Show that group ring QG is isomorphic to Q(e3)xQ.

Question

Let G be a group of order 3 and e3 - primitive root of unity of order 3. Show that group ring QG is isomorphic to Q(e3)xQ (Carthesian product).

Earlier I have shown that group ring Q(e3)G is isomorphic to 3 copies of Q(e3), but I don't know how to do this part. It seems counterintuitive bcs QG rather looks like only Q(e3), without Q on second carthesian coordinate.

Any hints?

Answer 1

Let $j$ be a primitive third root of $1$. Since $G$ has order $3$, it is cyclic.

I leave you as an exercise to prove the following sequence of isomorphisms of $\mathbb{Q}$-algebras:

$$\mathbb{Q}[G]\simeq \mathbb{Q}[X]/(X^3-1)\simeq \mathbb{Q}[X]/(X^2+X+1)\times \mathbb{Q}[X]/(X-1)\simeq \mathbb{Q}(j)\times\mathbb{Q}.$$