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 (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GLX. Beware of the difference between the letter 'O' and the digit '0'.