How to show $A\simeq \mathbb Z^{k-1}$?

Question

I know this might be a silly question, but I don't know module theory very well. Suppose I have an exact sequence $$0\longrightarrow A\longrightarrow \mathbb Z^k\longrightarrow \mathbb Z\longrightarrow 0,$$ where $A$ is a $\mathbb Z$-module. How to show $A\simeq \mathbb Z^{k-1}$?

Answer 1

You have a sequence of abelian groups, or $\mathbb{Z}$-modules. Clearly $\mathbb{Z}$ is a free $\mathbb{Z}$-module, hence a projective module.

A defining property of projective modules is that any short exact sequence whose third nonzero term is projective splits, so $A\oplus \mathbb{Z}\simeq\mathbb{Z}^k$. In particular, $A$ is (isomorphic to) a subgroup of the free abelian group $\mathbb{Z}^k$, hence free abelian. This is a result of Dedekind.

But rank is additive over short exact sequences, so $\operatorname{rank}(A)+\operatorname{rank}(\mathbb{Z})=\operatorname{rank}(\mathbb{Z}^k)$, so $\operatorname{rank}(A)=k-1$. The rank determines a free abelian group up to isomorphism, so here $A\simeq\mathbb{Z}^{k-1}$.

Answer 2

First lets examine what the nontrivial subgroups of $\mathbb{Z}$ actually are. We show that $A$ is isomorphic to $\mathbb{Z}^j$ for some $j \leq k$. Well, this follows the classification of finitely generated abelian groups and the fact that $\mathbb{Z}^k$ has no torsion elements.

Then one can show that with the generators of $A$ and a single generator of $\mathbb{Z}$, we can get all of $\mathbb{Z}^k$ and so $A \cong \mathbb{Z}^{k-1}$.

Try to work out this argument rigorously.

Answer 3

$\mathbb{Z}$ is obviously a free $\mathbb{Z}$-module; so it is projective, then the exact sequence $$0\longrightarrow A\longrightarrow \mathbb Z^k\longrightarrow \mathbb Z\longrightarrow 0$$ is splitting.

This implies $$\mathbb{Z}^k \cong A \oplus \mathbb{Z}$$ So if we consider the projection morphism we have $$A \cong \frac{\mathbb{Z}^k}{\mathbb{Z}} \cong \mathbb{Z}^{k-1}$$