Let \(R\) be a Principal Ideal Domain (PID) and let \(A \in M_n(R)\) be an \(n \times n\) matrix representing an \(R\)-module homomorphism \(A: R^n \to R^n\). Let \(M = R^n / \operatorname{im}(A)\) be the cokernel module. Prove that the following three conditions are equivalent: (1) \(M = R^n / \operatorname{im}(A)\) is a torsion module (every element of \(M\) is annihilated by some non-zero \(r \in R\)). (2) The matrix rank of \(A\) is \(\operatorname{rank}_R(A) = n\) (equivalently, \(\det(A) \ne 0\)). (3) The Smith Normal Form of \(A\) has exactly \(n\) non-zero invariant factors \(d_1, d_2, \dots, d_n \ne 0\).
Solution
Take Smith normal form \[ PAQ=\operatorname{diag}(d_1,\dots,d_n), \qquad P,Q\in\operatorname{GL}_n(R), \] with \(d_1\mid\cdots\mid d_n\), allowing some \(d_i=0\). Then \[ M=R^n/\operatorname{im}A \cong\bigoplus_{i=1}^n R/(d_i), \] where \(R/(0)=R\).
The module \(R/(d_i)\) is torsion exactly when \(d_i\ne0\): if \(d_i\ne0\), the element \(d_i\) annihilates the whole quotient; if \(d_i=0\), the summand is a copy of the torsion-free module \(R\). Hence \[ M\text{ is torsion} \iff d_i\ne0\text{ for every }i. \]
Let \(K=\operatorname{Frac}(R)\). Since \(P,Q\) remain invertible over \(K\), \[ \operatorname{rank}A =\operatorname{rank}_K\operatorname{diag}(d_1,\dots,d_n), \] which is the number of nonzero \(d_i\). Therefore \[ \operatorname{rank}A=n \iff d_i\ne0\text{ for every }i. \] Because \(R\) is a domain, \[ \det A\ne0 \iff \prod_i d_i\ne0 \iff d_i\ne0\text{ for every }i. \] Thus the three conditions are equivalent.