The Stacks project

Lemma 10.79.3. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime. Let $M$ be a finitely presented $R$-module. If $M_\mathfrak p$ is free, then there is an $f \in R$, $f \not\in \mathfrak p$ such that $M_ f$ is a free $R_ f$-module.

Proof. Choose a basis $x_1, \ldots , x_ n \in M_\mathfrak p$. We can choose an $f \in R$, $f \not\in \mathfrak p$ such that $x_ i$ is the image of some $y_ i \in M_ f$. After replacing $y_ i$ by $f^ m y_ i$ for $m \gg 0$ we may assume $y_ i \in M$. Namely, this replaces $x_1, \ldots , x_ n$ by $f^ mx_1, \ldots , f^ mx_ n$ which is still a basis as $f$ maps to a unit in $R_\mathfrak p$. Hence we obtain a homomorphism $\varphi = (y_1, \ldots , y_ n) : R^{\oplus n} \to M$ of $R$-modules whose localization at $\mathfrak p$ is an isomorphism. By Lemma 10.79.2 we can find an $f \in R$, $f \not\in \mathfrak p$ such that $\varphi _\mathfrak q$ is an isomorphism for all primes $\mathfrak q \subset R$ with $f \not\in \mathfrak q$. Then it follows from Lemma 10.23.1 that $\varphi _ f$ is an isomorphism and the proof is complete. $\square$


Comments (0)


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Your email address will not be published.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