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. $\square$

There are also:

• 2 comment(s) on Section 15.95: An operator introduced by Berthelot and Ogus

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).