The Stacks project

110.33 A finite flat module which is not projective

This is a copy of Algebra, Remark 10.78.4. It is not true that a finite $R$-module which is $R$-flat is automatically projective. A counter example is where $R = \mathcal{C}^\infty (\mathbf{R})$ is the ring of infinitely differentiable functions on $\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where $\mathfrak m = \{ f \in R \mid f(0) = 0\} $ and $I = \{ f \in R \mid \exists \epsilon , \epsilon > 0 : f(x) = 0\ \forall x, |x| < \epsilon \} $.

The morphism $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$ is also an example of a flat closed immersion which is not open.

Lemma 110.33.1. Strange flat modules.

  1. There exists a ring $R$ and a finite flat $R$-module $M$ which is not projective.

  2. There exists a closed immersion which is flat but not open.

Proof. See discussion above. $\square$

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Your email address will not be published.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 052H. Beware of the difference between the letter 'O' and the digit '0'.