Prove this is an exact sequence

Question

Here, o is a commutative integral domain. Maybe we can assume o is Noetherian.

$a,b \in o$. Prove the following sequence is exact. $$0 \longrightarrow ao \longrightarrow ao + bo \longrightarrow o/(ao : bo) \longrightarrow 0$$

The question comes from Nagata's paper "A General theory of Algebraic Geometry Over Dedekind Domains II". I know $0 \longrightarrow bo \longrightarrow ao + bo$ is exact. But the other positions are difficult.

Answer 1

If one of $a,b$ is 0, then the proof is easy. So assume $a,b\not=0$.

The last map in the sequence should be $$f: ao + bo \longrightarrow o/(ao:bo)$$ $$ as + br \mapsto r$$ To see $f$ is well-defined, if $as' + br' = as + br$, then $b(r'-r) = a(s-s')$. $r' - r \in (ao : bo)$. So $f(as' + br') = f(as + br)$.

Obviously, $f$ is surjective.

If $as + br \in \ker f$, then $r \in (ao:bo)$. Then $as + br \in ao$. So $\ker f = ao$.