Exercise A0.V.2.15

Proposition

Let $R$ be a Euclidean domain with valuation $v.$ Then $R$ admits a valuation $v'$ such that $v'(ab) \geq v'(b)$ for all non-zero $a,b \in R.$

Proof

For each non-zero $a \in R,$ define $v'(a)$ as the minimum of $v(ab)$ as $b$ ranges over non-zero elements of $R.$

We will now prove that $R$ is a Euclidean domain with respect to $v'.$

Suppose $a, b \in R$ are non-zero, and $b \nmid a.$ Define the set $S$ of pairs $(q',r')$ such that $a = bq' + r'.$ Let $(q, r)$ be the pair minimizing $v(r).$

Suppose, for contradiction, that $v'(r) \geq v'(b).$ Let $r'$ and $b'$ be such that $v'(r) = v(r'r)$ and $v'(b) = v(b'b).$

Note that $v(r'r) \leq v(r).$

Let $a = q_2(b'b) + r_2$ with $v(r_2) < v(b'b).$ But then $a = (q_2b')b + r_2.$ But note that

$$v(r_2) < v(b'b) \leq v(r'r) \leq v(r),$$

which contradicts minimality of $r$.

$\blacksquare$