Lemma 58.21.1. Let $(A, \mathfrak m)$ be a Noetherian local ring with $\dim (A) \geq 1$. Let $f \in \mathfrak m$. Then there exist a $\mathfrak p \in V(f)$ with $\dim (A_\mathfrak p) = 1$.
Proof. By induction on $\dim (A)$. If $\dim (A) = 1$, then $\mathfrak p = \mathfrak m$ works. If $\dim (A) > 1$, then let $Z \subset \mathop{\mathrm{Spec}}(A)$ be an irreducible component of dimension $> 1$. Then $V(f) \cap Z$ has dimension $> 0$ (Algebra, Lemma 10.60.13). Pick a prime $\mathfrak q \in V(f) \cap Z$, $\mathfrak q \not= \mathfrak m$ corresponding to a closed point of the punctured spectrum of $A$; this is possible by Properties, Lemma 28.6.4. Then $\mathfrak q$ is not the generic point of $Z$. Hence $0 < \dim (A_\mathfrak q) < \dim (A)$ and $f \in \mathfrak q A_\mathfrak q$. By induction on the dimension we can find $f \in \mathfrak p \subset A_\mathfrak q$ with $\dim ((A_\mathfrak q)_\mathfrak p) = 1$. Then $\mathfrak p \cap A$ works. $\square$
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