The Stacks project

15.95 An operator introduced by Berthelot and Ogus

In this section we discuss a construction introduced in [Section 8, Berthelot-Ogus] and generalized in [Section 6, BMS]. We urge the reader to look at the original papers discussing this notion.

Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. If $M$ is a $A$-module then by Lemma 15.88.3 following are equivalent

  1. $f$ is a nonzerodivisor on $M$,

  2. $M[f] = 0$,

  3. $M[f^ n] = 0$ for all $n \geq 1$, and

  4. the map $M \to M_ f$ is injective.

If these equivalent conditions hold, then (in this section) we will say $M$ is $f$-torsion free. If so, then we denote $f^ iM \subset M_ f$ the submodule consisting of elements of the form $f^ ix$ with $x \in M$. Of course $f^ iM$ is isomorphic to $M$ as an $A$-module. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules with differentials $d^ i : M^ i \to M^{i + 1}$. In this case we define $\eta _ fM^\bullet $ to be the complex with terms

\[ (\eta _ fM)^ i = \{ x \in f^ iM^ i \mid d^ i(x) \in f^{i + 1}M^{i + 1}\} \]

and differential induced by $d^ i$. Observe that $\eta _ fM^\bullet $ is another complex of $f$-torsion free $A$-modules. If $a^\bullet : M^\bullet \to N^\bullet $ is a map of complexes of $f$-torsion free $A$-modules, then we obtain a map of complexes

\[ \eta _ fa^\bullet : \eta _ fM^\bullet \longrightarrow \eta _ fN^\bullet \]

induced by the maps $f^ iM^ i \to f^ iN^ i$. The reader checks that we obtain an endo-functor on the category of complexes of $f$-torsion free $A$-modules. If $a^\bullet , b^\bullet : M^\bullet \to N^\bullet $ are two maps of complexes of $f$-torsion free $A$-modules and $h = \{ h^ i : M^ i \to N^{i - 1}\} $ is a homotopy between $a^\bullet $ and $b^\bullet $, then we define $\eta _ fh$ to be the family of maps $(\eta _ fh)^ i : (\eta _ fM)^ i \to (\eta _ fN)^{i - 1}$ which sends $x$ to $h^ i(x)$; this makes sense as $x \in f^ iM^ i$ implies $h^ i(x) \in f^ iN^{i - 1}$ which is certainly contained in $(\eta _ fN)^{i - 1}$. The reader checks that $\eta _ fh$ is a homotopy between $\eta _ fa^\bullet $ and $\eta _ fb^\bullet $. All in all we see that we obtain a functor

\[ \eta _ f : K(f\text{-torsion free }A\text{-modules}) \longrightarrow K(f\text{-torsion free }A\text{-modules}) \]

on the homotopy category (Derived Categories, Section 13.8) of the additive category of $f$-torsion free $A$-modules. There is no sense in which $\eta _ f$ is an exact functor of triangulated categories, see Example 15.95.1.

Example 15.95.1. Let $A$ be a ring. Let $f \in A$ be a nonzerodivisor. Consider the functor $\eta _ f : K(f\text{-torsion free }A\text{-modules}) \to K(f\text{-torsion free }A\text{-modules})$. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. Multiplication by $f$ defines an isomorphism $\eta _ f(M^\bullet [1]) \to (\eta _ fM^\bullet )[1]$, so in this sense $\eta _ f$ is compatible with shifts. However, consider the diagram

\[ \xymatrix{ A \ar[r]_ f & A \ar[r]_1 & A \ar[r] & 0 \\ 0 \ar[r] \ar[u] & 0 \ar[r] \ar[u] & A \ar[r]^{-1} \ar[u]^ f & A \ar[u] } \]

Think of each column as a complex of $f$-torsion free $A$-modules with the module on top in degree $1$ and the module under it in degree $0$. Then this diagram provides us with a distinguished triangle in $K(f\text{-torsion free }A\text{-modules})$ with triangulated structure as given in Derived Categories, Section 13.10. Namely the third complex is the cone of the map between the first two complexes. However, applying $\eta _ f$ to each column we obtain

\[ \xymatrix{ fA \ar[r]_ f & fA \ar[r]_1 & fA \ar[r] & 0 \\ 0 \ar[r] \ar[u] & 0 \ar[r] \ar[u] & A \ar[r]^{-1} \ar[u]^ f & A \ar[u] } \]

However, the third complex is acyclic and even homotopic to zero. Hence if this were a distinguished triangle, then the first arrow would have to be an isomorphism in the homotopy category, which is not true unless $f$ is a unit.

