Commutative integral domain does not finitely generate its field of fractions

Question

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.

Answer 1

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 $D=F$.

Answer 2

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.