Last time, we proved an important theorem. Namely, for a proper morphism of noetherian schemes , we showed that the cohomology of a coherent sheaf on , flat over the base, could be described as the cohomology of a finite complex of flat, finitely generated modules, and moreover that if we base-changed to some other scheme , we just had to compute the cohomology of 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 , for 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 be a proper morphism of noetherian schemes. Let be a coherent sheaf on , flat over . Then the function is upper semi-continuous on . Further, the function is locally constant on .
Proof: The assertions are local on , so we may assume that is affine, say . By yesterday’s discussion of the Grothendieck complex, there is a finite complex of finite flat -modules such that for the residue field at . 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 be a finite complex of finitely generated flat -modules, for noetherian. Then:
- For each , the function is upper semicontinuous on .
- The function is locally constant on .
Proof: Over a noetherian ring, a finitely generated flat module is projective, hence locally free. Since the assertion is local on , we may assume that 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 is always free of the same dimension, whatever may be.
where , and . Consequently, the rank of the cohomology equals
where is the kernel and the image. So we have an exact sequence , so by linear algebra its dimension equals
Moreover, by the next lemma, are lower semicontinuous as functions of . Meanwhile, is constant in . The difference is thus upper semicontinuous. The result is now clear in view of (1) above.
Lemma 8 Let be a noetherian ring, and finite free -modules. Let be a homomorphism. Then the function is lower semicontinuous.
Proof: Suppose ; we must show that the set of such that is closed.
Now to say that this dimension is less than is to say that the image of
is zero. This in turn is equivalent to the condition that the morphism be zero, which is a closed condition: if the morphism of free modules is represented by the matrix , then the condition is equivalent to having entries lying in the ideal associated to .