Local Ring with principal maximal ideal, powers of maximal ideal

Question

Let $(R,m)$ be a local ring such that $m = (x)$ is principal and $x^{k} \ne 0 , \forall k \geq 0$. How do I show that $(x^{k}) \ne (x^{k+1})$ for all $k \geq 0$? If I proceed by contradiction, I get that $x^{k} \in (x^{k+1}) \implies rx^{k+1} = x^{k} \implies x^{k}(rx-1) = 0$. If $R$ is a domain, I can get my desired contradiction, but that is not an assumption. There must be something I am not considering, any hint would be appreciated.

Answer 1

Sorry, this question is actually a duplicate. The argument is $1 + m \subset R^{*}$, because $1 + y : y \in m \implies 1 + y \notin m$, so $(1 + y) = R$ by the fact that $R$ is local. This implies $\forall a \in R$, $xa \mid a \iff a = 0$. Suppose $xa \mid a \implies a(1-rx) = 0$, but $1-rx$ is a unit so $a = 0$. If $(x^{k}) = (x^{k+1})$, then $x^{k}(1-rx) = 0$, but $1-rx \in R^{*} \implies x^{k} = 0$, which is a contradiction.