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I will use the notation and language of Stichtenoth, *Algebraic Function Fields and Codes*.

Let $F$ be a function field over a finite field $\mathbb F_q$, $S$ a non empty set of places (possibly infinite) and $O_S$ the holomorphy ring of $S$, i.e., $O_S:=\bigcap_{P\in S} O_P$ (being $O_P$ the valuation ring of the place $P$).

- Finite intersection of DVRs
- $R$ is a commutative integral ring, $R$ is a principal ideal domain imply $R$ is a field
- If R is a PID, is it true that $R/\ker \phi$ is also a PID?
- A subring of the field of fractions of a PID is a PID as well.
- Specific way of showing $\Bbb Z$ is not a Euclidean Domain when $d>2$
- Is $\mathbb{Z}$ a principal ideal domain?

Give necessary and sufficient conditions for $O_S$ to be a Principal Ideal Domain.

Many thanks in advance!

G.

- Finitely generated modules over PID
- Cardinal of quotient rings of gaussian integers.
- Is the localization of a PID a PID?
- An integral domain whose every prime ideal is principal is a PID
- Show that for a field $F$, the polynomial ring $F$ is not a PID for $n>1$.
- There are finitely many ideals containing $(a)$ in a PID
- Non-principal ideal in $K$?
- Ring of trigonometric functions with real coefficients

Over any ground field, when $S$ is finite, your holomorphy ring is the coordinate ring $k[C^{\circ}]$ of the corresponding nonsingular affine curve $C^{\circ}_{/k}$. This is a Dedekind domain, so it is a PID iff its ideal class group $\operatorname{Pic} k[C^{\circ}]$ vanishes. There is a canonical map

$(\operatorname{Pic}^0 C)(k) \rightarrow \operatorname{Pic} k[C^{\circ}]$

from the group of $k$-rational points on the Jacobian abelian variety to the ideal class group in question: you just remove the places at infinity of your degree zero divisor to get a divisor on the affine curve which need no longer have degree zero. The kernel is (clearly) the image in the Picard group of the subgroup of degree zero divisors supported on the set of infinite places. The cokernel is (less clearly) a cyclic group of order $d/i$, where $d$ is the least positive degree of a divisor supported on $S = C \setminus C^{\circ}$ and $i$ is the least positive degree of a divisor on $C$. In particular, this map is an isomorphism when $S$ consists of one place with residue field $k$. See Theorem 11 of this paper and the paper of Mike Rosen it refers to.

You also ask about what happens when the set of places is infinite. In general, when you pass from a Dedekind domain to an overring, then the ideal class group gets modded out by the subgroup generated by the classes of the prime ideals which you have “lost” in passing to the overring. (The language of “holomorphy rings” is in fact a very natural one here.) In your situation, you can take a finite nonempty subset $S’$ of your infinite set $S$ and understand the ideal class group of $O_S$ as a quotient of the ideal class group of $O_{S’}$. This is explained in the same paper: see in particular Theorem 5. In fact, the point of the paper is to use these two results — and transfinite induction — to give a proof of **Claborn’s Theorem** that every commutative group is the ideal class group of some Dedekind domain. The Dedekind domain can be taken to be the “holomorphy ring” of an elliptic curve over a field $k$ with respect to an infinite set $S$ of places (choosing $k$, $E$ and $S$ suitably, of course).

When $k$ is finite, then $(\operatorname{Pic}^0 C)(k)$ is the group of points on an algebraic variety over a finite field, so is finite. Moreover $i =1$ always in this case (F.K. Schmidt’s Theorem). It then follows easily from these results that for any $C$ you can get a PID by suitably enlarging $S$. If $S$ itself contains a place with residue field $k$, then you can get a PID by throwing in at most twice the genus of $C$ more places. And in fact, in this case if $S$ is infinite then the ideal class group will be the same as that of a sufficiently large finite subset.

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