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 R be a PID. We will start with the right implication:

\Rightarrow“: let d be the generator for the principal ideal generated by a and b, i.e.,

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

\Leftarrow“: reverse the previous proof.

Leave a Reply