Exercise A0.V.2.21

Definition (Dedekind-Hasse valuation)

A Dedekind-Hasse valuation is a valuation $v$ such that for all $a$ and all non-zero $b$ in $R,$ either $b \mid a$ or there exists $r \in (a,b)$ such that $v(r) < v(b).$

Proposition

An integral domain $R$ is a PID if and only if it admits a Dedekind-Hasse valuation.

Proof

Suppose $R$ admits a Dedekind-Hasse valuation $v.$ Let $I$ be an ideal of $R.$ If $I$ is the zero ideal, it is principal. Suppose $I$ is non-zero.

Let $b \in I$ be non-zero element minimizing $v.$ Take arbitrary $a \in I.$ If $b \mid a$ then $a \in (b).$

Suppose instead that $b \nmid a.$ Let $r \in (a,b)$ with $v(r) < v(b).$ Write $r = sa - qb,$ and observe that $sa \in I$ and $qb \in I,$ hence $r \in I.$ But this is impossible, as no element of $I$ has lower valuation than $v(b).$ Thus $b \mid a.$ This suffices to show that $(b) = I.$

Suppose conversely that $R$ is a PID. Let $v : R\setminus {0} \to \mathbb{Z}_{\geq 0}$ be the map sending $a$ to the size of the multiset of irreducible factors of $a.$

Fix arbitrary $a, b \in R$ such that $b$ is non-zero. If $b \mid a,$ we are done. Suppose $b \nmid a.$ Let $r = \gcd(a,b).$ We know that $r \in (a,b)$ and furthermore any irreducible factor of $r$ must be also a factor of $b,$ hence $v(r) \leq v(b).$ If $v(r) = v(b),$ then all factors of $b$ are factors of $r,$ which implies $b \mid a.$ But we have $b \nmid a,$ therefore $v(r) < v(b),$ as required.

$\blacksquare$