= 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 $N^3M=M$, 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 \b[there is exactly one simple <module> up to <isomorphism>], namely
$$
\boxed{A/(X)\cong k,\qquad X\cdot v=0.}
$$
This argument covers all simple <modules>, without assuming their finite dimensionality in advance.
Back to article page
