## Over a PID, isomorphic finitely generated torsion modules have the same elementary divisors

Let $R$ be a principal ideal domain and let $p \in R$ be a prime.

1. Let $M$ be a finitely generated torsion $R$-module, let $k \in \mathbb{N}$, and let $n_{p,k}$ denote the number of elementary divisors of $M$ of the form $p^\alpha$ with $\alpha > k$. Prove that $(p^k)M/(p^{k+1})M \cong_R (R/(p))^{n_{p,k}}$.
2. Suppose $M_1$ and $M_2$ are isomorphic finitely generated torsion $R$-modules. Use part (a) to prove that $M_1$ and $M_2$ have the same set of elementary divisors.

By the existence portion of FTFGMPID, $M \cong_R \bigoplus_i R/(p_i^{\alpha_i})$ where $p_i$ ranges over the set of primes in $R$ (up to associates). Now $(p^k)M/(p^{k+1})M \cong_R \left[ \bigoplus (p^k)(R/(p_i^{\alpha_i}) \right] / \left[ \bigoplus (p^{k+1})(R/(p_i^{\alpha_i}) \right]$ $\cong_R \bigoplus (p^k)(R/(p_i^{\alpha_i})/(p^{k+1})(R/(p_i^{\alpha_i})$ by this previous exercise. By this previous exercise, this is module-isomorphic to $\bigoplus T_i$ where $T_i = R/(p_i)$ if $p_i = p$ and $\alpha_i > k$ and 0 otherwise. Which, by construction, is isomorphic to $(R/(p))^{n_{p,k}}$.

Now suppose $M_1$ and $M_2$ are isomorphic finitely generated torsion $R$-modules.

We claim that $(p^k)M_1/(p^{k+1})M_1 \cong_R (p^k)M_2/(p^{k+1})M_2$ for all primes $p \in R$ and all natural numbers $k$. To see this, suppose $\theta : M_1 \rightarrow M_2$ is a module isomorphism. Certainly the relation $\{ (p^k \cdot m, p^k \cdot \theta(m)) \ |\ m \in M_1 \}$ is well defined, since $\theta$ is an isomorphism. So we have a surjective mapping $\psi : (p^k)M_1 \rightarrow (p^k)M_2$ given by $\psi(p^k \cdot m) = p^k \cdot \theta(m)$. Moreover, this is an $R$-module homomorphism, since $\psi(p^k \cdot m + r \cdot (p^k \cdot n)) = \psi(p^k \cdot (m+r \cdot n))$ $= p^k \cdot \psi(m) + r \cdot p^k \cdot \psi(n)$. Now the composite $\overline{\psi} = \pi \circ \psi : (p^k)M_1 \rightarrow (p^k)M_2/(p^{k+1})M_2$, where $\pi$ is the natural projection, is also a module homomorphism. We claim that $\mathsf{ker}\ \overline{\psi} = (p^{k+1})M_1$. It is certainly the case that $(p^{k+1})M_1 \subseteq \mathsf{ker}\ \overline{\psi}$. To see the other inclusion, suppose $p^k \cdot a \in \mathsf{ker}\ \overline{\psi}$. Then $p^k \theta(a) \in (p^{k+1})M_2$, so that $p^k \theta(a) = p^{k+1}b$ for some $b \in M_2$. (Note that if we took an element from $(p^k)$ with a non-unit factor other than $p^{k+1}$, then that factor can be attached to $b$.) Now $p^k(\theta(a) - pb) = 0$. Since $\theta$ is an isomorphism, we have $c \in M_1$ with $\theta(c) = b$. Then $p^k(\theta(a) - p\theta(c)) = 0$, so that $\theta(p^k(a - pc)) = 0$, and thus $p^k(a - pc) = 0$. So $p^ka = p^{k+1}c$, as desired. (There is probably a better way to do this.) By the First Isomorphism Theorem, we have $(p^k)M_1/(p^{k+1})M_1 \cong_R (p^k)M_2/(p^{k+1})M_2$.

We saw in Part 1 that $(p^k)M_1/(p^{k+1})M_1 \cong_R (R/(p))^n_{k+1}$, where $n_{k+1}$ is the number of elementary divisors of $M_1$ which are of the form $p^\alpha$ for some $\alpha > k$. In particular, we have that $m_k = n_k - n_{k+1}$ is the number of elementary divisors of $M_1$ of the form $p^k$. The isomorphism above demonstrates that these numbers are the same for $M_1$ and $M_2$. Note that $R/(p)$ is a field if $p$ is prime (since $R$ is a PID) and that if $F^a \cong F^b$ as $F$-vector spaces, then $a = b$ by this previous exercise.