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Let $R\subseteq S$ be two integral domains. $S$ is finitely generated over $R$. $Frac(S)$ is finitely generated over $Frac(R)$ (hence, finite by Zariski's lemma). Is it true that $S$ is finite (as module) over some $R[1/f]$ where $0\neq f\in R$?

I have $\mathbb{Z}$ and $\mathbb{Z}[1/2,\sqrt{2}]$ and $\mathbb{Z}$ and $\mathbb{Z}[e]$ in mind. The first can be taken as $\mathbb{Z}[1/2][\sqrt{2}]$ hence finite. Second is not finite but the fraction fields are not finite extension.

CO2
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1 Answers1

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This is false even if we assume that the unital homomorphism $R\rightarrow S$ induces an isomorphism on the fraction fields.

The issue is that if $S$ contains $R[1/f]$ for a non-unit $f$ then there is a maximal ideal of $R$ containing $f$ and no proper ideal (in particular, no prime ideal) of $S$ can lie over it. It is not too difficult to produce non-finite ring maps such that every prime ideal of $R$ has a prime ideal of $S$ lying over it. For example, take $R=\mathbb{k}[t^2-1, t^3-t]$, $S=\mathbb{k}[t, \frac{1}{t-1}]$ where $\mathbb{k}$ is an algebraically closed field; the function fields are both $\mathrm{Frac}(\mathbb{k}[t])$.

In the light of this issue I think that the question would be more reasonable if it asks $S\otimes_R R[1/f]$ to be a finite $R[1/f]$-module. I am not sure about this modified question at the moment.

bqe0
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  • @CO2 if $f$ is a unit then $S$ has to be finite over $R$ which is false. If $f$ is not a unit, we have to miss the maximal ideal containing $f$. – bqe0 Aug 08 '19 at 12:02
  • $S\otimes_RR[1/f]$ will work I guess. You can check that post too. I thought I proved for the original case. But it missed a final point that $1/f$ might not live in $S$. – CO2 Aug 08 '19 at 12:09
  • @CO2 consider accepting the answer if it accurately addresses your question. – bqe0 Aug 08 '19 at 12:10
  • Sure. A small question: geometrically, say $k=\mathbb{C}$, your $S$ will be the set of regular functions around the point $X=1$. And what will $R$ corresponds to? What geometric objects $\mathbb{C}[f_1(X),...,f_n(X)]$ will correspond to? – CO2 Aug 08 '19 at 12:26
  • @CO2 geometrically, $R$ is the coordinate ring of the affine nodal cubic, its normalization is the affine line which is injective everywhere except two points being mapped to the node (normalization is a finite map so that would not work as a counterexample). If we throw out one point, we get $S$, the coordinate ring of a punctured affine line mapped bijectely to the nodal cubic and this one is not a finite map. – bqe0 Aug 08 '19 at 12:28
  • I see, so you actually constructed proved there can not even be a map $R[1/f]\hookrightarrow S$? – CO2 Aug 08 '19 at 13:16
  • @CO2 if $f$ is not a unit, then there is no unital homomorphism $R[1/f]\rightarrow S$ whose composition with the canonical map $R\rightarrow R[1/f]$ is equal to the map $R\rightarrow S$. – bqe0 Aug 08 '19 at 13:23
  • But the proposition just states "$S$ is finite over $R[1/f]$" which is by definition just a finite map $\varphi:R[1/f]\to S$, i.e., a ring homomorphism with some extra conditions. There is no restriction on what it should be, why do we require the composition map to be $R\hookrightarrow S$? – CO2 Aug 08 '19 at 13:53
  • @CO2 the proposition states that "$S$ is finite over $R[1/f]$", which implies that the author has a particular map $R[1/f]\rightarrow S$ in mind (which, inferring from the context, could only be a map compatible with the composition). If the author meant to claim existence of a random map then it would be more clear to say just that. I do not know if there exists such a map without the condition on the composition. – bqe0 Aug 08 '19 at 14:09