Let $f$ be a monic polynomial in $\Bbb{Z}[X]$. Show that $\Bbb{Z}[X]/〈f〉$ is a finitely generated $\Bbb{Z}$-module.
I don't even know how to start. If $g\in\Bbb{Z}[X]/〈f〉$, we are trying to find $h_i\in\Bbb{Z}[X]/〈f〉$ and $a_i\in\Bbb{Z}$ such that $g+\langle f\rangle=a_1h_1+a_1\langle f\rangle +...+a_nh_n+a_n\langle f\rangle$. How do I use that $f$ is monic?
On
$\begin{eqnarray}{\bf Hint}\quad\ x^n &\in&\! \langle x^{n-1},\ldots,x,1\rangle\, =: M \\ \Rightarrow\ x^{n+1} &\in& \langle x^n,\,\ldots,x^2,x\rangle\, \subseteq\, M\ \ {\rm by}\ \ x^n\in M\\ &\vdots&\\ \Rightarrow\ x^{n+k} &\in& M \end{eqnarray}$
Remark $\, $ Instead of using the division algorithm, we interpret the congruence $\, 0\equiv f = x^n - g\,$ as a rewrite rule $\, x^n \to g\,$ used to inductively reduce all exponents on $\,x\,$ to be $\,< n.\, $ This breaks down if the leading coeff of $\,f\,$ is not $1$ (or a unit), e.g. we can't employ $\,2x\to 1\,$ to rewrite $\,x^2.$
For a monic $f$, we have division with remainder, that means for every $g\in \mathbb{Z}[X]$, there are uniquely determined $q,r\in\mathbb{Z}[X]$ with
$$g = q\cdot f + r$$
and $r = 0$ or $\deg r < \deg f$ (if you use the convenient convention $\deg 0 = -\infty$, the last distinction is not necessary).
That means that every element of $\mathbb{Z}[X]/\langle f\rangle$ is the residue class of a polynomial whose degree is less than $\deg f$. This should enable you to find a nice finite set of generators for $\mathbb{Z}[X]/\langle f\rangle$.
If you look at the structure closely, you will notice that the "natural" set of generators of $\mathbb{Z}[X]/\langle f\rangle$ is in fact a basis, and $\mathbb{Z}[X]/\langle f\rangle$ is a free $\mathbb{Z}$-module of rank $\deg f$, in symbols
$$\mathbb{Z}[X]/\langle f\rangle \cong \mathbb{Z}^{\deg f}$$
as $\mathbb{Z}$-modules.