I am going through theorem 2.14 in Eisenbud's Commutative Algebra.
Given a ring $R$ that is Noetherian, all of whose prime ideals are maximal, we want to prove that $R$ is Artinian. Assume that $R$ is not Artinian. Define an $I \subseteq R$ that is maximal such that $R/I$ is not of finite length. Eisenbud proves that $I$ must be maximal, so $R/I$ is a field. This is a contradiction.
It's not clear why this is a contradiction. As far as I can tell, $R/I$ is assumed to be not of finite length as an $R$-module in the proof. But fields do not need to be of finite length over other rings; for example $\mathbb{Q}$ is not a finite-length $\mathbb{Z}$-module. I am not sure where to go with this.
Hint, The key fact is: $R$-submodules of $R/I$ are of the forms $J/I$ where $J$ is an ideal containing $I$.