Every Euclidean Domain is a Principal Ideal Domain

Recall from the Euclidean Domains (EDs) page that if $(R, +, *)$ is an integral domain then $R$ is said to be a Euclidean domain if there exists a function $\delta : R \setminus \{ 0 \} \to \mathbb{N} \cup \{ 0 \}$ that satisfies the following properties:

• 1) For all $a, b \in R \setminus \{ 0 \}$ we have that:

(1)

\begin{align} \quad \delta (a) \leq \delta (ab) \end{align}

• 2) For all $a, b \in R \setminus \{ 0 \}$ there exists $q, r \in R$ such that:

(2)

\begin{align} \quad a = bq + r \end{align}

• and such that $r = 0$ or $\delta (r) < \delta (b)$.

We now prove an important result which says that every Euclidean domain is a principal ideal domain.

Theorem 1: Every Euclidean domain is a principal ideal domain.

• Proof: Let $(R, +, \cdot)$ be a Euclidean domain. Let $I \subseteq R$ be a nonzero ideal of $R$. Let $d \in I$ with $d \neq 0$ be any element such that $\delta (d)$ is minimum in $I$.

• We claim that $I = dR = \{ dx : x \in R \}$.

• Since $d \in I$ we have that:

(3)

\begin{align} \quad dR \subseteq I \quad (*) \end{align}

• Now let $a \in I$ with $a \neq 0$. Since $R$ is a Euclidean domain there exists elements $q, r \in R$ such that $a = dq + r$ and $r = 0$ or $\delta (r) < \delta (d)$. Now we have that $r = a - dq$ where $a \in I$ and $-dq \in I$, so $r = a - dq \in I$. From the minimality of $\delta (d)$ we have that:

(4)

\begin{align} \quad \delta (r) \geq \delta (d) \end{align}

• Hence, since $\delta (r) \not < \delta(d)$ we must have that $r = 0$. So $0 = a - dq$. Hence $a = dq$. So $a \in dR$. Thus:

(5)

\begin{align} \quad dR \supseteq I \quad (**) \end{align}

• From $(*)$ and $(**)$ we conclude that $I = dR$. In other words, $I = <d>$. So every Euclidean domain is a principal ideal domain. $\blacksquare$