Let \(R\) be a PID, \(n>0\), and \(A\in M_n(R)\). Put \[ M=R^n/\operatorname{im}A. \]
- Give a necessary and sufficient condition for \(M\) to be torsion.
- For \(R=F[x]\), give a square matrix \(A\) whose cokernel is \(R\times F\).
Solution
1. The cokernel is torsion exactly when \(\det A\ne0\).
By Smith normal form there are invertible \(P,Q\) such that \[ PAQ=\operatorname{diag}(d_1,\dots,d_r,0,\dots,0), \] with each \(d_i\ne0\). Hence \[ M\cong\bigoplus_{i=1}^rR/(d_i)\oplus R^{n-r}. \] This module is torsion exactly when \(r=n\), i.e. exactly when \(A\) has full rank over the fraction field of \(R\), equivalently when \(\det A\ne0\).
Alternatively, if \(d=\det A\ne0\), the adjugate identity \[ A\operatorname{adj}(A)=dI \] shows that \(d\) annihilates the entire cokernel.
2. Take \(R=F[x]\), \(n=2\), and\[ A=\begin{pmatrix}0&0\\0&x\end{pmatrix}. \]
Then \[ \operatorname{im}A=0\times xR, \] so \[ \operatorname{coker}A \cong R\times R/(x) \cong R\times F. \]