Past exam of the algebra course of the University of Oxford / 2007 / A / 3

Each $M_i$ is an ideal because the algebra is commutative. For $m\in M_i$, the product $m{e}_j$ belongs both to $M_i$ and $M_j$, so is zero when $j\ne i$. Multiplying $1={\sum }_j{e}_j$ by $m$ therefore gives $m=m{e}_i$. In particular $e_i^2=e_i$, different $e_i$ have zero product, andFor any idempotent $e$, the set $Ae$ is closed under multiplication and $e$ acts as its identity. With $f=1-e$ the map $a\mapsto (ae,af)$ is an algebra isomorphism $A\to Ae\times Af$: multiplication is componentwise because $ef=0$, and its inverse sends $(u,v)$ to $u+v$.

The $A$-submodules of $Ae$ are exactly its ideals as a unital algebra, since acting by $a$ on $Ae$ is the same as acting by $ae$. A nonzero commutative unital algebra is a simple module over itself exactly when every nonzero principal ideal is the whole algebra, which is equivalent to every nonzero element having an inverse. Thus a nonzero component $Ae$ is simple exactly when it is a field. If $e=0$, both properties fail under the nonzero-field convention.

A semisimple regular module is a finite direct sum of simple ideals. Applying the preceding decomposition gives a finite direct product of fields. Conversely the coordinate ideals of a finite product of fields are simple modules, so its regular module is semisimple. The fields are finite extensions of $K$, because $A$ has finite $K$-dimension.

For a generator $g$ of the cyclic group of order three, evaluation $X\mapsto g$ gives a surjection $\mathbb{Q}[X]\to \mathbb{Q}G$ with kernel $(X^3-1)$, by division and independence of $1,g,g^2$. The factors $X-1$ and $X^2+X+1$ are coprime, and the quadratic is irreducible over $\mathbb{Q}$. The Chinese remainder theorem yieldsThe explicit map sends $g$ to $(1,{\zeta }_3)$, where ${\zeta }_3=(-1+\sqrt{-3})/2$.