Lemma 10.125.8 (00QJ)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.125: Dimension of fibres
Lemma 10.125.8 (cite)

numberstags

Lemma 10.125.8. Let $R \to S$ be a ring homomorphism of finite presentation. Let $n \geq 0$ . The set

$V_ n = \{ \mathfrak q \in \mathop{\mathrm{Spec}}(S) \mid \dim _{\mathfrak q}(S/R) \leq n\}$

is a quasi-compact open subset of $\mathop{\mathrm{Spec}}(S)$ .

Proof. It is open by Lemma 10.125.6. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ be a presentation of $S$ . Let $R_0$ be the $\mathbf{Z}$ -subalgebra of $R$ generated by the coefficients of the polynomials $f_ i$ . Let $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ . Then $S = R \otimes _{R_0} S_0$ . By Lemma 10.125.7 $V_ n$ is the inverse image of an open $V_{0, n}$ under the quasi-compact continuous map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_0)$ . Since $S_0$ is Noetherian we see that $V_{0, n}$ is quasi-compact. $\square$

Comments (0)

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).

Comments (0)

Post a comment