Why is the decomposition of $\operatorname{Tor}(M)$ into cyclic modules a "subset" of the decomposition of $M$?

Question

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.

Answer 1

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 $M$.

Answer 2

1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.
2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.