Euclidean Domain

Definition ¹

In the domain $D$, if there exists a Euclidean Norm $\nu : D \setminus \left\{ 0 \right\} \to \mathbb{N}_{0}$ that satisfies the following two conditions, then $D$ is called a Euclidean Domain.

(i): For all $a,b \in D (b \ne 0 )$ $$ a = bq + r $$ There exist $q$ and $r$ that satisfy this. At this time, it must be either $r = 0$ or $\nu (r) < \nu (b)$.
(ii): For all $a,b \in D (b \ne 0 )$, $\nu ( a ) \le \nu ( ab )$

$\mathbb{N}_{0}$ represents the set of natural numbers including $0$.

Theorem

Let $0$ be the identity element of the Euclidean Domain $D$, $1$ be the unit element, and $\nu$ be the Euclidean norm.

[1]: Every ED is a PID.
[2]: Every ED is a UFD.
[3]: For all $d \in D$ that are not $0$, $\nu (1) \le \nu (d)$
[4]: $u \in D$ is a unit element $\iff$ $\nu ( u ) = \nu (1)$

PID refers to a Principal Ideal Domain, and UFD refers to a Unique Factorization Domain.
Unit Element is the multiplicative identity $1$, and Unit is an element that has a multiplicative inverse.

Explanation

The term ‘Euclidean Domain’ is not particularly long, but usually, the abbreviation ED is widely used.

Conditions (i) and (ii) are naturally satisfied in the integer ring $\mathbb{Z}$, due to the existence of a Euclidean norm $\nu ( n ) := | n |$ which makes $\mathbb{Z}$ a Euclidean Domain. The term ‘Euclidean Norm’ itself is derived from the Euclidean Algorithm in number theory.

On the other hand, considering ◁eq28◁ for a field $F$ by defining a Euclidean norm $\nu ( f(x) ) : = \deg ( f(x) )$, it becomes the Euclidean norm. The Division Algorithm falls under these conditions in the first place.

$20190107\_023727.png$ When diagramming various domains, it looks like the above, which easily shows how many good properties the ED possesses.

Proof

[1]

Let the ideal of $D$ be $N$.

$N = \left\{ 0 \right\} = \left< 0 \right>$ is naturally a Principal Ideal, so let’s consider $N \ne \left\{ 0 \right\}$.

Then for every $n \in N$ which is not $0$, $$ \nu (b) \le \nu (n) $$ we can find one $b \ne 0$ which satisfies it. Let’s denote this as $a \in N$, by condition (i), $$ a = b q + r $$ $q,r \in D$ must exist which satisfies this. Since $N = Nq$ is an ideal, $r = a - bq$ is also contained within $N$. Since $b$ was the element minimizing $\nu (b)$, by condition (i), $r=0$ must hold. That all elements $a \in N$ can be expressed as $a = bq$ implies that $N = \left< b \right>$, meaning every ideal $N$ is a principal ideal.

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[2]

Since ED is PID, and PID is UFD, ED is UFD.

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[3]

By condition (ii), $$ \nu (1) \le \nu ( 1 d) = \nu (d) $$

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[4]

$( \implies )$

Since $u$ is a unit, its inverse $u^{-1}$ exists, $$ \nu ( u ) \le \nu ( u u^{-1} ) = \nu (1) $$ and by theorem [3], $\nu (1) \le \nu (1)$, hence, $$ \nu ( u ) = \nu (1) $$

$( \impliedby )$

Let it be $1 = uq + r$. If $\nu (1) = \nu (u)$ then $\nu (1) < \nu (u)$ is not true, so by condition (i) of the definition, $r=0$ must hold. Hence, $1 = uq$, and thus $u$ is a unit.

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Euclidean Domain $\implies$ Principal Ideal Domain $\implies$ Unique Factorization Domain $\implies$ Domain
Euclidean Domain $\implies$ Principal Ideal Domain $\implies$ Noetherian Ring

Fraleigh. (2003). A first course in abstract algebra(7th Edition): p401. ↩︎