Classification of chain complexes of free abelian groups

Question

In Hatcher's book Algebraic Topology, the author claims (in the beginning of the Universal Coefficient Theorem for Cohomology section) "More generally, consider any chain complex of finitely generated free abelian groups. Such a chain complex always splits as the direct sum of elementary complexes of the forms $0 \to \mathbb{Z} \to 0$ and $0 \to \mathbb{Z} \overset{m}\to \mathbb{Z} \to 0$" referring to an exercise earlier in the book. But wouldn't that imply that every matrix with $\mathbb{Z}$ coefficients is diagonalizable, which is certainly not true? For a concrete example of what I have in mind, consider the chain complex $0\to \mathbb{Z}^2 \to \mathbb{Z}^2 \to 0$ where the mapping is given by the matrix $$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ What would Hatcher's statement for this mapping mean?

Answer 1

The point is that you are free to choose independent base changes for the domain and the codomain of the differential in your chain complex to bring it into simpler form. Hence, what Hatcher's claim implies in classical terms is that any matrix over ${\mathbb Z}$ is equivalent (not similar!) to a diagonal matrix - see Smith Normal Form.