Say I want to find a basis for the $\mathbb Z$-submodule of $\mathbb Z^3$ which consists of all the $(x,y,z)$ such that $x+2y+3z = 0$ and $x+4y + 9z=0$.
If $(x,y,z)$ is in that submodule, then we have $2y + 6z = 0$, giving (?) $y=-3z$. Also $x - 3z = 0$, i.e. $x = 3z$. Hence $(x,y,z) = (3z,-3z,z) = z(3,-3,1)$. This shows that $\{(3,-3,1)\}$ is a basis of that submodule.
This is suspiciously easy. In particular, I am wondering whether or not I am allowed to deduce $y = -3z$ from $2y+6z=0$ (perhaps I am not allowed to multiply by $2^{-1}$ both sides because $2^{-1} \notin \mathbb Z$; but then, why look at the module action? just look at the thing as an equation in $\mathbb Z$, which is a domain - from $2(y + 3z) = 0$, we get $y+3z =0$, so $y = -3z$).
Does this look right to you?
Thanks.
For reference, this is exercise $3$ in section $3.6$ of Basic Algebra $1$, N. Jacobson.