The Stacks project

Proof. Proof of (1). Choose generators $y_1, \ldots , y_ m$ of $M$ over $R$. For each $i$ we can write $y_ i = \sum z_{ij} y_ j$ with $z_{ij} \in I$ (since $M = IM$). In other words $\sum _ j (\delta _{ij} - z_{ij})y_ j = 0$. Let $f$ be the determinant of the $m \times m$ matrix $A = (\delta _{ij} - z_{ij})$. Note that $f \in 1 + I$ (since the matrix $A$ is entrywise congruent to the $m \times m$ identity matrix modulo $I$). By Lemma 10.15.5 (1), there exists an $m \times m$ matrix $B$ such that $BA = f 1_{m \times m}$. Writing out we see that $\sum _{i} b_{hi} a_{ij} = f \delta _{hj}$ for all $h$ and $j$; hence, $\sum _{i, j} b_{hi} a_{ij} y_ j = \sum _{j} f \delta _{hj} y_ j = f y_ h$ for every $h$. In other words, $0 = f y_ h$ for every $h$ (since each $i$ satisfies $\sum _ j a_{ij} y_ j = 0$). This implies that $f$ annihilates $M$.

By Lemma 10.19.1 an element of $1 + \text{rad}(R)$ is invertible element of $R$. Hence we see that (1) implies (2). We obtain (3) by applying (1) to $M/N$ which is finite as $N'$ is finite. We obtain (4) by applying (2) to $M/N$ which is finite as $N'$ is finite. We obtain (5) by applying (3) to $M$ and the submodules $\mathop{\mathrm{Im}}(N \to M)$ and $M$. We obtain (6) by applying (4) to $M$ and the submodules $\mathop{\mathrm{Im}}(N \to M)$ and $M$. We obtain (7) by applying (5) to the map $R^{\oplus n} \to M$, $(a_1, \ldots , a_ n) \mapsto a_1x_1 + \ldots + a_ nx_ n$. We obtain (8) by applying (6) to the map $R^{\oplus n} \to M$, $(a_1, \ldots , a_ n) \mapsto a_1x_1 + \ldots + a_ nx_ n$.

Part (9) holds because if $M = IM$ then $M = I^ nM$ for all $n \geq 0$ and $I$ being nilpotent means $I^ n = 0$ for some $n \gg 0$. Parts (10), (11), and (12) follow from (9) by the arguments used above. $\square$