Torsion is "multiplicative"

Question

Let $A$ be a PID and let $M$ be a finitely generated $A$-module. Then there exist $m\in \mathbb{N}$ and $d_1, \dotsc,d_n\in A$ such that $M\cong A^m\times (A/d_1A)\times\cdots\times (A/d_nA)$, with $d_1\mid d_2\mid\cdots \mid d_n$ unique up to associates. So the ideal $(d_1\cdots d_n)A$ of $A$ is well-defined, and we denote it by $\operatorname{det} M$. Now let $0\to M\to N\to P\to 0$ be an exact sequence of finitely generated $A$-modules. I want to prove that $\operatorname{det} N=(\operatorname{det} M)(\operatorname{det} P)$. My idea was to take projective resolutions of $M$ and $P$ and to apply the horseshoe lemma to find a projective resolution of $N$. The problem is that I don't think I can always get an adapted basis for the projective modules over $N$ doing that (if I always could, then the exact sequence obtained would always split, which is not true).

Answer 1

I'll asume you are talking about torsion modules.

If $M$ is a finitely generarted torsion module, then there is a some surjective map $A^n\to M$. Its kernel is also free and, since $M$ is torsion, has also rank $n$. There is then a short exact sequence $0\to A^n\xrightarrow{f} A^n\to M\to 0$. The Smith Normal Form theorem implies that there are maps $u$, $v$, $g:A^n\to A^n$ such that

          f
0 —> A^n —> A^n —> M —> 0
     |       |     |
     | u     | v   |
     V       V     V
0 —> A^n —> A^n —> M —> 0
         g

commutes and $g$ is diagonal with your $d_i$'s along the diagonal along with some $1$s, and $u$ and $v$ are isomorphisms. In particular $\det f$ and $\det g$ differ by a unit, and $\deg g$ generates the ideal $\det M$, so $$\det M=(\det f).$$

Now suppose that $M_1$, $M_2$ and $M_3$ are torsion f.g. modules, let $$0\to A^n\xrightarrow{f_1} A^n\to M_1\to 0, \qquad 0\to A^m\xrightarrow{f_3} A^m\to M_3\to 0$$ be two short exact sequences as above, and build a third one using the horseshoe lemma, which will be of the form $$0\to A^{n+m}\xrightarrow{f_2} A^{n+m}\to M_2\to 0$$ with the map $f_2$ given by an upper triangular by blocks, with the matrices of $f_1$ and of $f_3$ along the diagonal. It follows from this that $$\det M_2=\det f_2=\det f_1\cdot\det f_3=\det M_1\cdot\det M_3.$$

Answer 2

Let $K$ be the quotient of the abelian group generated by the isomorphism classes $[M]$ of f.g. torsion $A$-modules $M$ modulo the subgroup generated by all elements of the form $[N]-[M]-[P]$ such that there is a short exact sequence $0\to N\to M\to P\to0$. It is a good exercise to show that the classes $[A/(p)]$, one for each association class of prime elements $p$ in $A$, freely generated the abelian group $K$, so there is exactly one morphism of monoids $d:K\to I$, with $I$ the monoid of ideals in $A$ with their multiplication, such that $d([A/(p)])=(p)$ for all prime elements $p$.

Now if $M$ is a torsion module and and $M=\bigoplus_{i=1}^n A/(p_i^{r_i})$ is a decomposition of $M$ as a direct sum of primary cyclic modules, then clearly $d([M])=(p_1^{r_1}\cdots p_n^{r_n})$, so $d([M])=\det M$. The very fact that $d$ is defined on $K$ implies the multiplicativity you want.