**Proof.**
Proof of (1). If $\mathfrak p$ is minimal over $x$, then the only prime ideal of $R_\mathfrak p$ containing $x$ is the maximal ideal $\mathfrak p R_\mathfrak p$. This is true because the primes of $R_\mathfrak p$ correspond $1$-to-$1$ with the primes of $R$ contained in $\mathfrak p$, see Lemma 10.16.5. Hence Lemma 10.59.7 shows $\dim (R_\mathfrak p) = 1$ if $x$ is not nilpotent in $R_\mathfrak p$. Of course, if $x$ is nilpotent in $R_\mathfrak p$ the argument gives that $\mathfrak pR_\mathfrak p$ is the only prime ideal and we see that the height is $0$.

Proof of (2). By part (1) we see that $\mathfrak q/\mathfrak p$ is a prime of height $1$ or $0$ in $R/\mathfrak p$. This immediately implies there cannot be a prime strictly between $\mathfrak p$ and $\mathfrak q$.
$\square$

## Comments (2)

Comment #3527 by Jonas Ehrhard on

Comment #3666 by Johan on