Let $R$ be a commutative ring and let $a, b \in R\setminus\{0\}$. Suppose that $d$ is a gcd of $a$ and $b$.
Suppose further that $R$ is a Euclidean domain with valuation $\delta$. Let $c$ be another gcd of $a$ and $b$. Is it possible that $\delta(c) \ne \delta(d)$?
I am struggling with understanding Euclidean valuation. I understand that it is a function and its definition is:
$\delta : R\setminus\{0\} \to \mathbb{N}$ such that, for all nonzero $a, b \in R$,
• $\delta(a) \le \delta(ab)$, and
• $\exists q,r \in R$ such that $b = qa + r$ and either $r = 0$ or $\delta(r) < \delta(a)$.
Any hints as to how to approach this question?
Two gcd's may differ by a unit $b\in R^{\times}$ in a Euclidean ring $(R,d)$. However, we have the following result, see the notes of Keith Conrad, Theorem $3.1$.
Theorem $3.1.$ Let $(R,d)$ be a Euclidean domain where $d$ satisfies the $d$-inequality $d(a)\le d(ab)$ for all $a,b\in R^{\times}$. Then we have, for a unit $b$ in $R$, that $d(ab)=d(a)$ for all $a\in R$.
We may assume that the Euclidean function $d$ satisfies the $d$-inequality by Theorem $2.1$. You have already assume it.
Hence, if we have $gcd(a,b)=c$ and also $gcd(a,b)=c'$, then $c'=bc$ for a unit $b$, and $d(c)=d(c')$ by the Theorem.