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 $fI_ i(M^\bullet , f) = I_ i(N^\bullet , f)$.
$\square$

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