The Stacks project

Lemma 15.96.8. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet $ and $N^\bullet $ be two bounded complexes of finite free $A$-modules representing the same object in $D(A)$. Assume $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Then $J_ i(M^\bullet , f) = J_ i(N^\bullet , f)$ as ideals in $A/fA$.

Proof. Observe that the fact that $I_ i(M^\bullet , f)$ is a principal ideal implies that $I_ i(M^\bullet , f)$ is a principal ideal by Lemma 15.96.1 and hence the statement makes sense. As in the proof of Lemma 15.96.1 we may assume $N^\bullet = M^\bullet \oplus Q^\bullet $ for some trivial complex $Q^\bullet $, i.e.,

\[ Q^\bullet = \ldots \to 0 \to A \xrightarrow {1} A \to 0 \to \ldots \]

where $A$ is placed in degree $j$ and $j + 1$. Since $\eta _ f$ is compatible with direct sums, we see that the map

\[ (1, d^ i) : (\eta _ fN)^ i / f(\eta _ fN)^ i \longrightarrow f^ iN^ i/f^{i + 1}N^ i \oplus f^{i + 1}N^{i + 1}/f^{i + 2}N^{i + 1} \]

is the direct sum of the corresponding map for $M^\bullet $ and for $Q^\bullet $. By the universal property defining the ideals in question, we conclude that $J_ i(N^\bullet , f) = J_ i(M^\bullet , f) + J_ i(Q^\bullet , f)$. Hence it suffices to show that $J_ i(Q^\bullet , f) = 0$ for all $i$. This is a computation that we omit. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GSW. Beware of the difference between the letter 'O' and the digit '0'.