## 20.53 An operator introduced by Berthelot and Ogus

This section continuous the discussion started in More on Algebra, Section 15.95. We strongly encourage the reader to read that section first.

Lemma 20.53.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals. Consider the following two conditions

1. for every $x \in X$ there exists an open neighbourhood $U \subset X$ of $x$ and $f \in \mathcal{I}(U)$ such that $\mathcal{I}|_ U = \mathcal{O}_ U \cdot f$ and $f : \mathcal{O}_ U \to \mathcal{O}_ U$ is injective, and

2. $\mathcal{I}$ is invertible as an $\mathcal{O}_ X$-module.

Then (1) implies (2) and the converse is true if all stalks $\mathcal{O}_{X, x}$ of the structure sheaf are local rings.

Proof. Omitted. Hint: Use Modules, Lemma 17.24.4. $\square$

Situation 20.53.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals satisfying condition (1) of Lemma 20.53.11.

Lemma 20.53.3. In Situation 20.53.2 let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The following are equivalent

1. the subsheaf $\mathcal{F}[\mathcal{I}] \subset \mathcal{F}$ of sections annihilated by $\mathcal{I}$ is zero,

2. the subsheaf $\mathcal{F}[\mathcal{I}^ n]$ is zero for all $n \geq 1$,

3. the multiplication map $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{F}$ is injective,

4. for every open $U \subset X$ such that $\mathcal{I}|_ U = \mathcal{O}_ U \cdot f$ for some $f \in \mathcal{I}(U)$ the map $f : \mathcal{F}|_ U \to \mathcal{F}|_ U$ is injective,

5. for every $x \in X$ and generator $f$ of the ideal $\mathcal{I}_ x \subset \mathcal{O}_{X, x}$ the element $f$ is a nonzerodivisor on the stalk $\mathcal{F}_ x$.

Proof. Omitted. $\square$

In Situation 20.53.2 let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. If the equivalent conditions of Lemma 20.53.3 hold, then we will say that $\mathcal{F}$ is $\mathcal{I}$-torsion free. If so, then for any $i \in \mathbf{Z}$ we will denote

$\mathcal{I}^ i\mathcal{F} = \mathcal{I}^{\otimes i} \otimes _{\mathcal{O}_ X} \mathcal{F}$

so that we have inclusions

$\ldots \subset \mathcal{I}^{i + 1}\mathcal{F} \subset \mathcal{I}^ i\mathcal{F} \subset \mathcal{I}^{i - 1}\mathcal{F} \subset \ldots$

The modules $\mathcal{I}^ i\mathcal{F}$ are locally isomorphic to $\mathcal{F}$ as $\mathcal{O}_ X$-modules, but not globally.

Let $\mathcal{F}^\bullet$ be a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules with differentials $d^ i : \mathcal{F}^ i \to \mathcal{F}^{i + 1}$. In this case we define $\eta _\mathcal {I}\mathcal{F}^\bullet$ to be the complex with terms

\begin{align*} (\eta _\mathcal {I}\mathcal{F})^ i & = \mathop{\mathrm{Ker}}\left( d^ i, -1 : \mathcal{I}^ i\mathcal{F}^ i \oplus \mathcal{I}^{i + 1}\mathcal{F}^{i + 1} \to \mathcal{I}^ i\mathcal{F}^{i + 1} \right) \\ & = \mathop{\mathrm{Ker}}\left(d^ i : \mathcal{I}^ i\mathcal{F}^ i \to \mathcal{I}^ i\mathcal{F}^{i + 1}/ \mathcal{I}^{i + 1}\mathcal{F}^{i + 1} \right) \end{align*}

and differential induced by $d^ i$. In other words, a local section $s$ of $(\eta _\mathcal {I}\mathcal{F})^ i$ is the same thing as a local section $s$ of $\mathcal{I}^ i\mathcal{F}^ i$ such that its image $d^ i(s)$ in $\mathcal{I}^ i\mathcal{F}^{i + 1}$ is in the subsheaf $\mathcal{I}^{i + 1}\mathcal{F}^{i + 1}$. Observe that $\eta _\mathcal {I}\mathcal{F}^\bullet$ is another complex of $\mathcal{I}$-torsion free modules.

