Let \(A \in M_n(\mathbb{Z})\) be an \(n \times n\) matrix with integer entries, viewed as a \(\mathbb{Z}\)-module homomorphism \(A: \mathbb{Z}^n \to \mathbb{Z}^n\). (1) Prove that the cokernel \(\operatorname{coker}(A) = \mathbb{Z}^n / \operatorname{im}(A)\) is finite if and only if \(\det(A) \ne 0\). (2) Prove that when \(\det(A) \ne 0\), the order of the cokernel is given by: \[|\operatorname{coker}(A)| = |\det(A)|.\]
Solution
Take Smith normal form \[ UAV=\operatorname{diag}(d_1,\dots,d_n), \] with \(U,V\in\operatorname{GL}_n(\mathbb Z)\) and \(d_1\mid\cdots\mid d_n\), allowing some \(d_i=0\). Since \(U\) and \(V\) are automorphisms of \(\mathbb Z^n\), \[ \operatorname{coker}(A)\cong\bigoplus_{i=1}^n\mathbb Z/d_i\mathbb Z, \] where \(\mathbb Z/0\mathbb Z\) means \(\mathbb Z\).
Thus the cokernel is finite iff every \(d_i\neq0\), equivalently iff \[ \det A\neq0. \] When this holds, \[ |\operatorname{coker}(A)|=\prod_i |d_i|. \] Because \(\det U,\det V=\pm1\), \[ |\det A|=\prod_i|d_i|, \] so \[ |\operatorname{coker}(A)|=|\det A|. \]