Two ideals (a) and (b) in PID are comaximal iff gcd(a,b) = 1

How to prove that two ideals (a) and (b) in PID are comaximal iff gcd(a,b) = 1

The best way to tackle this problem is by using the definition of a comaximal.

Prove that any two ideal (a) and (b) in PID are comaximal iff gcd(a,b)=1

Proof: let RR be a PID. We will start with the right implication:

\Rightarrow“: let dd be the generator for the principal ideal generated by aa and bb, i.e.,

(d)=(a,b)={ax+by  x,yR}.\begin{equation*} (d) = (a,b) = \{ax+by \ | \ x,y\in R\}. \end{equation*}
We have given that (a)+(b)=R(a) + (b) = R since (a)(a) and (b)(b) are comaximal. So this means that 1(a)+(b)1 \in (a) + (b) and therefore we have an RR-linear combination ax+by=1ax + by = 1 for some x,yRx,y\in R. This means that gcd(a,b)=1gcd(a,b) = 1 since ax+by(a,b)ax + by \in (a,b).

\Leftarrow“: reverse the previous proof.

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