Why $\mathbb{Q}\mathbb{Z}_3 \cong \mathbb{Q(\varepsilon_3)} \times \mathbb{Q}$ as rings (or even algebras)?

Question

I don't see why the statement $\mathbb{Q}\mathbb{Z}_3 \cong \mathbb{Q(\varepsilon_3)} \times \mathbb{Q}$ holds (here $\varepsilon_3$ denotes the primitive $3$-root of unity). On one hand, $\mathbb{Q}\mathbb{Z}_3$ is semisimple, so it is a product of matrix algebras. It is clearly not the product of three copies of $\mathbb{Q}$, since the multiplication differs a lot. So what's that?

Answer 1

We have $\mathbb{Q}\mathbb{Z}_3\cong \mathbb{Q}[x]/(x^3-1)$ where $x$ corresponds to a generator of $\mathbb{Z}_3$. By the Chinese remainder theorem, $\mathbb{Q}[x]/(x^3-1)\cong\mathbb{Q}[x]/(x-1)\times\mathbb{Q}[x]/(x^2+x+1)$ where the first factor is just isomorphic to $\mathbb{Q}$ and the second factor is isomorphic to $\mathbb{Q}(\epsilon_3)$ since $x^2+x+1$ is the minimal polynomial of $\epsilon_3$.