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

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.

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 $$M$$.
• Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $\operatorname{Tor}(M) \cong R/(a_1) \oplus \cdots \oplus R/(a_2)$, wouldn't we have to show that $M \cong R^r \oplus \operatorname{Tor}(M)$ also? – Ovi Nov 18 '18 at 14:46
• But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $M\to R^k$ with kernel $T$. – Bernard Nov 18 '18 at 15:26