= Solution
For the <discrete valuation> $v$ corresponding to $|\mathord\cdot|$, the <valuation ring> and its maximal ideal are
$$
\mathcal O_K=\{x\in K:v(x)\geq0\},
\qquad
\mathfrak m=\{x\in K:v(x)>0\}.
$$
The first is a subring of the field $K$, hence an <integral domain>. An element of $\mathcal O_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak m$ and $\mathfrak m$ is the unique <maximal ideal>.
Choose a <uniformizer> $\pi$ with $v(\pi)=1$. If $I\ne0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u\pi^n$ for a unit $u$, so $(x)=(\pi^n)$. Every $y\in I$ has $v(y)\geq n$, hence $y/x\in\mathcal O_K$, and therefore $I=(x)$. Thus $\mathcal O_K$ is a <discrete valuation ring>, in particular a <principal ideal domain>.
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