Suppose that $D$ is an integral domain and $F$ its field of fractions, $D \neq F$. I want to show that $F$ as $D$-module is not Noetherian.
Attempt: By contradiction, suppose it is. Because $D \neq F$ I can find $d$ which is not in $F$. How to build an ascending chain that leads to a contradiction? I know that the different powers of $d$ are all different from 0 because $D$ is domain. But I get a descending chain, and I want an ascending one. Any ideas?
You were almost there.
Since $D$ is not a field, there is some nonzero $d \in D$ such that $d \notin D^\times$. Then $d^{-1} \in F \setminus D$.
Now consider the $D$-modules $d^{-n}D$, $n \in \Bbb N$.
Choose $f \in F \setminus D$, so $f = b/a$, with $b, a \in D$, and $a \ne 0$ not invertible in $D$. We have also $1/a \in F \setminus D$.
Consider the $D$-submodules of $F$ given by $$ \left\langle \frac{1}{a} \right\rangle \subseteq \left\langle \frac{1}{a^{2}} \right\rangle \subseteq \left\langle \frac{1}{a^{3}} \right\rangle \subseteq \dots $$ All inclusions are proper, because if $$ \left\langle \frac{1}{a^{i}} \right\rangle = \left\langle \frac{1}{a^{i+1}} \right\rangle $$ then there is $c \in D$ such that $$ \frac{c}{a^{i}} = \frac{1}{a^{i+1}} $$ or $c a^{i+1} = a^{i}$, which implies, since $D$ is a domain, $c a = 1$, thus $a$ invertible in $D$.
