Let \begin{align*} A=\begin{pmatrix} 1&0&-1&2\\ 1&2&1&0\\ 1&0&2&2\\ 1&2&2&0 \end{pmatrix} \in \mathbb{Z}^{4\times 4} \end{align*} Calculate the Smith normal form
\begin{align*} A &\leadsto \begin{pmatrix} 1 & 0 & -1 & 2 \\ 0 & 2 & 2 & -2 \\ 0 & 0 & 3 & 0 \\ 0 & 2 & 3 & -2 \end{pmatrix} \leadsto \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 2 & -2 \\ 0 & 0 & 3 & 0 \\ 0 & 2 & 3 & -2 \end{pmatrix} \leadsto \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \end{align*}
$3\nmid 1$, so what do I have to do next? I'm pretty sure I already messed up, because according to a calculator, I should have $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ at the end. Can somebody explain?
Your steps are fine, but your matrix is not in Smith normal form. We could further reduce your matrix as follows.
$$ \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 1 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0} \leadsto\\ \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0}. $$