Commutative integral domain does not finitely generate its field of fractions 
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.
 A: If the fractions $p_i/q$ ($i=1,2,\dots,n$) form a set of generators (it's not restrictive to assume the denominators are the same), then any element of $F$ can be written as
$$
\frac{x}{y}=\sum_{1\le i\le n}\frac{p_i}{q}d_i
=\frac{1}{q}\sum_{1\le i\le n}p_id_i
$$
which means $1/q$ is a generator. Since $F\ne D$, we have $1/q\notin D$. Then
$$
\frac{1}{q^2}=\frac{d}{q}
$$
for some $d\in D$, which means $dq=1$, a contradiction.
If $1\notin D$, we have that $\frac{q}{q^2}$ is a generator, but then
$$
\frac{q}{q^3}=\frac{qd}{q^2}
$$
and so
$$
\frac{q}{q^2}=q\frac{q}{q^3}=q\frac{qd}{q^2}=\frac{q^2d}{q^2}=d
$$
which is again a contradiction.
A: 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.
A: Let $\frac{a_i}{b_i}, 1\le i\le n$, be a generating set. Then 
$$\forall x\in D \ \ \exists c\in D \text{ such that } \frac{1}{x}=\frac{c}{a},$$
where $a=a_1\cdots a_n$, i.e.,
$$\forall x\in D \ \ \exists c\in D \text{ such that } cx=a.$$
For $x=a^2$ we have $ca=1$, so every $y\in D$ is invertible: if $ty=a$ then $cty=1$. Hence $D=F$.
