The Stacks project

Lemma 10.126.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \subset R$ be a multiplicative subset. Set $R' = S^{-1}(R/I) = S^{-1}R/S^{-1}I$.

  1. For any finite $R'$-module $M'$ there exists a finite $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$.

  2. For any finitely presented $R'$-module $M'$ there exists a finitely presented $R$-module $M$ such that $S^{-1}(M/IM) \cong M'$.

Proof. Proof of (1). Choose a short exact sequence $0 \to K' \to (R')^{\oplus n} \to M' \to 0$. Let $K \subset R^{\oplus n}$ be the inverse image of $K'$ under the map $R^{\oplus n} \to (R')^{\oplus n}$. Then $M = R^{\oplus n}/K$ works.

Proof of (2). Choose a presentation $(R')^{\oplus m} \to (R')^{\oplus n} \to M' \to 0$. Suppose that the first map is given by the matrix $A' = (a'_{ij})$ and the second map is determined by generators $x'_ i \in M'$, $i = 1, \ldots , n$. As $R' = S^{-1}(R/I)$ we can choose $s \in S$ and a matrix $A = (a_{ij})$ with coefficients in $R$ such that $a'_{ij} = a_{ij} / s \bmod S^{-1}I$. Let $M$ be the finitely presented $R$-module with presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$ where the first map is given by the matrix $A$ and the second map is determined by generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $M \to M'$, $x_ i \mapsto x'_ i$ induces an isomorphism $S^{-1}(M/IM) \cong M'$. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.126: Algebras and modules of finite presentation

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Your email address will not be published.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 05N5. Beware of the difference between the letter 'O' and the digit '0'.