Proposition 10.28.8 (05KE)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.28: A meta-observation about prime ideals
Proposition 10.28.8 (cite)

numberstags

Proposition 10.28.8. If $\mathcal{F}$ is an Oka family of ideals, then any maximal element of the complement of $\mathcal{F}$ is prime.

Proof. Suppose $I \not\in \mathcal{F}$ is maximal with respect to not being in $\mathcal{F}$ but $I$ is not prime. Note that $I \not= R$ because $R \in \mathcal{F}$ . Since $I$ is not prime we can find $a, b \in R - I$ with $ab \in I$ . It follows that $(I, a) \neq I$ and $(I : a)$ contains $b \not\in I$ so also $(I : a) \neq I$ . Thus $(I : a), (I, a)$ both strictly contain $I$ , so they must belong to $\mathcal{F}$ . By the Oka condition, we have $I \in \mathcal{F}$ , a contradiction. $\square$

Comments (0)

There are also:

6 comment(s) on Section 10.28: A meta-observation about prime ideals

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