Let \(R=\mathbb C[x]\) and \(M=\mathbb C^2\), with \(R\)-module structure defined by \[ p(x)\cdot v=p(A)v, \qquad A=\begin{pmatrix}1&2\\2&1\end{pmatrix}. \] Show that \(M\) is a cyclic \(R\)-module generated by \(e_1=(1,0)^t\), and show that \(M\) is a torsion \(R\)-module. Determine its annihilator.
Solution
Let \(e_1=(1,0)^t\). Since \[ Ae_1=(1,2)^t, \] the vectors \(e_1,Ae_1\) are linearly independent. Hence they form a basis of \(\mathbb C^2\), so every vector has the form \[ (s+tA)e_1 \] for suitable \(s,t\in\mathbb C\). Thus \(M\) is cyclic as a \(\mathbb C[x]\)-module, generated by \(e_1\).
The characteristic polynomial of \(A\) is \[ \chi_A(x)=x^2-2x-3=(x-3)(x+1). \] Because \(A\) has two distinct eigenvalues, its minimal polynomial is also \(x^2-2x-3\). Therefore \[ (x^2-2x-3)(A)=0, \] so every element of \(M\) is annihilated by this nonzero polynomial. Hence \(M\) is torsion, and since \(e_1\) is cyclic, \[ \operatorname{Ann}_{\mathbb C[x]}(M)=(x^2-2x-3). \] Equivalently, \[ M\cong \mathbb C[x]/(x^2-2x-3). \]