If $D$ is an integral domain then it is a PID iff it has a Dedekind-Hasse valuation, that is, a function $\nu:D-\{0\}\rightarrow \mathbb{Z}^+$ such that for any $a,b\in D-\{0\}$ either

• $a\in (b)$ or
• $\exists \alpha\in(a)\exists\beta\in(b)\left[0<\nu(\alpha+\beta)<\nu(b)\right]$

Proof: First, let $\nu$ be a Dedekind-Hasse valuation and let $I$ be an ideal of an integral domain $D$ Take some $b\in I$ with $\nu(b)$ minimal (this exists because the integers are well-ordered) and some $a\in I$ such that $a\neq 0$ $I$ must contain both $(a)$ and $(b)$ and since it is closed under addition, $\alpha+\beta\in I$ for any $\alpha\in(a),\beta\in(b)$

Since $\nu(b)$ is minimal, the second possibility above is ruled out, so it follows that $a\in (b)$ But this holds for any $a\in I$ so $I=(b)$ and therefore every ideal is princple.

For the converse, let $D$ be a PID. Then define $\nu(u)=1$ for any unit. Any non-zero, non-unit can be factored into a finite product of irreducibles (since every PID is a UFD), and every such factorization of $a$ is of the same length, $r$ So for $a\in D$ a non-zero non-unit, let $\nu(a)=r+1$ Obviously $r\in\mathbb{Z}^+$

Then take any $a,b\in D-\{0\}$ and suppose $a\notin (b)$ Then take the ideal of elements of the form $\{\alpha+\beta|\alpha\in (a), \beta\in(b)\}$ Since this is a PID, it is a principal ideal $(c)$ for some $r\in D-\{0\}$ and since $0+b=b\in(c)$ there is some non-unit $x\in D$ such that $xc=b$ Then $N(b)=N(xr)$ But since $x$ is not a unit, the factorization of $b$ must be longer than the factorization of $c$ so $\nu(b)>\nu(c)$ so $\nu$ is a Dedekind-Hasse valuation.