TITLE: A fraction field is not finitely generated over its subdomain
QUESTION [14 upvotes]: 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!

REPLY [7 votes]: 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 $$K=R[\frac {1}{\pi}]$$.