Lemma 10.60.12. Let $R$ be a Noetherian ring. Let $f_1, \ldots , f_ r \in R$.
If $\mathfrak p$ is minimal over $(f_1, \ldots , f_ r)$ then the height of $\mathfrak p$ is $\leq r$.
If $\mathfrak p, \mathfrak q \in \mathop{\mathrm{Spec}}(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, f_1, \ldots , f_ r)$, then every chain of primes between $\mathfrak p$ and $\mathfrak q$ has length at most $r$.
Proof. Proof of (1). If $\mathfrak p$ is minimal over $f_1, \ldots , f_ r$, then the only prime ideal of $R_\mathfrak p$ containing $f_1, \ldots , f_ r$ 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 Proposition 10.60.9 shows $\dim (R_\mathfrak p) \leq r$.
Proof of (2). By part (1) we see that $\mathfrak q/\mathfrak p$ is a prime of height $\leq r$. This immediately implies the statement about chains of primes between $\mathfrak p$ and $\mathfrak q$. $\square$
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