Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/Fraktur/Regular/Main.js

The Stacks project

Lemma 10.25.1. Let \mathfrak p be a minimal prime of a ring R. Every element of the maximal ideal of R_{\mathfrak p} is nilpotent. If R is reduced then R_{\mathfrak p} is a field.

Proof. If some element x of {\mathfrak p}R_{\mathfrak p} is not nilpotent, then D(x) \not= \emptyset , see Lemma 10.17.2. This contradicts the minimality of \mathfrak p. If R is reduced, then {\mathfrak p}R_{\mathfrak p} = 0 and hence it is a field. \square


Comments (0)

There are also:

  • 4 comment(s) on Section 10.25: Zerodivisors and total rings of fractions

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.