Euclidean implies Dedekind-Hasse

Statement

Verbal statement

Any Euclidean norm on a commutative unital ring is a Dedekind-Hasse norm.

Definitions used

Euclidean norm

Further information: Euclidean norm

Dedekind-Hasse norm

Further information: Dedekind-Hasse norm

Related facts

• Dedekind-Hasse not implies Euclidean

Proof

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 $s\neq 1$.

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 $b\neq 0$, either $a\in (b)$ or there exists $r\in (a,b)$ such that $N(r)<N(b)$.

Proof: Since $b\neq 0$ 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.