Lemma 15.95.2. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. There is a canonical isomorphism

\[ f^ i : H^ i(M^\bullet )/H^ i(M^\bullet )[f] \longrightarrow H^ i(\eta _ fM^\bullet ) \]

given by multiplication by $f^ i$.

Proof. Observe that $\mathop{\mathrm{Ker}}(d^ i : (\eta _ fM)^ i \to (\eta _ fM)^{i + 1})$ is equal to $\mathop{\mathrm{Ker}}(d^ i : f^ iM^ i \to f^ iM^{i + 1}) = f^ i\mathop{\mathrm{Ker}}(d^ i : M^ i \to M^{i + 1})$. This we get a surjection $f^ i : H^ i(M^\bullet ) \to H^ i(\eta _ fM^\bullet )$ by sending the class of $z \in \mathop{\mathrm{Ker}}(d^ i : M^ i \to M^{i + 1})$ to the class of $f^ iz$. If we obtain the zero class in $H^ i(\eta _ fM^\bullet )$ then we see that $f^ i z = d^{i - 1}(f^{i - 1}y)$ for some $y \in M^{i - 1}$. Since $f$ is a nonzerodivisor on all the modules involved, this means $f z = d^{i - 1}(y)$ which exactly means that the class of $z$ is $f$-torsion as desired. $\square$

Lemma 15.95.3. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. If $M^\bullet \to N^\bullet $ is a quasi-isomorphism of complexes of $f$-torsion free $A$-modules, then the induced map $\eta _ fM^\bullet \to \eta _ fN^\bullet $ is a quasi-isomorphism too.

Proof. This is true because the isomorphisms of Lemma 15.95.2 are compatible with maps of complexes. $\square$

Lemma 15.95.4. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. There is an additive functor1 $L\eta _ f : D(A) \to D(A)$ such that if $M \in D(A)$ is represented by a complex $M^\bullet $ of $f$-torsion free $A$-modules, then $L\eta _ fM = \eta _ fM^\bullet $ and similarly for morphisms.

Proof. Denote $\mathcal{T} \subset \text{Mod}_ A$ the full subcategory of $f$-torsion free $A$-modules. We have a corresponding inclusion

\[ K(\mathcal{T}) \quad \subset \quad K(\text{Mod}_ A) = K(A) \]

of $K(\mathcal{T})$ as a full triangulated subcategory of $K(A)$. Let $S \subset \text{Arrows}(K(\mathcal{T}))$ be the quasi-isomorphisms. We will apply Derived Categories, Lemma 13.5.8 to show that the map

\[ S^{-1}K(\mathcal{T}) \longrightarrow D(A) \]

is an equivalence of triangulated categories. The lemma shows that it suffices to prove: given a complex $M^\bullet $ of $A$-modules, there exists a quasi-isomorphism $K^\bullet \to M^\bullet $ with $K^\bullet $ a complex of $f$-torsion free modules. By Lemma 15.59.10 we can find a quasi-isomorphism $K^\bullet \to M^\bullet $ such that the complex $K^\bullet $ is K-flat (we won't use this) and consists of flat $A$-modules $K^ i$. In particular, $f$ is a nonzerodivisor on $K^ i$ for all $i$ as desired.

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

\[ K(\mathcal{T}) \xrightarrow {\eta _ f} K(\mathcal{T}) \to K(A) \to D(A) \]

which according to Lemma 15.95.3 sends quasi-isomorphisms to quasi-isomorphisms. Hence this functor factors over $S^{-1}K(\mathcal{T}) = D(A)$ by Categories, Lemma 4.27.8. $\square$

