If R is a Euclidean domain, then there are universal side divisors in R

The Mathematician — Fri, 02 Jun 2023 13:00:00 +0000

How to prove that a Euclidean Domain has universal side divisors

The best way to prove that is to use the definition of a Euclidean Domain straightly and by using the definition of universal side divisors.

Proof: let

R

be a Euclidean domain and let

b \in R - \widetilde{R} = R - R^{\times} \cup \{0\}

be a universal side divisor of minimal norm. This means that we took a universal side divisor which we are sure that

R

has no smaller norms of universal side divisors than the norm of

b

. By the definition of the Euclidean Domain, we can say

\begin{align*} x = qb + r \quad \text{where} \quad r = 0 \text{ or } N(r) < N(b). \end{align*}

Now, we need to show that

r \in R^{\times} \cup \{0\}

. It is obviously true that if

r = 0

, then

r \in R^{\times} \cup \{0\}

. If

N(r)

N(b)

, then we do know that

r

can't be a universal divisor since

b

is the smallest norm that is a universal divisor, which implicates that

r \not \in R - \widetilde{R}

. So

r

must be a unit and therefore

r \in R^{\times} \cup \{0\}

, which proves this proof.

It is crucial to notice that we are curious if there are universal side divisors, so showing one is perfectly fine.