Finite quotient ring of $\mathbb Z[X]$ 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.
The already-present answers show that $(f,g)$ necessarily contains a non-zero integer $n$, and so the quoitient is equal to $(\mathbb Z/n\mathbb Z)[x]/(f,g).$  So it remains to show that this quotient is finite.

There are various ways to proceed at this point.  Here is one.


First, suppose $n$ is a prime $p$.  Since $f$ and $g$ don't have a non-trivial common factor in $\mathbb Z[x]$, at least one of them is non-zero mod $p$, and the quotient of $\mathbb F_p[x]$ by any non-zero ideal is finite.
Now if $n$ is a prime power, say $p^e$, then by what we just proved, the ideal $(f,g)$ has non-zero image in $\mathbb F_p[X]$, hence it contains a polynomial which is monic when reduced mod $p$.  An induction proves that some power of this polynomial is actually monic mod $p^e$.  Thus $(f,g)$ contains a monic polynomial when reduced mod $p^e$, and so $(\mathbb Z/p^e\mathbb Z)[x]/(f,g)$ is finite.
Factor $n = p_1^{e_1}\cdots p_r^{e_r}$.  Then by CRT, there is an isomorphism


$$(\mathbb Z/n\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i
\mathbb Z)[x]/(f,g),$$
and so by what we have already proved, $(\mathbb Z/n\mathbb Z)[x]/(f,g)$ is a product of finite rings, hence finite.

Added: Now that some solutions from first principles have been put up, it may be worth describing a way to deduce this statement from general principles of commutative algebra.  

First, if $J$ is any ideal in a Noetherian ring $A$, and $I = $rad$(J)$, then since $J \subset I,$ we have that $A/I$ is a quotient of $A/J$.  Thus if $A/J$ is finite, so is $A/I$.  On the other hand, by Noetherianness (more precisely, by the fact that $I$ is f.g.), we have
$I^n \subset J$ for some $n$, and so $A/J$ is a quotient of $A/I^n$, which is filtered by $A\supset I \supset I^2 \supset \cdots \supset I^n.$ Each quotient $I^i/I^{i+1}$ is a f.g. $A/I$-module, and so if $A/I$ is finite, so is each $I^i/I^{i+1}$, hence so is $A/I^n$, and hence so is $A/J$.  Thus $A/J$ is finite if and only if $A/I$ is.

Now take $A = \mathbb Z[x]$.  This is a UFD of Krull dimension two, and so the prime ideals of $A$ are either $0$, height one and hence principal, or maximal.  Let $J = (f,g)$.  Since $f$ and $g$ have no non-trivial common divisor, the ideal $(f,g)$ cannot be contained in a principal prime ideal, and so each minimal prime of $J$ must be a maximal ideal of $A$.  Thus if $I$ is the radical of $J$, then $I = \mathfrak m_1 \cap \cdots \mathfrak m_r$ for some maximal
ideals $\mathfrak m_i$, and so $A/I$ embeds into the product $\prod_{i=1}^r A/\mathfrak m_i$.  Thus to show that $A/I$ is finite, it suffices to show that $A/\mathfrak m$ is finite for any maximal ideal of $A$.  This last fact follows from the general version of the Nullstellensatz for Jacobson rings applied to $\mathbb Z[x]$ (using that $\mathbb Z$ is Jacobson).
Consider $\mathbb{Z}[x]$ as a subring of $\mathbb{Q}[x]$ which is a PID. Since $\rm{gcd}(f,g) = 1$ in $\mathbb{Z}[x]$ the same is true in $\mathbb{Q}[x]$ so $\mathbb{Q}(f,g) = \mathbb{Q}[x]$ which means that $(f,g)$ contains an element that is invertible inside $\mathbb{Q}[x]$, so it contains some integer. Now we know that the quotient is actually a quotient of $(\mathbb{Z}/n\mathbb{Z})[x]$ which means that unless $(f,g)$ consists of just constants, we get a finite ring, as we get a bounded degree. But if this is the case, then both $f$ and $g$ are constants, and $(f,g) = \mathbb{Z}[x]$ since they have gcd equal to $1$.

Edit: The last part is missing some ingredient as noted in the comment.