Let $a^\bullet : \mathcal{F}^\bullet \to \mathcal{G}^\bullet$ be a map of complexes of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. Then we obtain a map of complexes

$\eta _\mathcal {I} a^\bullet : \eta _\mathcal {I}\mathcal{F}^\bullet \longrightarrow \eta _\mathcal {I}\mathcal{G}^\bullet$

induced by the maps $\mathcal{I}^ i\mathcal{F}^ i \to \mathcal{I}^ i\mathcal{G}^ i$. The reader checks that we obtain an endo-functor on the category of complexes of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules.

If $a^\bullet , b^\bullet : \mathcal{F}^\bullet \to \mathcal{G}^\bullet$ are two maps of complexes of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules and $h = \{ h^ i : \mathcal{F}^ i \to \mathcal{G}^{i - 1}\}$ is a homotopy between $a^\bullet$ and $b^\bullet$, then we define $\eta _\mathcal {I}h$ to be the family of maps $(\eta _\mathcal {I}h)^ i : (\eta _\mathcal {I}\mathcal{F})^ i \to (\eta _\mathcal {I}\mathcal{G})^{i - 1}$ which sends $x$ to $h^ i(x)$; this makes sense as $x$ a local section of $\mathcal{I}^ i\mathcal{F}^ i$ implies $h^ i(x)$ is a local section of $\mathcal{I}^ i\mathcal{G}^{i - 1}$ which is certainly contained in $(\eta _\mathcal {I}\mathcal{G})^{i - 1}$. The reader checks that $\eta _\mathcal {I}h$ is a homotopy between $\eta _\mathcal {I}a^\bullet$ and $\eta _\mathcal {I}b^\bullet$. All in all we see that we obtain a functor

$\eta _ f : K(\mathcal{I}\text{-torsion free }\mathcal{O}_ X\text{-modules}) \longrightarrow K(\mathcal{I}\text{-torsion free }\mathcal{O}_ X\text{-modules})$

on the homotopy category (Derived Categories, Section 13.8) of the additive category of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. There is no sense in which $\eta _\mathcal {I}$ is an exact functor of triangulated categories; compare with More on Algebra, Example 15.95.1.

Lemma 20.53.4. In Situation 20.53.2 let $\mathcal{F}^\bullet$ be a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. For $x \in X$ choose a generator $f \in \mathcal{I}_ x$. Then the stalk $(\eta _\mathcal {I}\mathcal{F}^\bullet )_ x$ is canonically isomorphic to the complex $\eta _ f\mathcal{F}^\bullet _ x$ constructed in More on Algebra, Section 15.95.

Proof. Omitted. $\square$

Lemma 20.53.5. In Situation 20.53.2 let $\mathcal{F}^\bullet$ be a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. There is a canonical isomorphism

$\mathcal{I}^{\otimes i} \otimes _{\mathcal{O}_ X} \left( H^ i(\mathcal{F}^\bullet )/H^ i(\mathcal{F}^\bullet )[\mathcal{I}] \right) \longrightarrow H^ i(\eta _\mathcal {I}\mathcal{F}^\bullet )$

of cohomology sheaves.

