Let \(\phi:\ZZ^4\to\ZZ^3\) be represented in the standard bases by \[ A= \begin{pmatrix} 1&2&0&3\\ 0&-3&3&1\\ -1&1&1&5 \end{pmatrix}. \] Compute \(\operatorname{coker}(\phi)=\ZZ^3/\operatorname{im}(A)\) using Smith normal form.
Solution
1. The first determinantal divisor is \(\Delta_1=1\).
The gcd of all entries of \(A\) is \(1\), since \(A\) contains the entry \(1\).
2. The second determinantal divisor is \(\Delta_2=1\).
Among the \(2\times2\) minors is \[ \det \begin{pmatrix} 1&3\\ 0&1 \end{pmatrix}=1, \] using rows \(1,2\) and columns \(1,4\). Hence the gcd of all \(2\times2\) minors is \(1\).
3. The third determinantal divisor is \(\Delta_3=1\).
The four \(3\times3\) minors obtained by deleting one column are \[ -12,\qquad -27,\qquad 23,\qquad 10. \] Their gcd is \(1\); for instance \[ \gcd(12,27,23,10)=1. \] Therefore \(\Delta_3=1\).
4. The Smith invariant factors are\[ d_1=1,\qquad d_2=1,\qquad d_3=1. \]
For a full-rank integer matrix, the Smith invariant factors satisfy \[ d_1=\Delta_1,\qquad d_1d_2=\Delta_2,\qquad d_1d_2d_3=\Delta_3. \] Using <1>1–<1>3 gives \(d_1=d_2=d_3=1\).
5. Hence \(\phi\) is surjective and\[ \operatorname{coker}(\phi)=0. \]
The Smith normal form is \[ \operatorname{diag}(1,1,1) \] as a \(3\times4\) matrix, with one additional zero column. Therefore \[ \ZZ^3/\operatorname{im}(A) \cong \ZZ/1\ZZ\oplus\ZZ/1\ZZ\oplus\ZZ/1\ZZ =0. \]