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For any domain $A$ let $A^\times$ be its group of units.

Let $A$ be a noetherian domain with only finitely many prime ideals, and field of fractions $K$.

Is the group $K^\times/A^\times$ finitely generated?

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up vote 3 down vote accepted

The answer is no in general.

First by an answer to the question Does every Noetherian domain have finitely many height 1 prime ideals?, we know that $A$ has dimension at most $1$. Let $m_1,\dots, m_n$ be the maximal ideals of $A$. It is easy to see that inside $K^\times$, we have $A^\times =\cap_i (A_{m_i})^\times$, so the canonical map $$ K^\times /A^\times \to \prod_i K^\times/(A_{m_i})^\times$$ is injective and we are reduced to the case where $A$ is local of dimension $\le 1$. If $A$ is integrally closed, then $K^\times /A^\times$ is generated by an uniformizing element of $A$.

So the first conclusion is $K^\times /A^\times$ is finitely generated if $A$ is integrally closed.

Now the question is what happens if $A$ is local, but not necessarily integrally closed. Let us consider an example. Let $F$ be a field and $c\in F$ an element which is not a square in $F$. Let $A$ be $F[x,y]/(y^2-cx^2)$ localized at the maximal ideal generated by $x,y$. It is integral, noetherian, with only two prime ideals. Let $L=F[\sqrt{c}]$. It can be identified to a subring of $K=\mathrm{Frac}(A)$ by sending $\sqrt{c}$ to $y/x$. It is easy to see that $L\cap A=F$. So $$ L^\times/F^\times \hookrightarrow K^\times/A^\times.$$ We have $L^\times=F^\times+ \sqrt{c}F^\times$, and, set-theoretically, $L^\times/F^\times$ is in bijection with $F^\times$. If we choose a uncountable field $F$, then $K^\times/A^\times$ is uncountable, hence not finitely generated. Concretly, we can take $F=\mathbb C(t)$ and $c=t$.

In this counterexample, if $B$ denotes the integral closure of $A$ in $K$, then $L\subset B$ and $B^\times/A^\times$ is not finitely generated.

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The article

D. D. Anderson, "Integral Domains with Finitely Generated Groups of Divisibility", Proceeding AMS 112 (3), 1991

most likely provides the information you seek for.

In particular theorem 0 in this article seems to show that in general the answer to your question is no: a necessary requirement seems to be that the integral closure $\overline{A}$ of $A$ in its field of fractions is a finite $A$-module, and is a Bezout domain. So e.g. if $\overline{A}$ is noetherian, it is a principal ideal domain.

Added after QiL's post:

Putting the information from the post of QiL and the cited article together, we get the following nice result:

Theorem: Let $A$ be a noetherian domain with finite spectrum and field of fractions $K$. Let $\overline{A}$ be the integral closure of $A$ in $K$ and let $f$ be the conductor ideal of the extension $A\subseteq\overline{A}$. Then the following statements are equivalent:

  1. $K^\times/A^\times$ is finitely generated.

  2. For every prime ideal $p\supseteq f$ the fraction field of $A/p$ is finite.

Proof. $1\Rightarrow 2$: By theorem 1 in Anderson's paper the ring $\overline{A}/f$ is finite. In particular the fraction fields of $\overline{A}/q$ are finite for all $f\subseteq q$. These fields contain the fraction fields of $A/p$, $p\supseteq f$.

2$\Rightarrow 1$: By QiL's post and Theorem 3 in Andersons's paper it suffices to prove that $\overline{A}/f$ is finite, which is equivalent to the finiteness of the fields $\overline{A}/q$, $f\subseteq q$. By Krull-Akizuki the extensions $A/q\cap A\subseteq \overline{A}/q$ are finite. Hence the assertion follows.

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If $\bar{A}$ is finite over $A$, it is Dedekind and semi-local, so is a PID. But this is not enough. – user18119 Aug 20 '12 at 9:34
Nice result ! Maybe we can add that prime ideals containing $f$ are these maximal ideals $p$ such that $A_p$ is not integrally closed. – user18119 Aug 21 '12 at 12:45

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