I am trying to solve the following exercise on basic module theory and I am stuck. Any help would be more than welcome!
So let $M\subseteq \mathbb{Z}^3$ the solutions to the following problem:
$-3x+5y+3z=0$
$-6x+20y+9y=0$
$-3x+25y+9z=0.$
What are the invariant factors of $M$ and $\mathbb{Z}^3/M$? Or, put in other words, what is the structure of $M$ and $\mathbb{Z}^3/M$?
What I have so far:
$M$ must be free since it is a submodule of a free module over a PID. But I think in order to answer anything else, I must be able to solve the above system. If I knew a basis for $M$ then I could calculate the map $f$ injecting $M$ in $\mathbb{Z}^3$ and then I would reduce the matrix of $f$ using Smith's algorithm to something simple. Is there any way to avoid solving the system? Also, since the "injection" matrix will be part of the identity matrix, in the end both $M$ and $\mathbb{Z^3}/M$ will be free of rank depending on the rank of the space of solutions.
NOTE: There is some similar problem which is basically to understand an abelian group given a representation, e.g., $<a,b|2a+4b,3a-8b>$. Are the solutions to these two problems in any way related?
Well, $M$ is the kernel of the map $$ f:\mathbb{Z}^3\to\mathbb{Z}^3,\;\;\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix}-3x+5y+3z\\-6x+20y+9z\\-3x+25y+9z\end{pmatrix} $$ Using the standard basis $\{e_1,e_2,e_3\}$ for $\mathbb{Z}^3$, we can represent $f$ above as a matrix $$ A=\begin{pmatrix}-3&5&3\\-6&20&9\\-3&25&9\end{pmatrix}. $$
If $U,V\in M_3(\mathbb{Z})$ are $3\times 3$-matrices with determinant $\pm1$, then $\{Ue_1,Ue_2,Ue_3\}$ is again a basis for the domain and $\{V^{-1}e_1,V^{-1}e_2,V^{-1}e_3\}$ is a basis for the codomain (both of which are $\mathbb{Z}^3$). The matrix for $f$ in these bases is $UAV$.
If $U$ and $V$ are as above, then they are a product of elementary matrices and these elementary matrices act by elementary row/column operations (since we are working over $\mathbb{Z}$, we can only rescale rows/columns by $\pm1$). Smith's algorithm describes how to find a basis for the domain ($\mathbb{Z}^3$) and the codomain ($\mathbb{Z}^3$) so that the matrix for $f$ is diagonal. My computation shows that we can find $U$ and $V$ such that $$ UAV=\begin{pmatrix}1&0&0\\0&3&0\\0&0&0\end{pmatrix} $$ I read from this that $M$ is 1-dimensional, $\mathrm{im}(f)\cong\mathbb{Z}^2$, and $\mathbb{Z}^3/\mathrm{im}(f)\cong\mathbb{Z}_3\times\mathbb{Z}$.