Finitely generated module $M$ over principal ideal domain splits into direct sum of $M_\text{tors}$ and $M' \subset M$.

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Let $R$ be a principal ideal domain and $M$ be an finitely generated $R$-module.

The torsion module $M_{\text{tors}}$ is defined as $$ M_{\text{tors}} := \left\{ m \in M \;|\; \text{there is}\; a\in R \setminus \{0\} \;\text{with}\; am = 0 \right\}. $$

I want to show that there is a $R$-submodule $M' \subset M$ with $M = M' \oplus M_{\text{tors}}$.

What I tried: I know that

$$ M \cong R^r \oplus R/(a_1) \oplus \cdots \oplus R/(a_s), $$

for some $r \in \mathbb{N}_0$ and $a_1, \dots a_s \in R \setminus \{0\}$.

It is also clear that $R/(a_i)$ are fields, since $(a_i)$ is maximal because we are in a principal ideal domain.

However, I don't know how to show that $R^r \cong M_\text{tors}$.