root of unity in quotient field Let R be an commutative ring, in which $n \ge 2$ is invertible. Show that the modulo class of $y$ in the quotient ring $ R'=R[y]/\langle y^n-1\rangle $ is not an primitiv $n$th-root of unity in R'.
Hi,
this is a problem which bugs me for quite a while. 
I don't know exactly where I shall begin. I know n is invertible which means there exists $n^{-1} \in R$ and n is invertible in R' too, and the modulo class of $y$ is $y~mod~y^n-1$. Can someone give me an hint how I can  proceed from here? Thank you.
Greets Bernd.
 A: I think you need $n$ to be even, and $R\neq0$ for this to be true.
As I indicated in the comments, I would first prove that the polynomials $y-1$ and $y^{n-1}+\cdots+1$ are relatively prime in $R[y]$, i.e., the ideal
$$
(y-1,y^{n-1}+\cdots+1)
$$
in $R[y]$ is the trivial ideal $(1)$. In order to show this, use transitivity of quotients:
$$
R[y]/(y-1,y^{n-1}+\cdots+1)=
(R[y]/(y-1))/(\bar y^{n-1}+\cdots+1)=
R/(1+\cdots+1)=R/n=0
$$
since $n$ is invertible in $R$, where $\bar y$ denotes the class of $y$ mod $y-1$, i.e., $\bar y=1$ in $R[y]/(y-1)=R$. Therefore
$$
(y-1,y^{n-1}+\cdots+1)=(1)
$$
in $R[y]$.
By the Chinese remainder theorem, the quotient morphisms
$$
R[y]\rightarrow R[y]/(y-1)=R
\quad\text{and}\quad
R[y]\rightarrow R[y]/(y^{n-1}+\cdots+1)
$$
induce an isomorphism of rings
$$
R'=R[y]/(y^n-1)\rightarrow R\times R[y]/(y^{n-1}+\cdots+1).
$$
This morphism send $y$ to the element $(1,\bar y)$, where $\bar y$ now denotes the class of $y$ modulo $y^{n-1}+\cdots+1$.
It is now clear that $y$ is not a primitive $n$-th root of unity in $R'$, i.e., that $y$ does not generate the subgroup of $(R')^\times$ of elements having order a divisor of $n$. Indeed, since $n$ is even $(-1,1)$ is a $n$-th root of unity but 
$$
(-1,1)\neq (1,\bar y)^m
$$
for all integers $m$, otherwise $1=-1$ in $R$, hence $2=0$ in $R$, and $n=0$, and then $R=0$, which was excluded. This proves that $y$ is not a primitive $n$-root of unity in $R'$ if $n$ is even and $R\neq0$.
If $n$ is odd then the statement is not true.
Take $n=3$ and $R=\mathbf Z$. One gets $R'=\mathbf Z\times\mathbf Z[j]$, where $j$ is a primitive cubic root of unity, and the class of $y$ in $R'$ corresponds to the element $(1,j)$. The group of cubic roots of unity in $\mathbf Z\times\mathbf Z[j]$ is $\{(1,1),(1,j),(1,j^2)\}$, and $(1,j)$ is a generator. Hence, $y$ is a primitive cubic root of unity this case.
