Question.
Let $R$ be a Noetherian ring. Let $M$ be a finitely generated $R$-module such that $\dim(M)>0$. Take a prime filtration $0=K_0\subsetneq K_1\subsetneq ...\subsetneq K_t = M$, and let $P_i\in\text{Spec}(R)$ such that $K_i/K_{i-1}\cong R/P_i$ for $1\leq i\leq t$.
I am trying to show that none of the $P_i$ can be maximal.
I am not positive that this is true (I am hoping it is for the sake of reading a paper). If $A$ is an $R-$algebra, use the notation $\dim_R(A)$ to denote the dimension of $A$ over $R$ ($=\dim(R/\text{ann}_R(A)$) and $\dim(A)$ to denote the Krull dimension as a ring (and omit this subscript $R$ when the context is clear).
Some ideas towards a proof.
Note that, if $m\in \text{max}(R)$, then $\dim_R(R/m) = \dim(R/m)=0$ (since $m=\text{ann}_R(R/m)$), and it is well known that if we have a SES (short exact sequence) $0\to M'\to M\to M''\to 0$ (over $R$) of finitely generated $R$-modules, then $\dim M = \max\{\dim M'', \dim M'\}$. Consider the exact sequences induced by the filtration $0\to K_i/K_{i-1}\to K_{i+1}/K_{i-1}\to K_{i+1}/K_i\to 0$. We have that $\dim(K_t/K_{0}) = \dim(M)>0$).
For $j\in \{0,...,t\}$ we have an exact sequence
$0\to K_{t-(j-1)}/K_{t-j}\to K_t/K_{t-j}\to K_t/K_{t-(j-1)}\to 0$
so that $\dim(K_t/K_{t-j})= \max \{\dim(K_{t-(j-1)}/K_{t-j}), \dim(K_t/K_{t-(j-1)})\}$
We have a short exact sequence
$K_{t-1}\to K_t=M\to K_t/K_{t-1}$
and so either $K_{t-1}$ or $K_t/K_{t-1}$ has positive dimension. If $\dim(K_t/K_{t-1})=0$, then $\dim(K_{t-1})>0$ so that $K_{t-2}$ or $K_{t-1}/K_{t-2}$ has positive dimension.
