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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 $m<n$. 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 $m<n$. It is easily seen that this condition holds if and only if $x$ is not nilpotent. Why is this the case?

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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.

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