Let \(D = \QQ[x]\) and let \(M\) be a \(\QQ[x]\dash\)module such that \[ M \cong \frac{\mathbb{Q}[x]}{(x-1)^{3}} \oplus \frac{\mathbb{Q}[x]}{\left(x^{2}+1\right)^{3}} \oplus \frac{\mathbb{Q}[x]}{(x-1)\left(x^{2}+1\right)^{5}} \oplus \frac{\mathbb{Q}[x]}{(x+2)\left(x^{2}+1\right)^{2}} .\]
Determine the elementary divisors and invariant factors of \(M\).
Solution
Set \[ a=x-1, \qquad b=x+2, \qquad q=x^2+1. \] These are pairwise coprime irreducibles in \(\mathbb Q[x]\). By the Chinese remainder theorem, \[ D/(aq^5)\cong D/(a)\oplus D/(q^5) \] and \[ D/(bq^2)\cong D/(b)\oplus D/(q^2). \] Therefore the elementary divisors of \(M\) are \[ a^3,\quad a,\quad b,\quad q^5,\quad q^3,\quad q^2, \] i.e. \[ (x-1)^3,\quad x-1,\quad x+2,\quad (x^2+1)^5,\quad(x^2+1)^3,\quad(x^2+1)^2. \]
To form invariant factors, align the powers of each irreducible from the right. For \(a\) the exponent lists are \(1,3\); for \(q\) they are \(2,3,5\); for \(b\) there is one exponent \(1\). This yields three invariant factors \[ d_1=q^2, \qquad d_2=a q^3, \qquad d_3=a^3bq^5. \] Explicitly, \[ \boxed{ \begin{aligned} d_1&=(x^2+1)^2,\\ d_2&=(x-1)(x^2+1)^3,\\ d_3&=(x-1)^3(x+2)(x^2+1)^5. \end{aligned}} \] They satisfy \(d_1\mid d_2\mid d_3\), and their product is the product of the original cyclic annihilators, confirming the invariant-factor decomposition.