$R[x]/(x^n-1)=R[G]$ as rings Let $R$ be a commmutative ring with $1$ and $G$ finite cyclic group of order $n$. 
Show that $R[x]/(x^n-1)=R[G]$ (isomorphic) as rings.
This is what I did. Suppose $G=\langle b\rangle $. Let $\psi\colon R[x] \to R[G]$ by $\psi(a_0+a_1x+\ldots+a_mx^m)=a_0e+a_1b+\ldots+a_mb^m$. Check it's well defined and an epimorphism. I can see $(x^n-1)\subseteq \ker\psi$, but why $\ker\psi\subseteq(x^n-1)$?
If $G=\mathbb{Z}$, the additive integer group, what's the map for $R[G]=R[x,x^{-1}]$?
Thank you.
 A: If $a_0+a_1x+\ldots+a_mx^m$ is mapped to zero, we have $a_i+a_{n+i}+a_{2n+i}+\ldots+a_{k_in+i}=0$ for all $i=0,1,\ldots,n-1$, where $k_i$ satisfies $k_in+i\le m<(k_i+1)n+i$.  It follows that all $n$th roots of unity are roots of the polynomial $a_0+a_1x+\ldots+a_mx^m$, so that $x^n-1$ is a factor.
A: Let $f\in\ker \psi$. By subtracting a multiple of $x^n-1$, you may assume that $\deg f<n$.
But if $a_0+\ldots + a_{n-1}x^{n-1}\in\ker\psi$, then $a_0e+\ldots +a_{n-1}b^{n-1}=0$, i.e. $a_0=\ldots =a_{n-1}=0$.
The corresponding isomorphism $R[\mathbb Z]\to R[x,x^{-1}]$ is given by $rn\mapsto rx^n$, of course.
A: POf course, you suppose $b^n=e$. Then every element from has an unique form $a_0e+a_1b+\ldots+a_{n-1}b^{n-1}$. Now let $a_0+a_1x+\ldots+a_mx^m\in \ker\psi$. Write it (uniquely!) as $a_0+a_1x+\ldots+a_mx^m= c_0+c_1x+\ldots+c_{n-1}x^{n-1}+ f(x)(x^n-1)$. Then $0=\psi (a_0+a_1x+\ldots+a_mx^m)=c_0+c_1b+\ldots+c_{n-1}b^{n-1}$. Hence $c_0=c_1b=\ldots=c_{n-1}=0$ and  $a_0+a_1x+\ldots+a_mx^m=  f(x)(x^n-1)$.
A: Let $A$ be an $R$-algebra (commutative with identity). Then there is a natural bijection $\mathrm{Hom}_{R-\mathrm{Alg}}(R[X]/(X^n-1),A)=\{a\in A:a^n=1\}$ given by $\varphi\mapsto\varphi(x)$, where $x=X+(X^n-1)$ is the image of $X$ in $R[X]/(X^n-1)$. (In fancier language, $\mathrm{Spec}(R[X]/(X^n-1))$ is the group scheme of $n$-th roots of unity over $R$.) This is a consequence of the universal property of the $R$-algebra $R[X]$.
There is also a natural bijection $\mathrm{Hom}_{R-\mathrm{Alg}}(R[\langle b\rangle],A)\rightarrow\mathrm{Hom}_{\mathrm{Grp}}(\langle b\rangle,A^\times)$ given by restricting along the canonical injection $\langle b\rangle\hookrightarrow R[\langle b\rangle ]$. This is a consequence of the universal property of the group algebra $R[\langle b \rangle]$. The latter set of group homomorphisms is in canonical bijection with $\{a\in A:a^n=1\}$ by sending a group homomorphism to its image on the generator $b$ (this is just because $\langle b\rangle $ is a cyclic group of order $n$). I should mention that these bijections are only pinned down by the choice of a generator, but once one fixes a generator, one gets canonical bijections.
Thus both $R$-algebras represent the same functor on the category of all (commutative) $R$-algebras (with identity), so by Yoneda's lemma there is a unique isomorphism $R[X]/(X^n-1)\cong R[\langle b\rangle]$ compatible with the above identifications. As can be checked, this isomorphism is the one that interchanges $X$ and $b$.
Similar reasoning applies to $R[X,X^{-1}]$ and $R[\mathbf{Z}]$, replacing $\{a\in A:a^n=1\}$ with $A^\times$, the unit group of $A$. (Again in fancier language, the spectra of these two $R$-algebras both represent the multiplicative group scheme over $R$.) 
I realize the OP might not be familiar with this language, but I still thought it might be useful to have this perspective recorded as an answer.  
