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.
By definition,
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
We know that for any finite -module
, the function
is upper semicontinuous on
by the Nakayama’s lemma.
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
.
May 29, 2012 at 11:07 pm
In the statement of Lemma 7 you wrote “lower semicontinuous”…
May 30, 2012 at 11:15 am
Whoops! Thanks for the correction.