The Stacks project

Proof. The proof is the same as that of Lemma 15.125.5, using Lemma 15.125.8 instead of Lemma 15.125.2 and Lemma 15.125.9 instead of Lemma 15.125.4. We can use Lemma 15.125.8 because $R$ is a catenary domain, so every height one prime ideal of $R$ has dimension $d - 1$, and hence the spectrum of $R/(f + h)$ is equidimensional. For the convenience of the reader we write out the details.

Let $f \in J$ be a nonzero element. We will modify $f$ slightly to obtain an element that generates a radical ideal. The localization $R_\mathfrak p$ of $R$ at each height one prime ideal $\mathfrak p$ is a discrete valuation ring, see Algebra, Lemma 10.119.7 or Algebra, Lemma 10.157.4. We denote by $\text{ord}_\mathfrak p(f)$ the corresponding valuation of $f$ in $R_{\mathfrak p}$. Let $\mathfrak q_1, \ldots , \mathfrak q_ s$ be the distinct height one prime ideals containing $f$. Write $\text{ord}_{\mathfrak q_ j}(f) = m_ j \geq 1$ for each $j$. Then we define $\text{div}(f) = \sum _{j = 1}^ s m_ j\mathfrak q_ j$ as a formal linear combination of height one primes with integer coefficients. The ring $R/fR$ is reduced if and only if $m_ j = 1$ for $j = 1, \ldots , s$. Namely, if $m_ j$ is $1$ then $(R/fR)\mathfrak q_ j$ is reduced and $R/fR \subset \prod (R/fR)_{\mathfrak q_ j}$ as $\mathfrak q_1, \ldots , \mathfrak q_ j$ are the associated primes of $R/fR$, see Algebra, Lemmas 10.63.19 and 10.157.6.

Choose and fix $g_2, \ldots , g_{d - 1}$ and $N$ as in Lemma 15.125.9. For a nonzero $y \in R$ denote $t(y)$ the number of primes minimal over $y$. Since $R$ is a normal domain, these primes are height one and correspond $1$-to-$1$ to the minimal primes of $R/yR$ (Algebra, Lemmas 10.60.11 and 10.157.6). For example $t(f) = s$ is the number of primes $\mathfrak q_ j$ occurring in $\text{div}(f)$. Let $h \in \mathfrak m^ N$. Because $R$ is catenary, for each height one prime $\mathfrak p$ of $R$ we have $\dim (R/\mathfrak p) = d$. Hence by Lemma 15.125.8 we have

see Algebra, Lemma 10.52.5 for the first equality. Therefore we see that $t(f + h)$ is bounded independent of $h \in \mathfrak m^ N$.

By the boundedness proved above we may pick $h \in \mathfrak m^ N \cap J$ such that $t(f + h)$ is maximal among such $h$. Set $f' = f + h$. Given $h' \in \mathfrak m^ N \cap J$ we see that the number $t(f' + h') \leq t(f + h)$. Thus after replacing $f$ by $f'$ we may assume that for every $h \in \mathfrak m^ N \cap J$ we have $t(f + h) \leq s$.

Next, assume that we can find an element $h \in \mathfrak m^ N \cap J$ such that for each $j$ we have $\text{ord}_{\mathfrak q_ j}(h) \geq 1$ and $\text{ord}_{\mathfrak q_ j}(h) = 1 \Leftrightarrow m_ j > 1$. Then $\text{ord}_{\mathfrak q_ j}(f + h) = 1$ for every $j$ by elementary properties of valuations. Thus

for some pairwise distinct height one prime ideals $\mathfrak r_1, \ldots , \mathfrak r_ v$ and $e_ k \geq 1$. However, since $s = t(f) \geq t(f + h)$ we see that $v = 0$ and we have found the desired element.

Now we will pick $h$ that satisfies the above criteria. By prime avoidance (Algebra, Lemma 10.15.2) for each $1 \leq j \leq s$ we can find an element $a_ j \in \mathfrak q_ j \cap J$ such that $a_ j \not\in \mathfrak q_{j'}$ for $j' \not= j$. Next, we can pick $b_ j \in J \cap \mathfrak q_1 \cap \ldots \cap q_ s$ with $b_ j \not\in \mathfrak q_ j^{(2)}$. Here $\mathfrak q_ j^{(2)} = \{ x \in R \mid \text{ord}_{\mathfrak q_ j}(x) \geq 2\} $ is the second symbolic power of $\mathfrak q_ j$. Prime avoidance applies because the ideal $J' = J \cap \mathfrak q_1 \cap \ldots \cap q_ s$ is radical, hence $R/J'$ is reduced, hence $(R/J')_{\mathfrak q_ j}$ is reduced, hence $J'$ contains an element $x$ with $\text{ord}_{\mathfrak q_ j}(x) = 1$, hence $J' \not\subset \mathfrak q_ j^{(2)}$. Then the element

is an element of $J$ with $\text{ord}_{\mathfrak q_ j}(c) = 1$ for all $j = 1, \ldots , s$ by elementary properties of valuations. Finally, we let

where $y \in \mathfrak m^ N$ is an element which is not contained in $\mathfrak q_ j$ for all $j$. $\square$