Let \(N\subseteq\ZZ^4\) be the submodule generated by the columns of \[ A= \begin{pmatrix} 3 & 12 & 3 & 6\\ 0 & 6 & 0 & 0\\ -3 & 6 & -3 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}. \] Compute \(\ZZ^4/N\) using Smith normal form.
Solution
1. The Smith normal form of \(A\) is\[ \operatorname{diag}(3,6,6,0). \]
The invariant factors may be recovered from determinantal divisors. The gcd of all entries is \[ \Delta_1=3. \] The gcd of all \(2\times2\) minors is \[ \Delta_2=18, \] while the gcd of all \(3\times3\) minors is \[ \Delta_3=108. \] Since \(\operatorname{rank}(A)=3\), the nonzero Smith invariants are \[ d_1=\Delta_1=3, \qquad d_2=\Delta_2/\Delta_1=6, \qquad d_3=\Delta_3/\Delta_2=6. \] Thus the Smith normal form is \(\operatorname{diag}(3,6,6,0)\).
2. Therefore\[ \ZZ^4/N \cong \ZZ/3\ZZ\oplus\ZZ/6\ZZ\oplus\ZZ/6\ZZ\oplus\ZZ. \]
Unimodular row and column operations do not change the isomorphism type of the cokernel. Hence \[ \operatorname{coker}(A) \cong \operatorname{coker}(\operatorname{diag}(3,6,6,0)), \] which is exactly the displayed direct sum.