Past exam of the algebra course of the University of Oxford / 2011 / 3 / c / i / Solution

A direct determinant computation givesThe zero eigenvalue has algebraic multiplicity two but a one-dimensional eigenspace, so it gives one Jordan block of size two. The other block is $J_1(4)$. For an explicit Jordan basis, takeThese are linearly independent and satisfy $Dv=0$, $Du=v$, $Dw=4w$. ThusThe first summand has the chain $u\mapsto v\mapsto 0$ and is an indecomposable module; the other is simple.