# A Noetherian module annihilated by a power of maximal ideal must has finite length.

Let $M$ be a Noetherian $R$-module and $P^kM=0$ from some maximal ideal $P$ of $R$ and some integer $k$. How to show that $M$ has finite length?

The length of a module is defined to be the maximum length of the chain of submodule: $$0=M_0<M_1<\cdots<M_{n-1}<M_n=M$$

I have tried following. We can assume there is no strict submodule between $M_i$ and $M_{i+1}$, and try to prove such $n$ is bounded.

Then we have $$M_i/M_{i+1}\cong R/Q_i$$

where $Q_i=\operatorname{ann}_R(M_i/M_{i+1})$ is maximal.

Then I cannot move on. I tried to look at the localization of the chain at $P$, but it seems to provide nothing.

Could anyone help?

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## 2 Answers

WLOG, choose $k$ to be the smallest possible. Consider the sequence of submodules

$$(*)\qquad0=MP^{k} \subsetneq MP^{k-1} \subsetneq \dots \subsetneq MP \subsetneq M \,.$$

The goal is to show that every consequtive factor of this sequence is of finite length.

For arbitrary $l \in \{0,1, \dots, k-1\},$ consider the factor $MP^l/MP^{l+1}$. Since $M$ is noetherian, this clearly is a finitely generated module. The annihilator $\mathrm{Ann}_R(MP^l/MP^{l+1})$ clearly contains the maximal ideal $P$. On the other hand, from the strictness of the inclusion $MP^{l+1} \subsetneq MP^{l}$ it follows that $1 \notin \mathrm{Ann}_R(MP^l/MP^{l+1})$, hence (since $P$ is maximal) $\mathrm{Ann}_R(MP^l/MP^{l+1})=P,$ a maximal ideal.

Now, since any module $N$ can be considered as $R/\mathrm{Ann}_R(N)$-module (with the multiplication defined by $n \cdot (r+\mathrm{Ann}_R(N)):=nr$ and the important property that the lattice of submodules does not change by this shift of perspective), we can see that $MP^l/MP^{l+1}$ is actually finitely-generated $R/P$-module, i.e. a vector space of finite dimension. Hence, it is of finite length.

Adding more details:

It is a well-known fact that a module $M$ is of finite length iff it has a finite composition series, i.e. a finite chain of submodules from $0$ to $M$ with the consecutive factors simple. Now, we have shown (for arbitrary $l$) that $MP^l/MP^{l+1}$ are of finite length, hence there exist a composition series $$0=MP^{l+1}/MP^{l+1}=N_0^{l} \subseteq N_{1}^{l} \subseteq \dots \subseteq N_{k_l}^{l}=MP^{l}/MP^{l+1}$$ with simple consecutive factors. After applying the correspondence theorem, we obtain a chain of submodules of $M$ $$MP^{l+1}=\overline{N_0^{l}}\subseteq \overline{N_1^{l}} \subseteq \dots \subseteq \overline{N_{k_l}^{l}}=MP^{l}$$ again with simple consecutive factors. Doing this for every $l$, we obtain a refinement of the series $(*)$ with simple consecutive factors of finite length, i.e. a composition series of module $M$ of finite length.

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Sorry but I think you only proved this specific chain has finite length. What about other chains? Isn't the length of a module the maximal length of all possible chains? –  hxhxhx88 Nov 17 '13 at 21:54
Jordan - Hölder. –  Martin Brandenburg Nov 17 '13 at 22:00
@hxhxhx88 Sorry, I thought this was clear. It uses the fact that module is of finite length if and only if it has a finite composition series. I will add a comment about it into the answer. –  PavelC Nov 17 '13 at 22:16

$P^kM=0$ implies $P^k\subset Ann_R(M)$. This shows that $R/Ann_R(M)$ is an artinian ring (its only prime ideal being $P/Ann_R(M)$), so $M$ is an artinian $R/Ann_R(M)$-module, that is, an artinian $R$-module and we are done.

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Sorry, your approach is too advanced. My commutative algebra course has't covered the Artinian ring yet..If it is possible, could you expand it a little bit? –  hxhxhx88 Nov 17 '13 at 21:56