Prove ideal of matrix is that set Given the set of matrices: 
$M=\left\{
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
\mid
a,b \in \Bbb Z
\right\}$.
It is easily seen that $M$ is a commutative subring of $M_{2,2}(\Bbb Z)$. If K is ideal of M prove that exits $m,n \in \Bbb Z$ to $K=\left\{
\begin{pmatrix}
ma & 0 \\
0 & nb
\end{pmatrix}
\mid
a,b \in \Bbb Z
\right\}$
My ideas is : if $\begin{pmatrix}
m & 0 \\
0 & n
\end{pmatrix}\in K$ then  $\left\{
\begin{pmatrix}
am & 0 \\
0 & bn
\end{pmatrix}
\mid
a,b \in \Bbb Z
\right\}\in M$.
I've tried to find $m: \mid m \mid < \mid a \mid, \begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix} \in K $ and  $n: \mid n \mid < \mid b \mid, \begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix} \in K$. But I'm not sure existence of both m, n: $\mid m \mid < \mid a \mid,\mid n \mid < \mid b \mid $.
Beside, I've known that $K=\Bbb Z$x$ \Bbb Z$, and A is subgroup of $\Bbb Z$ if and only if $A=m\Bbb Z$.
I have trouble proving it, is it possible?

Hint: Let $K_1$ be the set of $a$ such that $\pmatrix{a & 0\cr 0 & b\cr} \in K$ for some $b$, and $K_2$ the set of $b$ such that $\pmatrix{a & 0\cr 0 & b\cr} \in K$ for some $a$.  Show that $K_1$ and $K_2$ are ideals of $\mathbb Z$...
