Past exam of the algebra course of the University of Oxford / 2010 / A / 2

A simple module is a nonzero module with no submodules other than zero and itself. Equivalently, every nonzero element generates the entire module: $Am=M$ whenever $m\ne 0$.

A composition series of a module is a finite chainof submodules whose factors $M_j/M_{j-1}$ are simple modules. A finite-dimensional module has such a series by induction on its dimension, choosing a maximal proper submodule at each step.

The Jordan–Hölder theorem for modules states that any two composition series of a module have the same length and the same simple factors up to isomorphism and reordering. Thus the multiset of composition factors is an invariant of the module, even when the series itself is not unique.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

The finite-dimensional qualification is essential for this scalar conclusion. For example, the field $\mathbb{C}(t)$ is a complex algebra, simple as a module over itself, and its endomorphisms include multiplication by $t$, which is not a complex scalar.

Let $M$ be a nonzero simple module for $A=k[X]/(X^3)$ and let $N$ be the action of $X$. Since $X$ is central, $NM$ is a submodule, and $N^3=0$. If $NM=M$, iterating would give $N^{3}M=M$, impossible. Thus $NM=0$.

The action therefore factors through $A/(X)\cong k$. A simple module over a field is one-dimensional: the span of any nonzero vector is already a nonzero submodule. Conversely, $k$ with $X$ acting as zero is simple. Hence there is exactly one simple module up to isomorphism, namelyThis argument covers all simple modules, without assuming their finite dimensionality in advance.

On a finite-dimensional $A$-module $M$, the actions of $X$ and $Y$ are endomorphisms $P,Q$ satisfying $PQ-QP=I_M$. Taking traces and using $\mathrm{tr}(PQ)=\mathrm{tr}(QP)$ givesSince $k$ has characteristic zero, this forces ${\dim }_{k}M=0$. Thus the zero module is the only finite-dimensional module for this Weyl algebra. The obstruction is the Weyl algebra has no nonzero finite-dimensional modules.