Let $R$ be a Noetherian integral domain and $\omega\in R\backslash\{0\}$ be a non-unit. For the multiplicative set $S=\{\omega^n:n\in\mathbb{Z}^+\}$ of $R$, let $R_{\omega}=S^{-1}R$ be the ring of fractions of $R$ by $S$. Prove that $R_\omega$ is a Noetherian ring, but not a Noetherian $R$-module.
My attempt: Proving that $R_\omega$ is a Noetherian ring is easy, since $R_\omega \cong R[x]/(x\omega-1)$ and $R[x]$ is a Noetherian ring. However, I'm not sure if my proof of the latter statement is correct. To obtain a contradiction, I'd like to prove that $R_\omega$ is not finitely generated as an $R$-module. Below is my solution.
Suppose $R_\omega\cong R[x]/(x\omega-1)$ is a finitely generated $R$-module. Then there exists $f_1,...,f_n\in R[x]$ such that $\langle f_1,...,f_n\rangle=R[x]/(x\omega-1)$.(Here I'm abusing the notation; each $f_i$ means the coset $f_i+(x\omega-1)$.) Let $\deg(f_i)= m_i$. Now let $M=\max\{m_1,...,m_n\}$. Then $x^{M+1}\notin \langle f_1,...,f_n\rangle$. If it were, then there would be $r_1 ,..., r_n \in R$ and $g\in R[x]$ such that $x^{M+1}= \sum_{i=1}^{n}r_i f_i + (x\omega-1)g\in R[x]$. Comparing the coefficients of the both sides, the leading coefficient of $g$ must be $1/\omega$, which is absurd since $\omega$ is a non-unit.
Is my solution correct?
The proof that $R_\omega$ is Noetherian is good.
Let $\{x_1/\omega^k,\dots,x_n/\omega^k\}$ be a finite set in $R_\omega$. It's not restrictive to assume the denominators are the same $k>0$, with $x_i\in R$. Let's show that $$ \frac{1}{\omega^{k+1}}\notin\frac{x_1}{\omega^k}R+\dots+\frac{x_n}{\omega^k}R $$ Assume the contrary; then we can write $$ \frac{1}{\omega^{k+1}}=\frac{r}{\omega^k} $$ in $R_\omega$, for some $r\in R$, which in turn implies $$ \frac{r}{1}=\frac{1}{\omega} $$ As $R$ is an an integral domain, this implies $r\omega=1$.
Note that the assumption $R$ is a domain is necessary. For instance, if $\omega$ is nilpotent, then $R_\omega=\{0\}$. But this is not required: for $R=\mathbb{Z}/6\mathbb{Z}$ and $\omega=\bar{2}$, the $R$-module $R_\omega$ is generated by $1/\bar{4}$.