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.

 A: 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).  
A: Complete Answer (?)
Note that $\mathbb Z[X]$ is not euclidian, but $\mathbb Q[X]$ is.
Since gcd$(f,g)=1$ there exists $h_1,h_2 \in \mathbb Q[X]$ so that 
$$1=h_1f+h_2g$$
Let $a$ be the common denominator of the the coefficients of $h_1,h_2$. Then we get
$$a=(ah_1)f+(ah_2)g \,.$$
Since $ah_1, ah_2 \in \mathbb Z[X]$ we get $a \in (f,g)$.
Now, by the Third Isomorphism Theorem 
$$\frac{\mathbb Z[X]}{(f,g)} \sim \frac{\mathbb Z [X]/(a)}{(f,g)/(a)} \sim  \frac{\left(\mathbb Z/a\mathbb Z \right)[X]}{(f,g)} $$
Now we know that
$$(\mathbb Z/a\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i
\mathbb Z)[x]/(f,g),$$
We claim that $ (\mathbb Z/p^{e_i}_i\mathbb Z)[x]/(f,g)$ is finite.
Indeed, since gcd$(f,g)=1$ then, one of them has a coefficient not divisible by $p_i$. Lets say this is $f$.
Then, we can write $f=f_1-f_2$ such that all the coefficients of $f_1$ are relatively prime to $p_i$ and all the coefficients of $f_2$ are divisible by $p_i$.
Then, since $f_2^{e_i}=0$ we have
$$f_1^{e^i}=f_1^{e^i}-f_2^{e^i}=(f_1-f_2)(\mbox{junk})=f(\mbox{junk}) \in (f,g)$$
Let $b$ be the largest coefficient of $f_1$, then the largest coefficient of $f_1^{e_i}$ is $b^{e_i}$. Since this is invertible, we get
$$b^{-e_i}f_1^{e^i} \in (f,g)$$
Since we found a polynomial with leading coefficient $1$ in $(f,g)$, it is easy to conclude that $ (\mathbb Z/p^{e_i}_i\mathbb Z)[x]/(f,g)$ is finite.
A: 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. 
A: If it's OK I would like to post my rough notes on understanding this brilliant question. I don't have a solution but for beginners to understand the problem itself this may be useful.

Some theory of GCD in $\mathbb Z[X]$ that I was not aware of:
Definition Let $D|P$ and $D|Q$ then $D$ is a common divisor of $P$ and $Q$.
Definition The set of all common divisors of $P$ and $Q$ is a preorder order by the divisibility relation, the greatest element is the greatest common divisor of $P$ and $Q$.
Example $\{-1,1\}$ is the set of common divisors of $2$ and $X$. So $\gcd(2,X) = 1$ (by convention we don't pick $-1$).
Theorem The GCD is unique if it exists. proof: Let $D$,$D'$ be maximal elements of the divisibility preorder order on the set of common divisors of $P$ and $Q$ and suppose for contradiction that they are not associates, then $D \not | D'$ and $D' \not | D$. Since $D|P$ and $D'|P$ let $P=DA=D'A'$ and since $D'|DA$ we find $D|A$ hence $DD'|P$ contradicts the maximality proving that $D$ and $D'$ are associates.
Theorem The GCD exists. proof: factor uniquely into irreducible powers and take the intersection. Any larger element of this in the poset will not be a common divisor so it's maximal.

my idea for a proof and some explicit calculations.
I view $\mathbb Z[X]$ as an infinite dimensional space with axis being integers, X*integers, X^2*integers, ...
To prove that $\mathbb Z[X]/(F,G)$ is finite we need to show that some integer $i \in (F,G)$ (so there are at most $i^\infty$ points in the quotient) and that some $X^r \in (F,G)$ (so the hypercube has finite dimension therefore at most $i^r$ points in the quotient).
Example 1: $(1+X+X^2,3+5X)$. The polynomials have gcd 1. and we can do the following two calculations:
5+5X+5X^2
 -3X-5X^2
---------
5-2X

25-10X
 6-10X
------
31 in an integer in the ideal


 3+3X+3X^2
-3-5X
----------
  -2X+3X^2

  -6X+ 9X^2
   6X+10X^2
  ---------
      19X^2 is in the ideal

 and since (31,19)=1 we get X^2 in the ideal.


Example 2: $(A,B) = (X^3 + X^2 + X + 1, 9X^2 + 5X + 3)$ when we perform the same process as before we get $81A-(9+9X-5)B = 34x + 69$ instead of the integer and $9A - B(3-2X) = 27X^3 - 8X^2$ which is not monic! but.. $\gcd(34X+69,27X^3-8X^2) = 1$ and it does have smaller degree so we can do the process again (induction proof).

It seems like a proof could work along these lines but the hardest bit is showing both processes lead to polynomials which are coprime.
A: 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.
