Call a ring local if it has a unique maximal ideal. Let $R$ be a local PID. I wish to show that it is a discrete valuation ring. I've already shown that $R$ has a unique irreducible element, say $t$. I wish to show that (*) any non-zero element $r$ of frac$(R)$ can be expressed as $r=t^n\cdot u$, for some integer $n$ and a unit $u$. Once I have this, I will use it to define the discrete valuation.
I'm stuck at (*). I feel like I can use that $R$ is a PID, hence a UFD, where unique factorization into irreducibles is possible. However, $r$ is an element of frac$(R)$. Any help is appreciated. Thanks.
Since $R$ is a local PID, every element $r\in R$ has a factorization $r=up^n$, where $p$ is a generator of the unique maximal ideal (and, up to associates, the only prime element in $R$, because in a PID, every nonzero prime ideal is maximal) and $u$ invertible.
If $r,s\in R$, with $r,s\ne0$, we have $r=up^n$ and $s=vp^m$, so $$ \frac{r}{s}=\frac{up^n}{vp^m}=(uv^{-1})p^{n-m} $$