Let $R$ be a noetherian integral domain.
Suppose that every finitely generated $R$-module is a direct sum of cyclic $R$-modules. Show that $R$ is a PID.
Here's my attempt:
Let $I$ be an ideal. Since $R$ is noetherian, $I$ can be considered as a finitely generated $R$-module. By the assumption, $I$ is isomorphic to direct sum of cyclic $R$-modules, say $M_1 \otimes \cdots \otimes M_n$. Since $I$ is torsion-free (as $R$ is a domain), it can be shown that $M_j \cong R$ for all $j$, and so $I \cong R^n$.
Now, if I can show $n = 1$, then it is not hard to deduce that $I$ is a principal ideal. But, this is where I am stuck.
I would appreciate any hint.
Let $m_1,m_2$ be generators for $M_1,M_2$. We know $M_1 \cap M_2 = 0$, since they are two of the modules in the direct sum decomposition. But $m_1 m_2 \in M_1 \cap M_2$, and because $R$ is a domain, $m_1 m_2 \neq 0$.