The Stacks project

Lemma 15.64.4. Let $R \to R'$ be a flat ring map. Let $M$, $N$ be $R$-modules.

  1. If $M$ is a finitely presented $R$-module, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R R' = \mathop{\mathrm{Hom}}\nolimits _{R'}(M \otimes _ R R', N \otimes _ R R')$.

  2. If $M$ is $(-m)$-pseudo-coherent, then $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R R' = \mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N \otimes _ R R')$ for $i < m$.

In particular if $R$ is Noetherian and $M$ is a finite module this holds for all $i$.

Proof. By Algebra, Lemma 10.73.1 we have $\mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N \otimes _ R R') = \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R R')$. Combined with Lemma 15.64.3 we conclude (1) and (2) holds. The final statement follows from this and Lemma 15.63.17. $\square$

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Your email address will not be published.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 087R. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.64.4 as pdf