1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $M$.

$\endgroup$
1
$\begingroup$
  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.
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Ovi Nov 18 '18 at 14:46
  • $\begingroup$ 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 $T$. $\endgroup$ – Bernard Nov 18 '18 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.