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?
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2It holds iff $\gcd\left(a,b\right)$ is expressible as an $R$-linear combination of $a$ and $b$. It certainly holds when $R$ is a Euclidean domain, then, as the Euclidean algorithm furnishes you with such an expression. I know this is not an answer to your question, but I hope it helps towards finding one. – Sam Streeter Nov 19 '18 at 11:53
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For an application of this property, see https://math.stackexchange.com/questions/597543 – Watson Nov 19 '18 at 13:30
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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
Matthew Towers
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