Lemma 10.60.11. Let $R$ be a Noetherian ring. Let $x \in R$.

1. If $\mathfrak p$ is minimal over $(x)$ then the height of $\mathfrak p$ is $0$ or $1$.

2. If $\mathfrak p, \mathfrak q \in \mathop{\mathrm{Spec}}(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, x)$, then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$.

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 10.17.5. Hence Lemma 10.60.8 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 $\mathfrak q$. $\square$

Comment #3527 by Jonas Ehrhard on

I think in the proof of part (2) we want to mod out $\mathfrak{p}$ instead of $\mathfrak{q}$, because $\mathfrak{p} \subset \mathfrak{q}$, so $\mathfrak{p} / \mathfrak{q} = 0$.

The same holds for the proof of 0BBZ which is essentially the same.

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