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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 ?

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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

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

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  • $\begingroup$ Ah right exactly ... $R[\mathbb Z/(n)]\cong R[X]/(X^n-1)$ and the latter is isomorphic with $R^n$ when $X^n-1$ splits in $R$ and has no multiple roots i.e. $R$ contains all the $n$-th roots of unity and $char R\ne n$ right ? Also, I wanted to point out that if $R$ satisfies a.c.c. on radical ideals, then $Spec R$ has finitely many connected components ... $\endgroup$ – user495643 Jun 25 '18 at 21:01
  • $\begingroup$ @misao: you really do need the additional hypothesis that $n$ is invertible. For example, when $n = 2$, the ring $R = \mathbb{Z}$ has all second roots of unity and its characteristic is not $2$, but $\mathbb{Z}[x]/(x^2 - 1)$ is not isomorphic to $\mathbb{Z}^2$. $\endgroup$ – Qiaochu Yuan Jun 25 '18 at 21:05
  • $\begingroup$ ah I see that now ... I thought only having distinct roots would make the quotient split ... what I do need is that for every two roots $a,b \in R$ of $X^n-1$, $(X-a)R[X]$ and $(X-b)R[X]$ are co-maximal in $R[X]$ ... is that assured by invertiblty of $n$ in $R$ ? $\endgroup$ – user495643 Jun 25 '18 at 21:10
  • $\begingroup$ Umm ... I seem to have stuck on why invertiblity of $n$ along with existence of all $n$-th roots of unity is enough to conclude $R[X]/(X^n-1) \cong R^n$ ... could you please elaborate here .? $\endgroup$ – user495643 Jun 26 '18 at 10:48
  • $\begingroup$ @misao: you can write down an explicit isomorphism using the discrete Fourier transform (en.wikipedia.org/wiki/Discrete_Fourier_transform). $\endgroup$ – Qiaochu Yuan Jun 26 '18 at 23:23
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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.

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  • $\begingroup$ Thanks ... I didn't know $R[\mathbb Z/(n)] \cong R[X]/(X^n -1)$ ... could you please provide a proof for that ? $\endgroup$ – user495643 Jun 24 '18 at 10:19
  • $\begingroup$ Both rings are naturally free $R$-modules of rank $n$, one with basis $\overline{k}\in\Bbb{Z}/(n)$ the other with $X^k, 0\le k<n$. Map $\overline{k}$ to the coset of $X^k$. Modulo $X^n-1$ precisely takes care of $k\equiv k'\pmod n$ as in that case $X^k\equiv X^{k'}\pmod{X^n-1}$. Therefore you get an isomorphism of rings as well as of $R$-modules. $\endgroup$ – Jyrki Lahtonen Jun 24 '18 at 11:48
  • $\begingroup$ Thanks a lot ... so we have counterexamples ... unfortunately yours are not Noetherian ... but you still answer half of my question ... thank you for that .. $\endgroup$ – user495643 Jun 24 '18 at 12:00

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