is there a ring map from $\mathbb{Z}[e^{2\pi i/(p-1)}]$ to $\mathbb{Z}$ mod p where p prime? I have learned about splitting fields and finite fields and some related concepts from my maths courses and I got suspicious that $F_p$ can be though of as not just as splitting field for $x^{p-1} - 1$ where $1$ is in $F_p$ ( a bit cyclic there but ) or we first factorize $x^{p-1} - 1$ where $1$ is in $\mathbb{Z}$ then somehow roles of $1$s can be interchanged by a ring map. I think this is true and somehow must be well known but to verify and to get some reference for it I thought to write here. Could you advise some textbooks that deal with this or similar things as well. I suppose this is from algebraic number theory.
For what have I looked for, I know that if this is true this gives easy proof that $F_p - {0}$ is cyclic group and other kinds of results, and for example $F_5$ can be thought of $\mathbb{Z}[i]/(3 + i)$
Thank you
 A: The quotient ring $\mathbf Z[i]/(3+i)$ is not a field of order $5$. It is a ring of order $10$.  But $\mathbf Z[i]/(2+i)$ is a field of order $5$.
As you suspected, what you ask about is well known and is part of algebraic number theory. If $\alpha$ is an algebraic integer and $\mathfrak a$ is a nonzero ideal in $\mathbf Z[\alpha]$ then $\mathbf Z[\alpha]/\mathfrak a$ is a finite ring (just like $\mathbf Z/m\mathbf Z$ is a finite ring). In particular, if $\mathfrak p$ is a nonzero prime ideal in $\mathbf Z[\alpha]$ then $\mathbf Z[\alpha]/\mathfrak p$ is a finite field (finite integral domains are fields). The intersection $\mathfrak p \cap \mathbf Z$ is a nonzero prime ideal $p\mathbf Z$ in $\mathbf Z$ and $\mathbf Z[\alpha]/\mathfrak p$ has characteristic $p$ since $p \equiv 0 \bmod \mathfrak p$.
Example.
Let $\zeta_m = e^{2\pi i/m}$. In the ring $\mathbf Z[\zeta_m]$,
for each prime $p$ not dividing $m$ there are prime ideals
$\mathfrak p$ such that $\mathfrak p \cap \mathbf Z = p\mathbf Z$ and the size of $\mathbf Z[\zeta_m]/\mathfrak p$ is the order of $p \bmod m$. (The story when $p \mid m$ is more complicated and I omit it.) Properties of prime ideals in $\mathbf Z[\zeta_m]$ are related to how the $m$th cyclotomic polynomial $\Phi_{m}(x)$ factors modulo $p$. (You are incorrect in focusing on $x^{m}-1$, which for $m > 2$ is not the minimal polynomial of the roots of unity of order $m$.)
Taking $m = p-1$, so $p \equiv 1 \bmod m$, we see that $\mathbf Z[\zeta_{p-1}]/\mathfrak p$ has order $p$ when $\mathfrak p$ is a prime ideal containing $p$.
Therefore the natural ring homomorphism $\mathbf Z/p\mathbf Z \to \mathbf Z[\zeta_{p-1}]/\mathfrak p$ is an isomorphism.
Due to the background needed to fill in the details for this kind of material, I do not think it is really an "easy" proof of $(\mathbf Z/p\mathbf Z)^\times$ being cyclic if you have not already seen a certain amount of algebraic number theory. Others have observed this kind of approach before.  An example of such a proof that views $\mathbf Q(\zeta_{p-1})$ inside the $p$-adic numbers $\mathbf Q_p$ rather than using a prime ideal in the ring $\mathbf Z[\zeta_{p-1}]$ is in Matt Baker's blog post here.
