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?
Best Answer
Your steps are fine, but your matrix is not in Smith normal form. We could further reduce your matrix as follows.
$$ \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 1 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0} \leadsto\\ \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0} \leadsto \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0}. $$