TITLE: Why is the decomposition of $\operatorname{Tor}(M)$ into cyclic modules a "subset" of the decomposition of $M$?
QUESTION [1 upvotes]: In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then 
$$M \cong R^r \oplus R/(a_1) \oplus \cdots \oplus R/(a_n)$$
and that 
$$\operatorname{Tor}(M) \cong R/(a_1) \oplus \cdots \oplus R/(a_n)$$
I don't understand how the second part follows from the first. First, I don't see why $\operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.
Secondly, if we assume that $\operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get
$$\operatorname{Tor}(M) \cong R^k \oplus R/(b_1) \oplus \cdots \oplus R/(b_m)$$ 
I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) \oplus \cdots \oplus R/(b_m)$ part has to be the same as the $R/(a_1) \oplus \cdots \oplus R/(a_n)$ part.

REPLY [1 votes]: Instead of thinking of Tor$(M)$, just think of Tor$(R^k \oplus R/(b_1) \oplus \cdots \oplus R/(b_m))$. It is clear that this is just $R/(b_1) \oplus \cdots \oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.
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