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

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.

On the same basis, $Y=D^2$ kills $v,u$ and sends $w$ to $16w$. Explicitly,Thus all three basis lines are submodules after restriction of scalars, andEach summand is a simple module, hence an indecomposable module. Squaring has destroyed the length-two nilpotent Jordan chain.