In a PID show that $(ab)=(a)\cap(b)$ $\iff$ $(a)+(b)=1$

Question

In a PID show that $(ab)=(a)\cap(b)$ $\iff$ $(a)+(b)=1$

I showed that backwards direction. Which is just corollary of Chinese reminder theorem. I am not sure how to show the forward direction.

Answer 1

First, write a proof when the PID is $\mathbb Z$, using facts about divisibility, and then change the language:

Proof for $\mathbb Z$: Let $(a) \cap (b) = (l)$ and $(a) + (b) = (d)$. We may assume $d = \gcd(a, b)$ and $ab = l = \operatorname{lcm}(a, b)$. But $a$ and $b$ divide $ab/d$, and if $d > 1$ this is a strict divisor of $ab = l$, so that $l$ is not the lcm, a contradiction. Thus $d = 1$.

Proof for a general PID $A$:

Let $(a) \cap (b) = (l)$ and $(a) + (b) = (d)$. We claim that $(a) \cap (b) \supset (ab/d)$. Indeed, $b/d \in A$, so $ab/d \in (a)$, and $a/d \in A$, so $ab/d \in (b)$. Thus $(ab/d) \subset (l)$, so that $l = ab$ divides $ab/d$. I.e. $dab$ divides $ab$. When $ab \neq 0$, this contradicts that $A$ is an integral domain.