Finite length module over Noetherian ring

Question

Let $M$ be a finitely generated module over a noetherian ring $R$ and suppose that $\dim R\leq 1$. Then $M$ has finite length.

A preliminary lemma:

There exists an chain $\{0\}=M_0\subsetneq M_1\subsetneq\cdots\subsetneq M_r=M$ of submodules of $M$, such that $M_i/M_{i-1}\approx R/\mathfrak{p}_i$ for $i=1,\ldots, r$.

We have:

1) If $ht(\mathfrak p_i)=1$ then $M_i/M_{i-1}$ is simple (right?).

2) If $ht(\mathfrak p_i)=0$ then $\mathfrak p_i\subsetneq\mathfrak q_i$ with $\mathfrak q_i$ be a maximal ideal of $R$. Thus we obtain a submodule $M_i'$ such that, $M_{i-1}\subsetneq M_i'\subsetneq M_i$ with $M_i/M_{i}'\approx R/\mathfrak q_i$ (as $R$-modules) and $M_i'/M_{i-1}\approx \mathfrak q_i/\mathfrak p_i$ (as $R$-modules).

I don't see that $\mathfrak q_i/\mathfrak p_i$ is a simple $R$-module, if it is true.

Thanks for your help.

$\textbf{Edit:}$ Thanks to Eric Wofsey. Now if $N$ is a non-zero-submodule $M/N$ has finite length as $R$-module?

Thanks one more time.

Answer 1

This isn't true. For instance, if $R=\mathbb{Z}$, then since every simple module is finite, any finite length module must be finite. But there are finitely generated modules that are not finite (e.g., $\mathbb{Z}$ itself). For your followup question, you can take $M=\mathbb{Z}^2$ and $N=\mathbb{Z}$ to get a counterexample.