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 15.75.7 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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