How to prove the ring $R_J$ is Noetherian without use the theorem of I. S. Cohen?

Question

First define $R:=k\left[{\{X_i\}}_{i\in\mathbb{N}}\right]$ where $k$ is a field (it could be an integral domain as $\mathbb{Z}$ too for example). This ring is an integral domain and it is not Noetherian. The second thing to do is define the primes ideals that I want there exist in the new ring. For this I define ${J_1}:={I_1}:={\langle X_1 \rangle}_R$, ${\forall i>2}:\;I_{\{2,i\}}:={\langle X_2,X_i \rangle}_R$, $J_2 := \underset{i>2}{\bigcup}{I_{\{2,i\}}}$, and so, ${I_{\{{n,i_{n−1},i_{n−2}, \ldots, i_1}\}}}:={\langle{X_n, X_{i_{n-1}}, X_{i_{n-2}}, \ldots, X_{i_1}} \rangle}_R$ where $i_1 > i_2 > \ldots > i_{n−1} > n$ are naturals, and $J_n:=\underset{i_1 > i_2 > \dots > i_{n−1} > n}{\bigcup}I_{\{n,i_{n−1},i_{n−2},\ldots,i_1\}}$. Then I define $J:=\underset{k \in \mathbb{N}}{\bigcup}{J_k}$. All the ideals defined are primes, then $J$ is union of primes ideals of $R$, which are all of them finitely many generated as $R$-mod. By the theorem of I. S. Cohen is easy seeing that the localization $R_J$ is Noetherian (and his Krull dimension is infinite). But I want prove the Noetherianity of $R_J$ without the theorem of I. S. Cohen. We can use for example that each ascending chain has a finite length in a localization of $R_J$ on any $I\in J$ is equivalent to Noetherianity of $R_J$, but I do not know how to proceed. Someone has an idea?