Lemma 10.126.5 (05GJ)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.126: Algebras and modules of finite presentation
Lemma 10.126.5 (cite)

Lemma 10.126.5. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $M$ be an $R$-module.

If $M_{\mathfrak p}$ is a finite $R_{\mathfrak p}$-module then there exists a finite $R$-module $M'$ and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$.
If $M_{\mathfrak p}$ is a finitely presented $R_{\mathfrak p}$-module then there exists an $R$-module $M'$ of finite presentation and a map $M' \to M$ which induces an isomorphism $M'_{\mathfrak p} \to M_{\mathfrak p}$.

Proof. This is a special case of Lemma 10.126.4 $\square$

Comments (0)

There are also:

3 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Your email address will not be published.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).

All contributions are licensed under the GNU Free Documentation License.