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Let $K$ be a number field and $\mathcal O_K$ the ring of algebraic integers in $K$.

If $\mathfrak p$ is a prime ideal, then $\mathcal O_K/\mathfrak p$ is finite field.

My question is:

Finding a Dedekind domain $D$ and $\mathfrak p$ a prime ideal (in $D$) such that $D/\mathfrak p$ is infinite.

I appreciate any reference.

Thank you all.

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A very easy answer is to take $\mathfrak{p}=(0)$.

A nontrivial answer is $D=\mathbb{C}[T]$ and $\mathfrak{p}=(T-a)$ for any $a\in\mathbb{C}$.

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Hint: Consider PIDs of the form $F[x]$, $F$ a field.

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  • $\begingroup$ Alex Youcis; thank you. $\endgroup$ – user126033 Dec 6 '14 at 6:28
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By thinking about this geometrically, we can find huge families of examples. A Dedekind domain is an integrally closed, Noetherian domain with Krull dimension one. In particular, the ring of regular functions of any nonsingular affine algebraic curve $C$ over a field $K$ is a Dedekind domain. (For example, if $K$ is any field, then $K[x]$ is the ring of regular functions of the affine line over $K$.)

The residue field $\kappa(x)$ at each closed point $x$ of any such curve $C$ is a finite extension of $K$, so if $K$ is infinite, then $\kappa(x)$ has the same cardinality as $K$.

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You can actually do much better - Whenever $F_n, \; n \in \mathbb{N}$ is some list of at most countable fields such that every positive characteristic occurs at most finitely many times, there exists a PID $R$ such that all $F_n$'s are realized as residue fields of $R$.

In particular, one can have a PID with both finite and infinite residue fields (which seems not obvious to me).

See

Heitmann, Raymond C.: PID’s with specified residue fields. Duke Math. J. 41 (1974), no. 3, 565--582.

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All the counter examples given up till now are all quite geometric. Here a counterexample that is slightly more number theoretic in flavour:

Let $\mathbb Q_p^{unr}$ be the maximal abelian extension of $\mathbb Q_p$ unramified at $p$, and $\mathbb Z_p^{unr}$ its ring of integers, then $\mathbb Z_p^{unr}$ is Dedekind (in fact it is a D.V.R. and $p$ is it's unique maximal ideal). But $\mathbb Z_p^{unr}/p\mathbb Z_p^{unr} \cong \overline{\mathbb F}_p$.

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