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

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.