Theorem N29.1. Any finitely generated abelian group is isomorphic to a product of cyclic groups:
(Since (i.e., coprime), we can factor further, e.g., to arrange that all the are powers of primes.)
If is finitely generated by , we get a surjection , and
Proposition N29.2 (Prop 1). If is a finitely generated module (over ), then there exist and such that . Equivalently, there is an exact sequence
The proof rests on the following.
Lemma N29.3. Any submodule (subgroup) of is finitely generated (and free of rank , too).
Proof. By induction on .
- : the subgroups of are and , .
- Assume the lemma is true for , and consider a submodule .
- Consider the homomorphism
i.e., .
So is a submodule, hence finitely generated (free) by induction. And is a subgroup of , hence free of rank 0 or 1. So and are finitely generated (free), which gives that is free (finitely generated). The proof goes like the one for the rank theorem.
Let be generators (a basis) of , and let be generators (a basis) of . We claim that generate (are a basis of) .
For all we have , so there are such that
so there are such that , i.e.,
∎
Proof of Prop 1. If is finitely generated, with generators , then is surjective, and is a subgroup/submodule of , so it is finitely generated by the lemma. So (go over this) for some , which gives the exact sequence
So, given , we get . ∎
Divisibility
Divisibility of elements of , in a free module (over ).
Definition N29.4. The divisibility of , , is the largest such that there is with ; concretely, . We say is primitive if its divisibility is .
Lemma N29.5. An element of a free finitely generated module () can be part of a basis if and only if it is primitive.
Proof. An element of a basis must be primitive: with would contradict linear independence.
Conversely, let be a primitive element, written in a basis . Since , WLOG we can assume and
i.e., is minimal among the nonzero coefficients.
Guys, we’re changing basis now!! ∎
Proposition N29.6 (Prop 2). For every there exists a basis …