Generators of $\mathbb{Z} \times \mathbb{Z}$ Why is $\mathbb{Z} \times \mathbb{Z}$ generated by $(1,0)$ and $(0,1)$? I thought it should be $(1,1)$, but going through the solution of a certain exercise I am working on, it seems I am wrong. 
Can anyone explain this?

Just think about the group that is generated by $(1,1)$. Okay, so you certainly have $(1,1)$. You must also have the identity $(0,0)$. Also you have $2\cdot (1,1) = (2,2)$, and $3\cdot (1,1) = (3,3)$,...see the pattern?
The group generated by $(1,1)$ is the subgroup of $\mathbb{Z}\times\mathbb{Z}$ consisting of elements of the form $(n,n)$. This is clearly not the entirety of $\mathbb{Z}^2$.
Now what about the group generated by $(1,0)$ and $(0,1)$? Every element generated by this group has the form $a\cdot (1,0) + b\cdot (0,1) = (a,b)$, which clearly is all of $\mathbb{Z}^2$.
