Lemma 10.128.3. Let $R \to S$, $M$, $\Lambda $, $R_\lambda \to S_\lambda $, $M_\lambda $ be as in Lemma 10.127.13. Assume that $M$ is flat over $R$. Then for some $\lambda \in \Lambda $ the module $M_\lambda $ is flat over $R_\lambda $.

**Proof.**
Pick some $\lambda \in \Lambda $ and consider

See Remark 10.75.9. The right hand side shows that this is a finitely generated $S_\lambda $-module (because $S_\lambda $ is Noetherian and the modules in question are finite). Let $\xi _1, \ldots , \xi _ n$ be generators. Because $M$ is flat over $R$ we have that $0 = \mathop{\mathrm{Ker}}(\mathfrak m_\lambda R \otimes _ R M \to M)$. Since $\otimes $ commutes with colimits we see there exists a $\lambda ' \geq \lambda $ such that each $\xi _ i$ maps to zero in $\mathfrak m_{\lambda }R_{\lambda '} \otimes _{R_{\lambda '}} M_{\lambda '}$. Hence we see that

is zero. Note that $M_\lambda \otimes _{R_\lambda } R_\lambda /\mathfrak m_\lambda $ is flat over $R_\lambda /\mathfrak m_\lambda $ because this last ring is a field. Hence we may apply Lemma 10.99.14 to get that $M_{\lambda '}$ is flat over $R_{\lambda '}$. $\square$

## 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Your email address will not be published.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.

## Comments (0)