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
0 \to R \to \prod \nolimits _{i = 1}^ t R/\mathfrak p_ i \to Q \to 0
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
R_{\mathfrak p_ j} \to M_{\mathfrak p_ j} = \left(\prod \nolimits _{i=1}^ t R/\mathfrak p_ i\right)_{\mathfrak p_ j} = (R/\mathfrak p_ j)_{\mathfrak p_ j}
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
\xymatrix{ 0 \ar[r] & R \ar[r] \ar[d]^-{x^ n} & M \ar[r] \ar[d]^-{x^ n} & Q \ar[r] \ar[d]^-{x^ n} & 0 \\ 0 \ar[r] & R \ar[r] & M \ar[r] & Q \ar[r] & 0 }
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:
0 \to Q \to R/x^ nR \to M/x^ nM \to Q \to 0
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
n \text{length}_ R(R/xR) = \sum \nolimits _{i = 1}^ t n \text{length}_{R_ i}(R_ i/x R_ i)
Since \text{length}_{R_ i}(R_ i/x R_ i) \geq 1 the lemma is proved. \square
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