The Stacks project

Lemma 15.96.1. 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 of $D(A)$. Then

\[ f^ m I_ i(M^\bullet , f) = f^ n I_ i(N^\bullet , f) \]

as ideals of $A$ for integers $n, m \geq 0$ such that

\[ m + \sum \nolimits _{j \geq i} (-1)^{j - i}rk(M^ j) = n + \sum \nolimits _{j \geq i} (-1)^{j - i}rk(N^ j) \]

Proof. It suffices to prove the equality after localization at every prime ideal of $A$. Thus by Lemma 15.75.8 and an induction argument we omit 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 $j + 1$. If $j \not= i - 1, i, i + 1$ then we clearly have equality $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$ and $m = n$ and we have the desired equality. If $j = i + 1$ then the maps

\[ (f, d^ i) : M^ i \to M^ i \oplus M^{i + 1} \quad \text{and}\quad (f, d^ i, 0) : M^ i \to M^ i \oplus M^{i + 1} \oplus A \]

have the same nonzero minors hence in this case we also have $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$ and $m = n$. If $j = i$, then $I_ i(M^\bullet , f)$ is the ideal generated by the $r_ i \times r_ i$-minors of

\[ (f, d^ i) : M^ i \to M^ i \oplus M^{i + 1} \]

and $I_ i(N^\bullet , f)$ is the ideal generated by the $(r_ i + 1) \times (r_ i + 1)$-minors of

\[ (f \oplus f, d^ i \oplus 1) : (M^ i \oplus A) \to (M^ i \oplus A) \oplus (M^{i + 1} \oplus A) \]

With suitable choice of coordinates we see that the matrix of the second map is in block form

\[ T = \left( \begin{matrix} T_1 & 0 \\ 0 & T_2 \end{matrix} \right), \quad T_1 = \text{matrix of first map}, \quad T_2 = \left( \begin{matrix} f \\ 1 \end{matrix} \right) \]

With notation as in Lemma 15.8.1 we have $I_0(T_2) = A$, $I_1(T_2) = A$, $I_ p(T_2) = 0$ for $p \geq 2$ and hence $I_{r_ i + 1}(T) = I_{r_ i + 1}(T_1) + I_{r_ i}(T_1) = I_{r_ i}(T_1)$ which means that $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. We also have $m = n$ so this finishes the case $j = i$. Finally, say $j = i - 1$. Then we see that $m = n + 1$, thus we have to show that $fI_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. In this case $I_ i(M^\bullet , f)$ is the ideal generated by the $r_ i \times r_ i$-minors of

\[ (f, d^ i) : M^ i \to M^ i \oplus M^{i + 1} \]

and $I_ i(N^\bullet , f)$ is the ideal generated by the $(r_ i + 1) \times (r_ i + 1)$-minors of

\[ (f \oplus f, d^ i) : (M^ i \oplus A) \to (M^ i \oplus A) \oplus M^{i + 1} \]

With suitable choice of coordinates we see that the matrix of the second map is in block form

\[ T = \left( \begin{matrix} T_1 & 0 \\ 0 & T_2 \end{matrix} \right), \quad T_1 = \text{matrix of first map}, \quad T_2 = \left( \begin{matrix} f \end{matrix} \right) \]

Arguing as above we find that indeed $fI_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. $\square$


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 0GSS. Beware of the difference between the letter 'O' and the digit '0'.