# Rings of $S$-integers are finitely generated as rings

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.

• Does $S$ contain all archimedian valuations? – Mark Sapir Dec 6 '19 at 1:00
• I'm not sure "Archimedean valuation" makes sense. Archimedean norms on $K$ do not give rise to valuations. – YCor Dec 6 '19 at 1:52
• I am unsure why this question was downvoted. If there is anything I can do to improve upon it, please let me know. – Bib-lost Dec 6 '19 at 9:40
• Probably because it shows no work and it's just a reference request for an elementary statement. – John Samples Dec 6 '19 at 15:25
• Ok, I'll add some work. It's an elementary statement, I won't deny that. I've also tried quite some books already. – Bib-lost Dec 6 '19 at 23:11

$$\newcommand{\order}{\mathcal{O}} \newcommand{\Z}{\mathbb{Z}}$$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 $$X$$.
• 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 $\mathcal{O}$? – Bib-lost Dec 6 '19 at 12:54
• 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. – Aurel Dec 6 '19 at 13:19
• 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 $F[t]$, ...) – Bib-lost Dec 6 '19 at 13:40