The Stacks project

Lemma 10.126.2. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is of finite presentation if and only if $R' \to S'$ is of finite presentation.

Proof. It is clear that if $R \to S$ is of finite presentation then $R' \to S'$ is of finite presentation. Assume that $R' \to S'$ is of finite presentation. By Lemma 10.126.1 we see that $R \to S$ is of finite type. Write $S = R[x_1, \ldots , x_ n]/I$. By flatness $S' = R'[x_1, \ldots , x_ n]/R'\otimes I$. Say $g_1, \ldots , g_ m$ generate $R'\otimes I$ over $R'[x_1, \ldots , x_ n]$. Write $g_ j = \sum _ i a_{ij} \otimes f_{ji}$ for some $a_{ij} \in R'$ and $f_{ji} \in I$. Let $J \subset I$ be the ideal generated by the $f_{ij}$. By flatness we have $R' \otimes _ R J \subset R'\otimes _ R I$, and both are ideals over $R'[x_1, \ldots , x_ n]$. By construction $g_ j \in R' \otimes _ R J$. Hence $R' \otimes _ R J = R'\otimes _ R I$. By faithful flatness $J = I$. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.126: Algebras and modules of finite presentation

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Your email address will not be published.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 00QQ. Beware of the difference between the letter 'O' and the digit '0'.