Section 10.52 (00IU): Length—The Stacks project

10.52 Length

Definition 10.52.1. Let $R$ be a ring. For any $R$-module $M$ we define the length of $M$ over $R$ by the formula

\[ \text{length}_ R(M) = \sup \{ n \mid \exists \ 0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M, \text{ }M_ i \not= M_{i + 1} \} . \]

In other words it is the supremum of the lengths of chains of submodules. There is an obvious notion of when a chain of submodules is a refinement of another. This gives a partial ordering on the collection of all chains of submodules, with the smallest chain having the shape $0 = M_0 \subset M_1 = M$ if $M$ is not zero. We note the obvious fact that if the length of $M$ is finite, then every chain can be refined to a maximal chain. But it is not as obvious that all maximal chains have the same length (as we will see later).

slogan

Lemma 10.52.2. Let $R$ be a ring. Let $M$ be an $R$-module. If $\text{length}_ R(M) < \infty $ then $M$ is a finite $R$-module.

Proof. Omitted. $\square$

slogan

Lemma 10.52.3. If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules over $R$ then the length of $M$ is the sum of the lengths of $M'$ and $M''$.

Proof. Given filtrations of $M'$ and $M''$ of lengths $n', n''$ it is easy to make a corresponding filtration of $M$ of length $n' + n''$. Thus we see that $\text{length}_ R M \geq \text{length}_ R M' + \text{length}_ R M''$. Conversely, given a filtration $M_0 \subset M_1 \subset \ldots \subset M_ n$ of $M$ consider the induced filtrations $M_ i' = M_ i \cap M'$ and $M_ i'' = \mathop{\mathrm{Im}}(M_ i \to M'')$. Let $n'$ (resp. $n''$) be the number of steps in the filtration $\{ M'_ i\} $ (resp. $\{ M''_ i\} $). If $M_ i' = M_{i + 1}'$ and $M_ i'' = M_{i + 1}''$ then $M_ i = M_{i + 1}$. Hence we conclude that $n' + n'' \geq n$. Combined with the earlier result we win. $\square$

Lemma 10.52.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$. If $M$ is an $R$-module and $\mathfrak m^ n M \not= 0$ for all $n \geq 0$, then $\text{length}_ R(M) = \infty $. In other words, if $M$ has finite length then $\mathfrak m^ nM = 0$ for some $n$.

Proof. Assume $\mathfrak m^ n M \not= 0$ for all $n\geq 0$. Choose $x \in M$ and $f_1, \ldots , f_ n \in \mathfrak m$ such that $f_1f_2 \ldots f_ n x \not= 0$. The first $n$ steps in the filtration

\[ 0 \subset R f_1 \ldots f_ n x \subset R f_1 \ldots f_{n - 1} x \subset \ldots \subset R x \subset M \]

are distinct. For example, if $R f_1 x = R f_1 f_2 x$ , then $f_1 x = g f_1 f_2 x$ for some $g$, hence $(1 - gf_2) f_1 x = 0$ hence $f_1 x = 0$ as $1 - gf_2$ is a unit which is a contradiction with the choice of $x$ and $f_1, \ldots , f_ n$. Hence the length is infinite. $\square$

Lemma 10.52.5. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. We always have $\text{length}_ R(M) \geq \text{length}_ S(M)$. If $R \to S$ is surjective then equality holds.

Proof. A filtration of $M$ by $S$-submodules gives rise a filtration of $M$ by $R$-submodules. This proves the inequality. And if $R \to S$ is surjective, then any $R$-submodule of $M$ is automatically an $S$-submodule. Hence equality in this case. $\square$

Lemma 10.52.6. Let $R$ be a ring with maximal ideal $\mathfrak m$. Suppose that $M$ is an $R$-module with $\mathfrak m M = 0$. Then the length of $M$ as an $R$-module agrees with the dimension of $M$ as a $R/\mathfrak m$ vector space. The length is finite if and only if $M$ is a finite $R$-module.

