Lemma 15.125.5. Let $R$ be a Noetherian local normal domain of dimension $2$. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be pairwise distinct primes of height $1$. There exists a nonzero element $f \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ such that $R/fR$ is reduced.

**Proof.**
Let $f \in \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ 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. Note for later use that each of the primes $\mathfrak p_ i$ occurs among the primes $\mathfrak q_ j$. 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 $g$ and $N$ as in Lemma 15.125.4. 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 $h \in \mathfrak m^ N$. By Lemma 15.125.2 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 $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ such that $t(f + h)$ is maximal among such $h$. Set $f' = f + h$. Given $h' \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ we see that the number $t(f' + h') \leq t(f + h)$. Thus after replacing $f$ by $f'$ we may assume that for every $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$ we have $t(f + h) \leq s$.

Next, assume that we can find an element $h \in \mathfrak m^ N$ such that for each $j$ we have $\text{ord}_{\mathfrak q_ j}(h) \geq 1$ and $\text{ord}_{\mathfrak q_ j}(h) = 1 \Leftrightarrow m_ j > 1$. Observe that $h \in \mathfrak m^ N \cap \mathfrak p_1 \cap \ldots \cap \mathfrak p_ r$. 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 $a_ j \in \mathfrak q_ j$ such that $a_ j \not\in \mathfrak q_{j'}$ for $j' \not= j$ and $a_ j \not\in \mathfrak q_ j^{(2)}$. Here $\mathfrak q_ j^{(2)} = \{ x \in R \mid \text{ord}_{\mathfrak q_ j}(x) \geq 2\} $ is the second symbolic power of $\mathfrak q_ j$. Then we take

Then $h$ clearly satisfies the conditions on valuations imposed above. If $h \not\in \mathfrak m^ N$, then we multiply by an element of $\mathfrak m^ N$ which is not contained in $\mathfrak q_ j$ for all $j$. $\square$

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