Euclidean implies Dedekind-Hasse

Verbal statement
Any fact about::Euclidean norm on a commutative unital ring is a fact about::Dedekind-Hasse norm.

Related facts

 * Dedekind-Hasse not implies Euclidean

Key idea
The main idea here is that while the Euclidean condition allows us to find an element of the form $$a - bq$$ of smaller norm, the Dedekind-Hasse condition only requires some $$R$$-linear combination of $$a$$ and $$b$$ to have smaller norm. In particular, for the Dedekind-Hasse condition, we can pick $$sa - bq$$ with $$s \ne 1$$.

Proof details
Given: A commutative unital ring $$R$$ with a Euclidean norm $$N$$.

To prove: $$N$$ is a Dedekind-Hasse norm on $$R$$: given $$a,b \in R$$ with $$b \ne 0$$, either $$a \in (b)$$ or there exists $$r \in (a,b)$$ such that $$N(r) < N(b)$$.

Proof: Since $$b \ne 0$$ and $$N$$ is a Euclidean norm, we can write:

$$a = bq + r$$

where either $$r = 0$$ or $$N(r) < N(b)$$. Consider both cases:


 * $$r = 0$$: In this case, $$a = bq$$, so $$a \in (b)$$, so the Dedekind-Hasse condition is satisfied.
 * $$N(r) < N(b)$$: In this case, $$r = a - bq$$, so $$r \in (a,b)$$, with $$N(r) < N(b)$$, so the Dedekind-Hasse condition is satisfied.