The Stacks project

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.,

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

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

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

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

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

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

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

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