Remark 15.95.5. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. For every $i$ set $\overline{M}^ i = M^ i/fM^ i$. Denote $B^ i \subset Z^ i \subset \overline{M}^ i$ the boundaries and cocycles for the differentials on the complex $\overline{M}^\bullet = M^\bullet \otimes _ A A/fA$. We claim that there exists a commutative diagram

\[ \xymatrix{ 0 \ar[r] & B^{i + 1} \ar[r] \ar@{=}[d] & B^{i + 1} \oplus B^ i \ar[r] \ar[d]^{s, s'} & B^ i \ar[r] \ar[d] & 0 \\ 0 \ar[r] & B^{i + 1} \ar[r]^-s & (\eta _ fM)^ i /f(\eta _ fM)^ i \ar[r]^-t & Z^ i \ar[r] & 0 } \]

with exact rows. Here are the constructions of the maps

  1. If $x \in (\eta _ fM)^ i$ then $x = f^ ix'$ with $d^ i(x') = 0$ in $\overline{M}^{i + 1}$. Hence we can define the map $t$ by sending $x$ to the class of $x'$.

  2. If $y \in M^{i + 1}$ has class $\overline{y}$ in $B^{i + 1} \subset \overline{M}^{i + 1}$ then we can write $y = fy' + d^ i(x)$ for $y' \in M^{i + 1}$ and $x \in M^ i$. Hence we can define the map $s$ sending $\overline{y}$ to the class of $f^{i + 1}x$ in $(\eta _ fM)^ i /f(\eta _ fM)^ i$; we omit the verification that this is well defined.

  3. If $x \in M^ i$ has class $\overline{x}$ in $B^ i \subset \overline{M}^ i$ then we can write $x = fx' + d^{i - 1}(z)$ for $x' \in M^ i$ and $z \in M^{i - 1}$. We define the map $s'$ by sending $\overline{x}$ to the class of $f^ i d^{i - 1}(z)$ in $(\eta _ fM)^ i/f(\eta _ fM)^ i$. This is well defined because if $fx' + d^{i - 1}(z) = 0$, then $f^ ix'$ is in $(\eta _ fM)^ i$ and consequently $f^ id^{i - 1}(z)$ is in $f(\eta _ fM)^ i$.

We omit the verification that the lower row in the displayed diagram is a short exact sequence of modules. It is immediately clear from these constructions that we have commutative diagrams

\[ \xymatrix{ B^{i + 1} \oplus B^ i \ar[d]^{s, s'} \ar[r] & B^{i + 2} \oplus B^{i + 1} \ar[d]^{s, s'} \\ (\eta _ fM)^ i /f(\eta _ fM)^ i \ar[r] & (\eta _ fM)^{i + 1} /f(\eta _ fM)^{i + 1} } \]

where the upper horizontal arrow is given by the identification of the summands $B^{i + 1}$ in source and target. In other words, we have found an acyclic subcomplex of $\eta _ fM^\bullet / f(\eta _ fM^\bullet ) = \eta _ fM^\bullet \otimes _ A A/fA$ and the quotient by this subcomplex is a complex whose terms $Z^ i/B^ i$ are the cohomology modules of the complex $\overline{M}^\bullet = M^\bullet \otimes _ A A/fA$.

To explain the phenomenon observed in Remark 15.95.5 in a more canonical manner, we are going to construct the Bockstein operators. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. For every $i \in \mathbf{Z}$ there is a commutative diagram (with tensor products over $A$)

\[ \xymatrix{ 0 \ar[r] & M^\bullet \otimes f^{i + 1}A \ar[r] \ar[d] & M^\bullet \otimes f^ iA \ar[r] \ar[d] & M^\bullet \otimes f^ iA/f^{i + 1}A \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & M^\bullet \otimes f^{i + 1}A/f^{i + 2}A \ar[r] & M^\bullet \otimes f^ iA/f^{i + 2}A \ar[r] & M^\bullet \otimes f^ iA/f^{i + 1}A \ar[r] & 0 } \]

whose rows are short exact sequences of complexes. Of course these short exact sequences for different $i$ are all isomorphic to each other by suitably multiplying with powers of $f$. The long exact sequence of cohomology of the bottom sequence in particular determines the Bockstein operator

\[ \beta = \beta ^ i : H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \to H^{i + 1}(M^\bullet \otimes f^{i + 1}A/f^{i + 2}A) \]

