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

Every finite-dimensional $\mathbb{C}[X]$-module is a finitely generated torsion module over the principal ideal domain $\mathbb{C}[X]$. The structure theorem for finitely generated modules over a principal ideal domain giveswhere the $p_j$ are uniquely determined monic nonconstant polynomials. Equivalently, factoring over $\mathbb{C}$ and using the Chinese remainder theorem givesThere are finitely many summands, and the multisets of positive integers $m_{\lambda j}$ are uniquely determined. Each displayed primary cyclic summand is an indecomposable module.

Apply the decomposition to the module on which $X$ acts as $\alpha$. On $\mathbb{C}[X]/((X-\lambda {)}^m)$ put $t=X-\lambda$ and use the ordered basisMultiplication by $X=\lambda +t$ has diagonal entries $\lambda$ and superdiagonal entries $1$, so its matrix is the Jordan block $J_m(\lambda )$. Concatenating the bases of the direct sum gives a basis of $V$ in whichThis proves existence of Jordan normal form; uniqueness of the primary cyclic decomposition also proves uniqueness of its blocks up to order.

The universal property of the polynomial ring gives the unique complex algebra homomorphismDistinct monomials $X^{2j}$ are linearly independent, so $\iota$ is injective. Restriction of scalars defines $b\cdot m=\iota (b)m$ for $b\in B$ and $m\in M$; its module axioms follow from those of the original action and the algebra homomorphism. In particular $Y$ acts as the square of the operator defining the $X$ action.

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.

A two-dimensional indecomposable module over $\mathbb{C}[X]$ is exactly $A/((X-\lambda {)}^2)$ for some $\lambda \in \mathbb{C}$. Write its operator as $T=\lambda I+N$ with $N^2=0$ and $N\ne 0$. On restriction of scalars,For $\lambda =0$ this is the zero operator on a two-dimensional space, which splits as two one-dimensional submodules. For $\lambda \ne 0$, its nilpotent part is nonzero of square zero, so it is one Jordan block and is indecomposable. Therefore the full list isDifferent $\lambda$ give nonisomorphic $A$-modules; the restricted $B$-modules for $\lambda$ and $-\lambda$ are isomorphic.