Let $R$ be a Noetherian ring and $M$ a finite $R$-module. Show that $\ell(M)<\infty$ if and only if $\text{Supp}(M)\subset m\text{-Spec}(R)$

Question

Let $R$ be a Noetherian ring and $M$ a finite $R$--module. Show that $\ell(M)<\infty$ if and only if $\operatorname{Supp}(M)\subset m\operatorname{-Spec}(R)$.

What does $\ell(M)<\infty$ mean? Is it that $M$ is finitely generated? If so, how is it that I am supposed to use that to show $\operatorname{Supp}(M)\subset m\operatorname{-Spec}(R)$?

Answer 1

$l(M)<\infty$ means that $M$ has a composition series.we always assume $R$ is Noetherian in following:

definition1:let $N$ be an aribitray module,a prime ideal $P$ is called the associated prime ideal of $N$ if $p=ann(x)$ for some $X\in N$.and denote $Ass(N)$ be the set of all the associted prime ideals.

it is clear that $p\in Ass(N)$ iff $R/p$ is a submodule of $N$.

lemma1:for any module $N$,if $N$ is not zero module,then $Ass(N)$ is not empty.

if $Ass(N)$ is not empty,for $x\neq 0\in N$,$ann(x)$ is not prime,hence you can find $ab\in ann(x)$ but $a,b\notin ann(x)$.So $ann(x)\varsubsetneqq ann(ax)$. then as before,you can get a filtration.this is a contradiction.

lemma2:if $N$ is finite generated module,then there is a filtration $0\varsubsetneqq N_1 \varsubsetneqq N_2 \varsubsetneqq ...\varsubsetneqq N_k=N$ such that $N_i/N_{i-1}\cong R/p_i$,where $p_i$ are prime ideals.

the proof relies on lemma1.

lemma3:if $N$ is finite generated module,then $N_p\neq 0$ iff $ ann(N)\subset p$.

proof:it is trivial that if $N_p\neq 0$ ,then $ ann(N)\subset p$.Now suppove $ ann(N)\subset p$.denote $N=<n_1,...,n_k>$.if $N_p=0$,you can find $r_i\notin p$ such that $r_in_i=0$.denote $r=r_1r_2...r_k$,then $r\notin p$ and $rN=0$.contradiction.

by lemma3,we know $Supp(R/p)=V(p)$.

lemma4:if $0\rightarrow N_1\rightarrow N_2\rightarrow N_3\rightarrow 0$ is an short exact sequence,then $Supp(N_2)= Supp(N_1)\bigcup Supp(N_3)$

BY lemma2,3,4,you can get the proof of your question.