0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.