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 $y_ i = \sum (a_{ij}/s_{ij})x_ j$ in $S^{-1}M$ for some $a_{ij} \in R$ and $s_{ij} \in S$. Multiplying by the product $s \in S$ of the $s_{ij}$ we see that $sy_ i = \sum a'_{ij}x_ j$ in $S^{-1}M$ for some $a'_{ij} \in R$. This in turn means there exist $t_ i \in S$ such that $t_ isy_ i = \sum t_ ia'_{ij}x_ j$ in $M$. Thus if $t \in S$ is the product of the $t_ i$, then we see that $y_ i$ is in the $R_{st}$-submodule generated by $x_1, \ldots , x_ r$ of $M_{st}$. Hence $x_1, \ldots , x_ r$ generates $M_{st}$.
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 $f \in R$, $f \not\in \mathfrak p$.
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Comment #10017 by F. Aurelien on
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