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Is this true in any commutative rings? I.e. $$\gcd(a,b)=1\implies (a)+(b)=R$$ I think there must be some conditions on the ring to make this implication otherwise it does not work.

This may related to this question here.

CO2
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    This is not true in arbitrary rings. For (counter)example, in a polynomial ring in two variables $R=k[x,y]$, we have $\gcd(x,y)=1$, but $(x)+(y) \neq R$. – Zach Teitler Jun 21 '18 at 22:08
  • In which domain is it true? – CO2 Jun 21 '18 at 22:09
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    Gcds do not exist in general. Rings in which they exist are called gcd domains, and they are integrally closed domains. U.F.D.s are gcd domains, but the converse is false. – Bernard Jun 21 '18 at 22:09
  • The implication you mention is true in P.I.D.s or more generally in Bézout domains for the non-noetherian case. – Bernard Jun 21 '18 at 22:14

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This is not true in arbitrary rings. For (counter)example, in a polynomial ring in two variables $R = k[x,y]$, we have $\gcd(x,y)=1$, but $(x)+(y) \neq R$.

It is true in Bézout domains. I don’t know if it’s equivalent to being a Bézout domain.

Zach Teitler
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  • $K$ is a field? – CO2 Jun 21 '18 at 22:16
  • $k$ can be a field but it doesn’t have to be. For $R$ to be a domain we must have $k$ a domain. That’s all I need for the $k[x,y]$ example. If you want a simple, concrete example, then take $k=\mathbb{Z}$ or a field. – Zach Teitler Jun 21 '18 at 22:20