Proof. We define a map

$\mathcal{I}^{\otimes i} \otimes _{\mathcal{O}_ X} H^ i(\mathcal{F}^\bullet ) \longrightarrow H^ i(\eta _\mathcal {I}\mathcal{F}^\bullet )$

as follows. Let $g$ be a local section of $\mathcal{I}^{\otimes i}$ and let $\overline{s}$ be a local section of $H^ i(\mathcal{F}^\bullet )$. Then $\overline{s}$ is (locally) the class of a local section $s$ of $\mathop{\mathrm{Ker}}(d^ i : \mathcal{F}^ i \to \mathcal{F}^{i + 1})$. Then we send $g \otimes \overline{s}$ to the local section $gs$ of $(\eta _\mathcal {I}\mathcal{F})^ i \subset \mathcal{I}^ i\mathcal{F}$. Of course $gs$ is in the kernel of $d^ i$ on $\eta _\mathcal {I}\mathcal{F}^\bullet$ and hence defines a local section of $H^ i(\eta _\mathcal {I}\mathcal{F}^\bullet )$. Checking that this is well defined is without problems. We claim that this map factors through an isomorphism as given in the lemma. This we my check on stalks and hence via Lemma 20.53.4 this translates into the result of More on Algebra, Lemma 15.95.2. $\square$

Lemma 20.53.6. In Situation 20.53.2 let $\mathcal{F}^\bullet \to \mathcal{G}^\bullet$ be a map of complexes of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. Then the induced map $\eta _\mathcal {I}\mathcal{F}^\bullet \to \eta _\mathcal {I}\mathcal{G}^\bullet$ is a quasi-isomorphism too.

Proof. This is true because the isomorphisms of Lemma 20.53.5 are compatible with maps of complexes. $\square$

Lemma 20.53.7. In Situation 20.53.2 there is an additive functor2 $L\eta _\mathcal {I} : D(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ such that if $M$ in $D(\mathcal{O}_ X)$ is represented by a complex $\mathcal{F}^\bullet$ of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules, then $L\eta _\mathcal {I}M = \eta _\mathcal {I}\mathcal{F}^\bullet$. Similarly for morphisms.

Proof. Denote $\mathcal{T} \subset \textit{Mod}(\mathcal{O}_ X)$ the full subcategory of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. We have a corresponding inclusion

$K(\mathcal{T}) \quad \subset \quad K(\textit{Mod}(\mathcal{O}_ X)) = K(\mathcal{O}_ X)$

of $K(\mathcal{T})$ as a full triangulated subcategory of $K(\mathcal{O}_ X)$. Let $S \subset \text{Arrows}(K(\mathcal{T}))$ be the quasi-isomorphisms. We will apply Derived Categories, Lemma 13.5.7 to show that the map

$S^{-1}K(\mathcal{T}) \longrightarrow D(\mathcal{O}_ X)$

is an equivalence of triangulated categories. The lemma shows that it suffices to prove: given a complex $\mathcal{G}^\bullet$ of $\mathcal{O}_ X$-modules, there exists a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{G}^\bullet$ with $\mathcal{F}^\bullet$ a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. By Lemma 20.26.12 we can find a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{G}^\bullet$ such that the complex $\mathcal{F}^\bullet$ is K-flat (we won't use this) and consists of flat $\mathcal{O}_ X$-modules $\mathcal{F}^ i$. By the third characterization of Lemma 20.53.3 we see that a flat $\mathcal{O}_ X$-module is an $\mathcal{I}$-torsion free $\mathcal{O}_ X$-module and we are done.

With these preliminaries out of the way we can define $L\eta _ f$. Namely, by the discussion following Lemma 20.53.3 this section we have already a well defined functor

$K(\mathcal{T}) \xrightarrow {\eta _ f} K(\mathcal{T}) \to K(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$

which according to Lemma 20.53.6 sends quasi-isomorphisms to quasi-isomorphisms. Hence this functor factors over $S^{-1}K(\mathcal{T}) = D(\mathcal{O}_ X)$ by Categories, Lemma 4.27.8. $\square$

In Situation 20.53.2 let us construct the Bockstein operators. First we observe that there is a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathcal{I}^{i + 1} \ar[r] \ar[d] & \mathcal{I}^ i \ar[r] \ar[d] & \mathcal{I}^ i/\mathcal{I}^{i + 1} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2} \ar[r] & \mathcal{I}^ i/\mathcal{I}^{i + 2} \ar[r] & \mathcal{I}^ i/\mathcal{I}^{i + 1} \ar[r] & 0 }$

whose rows are short exact sequences of $\mathcal{O}_ X$-modules. Let $M$ be an object of $D(\mathcal{O}_ X)$. Tensoring the above diagram with $M$ gives a morphism

