A proper scheme with infinite-dimensional fppf cohomology

In algebraic geometry, very often one encounters theorems of the following flavor:

Theorem: Let $f : X \to S$ be a proper morphism of spaces. Then for every sheaf $\mathcal{F}$ on $X$ that is finite, so is its pushforward $Rf_\ast \mathcal{F}$ .

Notice how I was being deliberately vague in the theorem above. What are $X$ and $Y$ ? 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 $f : X \to S$ be a proper morphism of locally Noetherian schemes. Let $\textbf{D}^+_{\text{coh}}(X)$ denote the derived category of bounded above $\mathcal{O}_X$ -modules with coherent cohomology. Then for any $\mathcal{F} \in \textbf{D}^+_{\text{coh}}(X)$ , the direct image $Rf_\ast \mathcal{F} \in \textbf{D}^+_{\text{coh}}(S)$ .

Theorem 2: Let $f : X \to S$ be a proper morphism of locally Noetherian schemes. Let $\textbf{D}^+_{\text{c}}(X)$ denote the derived category of bounded above abelian sheaves (in the étale topology) with constructible cohomology. Then for any $\mathcal{F} \in \textbf{D}^+_{\text{c}}(X)$ , the direct image $Rf_\ast \mathcal{F} \in \textbf{D}^+_{\text{c}}(S)$ .

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 $X$ .

In today’s post, I will show that there is no such analogue of a finiteness theorem in the fppf topology:

Main Theorem: Let $k$ be an algebraically closed field of characteristic $p$ . Let $X/k$ be the cuspidal cubic. Then in the fppf topology,

$H^2(X, \mu_p) \simeq k^+\times \mathbf{Z}/p\mathbf{Z}.$

This is infinite-dimensional as an $\mathbf{F}_p$ -vector space!

Before proving this result, we remark that in the étale topology, the group $H^2(X,\mu_p)$ is zero, because $\mu_p = 0$ ! I claim that on any reduced, $\mathbf{F}_p$ -scheme $X$ , $\mu_p = 0$ . To see this, take an étale open $U \to X$ . Then $\mu_p(U)$ consists of sections $f$ such that $f^p = 1$ . But in characteristic $p$ , this says $(f-1)^p = 0$ , and hence $f = 1$ since $U$ 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 $k$ be an algebraically closed field (of any characteristic) and $X/k$ a (possibly non-reduced, singular) curve. Then $H^2(X, \mathbf{G}_m) = 0$ .

Proof: First I claim we may assume that $X$ is reduced. Indeed, for a closed subscheme $j : Y \subseteq X$ defined by an ideal $\mathcal{I}$ with $\mathcal{I}^2 = 0$ , we also have $H^2(X, \mathbf{G}_m) \hookrightarrow H^2(Y, \mathbf{G}_m)$ . Indeed, consider the exact sequence of étale sheaves

$0 \to 1+ \mathcal{I} \to \mathbf{G}_m \to j_\ast \mathbf{G}_m \to 0.$

Taking cohomology, we see that a sufficient condition for $H^2(X, \mathbf{G}_m) \to H^2(Y, \mathbf{G}_m)$ to be injective is that $H^2(X, 1 + \mathcal{I}) = 0$ . Observe that $1 + \mathcal{I} \simeq \mathcal{I}$ via the map $i \mapsto 1 +i$ . But now $\mathcal{I}$ is a coherent sheaf, and therefore the cohomology $H^2(X, \mathcal{I})$ may be taken to be coherent. Since $X$ is a curve, it follows that $H^2(X,\mathcal{I}) = 0$ . The claim now follows from the fact that the nilradical of $X$ admits a filtration with successive quotients square-zero ideals in $\mathcal{O}_X$ .

Ok, so now that $X$ is reduced, we may consider the normalization $\widetilde{X}$ , with normalization morphism $\pi : \widetilde{X} \to X$ . We have a Leray spectral sequence with $E_2^{p,q} := H^p(X, R^q \pi_\ast \mathbf{G}_m)$ converging to $E_\infty^{p+q} := H^{p+q}(\widetilde{X}, \mathbf{G}_m)$ .

Consider the second abutment $E_\infty^2 = H^2(\widetilde{X}, \mathbf{G}_m)$ . Then $\widetilde{X}$ , being a (possibly union) of smooth curve(s) over an algebraically closed field $k$ , must have

$E_\infty^2 = H^2(\widetilde{X},\mathbf{G}_m ) = 0$

by Tsen’s theorem. Now I claim there is an injection $E_2^{2,0} \hookrightarrow E_\infty^2$

, and consequently that $E_2^{2,0} = H^2(X, \pi_\ast \mathbf{G}_m)$

is zero. Indeed, since $\pi$

is finite, $R^1\pi_\ast \mathbf{G}_m$

is zero, and therefore $E_2^{0,1} = 0$

, i.e. $E_3^{2,0} = E_2^{2,0}/\text{Im}(E_2^{0,1} \to E_2^{2,0}) = E_2^{2,0}$

It is now easy to see that in fact

$E_2^{2,0} = E_3^{2,0} = \ldots = E_\infty^{2,0}$

and hence $E_2^{2,0} \hookrightarrow E_\infty^2$

. Finally, I claim that $H^2(X,\mathbf{G}_m) = H^2(\widetilde{X}, \pi_\ast \mathbf{G}_m)$

which is sufficient to prove the lemma. Indeed, the exact sequence

$0 \to \mathbf{G}_m \to \pi_\ast \mathbf{G}_m \to Q \to 0$

has cokernel $Q$ supported on the singular locus of $X$ . By TAG 056V, the singular locus is a proper closed subset of $X$ , and therefore consists of a finite union of $k$ -valued points. Hence $H^1(X,Q) = H^2(X,Q) = 0$ , proving that $H^2(X,\mathbf{G}_m) = H^2(\widetilde{X}, \pi_\ast \mathbf{G}_m)$ .

