The Stacks project

Proof. If $x$ is a unit then the result is true. Hence we may assume $x \in \mathfrak m$ the maximal ideal of $R$. Since $x$ is not zero and $R$ is a domain we have $\dim (R/xR) = 0$, and hence $R/xR$ has finite length. Consider $M \subset K^{\oplus r}$ as in the lemma. We may assume that the elements of $M$ generate $K^{\oplus r}$ as a $K$-vector space after replacing $K^{\oplus r}$ by a smaller subspace if necessary.

Suppose first that $M$ is a finite $R$-module. In that case we can clear denominators and assume $M \subset R^{\oplus r}$. Since $M$ generates $K^{\oplus r}$ as a vectors space we see that $R^{\oplus r}/M$ has finite length. In particular there exists an integer $c \geq 0$ such that $x^ cR^{\oplus r} \subset M$. Note that $M \supset xM \supset x^2M \supset \ldots $ is a sequence of modules with successive quotients each isomorphic to $M/xM$. Hence we see that

The same argument for $M = R^{\oplus r}$ shows that

By our choice of $c$ above we see that $x^ nM$ is sandwiched between $x^ n R^{\oplus r}$ and $x^{n + c}R^{\oplus r}$. This easily gives that

Hence in the finite case we actually get the result of the lemma with equality.

Suppose now that $M$ is not finite. Suppose that the length of $M/xM$ is $\geq k$ for some natural number $k$. Then we can find

with $N_ i \not= N_{i + 1}$ for $i = 0, \ldots k - 1$. Choose an element $m_ i \in M$ whose congruence class mod $xM$ falls into $N_ i$ but not into $N_{i - 1}$ for $i = 1, \ldots , k$. Consider the finite $R$-module $M' = Rm_1 + \ldots + Rm_ k \subset M$. Let $N'_ i \subset M'/xM'$ be the inverse image of $N_ i$. It is clear that $N'_ i \not=N'_{i + 1}$ by our choice of $m_ i$. Hence we see that $\text{length}_ R(M'/xM') \geq k$. By the finite case we conclude $k \leq r\text{length}_ R(R/xR)$ as desired. $\square$