# Regular local rings of Krull dimension $1$ are integral domains

I am reading a proof of the fact that any regular local ring $$R$$ of Krull dimension $$1$$ is an integral domain.

It was previously shown that the maximal ideal $$\mathfrak m$$ of $$R$$ is generated by some $$x\in R$$. Then, it is proven that any non-zero ideal of $$R$$ is of the form $$(x^n)$$ for some $$n\geq 0$$. One now argues that $$(x^n)$$ cannot be prime for $$n\geq 0$$, since $$x^m\not\in (x^n)$$ whenever $$m. This implies that $$(0)$$ is a prime ideal (since $$R$$ has Krull dimension $$1$$), hence $$R$$ is an integral domain.

It isn't clear to me why $$x^m$$ does not belong to $$(x^n)$$ when $$m. It is easily seen that this condition holds if and only if $$x$$ is not nilpotent. Why is this the case?

Recall that the nilradical is the intersection of all prime ideals. By hypothesis $$R$$ has positive dimension; in particular, there exists some non-maximal prime ideal $$p$$. If $$x$$ were to be nilpotent, then $$x \in \mathfrak N_{\mathsf{nil}} \subset p$$ and by maximality $$(x) = p$$; a contradiction.