Structure of Modules and Elementary Divisors Matrix Find the structure of the module $\mathbb{Z}^{3}/N$ where $N$ is spanned by $(1,1,1),(6,3,2), (4,1,0)$. 

I checked that the above vectors are not linearly independent. 
I have generally learned that for a basis of $N$ (not a generating set), we can apply the elementary divisors theorem to a basis $y_{1}, \dots, y_{m}$ of $N$ and a basis $x_{1}, \dots, x_{n}$ of $R^{n}$ (in the general case where we have a surjective homomorphism of $R$-modules $f:R^{n} \rightarrow M$ and $\mathrm{Ker}(f)=N$) to find the respective change of bases $x_{j}'$ and $y_{j}'$ respectively. 
If the exercise given were for a linearly independent set, one would form the matrix $A$ out from the linearly independent set and then perform row operations until the matrix of elementary divisors is found.
What would be a way to extend this idea to a generating set of $\mathrm{Ker}(f)$ instead of a basis of it? Thanks for the help.
 A: You have to find the Smith normal form of the matrix
$$\begin{pmatrix}1&6&4\\1&3&1\\1&2&0\end{pmatrix}.$$
There is an algorithm to do this, however the fastest way is using Fitting ideals. The Fitting ideals $I_i$ are the ideals generated by the $i \times i$-minors of the matrix, by computing them we find
$I_1 = (1), I_2= (1), I_3=(0).$
The Smith normal form is the unique diagonal matrix with the same Fitting ideals, which is
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$
and we find $\mathbb Z^3/N \cong \mathbb Z$.
It would be interesting to find the generator. It is the equivalence class of $\begin{pmatrix}3\\1\\0\end{pmatrix}$, since we have
$$\det \begin{pmatrix}1&4&3\\1&1&1\\1&0&0\end{pmatrix}=1.$$
So these $3$ vectors form a basis of $\mathbb Z^3$, i.e.
$$\mathbb Z^3 = \langle \begin{pmatrix}1\\1\\1\end{pmatrix} \rangle \oplus  \langle \begin{pmatrix}4\\1\\0\end{pmatrix} \rangle \oplus \langle \begin{pmatrix}3\\1\\0\end{pmatrix} \rangle = N \oplus \langle \begin{pmatrix}3\\1\\0\end{pmatrix} \rangle.$$
