# Motivation of the definition of Euclidean Domain

We know the definition of Euclidean Domain is

An Euclidean Domain is an Integral domain $$(E,+,\ast )$$ together with a function $$v: E\setminus \{0\} \to \mathbb{N} \cup \{0\}$$ such that

(i) for all $$a,b \in E$$ with $$b \neq 0$$, there exist $$q,r \in E$$ such that $$a = qb + r$$ ,where $$r=0$$ or $$v(r) \lt v(b)$$

(ii) for all $$a,b \in E \setminus \{0\}$$, $$v(a) \leq v(ab)$$.

What is the motivation behind the definition of Euclidean Domain. Is it a generalisation of something?

• It is a way to prove it is a PID, an algorithm to find $c$ such that $(a,b)=(c)$. In PIDs there is unique factorization in a way quite similar to integers. – reuns Oct 24 at 11:27
• $\mathbb Z$ and polynomial rings? – Thomas Shelby Oct 24 at 11:38

## 2 Answers

Yes, in fact it's generalization of division. In integers we can divide any integer $$a$$ to to any other like $$b$$ and write $$a=qb+r$$ that $$q$$ is quotient and $$r$$ is reminder that less than $$b$$ always. Now we generalize it to Euclidean domain by that definition and function.

The second criteria is generalization of $$a \leq ab$$.

Also the "order" is actually generalized before this.

The motivation comes from a strong property of natural (and integer) numbers: the Euclidean algorithm that allows to find a gcd between two numbers.

The function $$v$$ is then just a degree function that gives you a weight comparison between these "abstract" elements of the ring. For the integers $$v=|\cdot|$$, for the ring of polynomial is the degree function.

• The possibility of finding a gcd between elements is what really matters in proving that the ring is a PID. – blipgo Oct 24 at 11:40