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. $\square$

 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 $f \in R$, $f \not\in \mathfrak p$.

There are also:

• 6 comment(s) on Section 10.20: Nakayama's lemma

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).