Consider a Noetherian ring $R$ and an ideal $I$, whose associated primes are maximal ideals. I want to show that $R$ is Artinian (to conclude that $\frac{R}{I}$ is Artinian).
My attempt:
I assumed it's not and that there's a chain $J_1 \supset J_2 \supset J_3 \supset ...$ that doesn't stabilize. Since $R$ is Noetherian, it has finitely many prime ideals. Let's give that set of prime ideals a name, $S$. Every ideal admits a minimal primary decomposition. But that must mean that all of those $J_i$ are intersections of elements in $S$. But there are only $2^{|S|}$ ways of picking those elements, which is a finite number. So at some point, those $J_i$ have to repeat, thus the chain has to stabilize.
Now, obviously, that can't be true, since if this was the case, then every Noetherian ring would be Artinian (I didn't even use the maximality of $I$'s associated primes). But I'm not sure where my mistake is, and I don't find any other way to solve the problem.
Another thing I tried was starting from $\frac{R}{I}$, get a chain $J_1 \supset J_2 \supset J_3 \supset ...$ of ideals, and then use the inverted quotient map to get ideals $\pi^{-1}(J_1) \supset \pi^{-1}(J_2) \supset \pi^{-1}(J_3) \supset ...$ in $R$. The difference now being that those are ideals that contain $I$, so I wonder if I get somehow use that, since $I$ associated primes are maximal elements, so they could potentially upper bound the $J_i$, I just don't see how and how that would help further.