How do we compute $\mathbb{Z}^2/(n, m)$? I have been trying to compute quotients of this form (as modules). Does anyone have a quick method of doing this? I'm sure I'm missing something obvious.
Any help is appreciated! 
 A: Let $\gcd(n,m)=d$, with $n=da$ and $m=db$. Of course, $\gcd(a,b)=1$, so Bezout's identity implies that there exist integers $x$ and $y$ such that $ax+by=1$. 
The set $\{v_1,v_2\}=\{(a,b),(-y,x)\}$ is a basis for $\mathbb{Z}^2$ because
$$\det\begin{bmatrix} a & -y \\ b & \hphantom{-}x\end{bmatrix}=1\in\mathbb{Z}^\times.$$ 
Alternatively, you can see that it is a basis because its $\mathbb{Z}$-span includes $(1,0)$ and $(0,1)$:
$$\begin{align*}
xv_1-bv_2&=(ax+by,bx-bx)=(1,0)\\
yv_1+av_2&=(ay-ay,by+ax)=(0,1)
\end{align*}$$
In this basis, we have that $(n,m)=d\cdot v_1$, so that $\mathbb{Z}^2/\langle (n,m)\rangle\cong \mathbb{Z}/d\mathbb{Z}\oplus\mathbb{Z}$. 
To give a bit more detail, consider the map $\varphi:\mathbb{Z}^2\to\mathbb{Z}^2$ defined by $\varphi(e_1)=v_1$, $\varphi(e_2)=v_2$. Because $\{v_1,v_2\}$ is a basis, the map $\varphi$ is an isomorphism. Note that $\varphi((n,m))=(d,0)$, so the kernel of the composition
$$\mathbb{Z}^2\xrightarrow{\;\;\varphi\;\;}\mathbb{Z}^2\xrightarrow{\;\;q\;\;}\mathbb{Z}^2/\langle (d,0)\rangle$$
is $\langle (n,m)\rangle$, so the first isomorphism theorem implies that
$$\mathbb{Z}^2/\langle (n,m)\rangle\cong \mathbb{Z}^2/\langle (d,0)\rangle.$$
But it is obvious that
$$\mathbb{Z}^2/(d,0)\cong\mathbb{Z}/d\mathbb{Z}\oplus\mathbb{Z},$$
so we are done.
