A proper scheme with infinite-dimensional fppf cohomology
In algebraic geometry, very often one encounters theorems of the following flavor:
Theorem: Let be a proper morphism of spaces. Then for every sheaf
on
that is finite, so is its pushforward
.
Notice how I was being deliberately vague in the theorem above. What are and
? What does “finite” mean? Well, it turns out that this depends on the context. In the setting of coherent cohomology, “finite” should mean coherent. On the other hand, in étale cohomology “finite” should mean constructible. Now unfortunately (or fortunately?) I am not a number theorist, so to me I’ll take constructible to mean “a finite set”. So in this case, finite really means – as you guessed it – finite.
Enough with the handwaving. Let me now give two theorems (in the coherent and étale setting) that illustrate this:
Theorem 1: Let be a proper morphism of locally Noetherian schemes. Let
denote the derived category of bounded above
-modules with coherent cohomology. Then for any
, the direct image
.
Theorem 2: Let be a proper morphism of locally Noetherian schemes. Let
denote the derived category of bounded above abelian sheaves (in the étale topology) with constructible cohomology. Then for any
, the direct image
.
Remark 1: I do not know/remember if the Noetherian assumption is necessary in Theorem 2, but it certainly is for Theorem 1, because the proof of Theorem 1 is by dévissage on the category of coherent sheaves on .
In today’s post, I will show that there is no such analogue of a finiteness theorem in the fppf topology:
Main Theorem: Let be an algebraically closed field of characteristic
. Let
be the cuspidal cubic. Then in the fppf topology,
This is infinite-dimensional as an -vector space!
Before proving this result, we remark that in the étale topology, the group is zero, because
! I claim that on any reduced,
-scheme
,
. To see this, take an étale open
. Then
consists of sections
such that
. But in characteristic
, this says
, and hence
since
is reduced (étale over reduced = reduced).
The following lemma seemed very surprising to me when I first “discovered” it. All cohomology considered will be étale cohomology.
Lemma 1: Let be an algebraically closed field (of any characteristic) and
a (possibly non-reduced, singular) curve. Then
.
Proof: First I claim we may assume that is reduced. Indeed, for a closed subscheme
defined by an ideal
with
, we also have
. Indeed, consider the exact sequence of étale sheaves
Taking cohomology, we see that a sufficient condition for to be injective is that
. Observe that
via the map
. But now
is a coherent sheaf, and therefore the cohomology
may be taken to be coherent. Since
is a curve, it follows that
. The claim now follows from the fact that the nilradical of
admits a filtration with successive quotients square-zero ideals in
.
Ok, so now that is reduced, we may consider the normalization
, with normalization morphism
. We have a Leray spectral sequence with
converging to
.
Consider the second abutment . Then
, being a (possibly union) of smooth curve(s) over an algebraically closed field
, must have
It is now easy to see that in fact
has cokernel supported on the singular locus of
. By TAG 056V, the singular locus is a proper closed subset of
, and therefore consists of a finite union of
-valued points. Hence
, proving that
.
Proof of Main Theorem: Consider the Kummer sequence (in the fppf topology)
This gives rise to an exact sequence
But is smooth, and therefore by a theorem of Grothendieck, the cohomology of
in either the étale or fppf topology is the same. It follows by Lemma 1 that
The main theorem now follows from the following result, noting that is not
-divisible in characteristic
.
Proposition 1: Let be an algebraically closed field of characteristic
, and let
be the cuspidal cubic. There is an exact sequence of abelian groups
As in Lemma 1, let denote the normalization morphism. Consider the long exact sequence in cohomology obtained from
Since is supported at the cusp, to compute
it is enough to describe the stalk of
at the cusp
. We must now work in coordinates: Consider the commutative diagram
where are the strict henselizations of
respectively at the origin. The top row of this diagram is the stalk of (2) at
(in the étale topology). The maps
are the ones obtained from the universal property of the henselization (note the henselization and strict henselization are the same since the residue field is algebraically closed). The map
is the induced map on quotients, which exists since since all
(on the level of rings) are local homomorphisms. The projection map
is given by
(note the logarithmic derivative is well-defined since
).
It is sufficient to show that the map is an isomorphism. The key thing we must show is that
is surjective (injectivity is kinda clear). Now any
is hit by the polynomial
. But this polynomial lies in
, i.e. is algebraic. Why? It satisfies
, where
As a final remark, we note that the computation of the Picard group of the cuspidal cubic in Hartshorne assumes the hypothesis that the characteristic of is not
. This is not necessary, as our computation above shows.
Hi Ben,
Nice post!
But I think it is a bit easier to construct examples when
is infinite for some
. For example, your comment there shows that
is infinite for any supersingular elliptic curve
and infinite field
(for instance, for any alg. closed
). I think that actually
is infinite over an infinite field as long as it is non-zero because it should have a structure of a
-vector space (at least for a perfect field
).
Yeah that’s true. But I wanted an example with
because somehow like
is kinda messed-up and
is more interesting? If
is not equal to the characteristic then the cohomology is finite (by comparing with \'{e}tale cohomology).