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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 an $R$-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!

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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

3 Answers 3

up vote 12 down vote accepted

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 $\frak{m} \neq 0$. 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.

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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 6:19
    
@ZevChonoles Are we assuming here that are rings are unital so that our max ideals are prime? –  hmmmm Dec 8 '13 at 12:30

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. $R=R'$.

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Let $R$ be a commutative ring, $S\subset R$ a multiplicative set, and $\varphi_S:R\to S^{-1}R$ the canonical morphism, that is, $\varphi_S(a)=a/1$. If $S^{-1}R$ is a finitely generated $R$-module, then $\varphi_S$ is surjective.

This is rather easy to prove: if $S^{-1}R$ is a finitely generated $R$-module then it is cyclic generated by an element of the form $1/s$, $s\in S$. Since $1/s^2\in R(1/s)$ there is $a\in R$ such that $1/s^2=a/s$, and therefore $1/s\in\operatorname{Im}\varphi_S$, so $\varphi_S$ is surjective.

In the particular case of the question $\varphi_S$ is injective (this also happens in the case described by the answer of Martin Brandenburg), so $R\simeq K$. It follows that $R$ is a field, a contradiction.

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