Integral domains such that all proper factor rings are finite 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? 
 A: Such rings are called residually finite, or rings with the finite norm property (FNP). They have been studied at length 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)

A: 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.
