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

Let be a principal ideal domain and let be a prime.

- Let be a finitely generated torsion -module, let , and let denote the number of elementary divisors of of the form with . Prove that .
- Suppose and are isomorphic finitely generated torsion -modules. Use part (a) to prove that and have the same set of elementary divisors.

By the existence portion of FTFGMPID, where ranges over the set of primes in (up to associates). Now by this previous exercise. By this previous exercise, this is module-isomorphic to where if and and 0 otherwise. Which, by construction, is isomorphic to .

Now suppose and are isomorphic finitely generated torsion -modules.

We claim that for all primes and all natural numbers . To see this, suppose is a module isomorphism. Certainly the relation is well defined, since is an isomorphism. So we have a surjective mapping given by . Moreover, this is an -module homomorphism, since . Now the composite , where is the natural projection, is also a module homomorphism. We claim that . It is certainly the case that . To see the other inclusion, suppose . Then , so that for some . (Note that if we took an element from with a non-unit factor other than , then that factor can be attached to .) Now . Since is an isomorphism, we have with . Then , so that , and thus . So , as desired. (There is probably a better way to do this.) By the First Isomorphism Theorem, we have .

We saw in Part 1 that , where is the number of elementary divisors of which are of the form for some . In particular, we have that is the number of elementary divisors of of the form . The isomorphism above demonstrates that these numbers are the same for and . Note that is a field if is prime (since is a PID) and that if as -vector spaces, then by this previous exercise.

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