Decomposition of finitely generated Module over PID Suppose
Let $M = <x,y>/<2x-3y> $ where $<x,y>, <2x-3y>$ are $ \mathbb{Z} $-modules.
How can I find the decomposition of $M$ to its invariant factors?
 A: Let $A =\langle{x,y}\rangle$ and let $B$ be the cyclic subgroup of $A$ given by
$B=\langle{2x-3y}\rangle$.

Let $M=A/B$ and let $z\in M$ be given by $z=(x-y)+B$.

Consider the cyclic subgroup of $\langle{z}\rangle$ of $M$.

Suppose $nz=0\;$for some integer $n$.
\begin{align*}
\text{Then}\;\;&
nz = 0
\\[4pt]
\implies\;&
n(x-y)+B = 0
\\[4pt]
\implies\;&
n(x-y)\in B
\\[4pt]
\implies\;&
n(x-y)=k(2x-3y)\;\text{for some integer $k$}
\\[4pt]
\implies\;&
(n-2k)x-(n-3k)y=0
\\[4pt]
\implies\;&
n=2k\;\;\text{and}\;\;n=3k
\\[4pt]
\implies\;&
k=0
\\[4pt]
\implies\;&
n=0
\\[4pt]
\end{align*}
hence $\langle{z}\rangle$ is infinite.

It's clear that the set $\{x+B,y+B\}$ generates $M$, hence from
$$
\left\lbrace
\begin{align*}
&3(x-y)=3x-3y=x+(2x-3y)\equiv x\;(\text{mod}\; B)\\[4pt]
&2(x-y)=2x-2y=y+(2x-3y)\equiv y\;(\text{mod}\; B)\\[4pt]
\end{align*}
\right.
$$
if follows that $\langle{z}\rangle=M$.

Thus $M\cong \mathbb{Z}$.

As an alternate, easier solution, here's an outline . . .

Consider the homomorphism $f:A\to\mathbb{Z}$ given by $f(mx+ny)=3m+2n$.

It's not hard to show that

*

*$f$ is surjective.$\\[4pt]$

*$\ker(f)=B$.

hence $M\cong \mathbb{Z}$.
