The Stacks project

Lemma 10.6.2. The notions finite type and finite presentation have the following permanence properties.

  1. A composition of ring maps of finite type is of finite type.

  2. A composition of ring maps of finite presentation is of finite presentation.

  3. Given $R \to S' \to S$ with $R \to S$ of finite type, then $S' \to S$ is of finite type.

  4. Given $R \to S' \to S$, with $R \to S$ of finite presentation, and $R \to S'$ of finite type, then $S' \to S$ is of finite presentation.

Proof. We only prove the last assertion. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and $S' = R[y_1, \ldots , y_ a]/I$. Say that the class $\bar y_ i$ of $y_ i$ maps to $h_ i \bmod (f_1, \ldots , f_ m)$ in $S$. Then it is clear that $S = S'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m, h_1 - \bar y_1, \ldots , h_ a - \bar y_ a)$. $\square$


Comments (3)

Comment #584 by Wei Xu on

A in last sentence: "Then it is clear that .", should be and should be .

Comment #586 by Anfang on

There is also an index error. It should be .

There are also:

  • 2 comment(s) on Section 10.6: Ring maps of finite type and 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Your email address will not be published.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 00F4. Beware of the difference between the letter 'O' and the digit '0'.