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. $\square$

Proof. Special case $I = 0$. Let $y_1, \ldots , y_ s$ be generators for $M$ over $R$. Since $S^{-1}M$ is generated by $x_1, \ldots , x_ r$, for each $i$ we can write $y_ i = \sum (a_{ij}/s_{ij})x_ j$ for some $a_{ij} \in R$ and $s_{ij} \in S$. Let $s \in S$ be the product of all of the $s_{ij}$. Then we see that $y_ i$ is contained in the $R_ s$-submodule of $M_ s$ generated by $x_1, \ldots , x_ r$. Hence $x_1, \ldots , x_ r$ generates $M_ s$.

General case. By the special case, we can find an $s \in S$ such that $x_1, \ldots , x_ r$ generate $(M/IM)_ s$ over $(R/I)_ s$. By Lemma 10.20.1 we can find a $g \in 1 + I_ s \subset R_ s$ such that $x_1, \ldots , x_ r$ generate $(M_ s)_ g$ over $(R_ s)_ g$. Write $g = 1 + i/s'$. Then $f = ss' + is$ works; details omitted. $\square$

Proof. To show that $A \to B$ is surjective, we view it as a map of $A$-modules and apply Lemma 10.20.1 (6). We conclude it suffices to show that $A/\mathfrak m_ A \to B/\mathfrak m_ AB$ is surjective. As $A/\mathfrak m_ A = B/\mathfrak m_ B$ it suffices to show that $\mathfrak m_ AB \to \mathfrak m_ B$ is surjective. View $\mathfrak m_ AB \to \mathfrak m_ B$ as a map of $B$-modules and apply Lemma 10.20.1 (6). We conclude it suffices to see that $\mathfrak m_ AB/\mathfrak m_ A\mathfrak m_ B \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective. This follows from assumption (4). $\square$