Multi-stage Euclidean implies Bezout

Verbal statement
Any multi-stage Euclidean ring is a Bezout ring. This also shows that any multi-stage Euclidean domain is a Bezout domain.

Proof
Let $$R$$ be a $$n$$-stage Euclidean ring. We need to show that any finitely generated ideal in $$R$$ is principal. In other words, we need to demonstrate that given $$a_1, a_2, \ldots, a_l \in R$$ we can find a single element $$b$$ which generates the ideal spanned by the $$a_i$$s.

From the definition of multi-stage Euclidean norm, we can, for any two elements $$p,q$$ of $$R$$, find one of the following:


 * A pair $$(p',q')$$ with $$N(q') < N(q)$$, and which generates the same ideal as $$(p,q)$$
 * A single element $$r$$ which generate the same ideal as $$(p,q)$$

Now, starting with the $$a_i$$s, we can keep applying this procedure to pairs, replacing a pair $$(p,q)$$ by the corresponding $$(p',q')$$ or by the corresponding $$r$$. The procedure terminates in finitely many steps, and it can only terminate when there is a single element. This is the generator of the ideal.