The Stacks project

Lemma 10.136.9. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection (Definition 10.136.5)

  1. For any $R \to R'$ the base change $R' \otimes _ R S = R'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection.

  2. For any $g \in S$ which is the image of $h \in R[x_1, \ldots , x_ n]$ the ring $S_ g = R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, hx_{n + 1} - 1)$ is a relative global complete intersection.

  3. If $R \to S$ factors as $R \to R_ f \to S$ for some $f \in R$. Then the ring $S = R_ f[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection over $R_ f$.

Proof. By Lemma 10.116.5 the fibres of a base change have the same dimension as the fibres of the original map. Moreover $R' \otimes _ R R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) = R'[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Thus (1) follows. The proof of (2) is that the localization at one element can be described as $S_ g \cong S[x_{n + 1}]/(gx_{n + 1} - 1)$. Assertion (3) follows from (1) since under the assumptions of (3) we have $R_ f \otimes _ R S \cong S$. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 10.136: Syntomic morphisms

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Your email address will not be published.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



View Lemma 10.136.9 as pdf