Euclidean implies principal ideal

This article gives the statement and possibly, proof, of an implication relation between two commutative unital ring properties. That is, it states that every commutative unital ring satisfying the first commutative unital ring property (i.e., Euclidean ring) must also satisfy the second commutative unital ring property (i.e., principal ideal ring)
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Statement

Property-theoretic statement

The property of commutative unital rings of being a Euclidean ring is stronger than the property of being a principal ideal ring.

Verbal statement

Any Euclidean ring is a principal ideal ring. In particular, any Euclidean domain is a principal ideal domain.

Proof

The key idea behind the proof is to, given an ideal, exhibit an element that generates the ideal. The claim is as follows:

Consider the Euclidean norm as a function on the ideal. Pick any element with minimum norm (such an element exists because the set of possible norms is a well-ordered set). Then, the ideal is generated by that element.

(Note that when the ideal is the zero ideal, the element of smallest possible norm is zero, whose norm is assumed to be $\infty$).

Let's prove this claim. Suppose $R$ is a Euclidean ring with Euclidean norm $N$, and $I$ is an ideal inside $R$. Let $d$ be the smallest possible norm among elements of $I$, and let $a\in I$ be an element such that $N(a)=d$. Now pick and $c\in I$. We want to show that $c$ is a multiple of $a$.

By the division algorithm over $R$, there exist $q,r\in R$ such that:

$c=aq+r,\qquad N(r)<N(a)$ or $r=0$

Now note that $r=c-aq$, and since $c,a\in I$ and $I$ is an ideal, $r\in I$. Since $a$ was chosen with minimum norm, $N(r)<N(a)$ is impossible, so $r=0$, and thus $c$ is a multiple of $a$.