Isomorphism of $\mathbb Z[\xi]/\left<\xi-1\right>$ with $\xi$ root of unity. I am trying to prove the following isomorphism but I am not sure how to. 

Let $\xi=e^{2\pi i/p}$ and $p$ a prime. Show $$\mathbb Z[\xi]/\left<\xi-1\right> \cong \mathbb F_p .$$

I wanted to create a map $\phi : \mathbb Z \rightarrow \mathbb F_p $ with kernel $<\xi-1>$ but so for I have not have good ideas. Any hint? I know that any element in $<\xi-1>$ is of the form 
$$
(a_{p-1}-a_0)+(a_0-a_1)\xi+\cdots+(a_{p-2}-a_{p-1})\xi^{p-1}
$$
but I do not know how to use this information.
Thank you!
 A: For a different approach, you can use the Third Isomorphism Theorem.  Since
$$
\mathbb{Z}[\xi] \cong \frac{\mathbb{Z}[x]}{(x^{p-1} + x^{p-2} + \cdots + x + 1)}
$$
then
\begin{align*}
\frac{\mathbb{Z}[\xi]}{(\xi - 1)} \cong \frac{\mathbb{Z}[x]}{(x^{p-1} + x^{p-2} + \cdots + x + 1, x - 1)} \cong \frac{\mathbb{Z}[1]}{(1^{p-1} + 1^{p-2} + \cdots + 1 + 1)} \cong \frac{\mathbb{Z}}{(p)} = \mathbb{F}_p \, .
\end{align*}
A: A couple of hints:


*

*You should know that $\xi$ is a zero of the polynomial $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots+x+1.$$ Therefore $\Bbb{Z}[\xi]\simeq \Bbb{Z}[x]/\langle \Phi_p(x)\rangle$. Your mapping $\phi$ therefore must have the property
$$\xi^{p-1}+\cdots+\xi+1\mapsto 0.$$

*When your kernel is to contain the ideal $I=\langle \xi-1\rangle$, basically that means that $\phi$ should equate $\xi$ and $1$, or $\phi(1+I)=\phi(\xi+I)$.

*By induction the previous bullet implies that $\phi(\xi^k+I)=\phi(1+I)=1$ for all $k$.


Combining items 1 and 3 should get you there.
A: Hint: you need a homomorphism $\phi$ from $\Bbb{Z}[\xi]$ to $\Bbb{F}_p$ with $\xi - 1$ in its kernel, i.e., with $\phi(\xi - 1) = 0$. As $\phi(\xi - 1) = \phi(\xi) - \phi(1) = \phi(\xi) - 1$, what must $\phi(\xi)$ be? 