$\xymatrix{ M \otimes ^\mathbf {L} \mathcal{I}^{i + 1} \ar[r] \ar[d] & M \otimes ^\mathbf {L} \mathcal{I}^ i \ar[r] \ar[d] & M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 1} \ar[d]^{\text{id}} \\ M \otimes ^\mathbf {L} \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2} \ar[r] & M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 2} \ar[r] & M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 1} }$

of distinguished triangles. The long exact sequence of cohomology sheaves associated the bottom triangle in particular determines the Bockstein operator

$\beta = \beta ^ i : H^ i(M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 1}) \longrightarrow H^{i + 1}(M \otimes ^\mathbf {L} \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2})$

for all $i \in \mathbf{Z}$. For later use we record here that by the commutative diagram above there is a factorization

20.53.7.1
$$\label{cohomology-equation-factorization-bockstein} \vcenter { \xymatrix{ H^ i(M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 1}) \ar[r]_\delta \ar[rd]_\beta & H^{i + 1}(M \otimes ^\mathbf {L} \mathcal{I}^{i + 1}) \ar[d] \\ & H^{i + 1}(M \otimes ^\mathbf {L} \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2}) } }$$

of the Bockstein operator where $\delta$ is the boundary operator coming from the top distinguished triangle in the commutative diagram above. We obtain a complex

20.53.7.2
$$\label{cohomology-equation-complex-bocksteins} H^\bullet (M/\mathcal{I}) = \left[ \begin{matrix} \ldots \\ \downarrow \\ H^{i - 1}(M \otimes ^\mathbf {L} \mathcal{I}^{i - 1}/\mathcal{I}^ i) \\ \downarrow \beta \\ H^ i(M \otimes ^\mathbf {L} \mathcal{I}^ i/\mathcal{I}^{i + 1}) \\ \downarrow \beta \\ H^{i + 1}(M \otimes ^\mathbf {L} \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2}) \\ \downarrow \\ \ldots \end{matrix} \right]$$

i.e., that $\beta \circ \beta = 0$. Namely, we can check this on stalks and in this case we can deduce it from the corresponding result in algebra shown in More on Algebra, Section 15.95. Alternative proof: the short exact sequences $0 \to \mathcal{I}^{i + 1}/\mathcal{I}^{i + 2} \to \mathcal{I}^ i/\mathcal{I}^{i + 2} \to \mathcal{I}^ i/\mathcal{I}^{i + 1} \to 0$ define maps $b^ i : \mathcal{I}^ i/\mathcal{I}^{i + 1} \to (\mathcal{I}^{i + 1}/\mathcal{I}^{i + 2})[1]$ in $D(\mathcal{O}_ X)$ which induce the maps $\beta$ above by tensoring with $M$ and taking cohomology sheaves. Then one shows that the composition $b^{i + 1}[1] \circ b^ i : \mathcal{I}^ i/\mathcal{I}^{i + 1} \to (\mathcal{I}^{i + 1}/\mathcal{I}^{i + 2})[1] \to (\mathcal{I}^{i + 2}/\mathcal{I}^{i + 3})[2]$ is zero in $D(\mathcal{O}_ X)$ by using the criterion in Derived Categories, Lemma 13.27.7 using that the module $\mathcal{I}^ i/\mathcal{I}^{i + 3}$ is an extension of $\mathcal{I}^{i + 1}/\mathcal{I}^{i + 3}$ by $\mathcal{I}^ i/\mathcal{I}^{i + 1}$.

