Formally,
\(M\) is the quotient module
\(M=\mathbb{Z}^{4} / \operatorname{im}\left(t_{A}\right)\text{,}\) where
\(t_{A}: \mathbb{Z}^{3} \rightarrow \mathbb{Z}^{4}\) is defined by
\(t_{A}(v)=A v\text{.}\) Since
\(\mathbb{Z}^{4}\) is generated by its standard basis elements
\(\left\{e_{1}, e_{2}, e_{3}, e_{4}\right\}\text{,}\) we deduce as in
Lemma 13.10 that
\(M=\mathbb{Z}^{4} / \operatorname{im}\left(t_{A}\right)\) is generated by the cosets of the
\(e_{i}\text{.}\) To keep the notation short, we set
\(m_{i}=e_{i}+\operatorname{im}\left(t_{A}\right)\text{.}\)
Let \(N=\operatorname{im}\left(t_{A}\right)\) and note that \(N\) is the submodule of \(\mathbb{Z}^{4}\) generated by the columns of \(A\text{:}\)
\begin{equation*}
N=R\left\{\left[\begin{array}{l}
2 \\ 3 \\ 1 \\ 0
\end{array}\right],\left[\begin{array}{c}
1 \\ 9 \\ -2 \\ 1
\end{array}\right],\left[\begin{array}{l}
0 \\ 5 \\ 7 \\ 2
\end{array}\right]\right\}=R\left\{2 e_{1}+3 e_{2}+e_{3}, e_{1}+9 e_{2}-2 e_{3}+e_{4}, 5 e_{2}+7 e_{3}+2 e_{4}\right\}
\end{equation*}
Since \(N\) maps to 0 under the quotient map \(q: \mathbb{Z}^{4} \rightarrow M=\mathbb{Z}^{4} / N\text{,}\) the relations of \(M\) can be written as
\begin{equation*}
\begin{cases}2 m_{1}+3 m_{2}+m_{3} & =0 \\ m_{1}+9 m_{2}-2 m_{3}+m_{4} & =0 \\ 5 m_{2}+7 m_{3}+2 m_{4} & =0\end{cases}
\end{equation*}
We can now see that this is a rather inefficient presentation, since we can clearly use the first equation to solve for for \(m_{3}=-2 m_{1}-3 m_{2}\text{.}\) This implies that \(M\) can be generated using only \(m_{1}, m_{2}\) and \(m_{4}\text{,}\) that is
\begin{equation*}
M=R\left\{m_{1}, m_{2}, m_{3}, m_{4}\right\}=R\left\{m_{1}, m_{2}, m_{4}\right\}.
\end{equation*}
This eliminates the first equation and the latter two become
\begin{equation*}
\begin{cases}5 m_{1}+15 m_{2}+m_{4} & =0 \\ -14 m_{2}-16 m_{2}+2 m_{4} & =0\end{cases}
\end{equation*}
Now we can also eliminate \(m_{4}\text{,}\) i.e leaving just two generators \(m_{1}, m_{2}\) that satisfy
\begin{equation*}
-24 m_{1}-46 m_{2}=0.
\end{equation*}
Another way to do this is to look at the matrix \(A\) and use elementary row operations to "make zeros" on the 1st and 2nd columns, as follows:
\begin{equation*}
A=\left[\begin{array}{ccc}
2 & 1 & 0 \\ 3 & 9 & 5 \\ 1 & -2 & 7 \\ 0 & 1 & 2
\end{array}\right] \rightarrow\left[\begin{array}{ccc}
0 & 5 & -14 \\ 0 & 15 & -16 \\ 1 & -2 & 7 \\ 0 & 1 & 2
\end{array}\right] \rightarrow\left[\begin{array}{ccc}
0 & 0 & -24 \\ 0 & 0 & -46 \\ 1 & 0 & 13 \\ 0 & 1 & 0
\end{array}\right]
\end{equation*}
Eliminating the generators \(m_{3}\) and \(m_{4}\) amounts to dropping the first two columns (which are the 3rd and 4th standard basis vectors) as well as the last two rows. As we will prove soon, this shows that the \(\mathbb{Z}\)-module presented by \(A\) is isomorphic to the \(\mathbb{Z}\)-module presented by
\begin{equation*}
B=\left[\begin{array}{l} -24 \\ -46 \end{array}\right]
\end{equation*}
We can go further. Set \(m_{1}^{\prime}:=m_{1}+2 m_{2}\text{.}\) Then \(m_{1}^{\prime}\) and \(m_{2}\) also form a generating set of \(M\text{.}\) The relation on \(m_{1}, m_{2}\) translates to
\begin{equation*}
-24 m_{1}^{\prime}+2 m_{2}=0
\end{equation*}
given by the matrix
\begin{equation*}
C=E_{1,2}(-2) B=\left[\begin{array}{c} -24 \\ 2 \end{array}\right].
\end{equation*}
Note that we have done a row operation (subtract twice row 1 from row 2 ) to get from \(B\) to \(C\text{.}\) Continuing in this fashion by subtracting 12 row 2 from row 1 we also form
\begin{equation*}
D=E_{1,2}(12) C=\left[\begin{array}{l} 0 \\ 2 \end{array}\right],
\end{equation*}
The last matrix \(D\) presents the module \(M^{\prime}=\mathbb{Z}^{2} / \operatorname{im}\left(t_{D}\right)\) with generators \(a, b\text{,}\) where
\begin{equation*}
a=e_{1}+\operatorname{im}\left(t_{D}\right), \quad b=e_{2}+\operatorname{im}\left(t_{D}\right)
\end{equation*}
and relation \(2 a=0\text{.}\) This module \(M^{\prime}\) is isomorphic to our original module \(M\text{.}\) As we will see, this proves \(M \cong \mathbb{Z} \oplus \mathbb{Z} / 2\text{.}\) An explicit isomorphism between \(M^{\prime}\) and \(\mathbb{Z} \oplus \mathbb{Z} / 2\) is given by sending \(\mathbb{Z}^{2} \rightarrow \mathbb{Z} \oplus \mathbb{Z} / 2\) by the unique \(\mathbb{Z}\)-module homomorphism defined by
\begin{equation*}
e_{1} \mapsto(1,0) \text { and } e_{2} \mapsto\left(0,[1]_{2}\right)
\end{equation*}
Now notice that the kernel of this homomorphism is the submodule \(\left(2 e_{2}\right) \mathbb{Z}=\operatorname{im}\left(t_{D}\right)\text{.}\) Then the first isomorphism theorem gives \(M^{\prime}=\mathbb{Z}^{2} / \operatorname{im}\left(t_{D}\right) \cong \mathbb{Z} \oplus \mathbb{Z} / 2\text{.}\)
