If $v$ is a discrete valuation on a field $K$, then $A:=\left\{ a \in K : v(a) \geq 0 \right\}$ is a principal ideal domain.

Question

J.S. Milne's Algebraic Number Theory says

[Proposition 3.27] Let $v$ be a discrete valuation on $K$, then $$A:=\left\{ a \in K : v(a) \geq 0 \right\}$$ is a principal ideal domain with maximal ideal $$\mathfrak{m}:= \left\{ a \in K : v(a)>0 \right\}.$$

I see that $A$ is a valuation ring since $v(a^{-1})=v(1)-v(a)=-v(a)$, but why should $A$ be PID?

Answer 1

For completeness I will write the proof that is essentially in the comment by Mathmo123.

Let $I\subseteq A$ be an ideal. Then $v(I)$ is a subset of $\mathbb{Z}_{\geq 0}$. Take $x\in I$ such that $v(x)=\min v(I)$.

Of course we have $(x)\subseteq I$. Now if $y\in (x)$ we have $v(yx^{-1})=v(x)-v(y)\geq 0$ so $yx^{-1}\in A$ and then $y=(yx^{-1})x\in I$. Hence $I\subseteq (x)$ from where $I=(x)$.