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

15.95.5.1
\begin{equation} \label{more-algebra-equation-factorization-bockstein} \vcenter { \xymatrix{ H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \ar[r]_\delta \ar[rd]_\beta & H^{i + 1}(M^\bullet \otimes f^{i + 1}A) \ar[d] \\ & H^{i + 1}(M^\bullet \otimes f^{i + 1}A/f^{i + 2}A) } } \end{equation}

of the Bockstein operator where $\delta $ is the boundary operator coming from the top row in the commutative diagram above. Let us show that we obtain a complex

15.95.5.2
\begin{equation} \label{more-algebra-equation-complex-bocksteins} H^\bullet (M^\bullet /f) = \left[ \begin{matrix} \ldots \\ \downarrow \\ H^{i - 1}(M^\bullet \otimes f^{i - 1}A/f^ iA) \\ \downarrow \beta \\ H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \\ \downarrow \beta \\ H^{i + 1}(M^\bullet \otimes f^{i + 1}A/f^{i + 2}A) \\ \downarrow \\ \ldots \end{matrix} \right] \end{equation}

i.e., that $\beta \circ \beta = 0$2. Namely, using the factorization (15.95.5.1) we see that it suffices to show that

\[ H^{i + 1}(M^\bullet \otimes f^{i + 1}A) \to H^{i + 1}(M^\bullet \otimes f^{i + 1}A/f^{i + 2}A) \xrightarrow {\beta ^{i + 1}} H^{i + 2}(M^\bullet \otimes f^{i + 2}A/f^{i + 3}A) \]

is zero. This is true because the kernel of $\beta ^{i + 1}$ consists of the cohomology classes which can be lifted to $H^{i + 1}(M^\bullet \otimes f^{i + 1}A/f^{i + 3}A)$ and those in the image of the first map certainly can!

Lemma 15.95.6. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. There is a canonical map of complexes

\[ \eta _ fM^\bullet \otimes _ A A/fA \longrightarrow H^\bullet (M^\bullet /f) \]

which is a quasi-isomorphism where the right hand side is the complex (15.95.5.2).

Proof. Let $x \in (\eta _ fM)^ i$. Then $x = f^ ix' \in f^ iM$ and $d^ i(x) = f^{i + 1}y \in f^{i + 1}M^{i + 1}$. Thus $d^ i$ maps $x' \otimes f^ i$ to zero in $M^{i + 1} \otimes f^ iA/f^{i + 1}A$. All tensor products are over $A$ in this proof. Hence we may map $x$ to the class of $x' \otimes f^ i$ in $H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A)$. It is clear that this rule defines a map

\[ (\eta _ fM)^ i \otimes A/fA \longrightarrow H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \]

of $A/fA$-modules. Observe that in the situation above, we may view $x' \otimes f^ i$ as an element of $M^ i \otimes f^ iA/f^{i + 2}A$ with differential $d^ i(x' \otimes f^ i) = y \otimes f^{i + 1}$. By the construction of $\beta $ above we find that $\beta (x' \otimes f^ i) = y \otimes f^{i + 1}$ and we conclude that our maps are compatible with differentials, i.e., we have a map of complexes.

To finish the proof, we observe that the construction given in the previous paragraph agrees with the maps $(\eta _ fM)^ i \otimes A/fA \to Z^ i/B^ i$ discussed in Remark 15.95.5. Since we have seen that the kernel of these maps is an acyclic subcomplex of $\eta _ fM^\bullet \otimes A/fA$, the lemma is proved. $\square$

Lemma 15.95.7. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. For $i \in \mathbf{Z}$ the following are equivalent

  1. $\mathop{\mathrm{Ker}}(d^ i \bmod f^2)$ surjects onto $\mathop{\mathrm{Ker}}(d^ i \bmod f)$,

  2. $\beta : H^ i(M^\bullet \otimes _ A f^ iA/f^{i + 1}A) \to H^{i + 1}(M^\bullet \otimes _ A f^{i + 1}A/f^{i + 2}A)$ is zero.

These equivalent conditions are implied by the condition $H^{i + 1}(M^\bullet )[f] = 0$.

