A question about minimal primes in the integral group ring of a finite abelian group

Question

Let $G$ be a finite abelian group of order $n$ and let $R=\mathbb{Z}G$ denote the integral group ring of $G$. Let $R'$ denote the localization of $R$ with respect to the multiplicatively closed set generated by $n$. Let $P$ and $P'$ be distinct minimal primes of $R$. Then the following appears to be true: The extensions of $P$ and $P'$ to $R'$ are comaximal.

I can prove this in some simple cases by using an explicit construction of the minimal primes of $R$, facts about cyclotomic polynomials, and some basic algebraic number theory, but the extension to the general case seems messy. Does anyone have a simple proof, or at least a reference to a proof?

Answer 1

$R/P$ has characteristic zero, so $P$ is the kernel of a ring homomorphism $R\to \mathbb C$, which in turn corresponds to a group homomorphism $\varphi: G\to \mathbb C^\ast$. Let $H\subset G$ be the kernel of $\varphi$.

Likewise $P'$ is the kernel of a homomorphism $R\to \mathbb C$. Let $H'$ be the kernel of the corresponding $\varphi':G\to \mathbb C^\ast$

Suppose for contradiction that $P+P'$ is contained in a maximal proper ideal $M\subset R'$. Let $F$ be the (finite) field $R'/M$. Its characteristic, say $p$, does not divide $n$.

Since the $n$th cyclotomic polynomial has distinct roots in characteristic $p$, the map $(R/P)^\ast \to F^\ast$ is one to one on $n$th roots of $1$. It follows that $H$ is the kernel of the composed map $G\to (R/P)^\ast\to F^\ast$. By the same reasoning, $H'$ is the kernel of the (same) composed map $G\to (R/P')^\ast\to F^\ast$. So $H=H'$.

We have two isomorphisms from $G/H=G/H'$ to the same group of complex roots of $1$. Composing with an automorphism of the relevant cyclotomic field, we can take them to be the same isomorphism. But this shows that $P=P'$.