Theorem 11.2 in Matsumura

Question

I have a question about the proof of $(4) \implies (3)$ in Theorem 11.2 of Matsumura's Commutative Ring Theory.

The setup is as follows. Let $R$ be a normal Noetherian local ring of dimension $1$ with maximal ideal $\mathfrak{m}$. Choose $x\in\mathfrak{m}\setminus\mathfrak{m}^2$. Matsumura now says:

Since $\dim R=1$ the only prime ideals of $R$ are $(0)$ and $\mathfrak{m}$, so that $\mathfrak{m}$ must be a prime divisor of $xR$, and there exists $y\in R$ such that $(xR:y)=\mathfrak{m}$.

Why is the last statement true?

Answer 1

That is because $\;\mathfrak m/xR$ is in $\operatorname{Ass}(R/xR)$ – actually it is the sole element of $\operatorname{Ass}(R/xR)$.