The proof is the same as that of Lemma \ref{lemma-radical-element}, using Lemma \ref{lemma-minprimespolyhigher} instead of Lemma \ref{lemma-minprimespoly} and Lemma \ref{lemma-sysparhigher} instead of Lemma \ref{lemma-syspar}. We can use Lemma \ref{lemma-minprimespolyhigher} 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. \medskip\noindent 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 \ref{algebra-lemma-characterize-dvr} or Algebra, Lemma \ref{algebra-lemma-criterion-normal}. 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 \ref{algebra-lemma-zero-at-ass-zero} and \ref{algebra-lemma-normal-domain-intersection-localizations-height-1}. \medskip\noindent Choose and fix $g_2, \ldots, g_{d - 1}$ and $N$ as in Lemma \ref{lemma-sysparhigher}. 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 \ref{algebra-lemma-minimal-over-1} and \ref{algebra-lemma-normal-domain-intersection-localizations-height-1}). 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 \ref{lemma-minprimespolyhigher} we have \begin{align*} t(f + h) & \leq \text{length}_{R/(f + h)}(R/(f + h, g_1, \ldots, g_{d - 1})) \\ & = \text{length}_R(R/(f + h, g_1, \ldots, g_{d - 1})) \\ & = \text{length}_R(R/(f, g_1, \ldots, g_{d - 1})) \end{align*} see Algebra, Lemma \ref{algebra-lemma-length-independent} for the first equality. Therefore we see that $t(f + h)$ is bounded independent of $h \in \mathfrak m^N$. \medskip\noindent 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 $t(f + h) \leq s$. \medskip\noindent 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 $$ \text{div}(f + h) = \sum\nolimits_{j = 1}^s \mathfrak q_j + \sum\nolimits_{k = 1}^v e_k \mathfrak r_k $$ 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. \medskip\noindent Now we will pick $h$ that satisfies the above criteria. By prime avoidance (Algebra, Lemma \ref{algebra-lemma-silly}) 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 $$ c = \sum\nolimits_{j = 1, \ldots, s} b_j \times \prod\nolimits_{j' \not = j} a_{j'} $$ 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 $$ h = c \times \prod\nolimits_{m_j = 1} a_j \times y $$ where $y \in \mathfrak m^N$ is an element which is not contained in $\mathfrak q_j$ for all $j$.