Multi-stage Euclidean implies Bezout

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 must also satisfy the second commutative unital ring property
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Statement

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.