Quotient Ring of finite order with root of irreducible polynomial The question is as follows:
Suppose that $\alpha\in\mathbb{C}$ is a root of the irreducible polynomial $f(t) = t^d + \sum_{i=0}^{d-1}a_it^i$ , where $a_i\in\mathbb{Z}$ ($0\le i\le d-1$). Let $\mathcal{R} = \mathbb{Z}[\alpha]$ be the subring of $\mathbb{C}$ generated by $\alpha$. Show that if $m\in\mathbb{Z}$ is an integer, then $\mathcal{R}/m\mathcal{R}$ is finite, and calculate its order.
Attempt at solution:
I feel as if I do not fully grasp the question as to my knowledge, $\mathbb{Z}[\alpha] = \lbrace c + d\alpha : c,d\in\mathbb{Z} \rbrace$, hence I would believe that $\mathcal{R}/m\mathcal{R} = \lbrace c + d\alpha + m\mathbb{Z}[\alpha] : c,d \in \mathbb{Z} \rbrace $. Hence c and d could both take m different values, making the order $m^2$. I haven't used the irrecubile polynomial or the fact that $\alpha$ is a root at all, so I think this may not be correct.
Help is much appreciated.
 A: A good strategy for almost any problem to do with quotient rings is to try and "visualise" it in the following sensible way:
For a ring $R$, ideal $I\unlhd R$, the quotient $R/I$ can be viewed as a copy of $R$, in which the addition of any element of $I$ can be considered equivalent to "adding zero" (since, intuitively, this is the same thing as noting that $x+I = I \ \ \forall x\in I$).
In this case, this suggests that $\mathcal{R}/m\mathcal{R}$ has a realisation as the set of polynomials (w/ integer coefficients) of degree at most $d$ (which is where the irreducibility of $f$ comes in) with any multiple of $m$ considered as an additive zero*. From this, we can see that any term $b_k\alpha ^k$ of some polynomial in $\mathcal{R}$ with coefficient $b_k\geq m$ can be written as $b_k\alpha^k = (a+bm)\alpha^k = a \alpha^k + \underbrace{m(b\alpha^k)}_{"= \ 0"} = a\alpha ^k$ (some $a,b\in \mathbb{Z}$) by the division algorithm in $\mathbb{Z}$; and that we can continue this process until $a$ is the least such positive integer.
This suggests that $\mathcal{R}/m\mathcal{R}$ can be viewed as the set of polynomials in $\alpha$ of degree at most $d$, with integer coefficients strictly smaller than $m$; which is better known as $\mathbb{Z}_{m}[\alpha]$. To count the number of such polynomials, how many choices are there for the constant term? How many ways can you independently choose the next coefficient? And the next etc etc until you've covered all $d$ powers of $\alpha$. The product of these will give you the answer, and it is indeed finite.
[*Whilst this serves as intuition, it doesn't count as a proof that $\mathcal{R} /m\mathcal{R} \cong \mathbb{Z}_m[\alpha]$. To see this, consider the map $\theta_m:\mathcal{R} \to \mathbb{Z}_m[\alpha]$ s.t. $\displaystyle\sum _{i=0}^d b_i\alpha ^i \mapsto \displaystyle\sum _{i=0}^d (b_i \ \text{mod}\ m)\alpha ^i$. Is it well-defined? Homomorphism? Kernel? Surjective? Then consider the first isomorphism theorem.]
