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
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