proper morphism of noetherian schemes. Let ${\mathcal{F}}$ be a coherent sheaf on ${X}$, flat over ${Y}$. Then the function ${y \rightarrow \mathrm{dim} H^p(X_y, \mathcal{F}_y)}$ is upper semi-continuous on ${Y}$. Further, the function ${y \rightarrow \sum_p (-1)^p H^p(X_y, \mathcal{F}_y)}$ is locally constant on ${Y}$.

Proof: The assertions are local on ${y \in Y}$, so we may assume that ${Y}$ is affine, say ${Y = \mathrm{Spec} A}$. By yesterday’s discussion of the Grothendieck complex, there is a finite complex ${K^*}$ of finite flat ${A}$-modules such that ${H^p(X_y, \mathcal{F}_y) = H^p( K \otimes_A k(y))}$ for ${k(y)}$ the residue field at ${y}$. This is a special case of the base-change formula we proved yesterday.

Let us now prove a lemma that will imply the semicontinuity theorem:

Lemma 7 Let ${K}$ be a finite complex of finitely generated flat ${A}$-modules, for ${A}$ noetherian. Then:

1. For each ${p}$, the function ${y \rightarrow \dim H^p(K \otimes_A k(y))}$ is upper semicontinuous on ${\mathrm{Spec} A}$.
2. The function ${y \rightarrow \sum_p (-1)^p \dim H^p(K \otimes_A k(y))}$ is locally constant on ${\mathrm{Spec} A}$.

Proof: Over a noetherian ring, a finitely generated flat module is projective, hence locally free. Since the assertion is local on ${y}$, we may assume that ${K}$ consists of free modules.

Then the second assertion is obvious, as the Euler characteristic (i.e. alternating sum) of the cohomology is the same thing as the Euler characteristic of the complex itself, and the complex $K \otimes k(y)$ is always free of the same dimension, whatever $y$ may be.

By definition,

$\displaystyle H^p(K \otimes_A k(y)) = \frac{\ker(K^p \otimes_A k(y) \rightarrow K^{p+1} \otimes_A k(y))}{\mathrm{Im}(K^{p-1}\otimes_A k(y) \rightarrow K^{p}\otimes_A k(y))} = Z^p/B^p,$

where ${B^p = \mathrm{Im}( K^{p-1} \otimes_A k(y) \rightarrow K^p \otimes_A k(y))}$, and ${Z^p = \ker( K^{p-1} \otimes_A k(y) \rightarrow K^p \otimes_A k(y))}$. Consequently, the rank of the cohomology equals

$\displaystyle \dim_{k(y)} H^p(K \otimes_A k(y)) = \dim Z^p - \dim B^p,$

where ${Z^p}$ is the kernel and ${B^p}$ the image. So we have an exact sequence ${0 \rightarrow Z^p \rightarrow K^p \otimes_A k(y) \rightarrow B^{p+1} \rightarrow 0}$, so by linear algebra its dimension equals

$\displaystyle \dim Z^p = \dim K^p \otimes_A k(y)- \dim B^{p+1} .$

In total, we find that

$\displaystyle \dim_{k(y)} H^p(K \otimes_A k(y)) = \dim K^p \otimes_A k(y) - \dim B^p - \dim B^{p+1}.\ \ \ \ \ (1)$

We know that for any finite ${A}$-module ${M}$, the function ${y \rightarrow \mathrm{dim}_{k(y)} M \otimes_A k(y)}$ is upper semicontinuous on ${y}$ by the Nakayama’s lemma.

Moreover, by the next lemma, ${\dim B^p, \dim B^{p+1}}$ are lower semicontinuous as functions of ${y}$. Meanwhile, ${\dim K^p \otimes_A k(y)}$ is constant in ${y}$. The difference is thus upper semicontinuous. The result is now clear in view of (1) above.

Lemma 8 Let ${A}$ be a noetherian ring, and ${M,N}$ finite free ${A}$-modules. Let ${\phi: M \rightarrow N}$ be a homomorphism. Then the function ${y \rightarrow \dim ( \mathrm{Im}(M\otimes_A k(y) \rightarrow N \otimes_A k(y))}$ is lower semicontinuous.

Proof: Suppose ${r \in \mathbb{N}}$; we must show that the set of ${y}$ such that ${\dim_{k(y)}( \mathrm{Im}(M\otimes_A k(y) \rightarrow N \otimes_A k(y)) < r}$ is closed.

Now to say that this dimension is less than ${r}$ is to say that the image of

$\displaystyle \bigwedge^r M \otimes_A k(y) \rightarrow \bigwedge^r N \otimes_A k(y)$

is zero. This in turn is equivalent to the condition that the morphism ${\bigwedge^r M \otimes_A k(y) \rightarrow \bigwedge^r N \otimes_A k(y)}$ be zero, which is a closed condition: if the morphism of free modules ${\bigwedge^r \phi: \bigwedge^r M \rightarrow \bigwedge^r N}$ is represented by the matrix ${A}$, then the condition is equivalent to ${A}$ having entries lying in the ideal associated to ${y}$.