I am a bit unsure about my following "proof", in particular the rationality of "taking projection".
Problem: $R$ is a ring with identity, not necessarily commutative. If $\epsilon\in R$ satisfies the equation $1+\epsilon+\cdots+\epsilon^{p-1}=0$, where $p$ is a prime number, show that for all integer $0<c<p$, we have $1+\epsilon^c+\epsilon^{2c}+\cdots+\epsilon^{(p-1)c}=0$.
Proof: By considering roots in $\mathbb{C}[X]$, there exists $\phi(X)\in\mathbb{Q}[X]$ s.t. \begin{equation*} 1+X^c+X^{2c}+\cdots+X^{(p-1)c}= (1+X+\cdots+X^{p-1})\phi(X). \end{equation*} Moreover this $\phi(X)$ is actually in $\mathbb{Z}[X]$ since $1+X+\cdots+X^{p-1}$ is monic. Now we take projection $\pi: \mathbb{Z}\to R$, $1\mapsto 1_R$ in the above equation and yield an identity on $R[X]$. Plugging in $X=\epsilon$ implies $1+\epsilon^c+\epsilon^{2c}+\cdots+\epsilon^{(p-1)c}=0$. $\square$
My confusion is that taking projection $\pi: \mathbb{Z}\to R$ and seems a bit causual. Is this proof correct?
The key is that the projection induces a homomorphism of the polynomial rings. In general, if $\psi:A\to B$ is a ring homomorphism, then it induces a homomorphism $\bar\psi:A[X]\to B[X]$ given by $$ \bar\psi \left(\sum_{i=0}^n a_i x^i\right) := \sum_{i=0}^n \psi(a_i) x^i. $$ Thus, any polynomial identity in $A[X]$ can be carried to a polynomial identity in $B[X]$.
Maybe a down-to-earth example would make it more clear. In $\Bbb Z[X]$, we have $(3X + 2)^2 = 9X^2 +12X+4$. Using the "mod 4" homomorphism $\Bbb Z \to \Bbb Z/4\Bbb Z$, we get $$(\bar3X + \bar2)^2 = \bar9X^2 +\bar{12}X+\bar4 = X^2$$ in $(\Bbb Z/4\Bbb Z)[X]$.
So in your question, $\sum X^{kc}$ and $\sum X^{k}$ in $\Bbb Z[X]$ maps to $\sum 1_RX^{kc}$ and $\sum 1_RX^{k}$ in $R[X]$. $\phi$ maps to some polynomial $\phi'$ in $R[X]$ (we don't care what it is). The equation $\sum X^{kc} = \left(\sum X^{k}\right)\phi'(X)$ still holds in $R[X]$ because the operation is homomorphic.