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I'm looking for proofs of the following fact.

Suppose that $R$ is a domain which is not a field with fraction field $K$. Then $K$ is not finitely generated as $R$-module.

I know this fact is true, at least, when $R$ is Noetherian and I guess it is true in general. I know two proofs, one when $R$ is Noetherian and one (very indirect) when $R$ is Noetherian local. Do you know any direct proof for any arbitrary domain $R$?

Thanks!

user26857
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mr.bigproblem
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  • thanks for the comment. in some sense, the claim in that article is a special case because one restricts to PIDs only. – mr.bigproblem Feb 23 '13 at 15:59
  • This follows from http://math.stackexchange.com/questions/879731. – Watson Jan 24 '17 at 10:14
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    $R \subset K$ finite $\implies R \subset K$ integral $\implies \dim(R) = \dim(K)=0 \implies R$ is a field (since $R$ is a domain). – Watson Jan 24 '17 at 15:03

3 Answers3

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The fraction field $K$ has the structure of an $R$-module, and for any maximal ideal $\mathfrak{m}$ of $R$, it also naturally has the structure of an $R_\mathfrak{m}$-module. If $K$ is finitely generated as an $R$-module, then $K$ must also be finitely generated as an $R_\mathfrak{m}$-module because $R_\mathfrak{m}\supseteq R$. Thus we are reduced to the case when $R$ is local.

Let $R$ be a local domain with maximal ideal $\mathfrak{m} \neq 0$. Clearly, $\mathfrak{m}K=K$. But the Jacobson radical of $R$ is just $\mathfrak{m}$, so if $K$ is finitely generated as an $R$-module, then Nakayama's lemma implies that $K=0$, which is a contradiction.

user26857
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Zev Chonoles
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  • Thanks Zev. Didn't expect the proof would be that simple. I feel the presence of Nakayama's lemma but I can't never get it right. – mr.bigproblem Feb 23 '13 at 06:19
  • @ZevChonoles Are we assuming here that are rings are unital so that our max ideals are prime? – hmmmm Dec 08 '13 at 12:30
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If $K$ is finitely generated module over $R$, it is integral over $R$.
But given domains $R\subset K$ with $K$ integral over $R$, $K$ is a field if and only if $R$ is a field. (Atiyah-Macdonald, Prop 5.7)
So if $R$ is not a field but $K$ is, then $K$ is not integral over $R$ and thus not a finitely generated module over $R$.
Edit
It is perfectly possible for the fraction field $K$ of a domain $R$ that is not a field to be finitely generated as an algebra over $R$.
The simplest class of examples is that of discrete valuation rings.
Indeed if $R$ is such a ring and if $\pi$ is a generator of its unique maximal ideal , then we have $$K=R[\frac {1}{\pi}]$$.

Georges Elencwajg
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  • Sorry, may you elaborate on how your first line holds? I know if $R\subseteq R[x_1,\ldots, x_n]$ where each $x_i$ is integral over $R$, then the extension is integral. So at least we require $R$ is a UFD here? – Bryan Shih May 13 '18 at 07:10
  • Dear @Bryan: Atiyah-Macdonald, page 60, Remark just before Corollary 5.4. – Georges Elencwajg Feb 23 '20 at 00:13
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Zev's proof uses the existence of maximal ideals, and is therefore not constructive. But it is not hard to make the proof constructive, and actually more generally:

Let $R$ be a ring and $R'$ a localization of $R$. If $R'$ is finitely generated and torsionfree over $R$, then $R=R'$.

Proof. Let $s \in R$ such that $s^{-1} \in R'$. Then $R' = sR'$. Generalized Nakayama's Lemma implies that there is some $a \in R$ such that $(1-as)R'=0$. Since $R'$ is torsionfree, this means $1=as$. So $R'$ has no more units than $R$, i.e. $R=R'$.

Martin Brandenburg
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