Lemma 15.125.1. Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension one, and let $x\in \mathfrak m$ be an element not contained in any minimal prime of $R$. Then

1. the function $P : n \mapsto \text{length}_ R(R/x^ n R)$ satisfies $P(n) \leq n P(1)$ for $n \geq 0$,

2. if $x$ is a nonzerodivisor, then $P(n) = nP(1)$ for $n \geq 0$.

Proof. Since $\dim (R) = 1$, we have $\dim (R/x^ n R) = 0$ and so $\text{length}_ R(R/x^ n R)$ is finite for each $n$ (Algebra, Lemma 10.62.3). To show the lemma we will induct on $n$. Since $x^0 R = R$, we have that $P(0) = \text{length}_ R(R/x^0R) = \text{length}_ R 0 = 0$. The statement also holds for $n = 1$. Now let $n \geq 2$ and suppose the statement holds for $n - 1$. The following sequence is exact

$R/x^{n-1}R \xrightarrow {x} R/x^ nR \to R/xR \to 0$

where $x$ denotes the multiplication by $x$ map. Since length is additive (Algebra, Lemma 10.52.3), we have that $P(n) \leq P(n - 1) + P(1)$. By induction $P(n - 1) \leq (n - 1)P(1)$, whence $P(n) \leq nP(1)$. This proves the induction step.

If $x$ is a nonzerodivisor, then the displayed exact sequence above is exact on the left also. Hence we get $P(n) = P(n - 1) + P(1)$ for all $n \geq 1$. $\square$

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