Euclidean Domains (EDs)

Recall from the Principal Ideal Domains page that an integral domain $(R, +, \cdot)$ is said to be a Principal ideal domain if every ideal $I \subseteq R$ is a principal ideal ($I = <a>$ for some $a \in R$).

We will now look at another special type of integral domain known as a Euclidean domain.

Definition: Let $(R, +, \cdot)$ be an integral domain. Then $R$ is said to be a Euclidean Domain if there exists a function $\delta : R \setminus \{ 0 \} \to \mathbb{N} \cup \{0 \}$ which satisfies the following properties: 1) For all $a, b \in R \setminus \{ 0 \}$ we have that $\delta (a) \leq \delta (ab)$. 2) For all $a, b \in R \setminus \{ 0 \}$ there exists $q, r \in R$ such that $a = bq + r$ and where $r = 0$ or $\delta (r) < \delta (b)$.

For example, for any field $(F, +, \cdot)$, the field of polynomials over $F$, $F[x]$, is a Euclidean domain where $\delta = \deg$. This is because for any polynomials $f, g \in F[x] \setminus \{ 0 \}$ we have that:

(1)

\begin{align} \quad \delta (f) = \deg (f) \leq \deg(fg) = \delta (fg) \end{align}

And from the Division algorithm for any $f, g \in F[x] \setminus \{ 0 \}$ there exists polynomials $q, r \in F[x]$ such that $f(x) = g(x)q(x) + r(x)$ and either $r(x) = 0$ or $\delta(r) = \deg(r) < \deg(g) = \delta(g)$.

For another example of a Euclidean domain, consider the integral domain $(\mathbb{Z}[i], +, \cdot)$ where:

(2)

\begin{align} \quad \mathbb{Z}[i] = \{ m + ni : m, n \in \mathbb{Z} \} \end{align}

Let $\delta : \mathbb{Z}[i] \setminus \{ 0 \} \to \mathbb{N} \cup \{ 0 \}$ be defined for all $m + ni \in \mathbb{Z}[i]$ by:

(3)

\begin{align} \quad \delta(m + ni) = m^2 + n^2 \end{align}

We now verify that $\delta$ satisfies (1) and (2). Let $a +bi, c + di \in \mathbb{Z}[i] \setminus \{ 0 \}$. Then:

(4)

\begin{align} \quad \delta((a+bi)(c+di)) &= \delta ((ac -bd) + (ad + bc)i) \\ &= (ac - bd)^2 + (ad + bc)^2 \\ &= (ac)^2 -2acbd + (bd)^2 + (ad)^2 + 2adbc + (bc)^2 \\ &= (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2 \\ &= a^2(c^2 + d^2) + b^2(c^2 + d^2) \\ & \geq a^2 + b^2 \\ & \geq \delta (a + bi) \\ \end{align}

So (1) holds.