Dedekind-Hasse valuation

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

• $a\in (b)$

  or

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 $a\ne 0$. $I$ 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 non-zero, non-unit 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 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.