Let $r, p,q \in \mathbb{Z}$ such that $\gcd(p,q)=1$ and $r\geq 1$. What is the quotient $\mathbb{Z}^2/\mathbb{Z}(rp,-rq)$? Is it isomorphic to $\mathbb{Z}\times \mathbb{Z}_r$?

I tried to define a map

$\mathbb{Z}\times \mathbb{Z}_r \to \mathbb{Z}^2/\mathbb{Z}(rp,-rq)$ by $(1,0) \mapsto \overline{(1,0)}$ and $(0,\overline{1}) \mapsto \overline{(p,-q)}$. This map is injective. But I am not sure whether $\{(1,0),(p,-q)\}$ generates $\mathbb{Z}^2$? If it generates then the above map is surjective also.

## Best Answer

Instead of $\{(1, 0), (p, -q)\}$, you need to make use of Bezout's identity.

Since $p, q$ are prime to each other, there exist $u, v \in \Bbb Z$ such that $pu + qv = 1$. This can be restated as: the matrix $A = \begin{pmatrix}p & v\\-q & u\end{pmatrix}$ is in $\operatorname{GL}_2(\Bbb Z)$.

This allows us to view $A$ as an isomorphism $f$ from $\Bbb Z^2$ to itself, namely we define $f(v)=Av$ for any (column) vector $v \in \Bbb Z^2$. The inverse homomorphism is given by $v \mapsto A^{-1}v$.

Under this isomorphism, the image of the vector $(r, 0)$ is just $(rp, -rq)$. Therefore we have $\Bbb Z^2 / \Bbb Z(rp, -rq) \simeq \Bbb Z^2 / \Bbb Z(r, 0)$, and it should be quite clear that the latter is isomorphic to $\Bbb Z \times \Bbb Z/r\Bbb Z$.