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My question is concerned with vanishing cycles of a locally constant sheaf for a smooth morphism in the case $l = p$. In the case $l \neq p$ this is a statement in SGA7-II. See below for the precise question. First let me start with some


Let us assume that one has an affine scheme $X$ over a field $k$ of characteristic $p >0$ and a function $f: X \to \mathbb{A}^1_k$. In SGA7-II Deligne then introduces the so-called nearby and vanishing cycles of $f$ at $0$ (say).

This roughly goes as follows: Let $S$ be the strict Henselization of $k[x]_{(x)}$. Let $s$ be the closed point of $\Spec S$ and let $\bar{\eta}$ be the spectrum of a separable closure of the quotient field of $S$. Consider the cartesian diagram

$ \begin{array}{ccccccc} &X_s&\xrightarrow{i}&X_{S}&\xleftarrow{j}&X_{\bar{\eta}} \newline & \downarrow & &\downarrow && \downarrow\newline &s&\xrightarrow{}&S&\xleftarrow{}&\bar{\eta} \end{array} $

For a constructible sheaf $\mathcal{F}$ on $X_{\acute{e}t}$ the nearby cycles of $\mathcal{F}$ are defined as $R \Psi(\mathcal{F}) = i^\ast Rj_\ast j^\ast \mathcal{F}$ (in $D(X_s)$). There is a natural morphism $i^\ast \mathcal{F} \to R\Psi(\mathcal{F})$ whose mapping cone we denote by $R\Phi(\mathcal{F})$ -- this is called the vanishing cycles.

(Of course, Deligne's construction is much more general and there are some groups acting here which I swept under the rug. But I hope the above is sufficient to get an idea of my question).

Deligne then proves the following (SGA7-II, 2.1.5): If $\mathcal{F}$ is a locally constant sheaf of $\mathbb{Z}/l$-modules (with $l$ a prime $\neq p$) and $f: X \to S$ defines a smooth morphism, then $R\Phi(\mathcal{F}) = 0$. My question is the following: Does this statement also hold for $l = p$? I am content to make additional assumptions on $X$ (e.g. of finite type over $k$) or my base field (take a finite field if you want). In particular, I am looking for a reference where this is proved or a counterexample.

Edit: My previous definition of Nearby/Vanishing cycles contained an error. I am indebted to Brian Conrad for pointing this out to me.

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The proof of the result of Deligne relies on the smooth base-change theorem, which requires $l\not= p$. So I guess that a counterexample could likely be constructed from a counterexample to the smooth base-change theorem, which should be classical (but I don't have one at hand). – Damian Rössler Feb 6 '13 at 14:05
Another remark: vanishing cycles when $l=p$ is an object studied in the framework of the p-adic Hodge-Tate decomposition. See for instance the paper by Bloch and Kato "p-adic étale cohomology" (which probably gives a whole class of counterexamples to the statement above when $l=p$). – Damian Rössler Feb 6 '13 at 14:13
up vote 5 down vote accepted

$\DeclareMathOperator{\ord}{ord}$ $\DeclareMathOperator{\Spec}{Spec}$

This negative answer to my question is a slight expansion of an email sent to me by Brian Conrad. Since I merely added some details for my own benefit this post is community wiki.

As Damian Rössler pointed out in his comment the essential ingredient is the failure of the smooth base change theorem in the $l = p$ case. This is can be done via Artin-Schreier examples (cf. Milne's book on étale cohomology VI, Remark 4.4).

Let $S$ be the spectrum of $R = k[[t]]$ where $k$ is an algebraically closed field of characteristic $p > 0$ and write $K$ for the quotient field of $R$. Let now $X = \mathbb{A}^1_{R} = \Spec k[[t]][y]$. Let us consider the constant sheaf $\mathcal{F} = \mathbb{Z}/p\mathbb{Z}$ on $X$ and its vanishing cycles complex at the origin $0$ of the special fiber. Denote by $A$ the strict henselization of $R[y]_{(t,y)}$.

