Lemma 15.125.2. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $1$. Let $x \in \mathfrak m$ be an element not contained in any minimal prime of $R$. Let $t$ be the number of minimal prime ideals of $R$. Then $t \leq \text{length}_ R(R/xR)$.

**Proof.**
Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the minimal prime ideals of $R$. Set $R' = R/\sqrt{0} = R/(\bigcap _{i = 1}^ t \mathfrak p_ i)$. We claim it suffices to prove the lemma for $R'$. Namely, it is clear that $R'$ has $t$ minimal primes too and $\text{length}_{R'}(R'/xR') = \text{length}_ R(R'/xR')$ is less than $\text{length}_ R(R/xR)$ as there is a surjection $R/xR \to R'/xR'$. Thus we may assume $R$ is reduced.

Assume $R$ is reduced with minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ t$. This means there is an exact sequence

Here $Q$ is the cokernel of the first map. Write $M = \prod _{i = 1}^ t R/\mathfrak p_ i$. Localizing at $\mathfrak p_ j$ we see that

is surjective. Thus $Q_{\mathfrak p_ j} = 0$ for all $j$. We conclude that $\text{Supp}(Q) = \{ \mathfrak m\} $ as $\mathfrak m$ is the only prime of $R$ different from the $\mathfrak p_ i$. It follows that $Q$ has finite length (Algebra, Lemma 10.62.3). Since $\text{Supp}(Q) = \{ \mathfrak m\} $ we can pick an $n \gg 0$ such that $x^ n$ acts as $0$ on $Q$ (Algebra, Lemma 10.62.4). Now consider the diagram

where the vertical maps are multiplication by $x^ n$. This is injective on $R$ and on $M$ since $x$ is not contained in any of the $\mathfrak p_ i$. By the snake lemma (Algebra, Lemma 10.4.1), the following sequence is exact:

Hence we find that $\text{length}_ R(R/x^ nR) = \text{length}_ R(M/x^ nM)$ for large enough $n$. Writing $R_ i = R/\mathfrak p_ i$ we see that $\text{length}(M/x^ nM) = \sum _{i = 1}^ t \text{length}_ R(R_ i/x^ nR_ i)$. Applying Lemma 15.125.1 and the fact that $x$ is a nonzerodivisor on $R$ and $R_ i$, we conclude that

Since $\text{length}_{R_ i}(R_ i/x R_ i) \geq 1$ the lemma is proved. $\square$

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