Finite quotient ring of $\mathbb Z[X]$

Solution 1:

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).

Solution 2:

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.