$\mathbb{Z}^k/A\mathbb{Z}^k \cong\mathbb{Z}^k/A'\mathbb{Z}^k$

Question

I am working on Ch 14 problem 4.6 from Artin's Algebra textbook.

Let $φ:\mathbb{Z}^k\to\mathbb{Z}^k$ be a homomorphism defined by $φ(x)=Ax$, where $A$ is some $k\times k$ integer matrix. Show that the image of $φ$ is of finite index iff $A$ is nonsingular and that if so, the index is equal to $|detA|$.

Now, I have completed the proof of this problem, but with the exception of an intermediate step which is not obvious to me.

I need to show $\mathbb{Z}^k/A\mathbb{Z}^k \cong \mathbb{Z}^k/A'\mathbb{Z}^k$ where $A'$ is the Smith normal form of $A$ and $A\neq0.$

I would prefer to find an explicit isomorphism $ψ:\mathbb{Z}^k/A\mathbb{Z}^k\to\mathbb{Z}^k/A'\mathbb{Z}^k$. But defining $ψ(x+A\mathbb{Z}^k)=x+A'\mathbb{Z}^k$ doesn't seem to work. What will work?

Answer 1

Try $ψ(x+A\mathbb{Z}^k)=Q^{-1}x+A'\mathbb{Z}^k$, where $A'$=$Q^{-1}AP$.

Answer 2

We have $A'= P A Q$, where $P,Q$ are invertible $k \times k$ integer matrices. They correspond to change of basis in the domain and codomain of $\varphi$. Draw a commutative diagram matching $\varphi$ and $\varphi': \mathbb{Z}^k\to\mathbb{Z}^k$ given by $x \mapsto A'x$.