Euclidean Geometry

Definitions ¹

In an integral 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 $a,b \in D (b \ne 0 )$, there exists $q$ and $r$ such that $$ a = bq + r $$ is satisfied. In this case, either $r = 0$ or $\nu (r) < \nu (b)$ must hold.
(ii): For all $a,b \in D (b \ne 0 )$, $\nu ( a ) \le \nu ( ab )$

$\mathbb{N}_{0}$ denotes the set including the natural numbers along with $0$.

Theorems

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

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

PID refers to a principal ideal domain, while UFD indicates a unique factorization domain.
The unit element is the identity for multiplication $1$, and a unit is an element with a multiplicative inverse.

Explanations

The term “Euclidean domain” may not be very long, but the abbreviation ED is often used.

Conditions (i) and (ii) are naturally satisfied in the ring of integers $\mathbb{Z}$, making it a Euclidean domain because there exists a Euclidean norm $\nu ( n ) := | n |$. Originally, the term Euclidean norm is derived from the Euclidean algorithm in number theory.

Considering a field $F$ and $F [ x ]$, it becomes a Euclidean domain by defining a Euclidean norm $\nu ( f(x) ) : = \deg ( f(x) )$. The division theorem applies to this condition.

$20190107\_023727.png$ Diagramming various domains like above easily shows how many favorable properties ED has.

Proofs

[1]

Let’s refer to an ideal of $D$ as $N$.

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

For all $n \in N$ not equal to $0$, one can find $b \ne 0$ that satisfies $$ \nu (b) \le \nu (n) $$ . If we call this $a \in N$, then by condition (i) $$ a = b q + r $$ must exist $q,r \in D$. Since $N = Nq$ is an ideal, $r = a - bq$ is also an element existing in $N$. $b$ being the element that minimizes $\nu (b)$ implies that according to condition (ii), it must be $r=0$. That all elements $a \in N$ are represented as $a = bq$ means $N = \left< b \right>$, hence all ideals $N$ are principal ideals.

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[2]

Since ED is a PID, and PID is UFD, hence 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 so $$ \nu ( u ) \le \nu ( u u^{-1} ) = \nu (1) $$ and by theorem [3], $\nu (1) \le \nu (1)$ thus $$ \nu ( u ) = \nu (1) $$

$( \impliedby )$

If we let $1 = uq + r$, then by theorem [2], $\nu ( u) = \nu (1)$ is the smallest except for $\nu (0)$. Since theorem [3] states that $\nu ( r) < \nu (u)$ satisfies only if $r=0$, $1 = uq$, and $u$ becomes a unit.

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