**Proof plan.** In a Euclidean domain with [function](/page/Function) $\varphi$, every non-zero ideal contains an element of minimal $\varphi$-value; the division algorithm forces every other element to be a multiple of this minimal element. **Step 1: Pick a minimal element.** Let $R$ be a Euclidean domain with Euclidean function $\varphi : R \setminus \{0\} \to \mathbb{Z}_{\geq 0}$. Let $I \trianglelefteq R$ be a non-zero ideal. Choose $b \in I \setminus \{0\}$ such that $\varphi(b)$ is minimal among all non-zero elements of $I$. **Step 2: Every element of $I$ is divisible by $b$.** [claim: All Elements Are Multiples of b] For any $a \in I$, we have $b \mid a$. [/claim] [proof] By the division algorithm in $R$, write $a = bq + r$ with $r = 0$ or $\varphi(r) < \varphi(b)$. Then $r = a - bq \in I$ (since $a, bq \in I$). The minimality of $\varphi(b)$ among non-zero elements of $I$ forces $r = 0$. So $a = bq$, i.e. $b \mid a$. [/proof] **Step 3: $I = (b)$.** Since $b \in I$ and $I$ is an ideal, $(b) \subseteq I$. By Step 2, $a \in (b)$ for all $a \in I$, so $I \subseteq (b)$. Thus $I = (b)$, a principal ideal. $\square$