When does $(a,b)=(\gcd(a,b))$ hold?

Question

I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) \neq (1) = \gcd(x,y)$, but I thought that $(a,b)=\gcd(a,b)$ was always true so obviously I was wrong. But when exactly does this relation hold then, if $ a,b \in R$ and $R$ is a ring?

Answer 1

For commutative rings $R$, it holds if and only if $\gcd(a,b) = \alpha a + \beta b$ for some $\alpha,\beta \in R$. For if $\gcd(a,b) = \alpha a + \beta b$ then clearly $(\gcd(a,b)) \subseteq (a,b)$, and on the other hand anything in $(a,b)$ has the form $ra + sb$ which is a multiple of $\gcd(a,b)$ so lies in $(\gcd(a,b))$.

Conversely if $(\gcd(a,b)) = (a,b)$ then $\gcd(a,b) \in (a,b)$, so $\gcd(a,b) = \alpha a + \beta b$ for some $\alpha,\beta \in R$.

Every two elementa $a,b$ having a GCD which can be expressed as a linear combination of $a,b$ is equivalent to $R$ being a Bézout domain