Lemma 4

Let $A$ be a principal ideal domain and let $0 \to N \to M \to P \to 0$ be an exact sequence of $A$-modules. Suppose that there exist coprime elements $a, b \in A$ such that $aN = 0$ and $bP = 0$. Then the exact sequence splits. If in addition $N$ and $P$ are both cyclic, then $M$ is cyclic.

The module $M$ is torsion, since $abM = 0$. The first assertion follows from Cor. 3 of VII, p. 9. If $N$ and $P$ are cyclic, then they are finitely generated, and hence so is $M$ (II, p. 17, Cor. 5); since each $p$-primary component of $M$ is isomorphic to a $p$-primary component either of $N$ or of $P$, it follows from Remark 2 of VII, p. 10, that $M$ is cyclic.