Proof. The equivalence of (1) and (2) follows from the definition of $\beta $ as the boundary map on cohomology of a short exact sequence of complexes isomorphic to the short exact sequence of complexes $0 \to fM^\bullet /f^2M^\bullet \to M^\bullet /f^2M^\bullet \to M^\bullet /fM^\bullet \to 0$. If $\beta \not= 0$, then $H^{i + 1}(M^\bullet )[f] \not= 0$ because of the factorization (15.95.5.1). $\square$

Lemma 15.95.8. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. If $\mathop{\mathrm{Ker}}(d^ i \bmod f^2)$ surjects onto $\mathop{\mathrm{Ker}}(d^ i \bmod f)$, then the canonical map

\[ (1, d^ i) : (\eta _ fM)^ i / f(\eta _ fM)^ i \longrightarrow f^ iM^ i/f^{i + 1}M^ i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1} \]

identifies the left hand side with a direct sum of submodules of the right hand side.

Proof. With notation as in Remark 15.95.5 we define a map $t^{-1} : Z^ i \to (\eta _ fM)^ i / f(\eta _ fM)^ i$. Namely, for $x \in M^ i$ with $d^ i(x) = f^2y$ we send the class of $x$ in $Z^ i$ to the class of $f^ ix$ in $(\eta _ fM)^ i / f(\eta _ fM)^ i$. We omit the verification that this is well defined; the assumption of the lemma exactly signifies that the domain of this operation is all of $Z^ i$. Then $t \circ t^{-1} = \text{id}_{Z^ i}$. Hence $t^{-1}$ defines a splitting of the short exact sequence in Remark 15.95.5 and the resulting direct sum decomposition

\[ (\eta _ fM)^ i / f(\eta _ fM)^ i = Z^ i \oplus B^{i + 1} \]

is compatible with the map displayed in the lemma. $\square$

Lemma 15.95.9. Let $A$ be a ring and let $f, g \in A$ be nonzerodivisors. Let $M^\bullet $ be a complex of $A$-modules such that $fg$ is a nonzerodivisor on all $M^ i$. Then $\eta _ f\eta _ gM^\bullet = \eta _{fg}M^\bullet $.

Proof. The statement means that in degree $i$ we obtain the same submodule of the localization $M^ i_{fg} = (M^ i_ g)_ f$. We omit the details. $\square$

Lemma 15.95.10. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $A \to B$ be a flat ring map and let $g \in B$ the image of $f$. Let $M^\bullet $ be a complex of $f$-torsion free $A$-modules. Then $g$ is a nonzerodivisor, $M^\bullet \otimes _ A B$ is a complex of $g$-torsion free modules, and $\eta _ fM^\bullet \otimes _ A B = \eta _ g(M^\bullet \otimes _ A B)$.

Proof. Omitted. $\square$

[1] Beware that this functor isn't exact, i.e., does not transform distinguished triangles into distinguished triangles. See Example 15.95.1.
[2] An alternative is to argue that $\beta $ occurs as the differential for the spectral sequence for the complex $(M^\bullet )_ f$ filtered by the subcomplexes $f^ iM^\bullet $. Yet another argument, which proves something stronger, is to first consider the case $M^\bullet = A$. Here the short exact sequences $0 \to f^{i + 1}A/f^{i + 2}A \to f^ iA/f^{i + 2}A \to f^ iA/f^{i + 1}A \to 0$ define maps $\beta ^ i : f^ iA/f^{i + 1}A \to f^{i + 1}A/f^{i + 2}A[1]$ in $D(A)$. Then one computes (arguing similarly to the text) that the composition $f^ iA/f^{i + 1}A \to f^{i + 1}A/f^{i + 2}A[1] \to f^{i + 2}A/f^{i + 3}A[2]$ is zero in $D(A)$. Since $M^\bullet \otimes f^ iA/f^{i + 1}A = M^\bullet \otimes ^\mathbf {L} f^ iA/f^{i + 1}A$ by our assumption on $M^\bullet $ having $f$-torsion free terms, we conclude the compostion
\[ (M^\bullet \otimes f^ iA/f^{i + 1}A) \to (M^\bullet \otimes f^{i + 1}A/f^{i + 2}A)[1] \to (M^\bullet \otimes f^{i + 2}A/f^{i + 3}A)[2] \]
in $D(A)$ is zero as well.

Comments (2)

Comment #6323 by Arthur Ogus on

I think it is worth pointing out that is , where is the -adic filtration of and is Deligne's décalée operation.

Comment #6429 by on

Yes, we should do this. Does anybody want to write a section on Deligne's decalee operation?

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