Isomorphism of group rings over finite cyclic groups

Question

Let $R$ be a commutative ring with unity. For any group $G$, let the group ring be denoted by $R[G]$. If $R[\mathbb Z/(n) ] \cong R [\mathbb Z / (m)]$ as rings , then is it true that $m=n$ ? If that is not true in general, then is it true if we put some extra condition on $R$, like say Noetherian ?

Answer 1

Fix any two distinct natural numbers $n,m$. Because $R[\Bbb{Z}/(n)]\simeq R[x]/(x^n-1)$ we do get counterexamples like $$ R=K[x_1,x_2,\ldots,y_1,y_2,\ldots]/(x_1^n-1,x_2^n-1,\ldots,y_1^m-1,y_2^m-1,\ldots), $$ when $R[x]/(x^n-1)\simeq R\simeq R[x]/(x^m-1)$. Here $K$ is a field.

I'm afraid I'm too ignorant about commutative algebra to guess what extra condition would make it true.

Answer 2

Suppose that $R$ has all $n^{th}$ roots of unity and that $n$ is invertible in $R$: for example, this holds if $R$ is a $\mathbb{C}$-algebra. Then you can show that

$$R[C_n] \cong R^n$$

as $R$-algebras, so the problem reduces to asking whether $R^n \cong R^m$ implies $n = m$.

The answer is yes if $\text{Spec } R$ has finitely many connected components, or equivalently if $R$ has finitely many idempotents, which is in particular the case if $R$ is either Noetherian or an integral domain, since then we can just count connected components (or idempotents) on both sides. It is no in general since, for example, we could have $R \cong k^{\mathbb{N}}$ for some other commutative ring $k$ satisfying the above conditions (and then $\text{Spec } R$ has infinitely many connected components).

More generally, the answer is yes if you can find any other commutative ring $S$ such that there are at least two but at most finitely many homomorphisms $S \to R$ (for idempotents take $S = \mathbb{Z}[x]/(x^2 - x) \cong \mathbb{Z}^2$, the free commutative ring on an idempotent), since then $R^n \cong R^m$ implies

$$\text{Hom}(S, R^n) \cong \text{Hom}(S, R)^n \cong \text{Hom}(S, R^m) \cong \text{Hom}(S, R)^m.$$