Let $A$ be a domain, $x \in A -\{0\}$ and $M$ be the $A$-submodule of $\text{Frac}(A)$ generated by the elements $\{1,x^{-1},x^{-2}, \dots,\}$. My question is: If $M$ is finitely generated, then it is true that $M=A$?
Obs: $A$ is an commutative ring with unity.
My attempt:
I tried interpret $A$ as a subset of $M$ by writing $a=\frac{a}{1}$ and tried to find an ideal $I$ of $A$ contained in the Jacobson radical of $A$ such that $M=IM+A$. The result then would follow by the application of the Nakayama lemma, but I don't know how to find such ideal $I$.
Suppose $M$ is finitely generated. Then $\{1, x^{-1}, \dots, x^{-n}\}$ generates $M$ for some $n \in \mathbb{N}$. Now $$x^{-n-1} = a_0 + a_1 x^{-1} + \dots + a_n x^{-n}$$ for some $a_0, \dots, a_n \in A$. Multiplying through by $x^n$ yields $$x^{-1} = a_0 x^n + a_1 x^{n-1} + \dots + a_n.$$ Thus, $x^{-1} \in A$, so $\{1, x^{-1}, \dots, x^{-n}\} \subseteq A$, so $M \subseteq A$. Also, since $1 \in M$, we have $A \subseteq M$. Thus, $A = M$. $\square$