Let $\alpha \in \mathbb Z [\sqrt{-1}]$. Is it true that $\mathbb Z[\alpha]/I$ is finite for any non-zero ideal $I$ of $\mathbb Z[\alpha]$?

Question

Let $\alpha \in \mathbb Z [\sqrt{-1}]$. Is it true that $\mathbb Z[\alpha]/I$ is finite for any non-zero ideal $I$ of $\mathbb Z[\alpha]$? I ask because on an old exam paper it asks me to prove this for all non-zero prime ideals $P$, but I haven't actually used that the ideal is prime, so I am confused.

Answer 1

You are right that it works for arbitrary $I$, though with a bit of knowledge on the prime ideals of $\mathbb{Z}[X]$ the proof is somewhat easier (but not that much) in the prime case.

$\alpha = a + ib$, hence $\mathbb{Z}[\alpha] = \mathbb{Z}[ib] \simeq \mathbb{Z}[X]/(X^2+b^2)$. Let $\pi: \mathbb{Z}[X] \to \mathbb{Z}[ib]$ be the associated projection map.

Let $I$ be an ideal of $\mathbb{Z}[ib]$. Then $\pi^{-1}(I)$ is an ideal of $\mathbb{Z}[X]$ that contains $X^2+b^2$. If $I$ is prime, so is $\pi^{-1}(I)$ and so the latter is of the form $(p, Q)$ where $p=0$ or is prime and $Q\in \mathbb{Z}[X]$ is an irreducible polynomial or $0$.

But if $p=0$ then $X^2+b^2 \in (Q)$ hence $Q=X^2+b^2$ (up to a unit, i.e. $1$ or $-1$), so $I=0$.

Hence $p\neq 0$, and so the quotient is a finite extension of $\mathbb{F}_p$. So when $I$ is prime, this works.

The problem with this proof for general $I$ is of course that the structure of ideals of $\mathbb{Z}[X]$ is very complicated otherwise.

But here we know that $X^2+b^2\in \pi^{-1}(I)$ so this helps: this is a monic polynomial so we can divide by it. If $P\in \pi^{-1}(I)$, then we may write $P=(X^2+b^2)Q + R$ with $\deg R < 2$ hence either $\pi^{-1}(I) = (X^2+b^2)$ or it contains a polynomial of degree $1$. In the first case, this corresponds to $I=0$.

In the second case, we have $cX+D \in \pi^{-1}(I)$ for some $c,d$. Now $cX+d$ and $X^2+b^2$ are coprime in $\mathbb{Q}[X]$ so there are integer polynomials $U,V$ and an integer $n$ such that $(cX+d)U+(X^2+b^2)V = n$. Hence $n\in \pi^{-1}(I)$, and so $n\in I$. Hence our quotient is actually a quotient of $\mathbb{Z}/n\mathbb{Z}[X]$ by an ideal $J$ containing $X^2+b^2$: hence all polynomials can be written as $eX+f$ modulo $J$: there are at most $n^2$ classes, and so the quotient is indeed finite.

Answer 2

There is a very simple proof of this that uses nothing but the abelian group structure. Note that $\mathbb{Z}[\sqrt{-1}]\cong\mathbb{Z}^2$ as an abelian group, so for any $\alpha\in\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\alpha]$ is a free abelian group of rank either $1$ or $2$. Note moreover that if $I\subseteq\mathbb{Z}[\alpha]$ is a nonzero ideal, then $I$ is a free abelian group of the same rank as $\mathbb{Z}[\alpha]$ (it must have the same rank because if $x\in I$ is nonzero then the ideal generated by $x$ is isomorphic as a group to $\mathbb{Z}[\alpha]$ via multiplication by $x$).

By the theory of finitely generated abelian groups, this implies the quotient $\mathbb{Z}[\alpha]/I$ is finite. For instance, we could tensor the exact sequence $0\to I\to\mathbb{Z}[\alpha]\to \mathbb{Z}[\alpha]/I\to 0$ with $\mathbb{Q}$. Since $I$ and $\mathbb{Z}[\alpha]$ have the same rank, the map $I\otimes\mathbb{Q}\to\mathbb{Z}[\alpha]\otimes\mathbb{Q}$ is an injection between $\mathbb{Q}$-vector spaces of the same dimension, and hence is surjective. It follows that $\mathbb{Z}[\alpha]/I\otimes\mathbb{Q}=0$. This implies $\mathbb{Z}[\alpha]/I$ is torsion, and hence finite since it is finitely generated.

More generally, this argument shows that if $R$ is any integral domain which is finitely generated as an abelian group and $I\subseteq R$ is a nonzero ideal, then $R/I$ is finite.