In a PID $R$ with $ a\in R $ irreducible and $\langle a \rangle$ the ideal generated by $a$, and $b \in R\setminus \langle a \rangle$, is $1\in \gcd(a,b)$.
Why is it sufficient to show that $g \in \gcd(a,b)$ is a unit in R to show the above?
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If $x$ is a gcd of $a$ and $b$ and $u$ is a unit, then $ux$ is a gcd of $a$ and $b$ – diracdeltafunk Mar 11 '21 at 01:20
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Is $\gcd(a,b)$ an element, or is it set of elements (and if so, what set)? Because otherwise $1\in\gcd(a,b)$ does not make sense. – Arturo Magidin Mar 11 '21 at 01:23
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1its the set of greatest common divisors – Shadesmar Mar 11 '21 at 01:25