Proof of Main Theorem: Consider the Kummer sequence (in the fppf topology)

$0 \to \mu_p \to \mathbf{G}_m \to \mathbf{G}_m \to 0.$

This gives rise to an exact sequence

$0 \to \frac{H^1_{\text{fppf}}(X, \mathbf{G}_m)}{p\cdot H^1_{\text{fppf}}(X, \mathbf{G}_m)} \to H^2_{\text{fppf}}(X,\mu_p) \to H^2_{\text{fppf}}(X, \mathbf{G}_m)[p] \to 0.$

But $\mathbf{G}_m$ is smooth, and therefore by a theorem of Grothendieck, the cohomology of $\mathbf{G}_m$ in either the étale or fppf topology is the same. It follows by Lemma 1 that

$H^2_{\text{fppf}}(X \mu_p) \simeq \frac{H^1_{\text{fppf}}(X, \mathbf{G}_m)}{p\cdot H^1_{\text{fppf}}(X, \mathbf{G}_m)} \simeq \frac{\text{Pic}(X)}{p\cdot \text{Pic}(X)}.$

The main theorem now follows from the following result, noting that $k^+$ is not $p$ -divisible in characteristic $p$ .

Proposition 1: Let $k$ be an algebraically closed field of characteristic $p$ , and let $X/k$ be the cuspidal cubic. There is an exact sequence of abelian groups

$0 \to k^+ \to \operatorname{Pic}(X) \to \mathbf{Z} \to 0.$

As in Lemma 1, let $\pi : \widetilde{X} \to X$ denote the normalization morphism. Consider the long exact sequence in cohomology obtained from

(2) $\begin{equation*} 0 \to \mathbf{G}_m \to \pi_\ast \mathbf{G}_m \to Q \to 0.\end{equation*}$

In other words,

$1\to k^\times \to k^\times \to H^0(X, Q) \to \operatorname{Pic}(X) \to H^1(X, \pi_\ast \mathbf{G}_m) \to 0.$

By a similar argument using the Leray spectral sequence in Lemma 1, and using the fact that $\widetilde{X} \simeq \mathbf{P}^1$

, we get

$H^1(X,\pi_\ast \mathbf{G}_m) = \operatorname{Pic}(\widetilde{X}) = \mathbf{Z}.$

Therefore, it remains to show $H^0(X,Q) = k^+$

. Why? Because the map $k^\times \to k^\times$

is simply the map on constants, and hence is the identity!

Since $Q$ is supported at the cusp, to compute $H^0(X, Q)$ it is enough to describe the stalk of $Q$ at the cusp $x_0 \in X$ . We must now work in coordinates: Consider the commutative diagram

$\begin{tikzcd} 0\ar{r} & k\{t^2, t^3\}^\times \ar{d}{\alpha} \ar{r} & k\{t\}^\times \ar{d}{\beta} \ar{r} & Q_{x_0} \ar{r} \ar[dashed]{d}{\gamma} \ar{r} & 0 \\ 0\ar{r} & k\llbracket t^2, t^3\rrbracket^\times \ar{r} & k\llbracket t\rrbracket^\times \ar{r} & k^+ \ar{r} & 0 \end{tikzcd}$

where $k\{t\}, k\{t^2, t^3\}$ are the strict henselizations of $k[t], k[t^2, t^3]$ respectively at the origin. The top row of this diagram is the stalk of (2) at $x_0 \in X$ (in the étale topology). The maps $\alpha, \beta$ 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 $\gamma$ is the induced map on quotients, which exists since since all $\alpha, \beta$ (on the level of rings) are local homomorphisms. The projection map $k\llbracket t \rrbracket^\times \to k^+$ is given by $f(t) \mapsto \frac{d}{d}\log f(t)\big|_{t = 0}$ (note the logarithmic derivative is well-defined since $f(0) \neq 0$ ).

It is sufficient to show that the map $\gamma$ is an isomorphism. The key thing we must show is that $\gamma$ is surjective (injectivity is kinda clear). Now any $a_1 \in k^+$ is hit by the polynomial $f(t) := 1 + a_1t \in k\llbracket t \rrbracket^\times$ . But this polynomial lies in $k\{t \}^\times$ , i.e. is algebraic. Why? It satisfies $p(t, f(t)) = 0$ , where

$p(x,y) = 1 + a_1x - y.$

Hence $\gamma$

is surjective and we win.

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 $k$ is not $2$ . This is not necessary, as our computation above shows.

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Bogdan Zavyalov

2 years ago

Hi Ben,

Nice post!

But I think it is a bit easier to construct examples when $\mathrm{H}^i_{fppf}(X, \alpha_p)$ is infinite for some $i$ . For example, your comment there shows that $\mathrm{H}^1_{fppf}(E, \alpha_p)=k$ is infinite for any supersingular elliptic curve $E$ and infinite field $k$ (for instance, for any alg. closed $k$ ). I think that actually $\mathrm{H}^i_{fppf}(X, \alpha_p)$ is infinite over an infinite field as long as it is non-zero because it should have a structure of a $k$ -vector space (at least for a perfect field $k$ ).

Last edited 2 years ago by Bogdan Zavyalov

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David Benjamin Lim

Reply to Bogdan Zavyalov

Yeah that’s true. But I wanted an example with $\mu_p$ because somehow like $\alpha_p$ is kinda messed-up and $\mu_p$ is more interesting? If $p$ is not equal to the characteristic then the cohomology is finite (by comparing with \'{e}tale cohomology).