Lemma 10.20.2 (0GLX)—The Stacks project

Lemma 10.20.2. Let $R$ be a ring, let $S \subset R$ be a multiplicative subset, let $I \subset R$ be an ideal, and let $M$ be a finite $R$ -module. If $x_1, \ldots , x_ r \in M$ generate $S^{-1}(M/IM)$ as an $S^{-1}(R/I)$ -module, then there exists an $f \in S + I$ such that $x_1, \ldots , x_ r$ generate $M_ f$ as an $R_ f$ -module.¹

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. $\square$

[1] Special cases: (I)

$I = 0$ . The lemma says if

$x_1, \ldots , x_ r$ generate

$S^{-1}M$ , then

$x_1, \ldots , x_ r$ generate

$M_ f$ for some

$f \in S$ . (II)

$I = \mathfrak p$ is a prime ideal and

$S = R \setminus \mathfrak p$ . The lemma says if

$x_1, \ldots , x_ r$ generate

$M \otimes _ R \kappa (\mathfrak p)$ then

$x_1, \ldots , x_ r$ generate

$M_ f$ for some

$f \in R$ ,

$f \not\in \mathfrak p$ .

Comments (1)

Comment #10017 by F. Aurelien on February 28, 2025 at 03:46

The proof of the special case $I=0$ of Lemma 0GLX is not true as written. An equality in $S^{-1}M$ does not necessarily imply the same equality in $M_s$ . A small modification solves the problem (see David Lui's comment in https://math.stackexchange.com/questions/3895881/geometric-nakayamas-lemma) : for each $i$ , we can write $y_i = \sum (a_{i,j}'/s)x_j$ wich implies the existence of $t_i \in S$ such that $t_i \cdot (s \cdot y - \sum a'_{i,j} \cdot x_j) = 0$ . Let $t$ be the product of the $t_i$ ; then this time $s' = st$ works.

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