Lemma 20.53.8. In Situation 20.53.2 let $M$ be an object of $D(\mathcal{O}_ X)$. There is a canonical isomorphism

$L\eta _\mathcal {I}M \otimes ^\mathbf {L} \mathcal{O}_ X/\mathcal{I} \longrightarrow H^\bullet (M/\mathcal{I})$

in $D(\mathcal{O}_ X)$ where the right hand side is the complex (20.53.7.2).

Proof. By the construction of $L\eta _\mathcal {I}$ in Lemma 20.53.6 we may assume $M$ is represented by a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules $\mathcal{F}^\bullet$. Then $L\eta _\mathcal {I}M$ is represented by the complex $\eta _\mathcal {I}\mathcal{F}^\bullet$ which is a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules as well. Thus $L\eta _\mathcal {I}M \otimes ^\mathbf {L} \mathcal{O}_ X/\mathcal{I}$ is represented by the complex $\eta _\mathcal {I}\mathcal{F}^\bullet \otimes \mathcal{O}_ X/\mathcal{I}$. Similarly, the complex $H^\bullet (M/\mathcal{I})$ has terms $H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$.

Let $f$ be a local generator for $\mathcal{I}$. Let $s$ be a local section of $(\eta _\mathcal {I}\mathcal{F})^ i$. Then we can write $s = f^ is'$ for a local section $s'$ of $\mathcal{F}^ i$ and similarly $d^ i(s) = f^{i + 1}t$ for a local section $t$ of $\mathcal{F}^{i + 1}$. Thus $d^ i$ maps $f^ is'$ to zero in $\mathcal{F}^{i + 1} \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1}$. Hence we may map $s$ to the class of $f^ is'$ in $H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$. This rule defines a map

$(\eta _\mathcal {I}\mathcal{F})^ i \otimes \mathcal{O}_ X/\mathcal{I} \longrightarrow H^ i(\mathcal{F}^\bullet \otimes \mathcal{I}^ i/\mathcal{I}^{i + 1})$

of $\mathcal{O}_ X$-modules. A calculation shows that these maps are compatible with differentials (essentially because $\beta$ sends the class of $f^ is'$ to the class of $f^{i + 1}t$), whence a map of complexes representing the arrow in the statement of the lemma.

To finish the proof, we observe that the construction given in the previous paragraph agrees on stalks with the maps constructed in More on Algebra, Lemma 15.95.6 hence we conclude. $\square$

Lemma 20.53.9. In Situation 20.53.2 let $\mathcal{F}^\bullet$ be a complex of $\mathcal{I}$-torsion free $\mathcal{O}_ X$-modules. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then $\eta _\mathcal {I}(\mathcal{F}^\bullet \otimes \mathcal{L}) = (\eta _\mathcal {I}\mathcal{F}^\bullet ) \otimes \mathcal{L}$.

Proof. Immediate from the construction. $\square$

Lemma 20.53.10. In Situation 20.53.2 let $M$ be an object of $D(\mathcal{O}_ X)$. Let $x \in X$ with $\mathcal{O}_{X, x}$ nonzero. If $H^ i(M)_ x$ is finite free over $\mathcal{O}_{X, x}$, then $H^ i(L\eta _\mathcal {I}M)_ x$ is finite free over $\mathcal{O}_{X, x}$ of the same rank.

Proof. Namely, say $f \in \mathcal{O}_{X, x}$ generates the stalk $\mathcal{I}_ x$. Then $f$ is a nonzerodivisor in $\mathcal{O}_{X, x}$ and hence $H^ i(M)_ x[f] = 0$. Thus by Lemma 20.53.5 we see that $H^ i(L\eta _\mathcal {I}M)_ x$ is isomorphic to $\mathcal{I}^ i_ x \otimes _{\mathcal{O}_{X, x}} H^ i(M)_ x$ which is free of the same rank as desired. $\square$

[1] The discussion in this section can be generalized to the case where all we require is that $\mathcal{I}$ is an invertible $\mathcal{O}_ X$-module as defined in Modules, Section 17.24.
[2] Beware that this functor isn't exact, i.e., does not transform distinguished triangles into distinguished triangles.

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