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. $\square$

There are also:

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

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).