How can we write this $\mathbb{Q}[x]$-module as a direct sum of cyclic $\mathbb{Q}[x]$-modules? If $L$ is the submodule of $\mathbb{Q}[x]^{(3)}$ generated by $(2x-1,x,x),(x,x,x),(x+1,2x,x)$. How do we write $\mathbb{Q}[x]^{(3)}/L$ as a direct sum of cyclic modules?
 A: Yes, you can simply take the matrix that has the vectors as rows and use Smith Normal Form; if the Smith normal form is
$$\left(\begin{array}{ccc}
\alpha & 0 & 0\\
0 & \beta & 0\\
0 & 0 & \gamma
\end{array}\right)$$
where $\alpha,\beta,\gamma\in\mathbb{Q}[x]$, then $\mathbb{Q}[x]^3/L$ will be isomorphic to $(\mathbb{Q}[x]/(\alpha)) \oplus (\mathbb{Q}[x]/(\beta)) \oplus (\mathbb{Q}[x]/(\gamma))$, which is a sum of cyclic $\mathbb{Q}[x]$-modules. 
Here, you can actually save a bit of work by noting that 
$$
\left(\begin{array}{ccc}
2x-1 & x & x \\
x & x & x\\
x+2 & 2x & x
\end{array}\right) \to \left(\begin{array}{ccc}
x & x & x\\
x+2 & 2x & x\\
2x-1 & x & x
\end{array}\right) \to \left(\begin{array}{ccc}
x& x & x\\
2 & x & 0\\
x-1 & 0 & 0
\end{array}\right)$$
so $L$ is generated by $(x-1,0,0)$, $(2,x,0)$, and $(x,x,x)$.
We can consider $\mathbb{Q}[x]^3$ as having basis by $(1,1,1)$, $(1,0,0)$ and $(0,1,0)$. If we do that, then the third generator of $L$ is just $x$ times the first generator of $\mathbb{Q}[x]^3$, and the submodule generated by the other two are in the direct summand generated by the second and third elements of the basis. So
$$\frac{\mathbb{Q}[x]^3}{L} \cong \frac{\langle (1,1,1)\rangle}{\langle x(1,1,1)\rangle} \oplus \frac{\langle (1,0,0),(0,1,0)\rangle}{\langle (x-1,0,0), (2,x,0)\rangle}.$$
The first summand is isomorphic to $\mathbb{Q}[x]/(x) \cong \mathbb{Q}$ via the map that sends $x$ to $0$ (and $\mathbb{Q}$ is a cyclic $\mathbb{Q}[x]$-module, where $x$ acts as multiplication by $0$). 
This reduces the problem to just trying to write $\mathbb{Q}[x]^2/M$ as a direct sum of cyclic modules, where $M$ is the submodule generated by $(x-1,0)$ and $(2,x)$. If nothing obvious suggests itself at this point, you can use the Smith Normal Form to see what this quotient looks like.
