TITLE: Finite quotient ring of $\mathbb Z[X]$
QUESTION [14 upvotes]: Since userxxxxx (I don't remember the numbers) deleted his own question which I find interesting, let me repost it: 

Let $f,g\in\mathbb Z[X]$ with $\mathrm{gcd}(f,g)=1$. Prove that the ring $\mathbb Z[X]/(f,g)$ is finite.

REPLY [3 votes]: Proposition. Let $f,g\in\mathbb Z[X]$ with $\mathrm{gcd}(f,g)=1$. Then $\mathbb Z[X]/(f,g)$ is finite.
Proof. If $(f,g)=\mathbb Z[X]$, then there is nothing to prove. Assume $(f,g)\neq\mathbb Z[X]$. Since $\mathrm{gcd}(f,g)=1$, the elements $f,g$ form an $\mathbb Z[X]$-sequence and therefore $\dim\mathbb Z[X]/(f,g)=0$, i.e. the ring $\mathbb Z[X]/(f,g)$ is artinian. In particular, there are only finitely many maximal ideals of $\mathbb Z[X]$ containing $(f,g)$ and their product at some power coincides with $(f,g)$. This shows that $\mathbb Z[X]/(f,g)$ is isomorphic to a finite product of rings of the form 
$$\mathbb Z[X]/(M^k+(f,g))\simeq\frac{\mathbb Z[X]/M^k}{(\hat{f},\hat{g})},$$ 
where $M\subset\mathbb Z[X]$ is a maximal ideal. Thus it is enough to show that $\mathbb Z[X]/M^k$ is a finite ring. In order to do this let's recall the form of the maximal ideals of $\mathbb Z[X]$: $M=p\mathbb Z[X]+u\mathbb Z[X]$ where $p\in\mathbb Z$ is a prime number and $u\in\mathbb Z[X]$ is a polynomial irreducible modulo $p$. Obviously 
$$\mathbb Z[X]/M\simeq\mathbb Z/p\mathbb Z[X]/(\overline{u})$$ 
is a finite field and $M^{i-1}/M^i$ are $\mathbb Z[X]/M$-vectorspaces of finite dimension. This shows that $\mathbb Z[X]/M^k$ is a finite ring.