Proof. The first part is a special case of Lemma 10.52.5. Thus the length is finite if and only if $M$ has a finite basis as a $R/\mathfrak m$-vector space if and only if $M$ has a finite set of generators as an $R$-module. $\square$

Lemma 10.52.7. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset. Then $\text{length}_ R(M) \geq \text{length}_{S^{-1}R}(S^{-1}M)$.

Proof. Any submodule $N' \subset S^{-1}M$ is of the form $S^{-1}N$ for some $R$-submodule $N \subset M$, by Lemma 10.9.15. The lemma follows. $\square$

Lemma 10.52.8. Let $R$ be a ring with finitely generated maximal ideal $\mathfrak m$. (For example $R$ Noetherian.) Suppose that $M$ is a finite $R$-module with $\mathfrak m^ n M = 0$ for some $n$. Then $\text{length}_ R(M) < \infty $.

Proof. Consider the filtration $0 = \mathfrak m^ n M \subset \mathfrak m^{n-1} M \subset \ldots \subset \mathfrak m M \subset M$. All of the subquotients are finitely generated $R$-modules to which Lemma 10.52.6 applies. We conclude by additivity, see Lemma 10.52.3. $\square$

Definition 10.52.9. Let $R$ be a ring. Let $M$ be an $R$-module. We say $M$ is simple if $M \not= 0$ and every submodule of $M$ is either equal to $M$ or to $0$.

Lemma 10.52.10. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

$M$ is simple,
$\text{length}_ R(M) = 1$, and
$M \cong R/\mathfrak m$ for some maximal ideal $\mathfrak m \subset R$.

Proof. Let $\mathfrak m$ be a maximal ideal of $R$. By Lemma 10.52.6 the module $R/\mathfrak m$ has length $1$. The equivalence of the first two assertions is tautological. Suppose that $M$ is simple. Choose $x \in M$, $x \not= 0$. As $M$ is simple we have $M = R \cdot x$. Let $I \subset R$ be the annihilator of $x$, i.e., $I = \{ f \in R \mid fx = 0\} $. The map $R/I \to M$, $f \bmod I \mapsto fx$ is an isomorphism, hence $R/I$ is a simple $R$-module. Since $R/I \not= 0$ we see $I \not= R$. Let $I \subset \mathfrak m$ be a maximal ideal containing $I$. If $I \not= \mathfrak m$, then $\mathfrak m /I \subset R/I$ is a nontrivial submodule contradicting the simplicity of $R/I$. Hence we see $I = \mathfrak m$ as desired. $\square$

Lemma 10.52.11. Let $R$ be a ring. Let $M$ be a finite length $R$-module. Choose any maximal chain of submodules

\[ 0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_ n = M \]

with $M_ i \not= M_{i-1}$, $i = 1, \ldots , n$. Then

$n = \text{length}_ R(M)$,
each $M_ i/M_{i-1}$ is simple,
each $M_ i/M_{i-1}$ is of the form $R/\mathfrak m_ i$ for some maximal ideal $\mathfrak m_ i$,
given a maximal ideal $\mathfrak m \subset R$ we have

\[ \# \{ i \mid \mathfrak m_ i = \mathfrak m\} = \text{length}_{R_{\mathfrak m}} (M_{\mathfrak m}). \]

Proof. If $M_ i/M_{i-1}$ is not simple then we can refine the filtration and the filtration is not maximal. Thus we see that $M_ i/M_{i-1}$ is simple. By Lemma 10.52.10 the modules $M_ i/M_{i-1}$ have length $1$ and are of the form $R/\mathfrak m_ i$ for some maximal ideals $\mathfrak m_ i$. By additivity of length, Lemma 10.52.3, we see $n = \text{length}_ R(M)$. Since localization is exact, we see that

\[ 0 = (M_0)_{\mathfrak m} \subset (M_1)_{\mathfrak m} \subset (M_2)_{\mathfrak m} \subset \ldots \subset (M_ n)_{\mathfrak m} = M_{\mathfrak m} \]

is a filtration of $M_{\mathfrak m}$ with successive quotients $(M_ i/M_{i-1})_{\mathfrak m}$. Thus the last statement follows directly from the fact that given maximal ideals $\mathfrak m$, $\mathfrak m'$ of $R$ we have

\[ (R/\mathfrak m')_{\mathfrak m} \cong \left\{ \begin{matrix} 0 & \text{if } \mathfrak m \not= \mathfrak m', \\ R_{\mathfrak m}/\mathfrak m R_{\mathfrak m} & \text{if } \mathfrak m = \mathfrak m' \end{matrix} \right. \]

