MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is a follow-up of the question I asked here.

Can you write down an explicit example of a monomorphism of finitely generated Abelian groups which is an indecomposable object in the category of two-step complexes of Abelian groups such that one of the two entries is not an indecomposable Abelian group?

share|cite|improve this question
up vote 1 down vote accepted

How about the map $f:\mathbb{Z}\to\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$ given by $f(n)=(2n,n)$?

Or for a finite example, the same idea works with a map $\mathbb{Z}/4\mathbb{Z}\to(\mathbb{Z}/8\mathbb{Z})\oplus(\mathbb{Z}/2\mathbb{Z})$.

Or if you want neither term to be indecomposable, the map $(m,n)\mapsto(2m,m+2n,n)$ from $(\mathbb{Z}/16\mathbb{Z})\oplus(\mathbb{Z}/4\mathbb{Z})$ to $(\mathbb{Z}/32\mathbb{Z})\oplus({\mathbb{Z}/8\mathbb{Z}})\oplus(\mathbb{Z}/2\mathbb{Z})$.

For example, in the third case, the cokernel of the map is cyclic of order 8, and in particular indecomposable. So if the complex decomposed, the map in one summand would have to be an isomorphism, which is impossible since the two terms of the original complex have no summands in common.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.