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

•
$a\in (b)$
or

•
$$
Proof: First, let $\nu $ be a DedekindHasse 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 wellordered) and some $a\in I$ such that $a\ne 0$. $I$ 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 nonzero, nonunit can be factored into a finite product of irreducibles^{} (since http://planetmath.org/node/PIDsareUFDsevery PID is a UFD), and every such factorization of $a$ is of the same length, $r$. So for $a\in D$, a nonzero nonunit, 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 nonunit $x\in D$ 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 DedekindHasse valuation.
Title  DedekindHasse valuation 

Canonical name  DedekindHasseValuation 
Date of creation  20130322 12:51:16 
Last modified on  20130322 12:51:16 
Owner  Henry (455) 
Last modified by  Henry (455) 
Numerical id  5 
Author  Henry (455) 
Entry type  Definition 
Classification  msc 13G05 
Related topic  EuclideanValuation 
Defines  DedekindHasse norm 
Defines  DedekindHasse valuation 