If $R$ is a Noetherian ring and $I$ a proper ideal, then the $R$-module $R/I$ has finite length iff the associated primes of $I$ are maximal

Question

Let $R$ be a Noetherian ring and $I$ a proper ideal in $R$.
We need to show that the $R$-module $R/I$ has finite length if and only if the set $Ass(I)$ of the associated primes of the ideal $I$ consists only of maximal ideals.
I'm completely stuck.
Any help would be appreciated.

Answer 1

user26857 has already answered your question (+1), but let me fill in just a few details. For an $R$-module $M$, write $\lambda(M)$ for the length of $M$. First note that, for any prime ideal $P$, if $P$ is not maximal then $\lambda(R\big/P)=\infty$. Indeed, then $R\big/P$ is an integral domain that is not a field, so there exists $\bar{x}\in R\big/P$ that is neither a unit nor zero, and then $$0\subsetneq(\bar{x}^n)\subsetneq (\bar{x}^{n-1})\subsetneq\dots\subsetneq(\bar{x})\subsetneq R\big/P$$ is a strictly ascending (why?) chain of submodules of length $n+1$ for any $n\in\mathbb{N}$. This shows the first direction of your desired result: if $I$ has an associated prime $P$ that is not maximal, then in particular $I$ is contained in $P$, so $\lambda(R\big/I)\geqslant\lambda(R\big/P)$ is thus infinite.

For the other direction, suppose every associated prime of $I$ is maximal. Since $\sqrt{I}$ is the intersection of the associated primes of $I$, and there are finitely many associated primes of $I$, this means that every prime lying above $I$ is maximal. Let $P_1,\dots,P_n$ be these maximal ideals. The nilradical of $R\big/I$ is then $N:=\bigcap_{i=1}^nP_i\big/I$, and – since $R$ is Noetherian and so $N$ is finitely generated – there exists some $k\in\mathbb{N}$ such that $N^k=\{\bar{0}\}$. On the other hand, the $P_i\big/I$ are pairwise coprime, so the $\left(P_i\big/I\right)^k$ are also pairwise coprime (why?), and now by the Chinese Remainder Theorem we have $$\frac{R}{I}\cong\prod_{i=1}^n\frac{R\big/I}{\left(P_i\big/I\right)^k}.$$ Now, every ideal of a finite product of rings is a product of the ideals of the rings, and so to show that $R\big/I$ has finite length it suffices to show that $(R/I)\big/(P_i/I)^k$ has finite length for each $i$. Thus the desired result follows from the following two lemmas:

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Lemma: If $R$ is any commutative ring and $A\leqslant B$ are $R$-modules, then $\lambda(B)=\lambda(A)+\lambda(B\big/A)$.

Proof: The inequality $\lambda(B)\geqslant\lambda(A)+\lambda(B\big/A)$ is hopefully clear. For the other direction, let $$0=B_0\subsetneq B_1\subsetneq\dots\subsetneq B_k=B$$ be any chain of submodules of $B$. For each $i\leqslant k$, define $A_i=A\cap B_i$ and $C_i=(A+B_i)\big/A$. This gives two ascending chains \begin{align}&0=A_0\subseteq A_1\subseteq\dots\subseteq A_k=A\\&0=C_0\subseteq C_1\subseteq\dots\subseteq C_k=B\big/A, \end{align}and so, to show $k\leqslant \lambda(A)+\lambda(B\big/A)$, it suffices to show that, for each $i<k$, at least one of $A_i\subsetneq A_{i+1}$ and $C_i\subsetneq C_{i+1}$ holds. Suppose otherwise; then $B_i\cap A=B_{i+1}\cap A$ and $$(A+B_i)\big/A=(A+B_{i+1})\big/A,$$ ie $A+B_i=A+B_{i+1}$. But now \begin{align}B_{i+1}&=B_{i+1}\cap (A+B_{i+1})=B_{i+1}\cap (A+B_i) \\ &=(B_{i+1}\cap A)+B_i=(B_i\cap A)+B_i=B_i, \end{align} where the third equality follows from the "modular law" and that fact that $B_i\subseteq B_{i+1}$. But this is a contradiction, since $B_{i+1}\supsetneq B_i$ by hypothesis, and so we are done. $\blacksquare$

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Lemma: If $R$ is any commutative Noetherian ring, and $M<R$ is a maximal ideal, then $\lambda(R\big/M^k)<\infty$ for any $k\geqslant 1$.

Proof: By applying the above lemma repeatedly, we see $$\lambda(R\big/M^k)=\lambda(R\big/M)+\lambda(M\big/M^2)+\dots+\lambda(M^{k-1}\big/M^k).$$ But each $M^{i}\big/M^{i+1}$ is a vector space over the field $R\big/M$, so its length over $R$ is equal to its dimension over $R\big/M$. On the other hand, since $R$ is Noetherian, each $M^i\big/M^{i+1}$ is finitely generated as an $R$-module, and hence finite-dimensional as a $R\big/M$-vector space, so each $\lambda(M^i\big/M^{i+1})<\infty$ and thus we are done. $\blacksquare$

Answer 2

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module. Then $M$ has finite length iff $\mathrm{Ass}(M)\subseteq\mathrm{Max}(R)$.

If $M$ has finite length, then it is artinian, so the ring $R/\mathrm{Ann}(M)$ is artinian. It follows that every prime ideal in $\mathrm{Supp}(M)$ is maximal. In particular, all asociated primes of $M$ are maximal.

For the converse it is enough to show that $\mathrm{Supp}(M)\subseteq\mathrm{Max}(R)$. Let $P$ be a prime minimal in $\mathrm{Supp}(M)$. Then it belongs to $\mathrm{Ass}(M)$ and therefore it is maximal.