Prove that if $\text{gcd}(v, w) \simeq av + bw$, then $\text{gcd}(a, b) \simeq 1$.

Question

Let $R$ be a principal ideal domain and $g, v, w \in R \setminus \{ 0 \}$, with $\text{gcd}(v, w) \simeq g$.

What needs to be proven is the following: If $a, b \in R$ with $av + bw \simeq g$, then $\text{gcd}(a, b) \simeq 1$.

My idea was to use Bézout's identity, since $R$ is a principal ideal domain, so there are $x, y \in R$ such that $g \simeq xv + yw \simeq av + bw$, but I do not see what to conclude from this, so if anybody could maybe give me a little nudge in the right direction, I would be very thankful.

Answer 1

As a guiding idea, consider the following. Bezout's lemma (and its converse) says that all linear combinations $av+bw$ (using elements $a,b\in R$) of $v,w$ are multiples of $g$, i.e. things of the form $dg$ with $d\in R$. What we should think is that $d$ is capturing common factors of the coefficients $a,b$. In particular, $a,b$ should be relatively prime if and only if $d=1$.

As a hint for your question, try proving your desired claim by contradiction, i.e. assume that $\gcd(a,b)\neq 1$ (i.e. $\gcd(a,b)$ is not a unit). Of course, you'll want to use $av+bw=g$ somehow.