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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?

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  • $\begingroup$ 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. $\endgroup$ – reuns Oct 24 at 11:27
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    $\begingroup$ $\mathbb Z$ and polynomial rings? $\endgroup$ – Thomas Shelby Oct 24 at 11:38
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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.

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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.

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

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