Reduction modulo: $\mathbb{Z}[x]\to\mathbb{Z}/p\mathbb{Z}[x]$ Let
$$f(x)=\sum_{i}a_ix^i\in\mathbb{Z}[x],$$
and
$$[f]_{p}(x)=\sum_{i}[a_i]_{p}x^i\in\mathbb{Z}/p\mathbb{Z}[x].$$
Is it true that:
$$f(\xi)=0\Rightarrow [f]_{p}(\xi)=0,$$
where $\xi$ is some root of unity.
i.e. is it true proof:
$$[f]_{p}(\xi)=\sum_{i}[a_i]_{p}{\xi}^i=\sum_{i}{a}_i{\xi}^i-\sum_{i}(a_i-[a_i]_{p}){\xi}^i=0-p\cdot\alpha=0-0\cdot\alpha=0.$$
Thanks.

If $\xi$ is any algebraic integer, then there is a ring homomorphism $\mathbb{Z}[\xi] \rightarrow \bar{\mathbb{F}}_p$ (actually taking values in $\mathbb{F}_{p^n}$ for some $n$).
Just take a maximal ideal $\mathfrak{m}$ of $\mathbb{Z}[\xi]$ containing $p$ (there is at least one thanks to Zorn's lemma because $p$ is not invertible), then $\mathbb{Z}[\xi]/\mathfrak{m} \simeq \mathbb{F}_{p^n}$ for some $n$.
This ring homomorphism lifts the one you described ($\mathbb{Z} \rightarrow \mathbb{F}_p$), and allows to give a meaning to your computation.
Note that there can be more than one such lift (i.e. more than one maximal ideal containing $p$).
In the special case that $\xi$ is a $p^{\text{something}}$-th or $k$-th ($k$ dividing $p-1$) root of unity, $n$ is equal to $1$.
In the first case, there is a unique lift, but in the second case there are several if $k>1$.
