Lemma 10.36.22 (00GU)—The Stacks project

Loading web-font TeX/Math/Italic

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.36: Finite and integral ring extensions
Lemma 10.36.22 (cite)

numberstags

Lemma 10.36.22. Let $R \to S$ be a ring map such that $S$ is integral over $R$ . Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q$ be a prime of $S$ mapping to $\mathfrak p$ . Then there exists a prime $\mathfrak q'$ with $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p'$ .

Proof. We may replace $R$ by $R/\mathfrak p$ and $S$ by $S/\mathfrak q$ . This reduces us to the situation of having an integral extension of domains $R \subset S$ and a prime $\mathfrak p' \subset R$ . By Lemma 10.36.17 we win. $\square$

Comments (0)

There are also:

2 comment(s) on Section 10.36: Finite and integral ring extensions

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