Dedekind domain in which quotient rings are not all finite? In rings of integers $O_K$, which are free $Z$-modules of finite rank (say n), we have that if m is an integer in the prime ideal $\mathfrak{p},$ then there is a surjection $$\frac{O_K}{(m)} \rightarrow \frac{O_K}{\mathfrak{p}}.$$The size of the first ring is $m^{n},$ which is finite. Therefore , the surjection of a finite ring into $\frac{O_K}{\mathfrak{p}}$ implies both rings are finite. But $\frac{O_K}{\mathfrak{p}}$ is a finite integral domain, therefore a field, therefore $\mathfrak{p}$ is a maximal ideal.
In short, every quotient of $O_K$ by an ideal is finite, so prime ideals must be maximal.
Since the only Dedekind domains I am familiar are rings of integers, is there an example of a Dedekind domain where this rule does not apply,i.e. where the quotient rings are not all finite?
 A: Any PID (that is not a field) is a Dedekind domain. So you can just take $K[X]$ where $K$ is an infinite field. The quotient $K[X]/\mathfrak m$ for some maximal ideal is a finite field extension of $K$, in particular again infinite.
A: Another class of examples is the coordinate ring $k[C]$ of a nonsingular affine plane curve $C$ over a field $k$, $k$ taken to be infinite to satisfy your stipulation.  Nonsingularity implies that $k[C]$ is integrally closed, and we take a curve so that $k[C]$ has Krull dimension $1$.
If we take $k$ to be algebraically closed, then a nonzero prime ideal $\mathfrak{p}$ corresponds to a point $(a,b)$ lying on $C$.  Since $k[C]$ is $1$-dimensional, then we must have $\mathfrak{p} = (x - a, y - b)$, so $k[C]/\mathfrak{p} \cong k$, which we assumed is infinite.
For instance, the affine plane curve $C: y^2 = x^3 - x$ over $\mathbb{C}$ has coordinate ring
$$
\mathbb{C}[C] = \frac{\mathbb{C}[x,y]}{(y^2 - (x^3 - x))}
$$
which is a Dedekind domain.  One can show that $\mathbb{C}[C]$ is not a UFD, hence not a PID.
One recovers the example given by MooS by taking $C = \mathbb{A}^1$, the affine line.
