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. $\square$

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).