The Stacks project

Lemma 10.78.7. Let $R$ be a semi-local ring. Let $M$ be a finite locally free module. If $M$ has constant rank, then $M$ is free. In particular, if $R$ has connected spectrum, then $M$ is free.

Proof. Omitted. Hints: First show that $M/\mathfrak m_ iM$ has the same dimension $d$ for all maximal ideal $\mathfrak m_1, \ldots , \mathfrak m_ n$ of $R$ using the rank is constant. Next, show that there exist elements $x_1, \ldots , x_ d \in M$ which form a basis for each $M/\mathfrak m_ iM$ by the Chinese remainder theorem. Finally show that $x_1, \ldots , x_ d$ is a basis for $M$. $\square$


Comments (0)

There are also:

  • 4 comment(s) on Section 10.78: Finite projective modules

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Your email address will not be published.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 02M9. Beware of the difference between the letter 'O' and the digit '0'.