Jacobson radical of the integral group ring

Question

I am trying to prove that the Jacobson radical of the integral group ring $\mathbb{Z}G$ for a finite group is zero. Most of what I find on semisimplicity, Jacobson semisimplicity, has to do with group algebras $KG$ where $K$ is a field. I did come across this,

Jacobson radical of a ring finitely generated over $\mathbb Z$

but it does not seem to apply here, as the Jacobson radical would only vanish for PID an Dedekind rings.

Thanks for the help!

Answer 1

I have located a proof as lemma 4.1 in the following article:

Akasaki, T. (1972). Idempotent ideals in integral group rings. Journal of Algebra, 23(2), 343–346.

The article is available online, but the proof is short, so I reproduce it here:

$\rm L{\small EMMA}$ 4.1. Let $\pi$ be any finite group. Then the Jacobson radical of $\mathbb{Z}\pi$ is zero.

Proof. For a given ring $R$, let its Jacobson radical be written as $J(R)$. Let $p$ be a prime in $\mathbb{Z}$ and $\eta_p: \mathbb{Z}\pi \to \mathbb{Z}_p \pi$ be coefficient reduction. Since $\eta_p$ is a ring epimorphism, $J(\mathbb{Z}\pi)$ is contained in the inverse image $\eta_p^{-1}(J(\mathbb{Z}_p \pi))$. Thus if $p \nmid |\pi|$ then, by Maschke’s theorem, $J(\mathbb{Z}\pi) \subset \ker(\eta_p)$. Therefore, if $\lambda \in J(\mathbb{Z}\pi)$, $\lambda \in \ker(\eta_p)$ for infinitely many primes $p \nmid |\pi|$. This means that $\lambda = 0$ and hence $J(\mathbb{Z}\pi) = (0)$.