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Let $\mathbb Z$ be the ring of rational integers. If $a\in\mathbb Z$ is a non-zero element, then the factor ring $\mathbb Z/(a)$ is finite and has order $|a|$. If $\mathbb Z[i]$ is the ring of Gaussian integers and $w\in\mathbb Z[i]$ is non-zero, we also have that $\mathbb Z[i]/(w)$ is finite and has order $w\bar w=N(w)$, that is, the norm of $w$.

I suspect that this is true for any ring of algebraic integers in a number field (in the case that the ring is Euclidean). Am I right?

Another question: Is there any classification (or name) for the integral domains $D$ that that satisfy property that for nonzero $a\in D$ the factor ring $D/(a)$ is finite? I think that all rings of algebraic integers (including non-Euclidean) in a number field satisfy this property. Is this true?

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What is "this" in your third line? That the quotient ring of an ID is always finite? Because, of course, it is not as you can easily check with some polynomial rings' quotients – DonAntonio Mar 14 '13 at 13:09
Yes, you are right, even if the ring is not Euclidean. – awllower Mar 14 '13 at 13:11
There is no need to restrict the question to principal ideals: If all $D/(a)$ with $a\neq 0$ are finite, then also all $D/I$ with $I\neq\{0\}$ are finite. – azimut Mar 14 '13 at 13:38
DonAntonio, "this" is for rings of integers in a number field – zacarias Mar 14 '13 at 14:14
up vote 5 down vote accepted

Such rings are called residually finite, or rings with the finite norm property. They have been studied, e.g. see the paper reviewed below.

Levitz, Kathleen B.; Mott, Joe L.
Rings with finite norm property.
Canad. J. Math. 24 (1972), 557--565.

Let $A$ be a ring with $A^2 \ne 0 ,$ and $A^+$ the additive group of $A$ . If each non-zero homomorphic image of $A$ is finite, then $A$ is said to be a ring with finite norm property (FNP ring). K. L. Chew and S. Lawn studied FNP rings with identity, which they called residually finite rings [same J. 22 (1970), 92--101; MR0260773 (41 #5396)]. In the paper under review, the authors extend the results of Chew and Lawn to arbitrary FNP rings. They also prove the following results:

$(1)\ $ If $A$ is an FNP ring then $A^+$ is torsion and bounded, or torsion-free and reduced, or torsion-free and divisible. Henceforth, $A$ will be a commutative integral domain with $1$ and with quotient field $K$ .

$(2)\ $ Let L be a finite extension of $K$ ; if $A$ is an FNP ring, then so is every intermediate ring of $L/A$ .

$(3)\ $ Let $A'$ be the integral closure of $A$ in $K$ ; then, $A$ is an FNP ring if and only if $A'$ is a Dedekind domain and $A_P$ is an FNP ring for every maximal ideal $P$ .

$(4)\ $ Let $K$ be of characteristic $0,$ then, every subring of $A$ is an FNP ring iff $K$ is a finite extension of the field of rational numbers.

$(5)\ $ Let $K \ne A$ be of prime characteristic; then, every subring of $A$ is an FNP ring iff $K$ is a finite extension of some $F(x),$ where is the prime field of $K$ and $x$ is transcendental over $F$ .

Review by H. Tominaga (AMS MR 45 #6872)

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If you copy text wholesale from Math Reviews, please do us the courtesy of at least including a link to the original. – Willie Wong Mar 14 '13 at 16:27
@Willie The review is copied from the paper. Please feel welcome to add a link to the review if you so desire. I thought the link to the paper was far more pertinent. – Math Gems Mar 14 '13 at 16:31
It is basic scholarship to acknowledge where you've taken information, especially if you take it verbatim. – Willie Wong Mar 14 '13 at 16:45
Also, for what it is worth: here's a link to the journal's website with the article in question. I won't edit it into your answer since I am not sure if it would be more or less accessible for those not affiliated with an academic institution than the Google-books link. – Willie Wong Mar 14 '13 at 16:48
@Willie In the rare cases that I hear complaints about links, usually they are about giving too many links, not too few! This case was an oversight. I didn't notice that the review number was not included in the text I extracted from my files on this topic. Thanks for updating it. – Math Gems Mar 14 '13 at 16:55

Yes this is always going to be true. In fact I claim that the quotient by any ideal $\neq 0$ is going to be finite. Let $K$ be a number field; we know that $\mathcal{O}_K$ is free abelian of rank $n = [K:\Bbb{Q}]$. This comes from using the fact that the trace as a bilinear form on $K$ is non-degenerate because any finite extension of $\Bbb{Q}$ is separable.

Now let $I$ be an ideal of $\mathcal{O}_K$ and choose $0 \neq a \in I$. Consider the principal ideal $(a)$. Then $(a) \cong \mathcal{O}_K$ and thus is free abelian of rank $n$. Then we get that

$$(a) \subseteq I \subseteq \mathcal{O}_K$$ and so $I$ itself has rank $n$. Now we have the ses

$$0 \to I \to \mathcal{O}_K \to \mathcal{O}_K/I \to 0.$$

If we apply the exact functor $-\otimes_{\Bbb{Z}} \Bbb{Q}$ then we see that the free part of the quotient is necessarily zero and so $ \mathcal{O}_K/I$ is torsion.

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