Consider $R=\mathbb{Z}[\{\frac{X}{p_n}\}_{n\ge1}]$ where $p_n$ is $n$'th prime. Suppose $f$ is an irreducible polynomial in $\mathbb{Q}[X]$. Then I need to show $R_{R\cap f\mathbb{Q}[X]}$ is noetherian.
I guess it might be some localization of $\mathbb{Q}[X]$ but can't figure out which localization. Any help is appreciated.
I think I've found an argument: Contraction of $R\cap f\mathbb{Q}[X]$ to $\mathbb{Z}[X]$ is $f\mathbb{Z}[X]$ (WLOG assuming $f\in \mathbb{Z}[X]$ with content one). Now $\mathbb{Z}[X]_{f\mathbb{Z}[X]}$ is a Dedekind domain and since $R_{R\cap f\mathbb{Q}[X]}$ is lying in between $\mathbb{Z}[X]_{f\mathbb{Z}[X]}$ and its fraction field $\mathbb{Q}(X)$ so its itself Dedekind and so is noetherian. Anybody care to check this argument or suggest any other argument?
Lemma. Let $A \subseteq B \subseteq K$ be rings, where $A$ is a discrete valuation ring, $B$ is a local ring, $K$ is the fraction field of $A$, and $A \to B$ is a local homomorphism. Then $A = B$.
Proof. By contradiction there exists $b \in B \setminus A$. Let $v \colon K^* \to \mathbb{Z}$ be the valuation associated to $A$. Since $b \notin A$, we have $v(b) < 0$. Then $v(b^{-1}) > 0$, therefore $b^{-1}$ is in the maximal ideal of $A$. As the homomorphism $A \to B$ is local, $1 = bb^{-1}$ lies in the maximal ideal of $B$, which is impossible. QED
Now consider the following case (which is slightly more general than yours): $f \in \mathbb{Z}[x]$ is primitive and irreducible and $\mathbb{Z}[x] \subseteq R \subseteq \mathbb{Q}[x]$ is any ring. So $f \mathbb{Q} [x] \cap \mathbb{Z}[x] = f \mathbb{Z}[x]$. Thus, if $P := f \mathbb{Q}[x] \cap R$, then $P \cap \mathbb{Z}[x] = f \mathbb{Z}[x]$. The ideal $f \mathbb{Z}[x]$ is prime of height $1$, hence the localisation $A = \mathbb{Z}[x]_{f \mathbb{Z}[x]}$ is a DVR. Moreover, the inclusion $A \hookrightarrow R_P$ is local. By the lemma above, $R_P = A$.