TITLE: Commutative integral domain does not finitely generate its field of fractions
QUESTION [6 upvotes]: I want to prove that if we have a commutative integral domain $D$ with field of fractions $F\neq D$ then $F$ is not finitely generated as a $D$-module. (In this question it may be the case that $1\not\in D$.) 

My original plan was to localize at a maximal ideal and then use Nakayama's lemma but as we don't have a $1$ our localization may not be local (in fact we need not have maximal ideals).
So I'm a bit lost as to where to go?
Thanks for any help.

REPLY [3 votes]: Since you already have two very good answers, I'd like to (just for fun) note that this also follows from very basic commutative algebra.
You have an inclusion $D\hookrightarrow F$ which, by assumption, is finite. Then, by the Lying Over Theorem, we'd have a surjection $\text{Spec}(F)\to\text{Spec}(D)$ which implies that $\text{Spec}(D)$ has only one prime. This prime must be $(0)$, and since it's the only prime, it must be maximal. So, $D$ is a field, and so $D=F$.
EDIT: I just noticed the strange assumption that $D$ needn't haven't a unit. It seems strange then to call $D$ a 'domain' and $F$ a 'field', since these usually have (definitionally) unit. So, the above only works if $D$ is unital.