Using SGA7-I, Exposé I, Proposition 2.3 we have $$(1) \quad (i^\ast R^n j_\ast j^\ast \mathcal{F})_0 = H^n(A_{K_s}, \mathcal{F}) = \varinjlim H^n(A_F, \mathcal{F}) ,$$ where $F$ varies over the finite extensions of $K$ in $K_s$. Since strict henselization is compatible with finite base change each $A_F$ is the analogue of $A_K$ with $R$ replaced by its normalization in $F$ (which is again local since $R$ is henselian).

We claim that $(1)$ is non-zero (in fact infinite). Admitting this for the moment we obtain a counterexample to my question. Indeed, part of the long exact sequence associated to the exact triangle $i^\ast \mathcal{F} \to R\Psi \mathcal{F} \to R\Phi \mathcal{F}$ is $$i^\ast R^1 j_\ast j^\ast \mathcal{F} \to R^1\Phi \mathcal{F} \to H^1(i^\ast\mathcal{F}),$$ but $i^\ast$ is exact so that $H^1(i^\ast \mathcal{F})$ is zero.

In order to show that $(1)$ is nonzero we consider the Artin-Schreier sequence $$0 \to \mathcal{F} \to \mathcal{O}_{A_{K_s}} \to \mathcal{O}_{A_{K_s}} \to 0.$$ Since $\Spec A_{K_s}$ is affine the higher étale cohomology groups of $\mathcal{O}_{A_{K_s}}$ vanish. It is therefore sufficient to show that the cokernel of $f \mapsto f^p - f$ on $A_{K_s}$ is infinite.

Claim: For any non-zero elements $c_i$ in the maximal ideal of the valuation ring of $K_s$ with pairwise distinct valuation the elements $y/c_i$ represent $\mathbb{F}_p$-linearly independent Artin-Schreier classes.

It suffices to proves this in each $H^1(A_F, \mathcal{F})$ separately (assuming all $c_i$ are contained in $A_F$). Hence, by replacing $R$ with with its normalization in $F$ we may assume that $F = K$ and all $c_i$ contained in $K$.

A non-trivial $\mathbb{F}_p$-linear combination of such $1/c_i$'s is an element of $K^\times$ with negative valuation. So it suffices to show that for nonzero $c \in R$ with $\ord(c) > 0$ the element $y/c \in A_K$ is not of the form $f^p - f$ for any $f \in A_K$. Here the valuation function $\ord$ is induced by $t^n \mapsto n$. In particular, $\ord(y/c) = -\ord(c) < 0$. If such $f$ exists then we must have $$0 > \ord(f^p - f) \geq \min(p \ord(f), \ord(f)),$$ so $\ord(f) < 0$. We therefore have $f = h/t^e$ with $e > 0$ and $h \in A$ not divisible by $t$.

Thus we get $$-\ord(c) = \ord(y/c) = \ord(f^p - f) = p \ord(f) = -pe. $$ This means that $c$ is contained in $t^{ep}R^\times$ and by changing $y$ by a unit we may assume that $c = t^{ep}$. Hence, $$\frac{y}{t^{ep}} = \frac{y}{c} = f^p - f = \frac{(h^p - t^{(p-1)e} h)}{t^{ep}}.$$ Equivalently we obtain the following equation in $A$ $$y = h^p - t^{(p-1)e}h = h(h^{p-1} - t^{(p-1)e}).$$

But $A$ is a regular local ring, so in particular a UFD, in which $y$ is irreducible. It follows that precisely one of the factors on the right hand side is a unit. Since $h^p = y + t^{(p-1)e}h$ lies in the maximal ideal of $A$ so does $h$. But then, since $e > 0$, $h^{p-1} - t^{(p-1)e}$ is also contained in the maximal ideal. This is a contradiction.

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