one dimensional regular local rings are dvrs

Suppose $R$ is regular. Lift a basis of the one-dimensional vector space $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ to $t\in \mathfrak{m}$. Then $\mathfrak{m}=(t)+{\mathfrak{m}}^2$. The finite module $Q=\mathfrak{m}\mathrm{/}(t)$ satisfies $\mathfrak{m}Q=Q$, so [F2] gives $Q=0$ and $\mathfrak{m}=(t)$. Since $\dim R=1$, $\mathfrak{m}\mathrm{\ne }0$ and $t$ is a nonzero nonunit. Also $t$ is not nilpotent: if $t^r=0$, every prime contains $t$ and hence $\mathfrak{m}$, making $\mathfrak{m}$ the only prime and forcing dimension zero.

By [F4], ${\bigcap }_{n\ge 0}(t^n)=0$. Thus for every nonzero $a\in R$ there is a largest $n$ with $a\in (t^n)$. Write $a=t^{n}u$. If $u\in \mathfrak{m}=(t)$ then $a\in (t^{n+1})$, a contradiction; therefore $u$ is a unit. Products of nonzero elements $t^{n}u$ and $t^{m}v$ are nonzero because $t$ is not nilpotent, so $R$ is a domain. In a nonzero ideal choose an element with least such exponent $n$; it generates the ideal. The zero ideal is principal too. Thus $R$ is a local PID with nonzero maximal ideal, hence a DVR by [F3].

Conversely, a DVR is a nonfield local PID of dimension one by [F3]. Its maximal ideal $(t)$ is nonzero, and $t\mathrm{\notin }(t^2)$ by cancellation in a domain, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has basis the class of $t$. Its embedding dimension is one, and it is regular by [F1].