Let $K$ be a global field (number field or algebraic function field over a finite field), $\mathcal{V}$ the set of $\mathbb{Z}$-valuations on $K$, $S \subseteq \mathcal{V}$ a finite set. The ring of $S$-integers is the subring of $K$ defined as $$ \mathcal{O}_S = \lbrace x \in K \mid \forall v \in \mathcal{V} \setminus S : v(x) \geq 0 \rbrace. $$ I am looking to puzzle together references for the following statement:

Let $R$ be a subring of $K$ such that $K$ is the fraction field or $R$. Then $R$ is finitely generated as a ring if and only if it is contained in some ring of $S$-integers.

A reference for the full statement would be amazing. I have been able to piece together pieces from different references, but the part which is generally missing is that $\mathcal{O}_S$ is actually finitely generated as a ring (equivalently, a $\mathbb{Z}$-algebra) for any choice of $S$.

Where should I look to find a reference for this statement? It feels like a statement from commutative algebra, but a minimal amount of number theory (respectively algebraic geometry) is needed to prove it, at least in the proofs I know of. On the other hand, there is no mention of the statement in any algebraic number theory books I consulted.

Alternatively, if someone has a very short proof, that is also very welcome. I need the statement for an article, but writing out all the details of the number theoretic proof would fall outside of the scope of the article.

  • $\begingroup$ Does $S$ contain all archimedian valuations? $\endgroup$
    – user6976
    Dec 6, 2019 at 1:00
  • $\begingroup$ I'm not sure "Archimedean valuation" makes sense. Archimedean norms on $K$ do not give rise to valuations. $\endgroup$
    – YCor
    Dec 6, 2019 at 1:52
  • $\begingroup$ I am unsure why this question was downvoted. If there is anything I can do to improve upon it, please let me know. $\endgroup$
    – Bib-lost
    Dec 6, 2019 at 9:40
  • 2
    $\begingroup$ Probably because it shows no work and it's just a reference request for an elementary statement. $\endgroup$ Dec 6, 2019 at 15:25
  • $\begingroup$ Ok, I'll add some work. It's an elementary statement, I won't deny that. I've also tried quite some books already. $\endgroup$
    – Bib-lost
    Dec 6, 2019 at 23:11

1 Answer 1


$\newcommand{\order}{\mathcal{O}} \newcommand{\Z}{\mathbb{Z}} $Here is a short proof, assuming that the class group is torsion (a result for which you should easily find a reference).

First, $\order = \order_\emptyset$ is finitely generated as a $\Z$-module, hence also as a $\Z$-algebra; let $a_1,\dots,a_k$ be generators. In the function field case, pick $v_0\in S$; then $\order_\emptyset = \order_{v_0}$. For every valuation $v\in S$, with $v\neq v_0$ in the function field case, let $x_v\in K$ be such that $v(x_v)<0$ and $w(x_v)=0$ for all $w\neq v$ ($w\neq v,v_0$ in the function field case). Such an element exists: in the number field case since the class group is torsion, and in the function field case by Riemann-Roch.

Claim: $X = \{a_1,\dots,a_k\}\cup \{x_v : v\in S\}$ generates $\order_S$.

Proof: Let $0\neq x\in \order_S$. By definition of $\order_S$, $x$ can only have negative valuation for $v\in S$, so there exists a product $y$ of the $x_v$'s such that $x/y$ has nonnegative valuation everywhere (except possibly at $v_0$), hence belongs to $\order$. So $x/y$ is a polynomial in the $a_i$ and therefore $x$ is a polynomial in the elements of $X$.

  • $\begingroup$ Thx; this is more or less the proof I had in mind in case $K$ is a number field, although I am still hoping to find a reference to replace it. If $K$ is a global function field, what is your $\mathcal{O}$? $\endgroup$
    – Bib-lost
    Dec 6, 2019 at 12:54
  • $\begingroup$ I had in mind the case of number fields when I wrote the answer, but in any case $\mathcal{O} = \mathcal{O}_\emptyset$ so in the function field case it is the field of constants. $\endgroup$
    – Aurel
    Dec 6, 2019 at 13:19
  • $\begingroup$ Careful there; the argument which you sketch certainly does not go through if $\mathcal{O}$ is the field of constants $F$, as then the valuations of $K$ are certainly not visible as ideals of $\mathcal{O}$. I am not saying that you need a fundamentally different argument for global function fields, but I fear you need to twist it a bit (e.g. use Strong Approximation Theorem to fix a transcendental $t$ which has negative valuation precisely at the primes of $S$, then taking the integral closure of $F[t]$, ...) $\endgroup$
    – Bib-lost
    Dec 6, 2019 at 13:40

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