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.

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