
Lemma 10.134.10. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection over $R$.

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 10.115.5 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 $R_ f \otimes _ R S \cong S$. $\square$

There are also:

• 2 comment(s) on Section 10.134: Syntomic morphisms

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).