# The Stacks Project

## Tag 00EU

Lemma 10.24.1. 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 10.16.2. This contradicts the minimality of $\mathfrak p$. If $R$ is reduced, then ${\mathfrak p}R_{\mathfrak p} = 0$ and hence it is a field. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 4280–4286 (see updates for more information).

\begin{lemma}
\label{lemma-minimal-prime-reduced-ring}
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.
\end{lemma}

\begin{proof}
If some element $x$ of ${\mathfrak p}R_{\mathfrak p}$
is not nilpotent, then $D(x) \not = \emptyset$, see
the minimality of $\mathfrak p$. If $R$ is reduced,
then ${\mathfrak p}R_{\mathfrak p} = 0$ and
hence it is a field.
\end{proof}

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