Applying the fundamental theorem of finitely generated abelian groups to the group $\Bbb Z^3/((1,0,3),(-1,2,1))$ Consider two vectors $v_1=(1,0,3), v_2=(-1,2,1)$ in $\Bbb Z^3$. Let $A$ be the subgroup generated by $v_1,v_2.$ Then $\Bbb Z^3/A$ would be a finitely generated abelian group, so by the fundamental theorem of finitely generated abelian groups, $\Bbb Z^3/A$ can be expressed as a finite direct sum of $\Bbb Z$'s and $\Bbb Z_m$'s. Since $v_1,v_2$ is linearly independent, $A$ should be free of rank $2$. So $\Bbb Z^3/A$ should have rank $1$. But how should I find the torsion part?
 A: Lets generalize a little. Let $\vec{v_1}, \ldots, \vec{v_m}$ be vectors in $\mathbb{Z}^n$. To compute the quotient group $\mathbb{Z}^n / \langle \vec{v_1}, \ldots, \vec{v_m}\rangle$, make the $m \times n$ matrix whose rows are the $\vec{v_i}$. 
Now swapping two rows doesn't change the isomorphism class of the quotient; that just corresponds to relabeling the vectors $\vec{v_i}$. Similarly, you can multiply any row by $\pm 1$ without changing the isomorphism class of the quotient, and (most usefully) you can replace any row by itself plus an arbitrary scalar multiple of another; that is, the isomorphism class of the quotient is preserved by $\mathbb{Z}$-linear elementary row operations on the matrix. 
Similarly, we can do change of basis operations in the ambient space $\mathbb{Z}^n$, which correspond to column operations on your matrix. 
Now you can get the matrix into "Smith normal form," which means that it is diagonal with non-decreasing entries (along the diagonal) with the elementary row operations, and then it is easy to read off the isomorphism class of the quotient.
In your example, we have
\begin{align*}
\begin{pmatrix}1 & 0 & 3 \\ -1 & 2 & 1 \end{pmatrix} & \sim \begin{pmatrix}1 & 0 & 3 \\ 0 & 2 & 4 \end{pmatrix} \\
&\sim \begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \end{pmatrix},
\end{align*}
using a row operation at step one and two column operations at step two. We see the quotient is isomorphic to 
$$
\frac{\mathbb{Z}}{\mathbb{Z}} \oplus \frac{\mathbb{Z}}{2\mathbb{Z}} \oplus \frac{\mathbb{Z}}{0\mathbb{Z}} = \mathbb{Z} \oplus \frac{\mathbb{Z}}{2\mathbb{Z}}
$$
