Lemma 10.116.2 (06RP)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.116: Dimension of finite type algebras over fields, reprise
Lemma 10.116.2 (cite)

numberstags

Lemma 10.116.2. Let $k$ be a field. Let $S$ be a finite type $k$ -algebra. Let $\mathfrak q \subset \mathfrak q' \subset S$ be distinct prime ideals. Then $\text{trdeg}_ k\ \kappa (\mathfrak q') < \text{trdeg}_ k\ \kappa (\mathfrak q)$ .

Proof. By Lemma 10.116.1 we have $\dim V(\mathfrak q) = \text{trdeg}_ k\ \kappa (\mathfrak q)$ and similarly for $\mathfrak q'$ . Hence the result follows as the strict inclusion $V(\mathfrak q') \subset V(\mathfrak q)$ implies a strict inequality of dimensions. $\square$

Comments (0)

There are also:

2 comment(s) on Section 10.116: Dimension of finite type algebras over fields, reprise

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