📂Abstract Algebra

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 $a,b \in D (b \ne 0 )$, $\nu ( a ) \le \nu ( ab )$

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• $\mathbb{N}_{0}$ 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 $\iff$ $\nu ( u ) = \nu (1)$

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• 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.

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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Related Information

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