Suppose the elementary divisors of a finitely generated torsion \(F[x]\)-module are \[ (x-1)^3,\quad (x-1),\quad (x^2+1)^4,\quad (x^2+1)^2,\quad (x^2+1)^2,\quad (x+2). \] Determine the invariant factors.
Solution
Group the elementary divisors by irreducible polynomial and order the powers increasingly: \[ \begin{array}{c|ccc} x-1 & 1 & (x-1) & (x-1)^3\\ x^2+1 & (x^2+1)^2 & (x^2+1)^2 & (x^2+1)^4\\ x+2 & 1 & 1 & (x+2). \end{array} \] There are three invariant factors because the largest number of elementary divisors belonging to a single irreducible is three.
Multiplying down the columns gives \[ d_1=(x^2+1)^2, \] \[ d_2=(x-1)(x^2+1)^2, \] and \[ d_3=(x-1)^3(x^2+1)^4(x+2). \] They satisfy \[ d_1\mid d_2\mid d_3, \] as required.
Thus the invariant-factor decomposition is \[ F[x]/(d_1)\oplus F[x]/(d_2)\oplus F[x]/(d_3). \]