Here, $\mathbb{Z}[\alpha]$ denotes the ring generated over $\mathbb{Z}$ by $\alpha$. The first thing that comes to mind is considering the homomorphism that takes a polynomial $p(\alpha)$ to the polynomial with coefficients $ \bmod m$. This is a surjective homomorphism with kernel the set of polynomials in $R$ with coefficients that are multiples of $m$ which is precisely $mR$. So the quotient $ R/mR$ is isomorphic to the set of polynomials with coefficients $ \bmod m $. But there is an infinite number of these polynomials because for all $n$ in $\Bbb Z$ there is always a polynomial of degree $d\ge n$.
Let $\alpha$ be an algebraic integer, $R=\Bbb Z[\alpha]$, and $m \in \mathbb{Z}$. Prove that $ R/mR$ is finite and find $|R/mR|$
abstract-algebrapolynomialsring-theory
Best Answer
Since $\alpha$ is algebraic, let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial of minimal degree with $f(\alpha) = 0$. (Such an $f(X)$ necessarily exists by the well-ordering of the natural numbers.) There is a surjective ring homomorphism $\varphi \colon \mathbb{Z}[X] \to R$ sending $g(X)$ to $g(\alpha)$. I will describe the rest of the problem in exercises, with hints hidden behind spoilers; I encourage you to try these exercises yourself first before looking.
Exercise $1$: show that $\ker(\varphi) = \langle f(X) \rangle$.
Hence, $R \cong \mathbb{Z}[X]/\langle f(X) \rangle$, and so $R/mR \cong (\mathbb{Z}/m\mathbb{Z})[X]/\langle \overline{f}(X) \rangle$, where $\overline{f}(X)$ denotes the class of $f(X)$ under the natural reduction map $\mathbb{Z}[X] \to (\mathbb{Z}/m\mathbb{Z})[X]$. Note that $\overline{f}(X)$ is also monic.
Exercise $2$: show that $(\mathbb{Z}/m\mathbb{Z})[X]/\langle \overline{f}(X) \rangle$ is a free $(\mathbb{Z}/m\mathbb{Z})$-module of rank $\deg(f)$. Conclude that $|R/mR| = m^{\deg(f)}$.