This we leave to the reader. $\square$

Lemma 10.52.12. Let $A$ be a local ring with maximal ideal $\mathfrak m$. Let $B$ be a semi-local ring with maximal ideals $\mathfrak m_ i$, $i = 1, \ldots , n$. Suppose that $A \to B$ is a homomorphism such that each $\mathfrak m_ i$ lies over $\mathfrak m$ and such that

\[ [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m)] < \infty . \]

Let $M$ be a $B$-module of finite length. Then

\[ \text{length}_ A(M) = \sum \nolimits _{i = 1, \ldots , n} [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m)] \text{length}_{B_{\mathfrak m_ i}}(M_{\mathfrak m_ i}), \]

in particular $\text{length}_ A(M) < \infty $.

Proof. Choose a maximal chain

\[ 0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_ m = M \]

by $B$-submodules as in Lemma 10.52.11. Then each quotient $M_ j/M_{j - 1}$ is isomorphic to $\kappa (\mathfrak m_{i(j)})$ for some $i(j) \in \{ 1, \ldots , n\} $. Moreover $\text{length}_ A(\kappa (\mathfrak m_ i)) = [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m)]$ by Lemma 10.52.6. The lemma follows by additivity of lengths (Lemma 10.52.3). $\square$

Lemma 10.52.13. Let $A \to B$ be a flat local homomorphism of local rings. Then for any $A$-module $M$ we have

\[ \text{length}_ A(M) \text{length}_ B(B/\mathfrak m_ AB) = \text{length}_ B(M \otimes _ A B). \]

In particular, if $\text{length}_ B(B/\mathfrak m_ AB) < \infty $ then $M$ has finite length if and only if $M \otimes _ A B$ has finite length.

Proof. The ring map $A \to B$ is faithfully flat by Lemma 10.39.17. Hence if $0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$ is a chain of length $n$ in $M$, then the corresponding chain $0 = M_0 \otimes _ A B \subset M_1 \otimes _ A B \subset \ldots \subset M_ n \otimes _ A B = M \otimes _ A B$ has length $n$ also. This proves $\text{length}_ A(M) = \infty \Rightarrow \text{length}_ B(M \otimes _ A B) = \infty $. Next, assume $\text{length}_ A(M) < \infty $. In this case we see that $M$ has a filtration of length $\ell = \text{length}_ A(M)$ whose quotients are $A/\mathfrak m_ A$. Arguing as above we see that $M \otimes _ A B$ has a filtration of length $\ell $ whose quotients are isomorphic to $B \otimes _ A A/\mathfrak m_ A = B/\mathfrak m_ AB$. Thus the lemma follows. $\square$

Lemma 10.52.14. Let $A \to B \to C$ be flat local homomorphisms of local rings. Then

\[ \text{length}_ B(B/\mathfrak m_ A B) \text{length}_ C(C/\mathfrak m_ B C) = \text{length}_ C(C/\mathfrak m_ A C) \]

Proof. Follows from Lemma 10.52.13 applied to the ring map $B \to C$ and the $B$-module $M = B/\mathfrak m_ A B$ $\square$

Comments (3)

Comment #5519 by Alex on September 27, 2020 at 03:16

In the proof of Lemma 51.12, I believe the length of the maximal chain should not use the index n, where n is the number of maximal ideals of B (or otherwise one should remove the Length_{B_{m_i}}(M_{m_i}) which encodes the multiplicities which would then necessarily be 1).

Comment #5712 by Johan on November 15, 2020 at 22:52

OK, I tried to improve the numbering. Thanks. See changes.

Comment #11030 by thesnakefromthelemma on December 14, 2025 at 13:44

Consider the following lemma, a stronger form of Jordan-Holder for modules over commutative rings (but not in general; there are easy counterexamples) whereby every module of finite length (naturally) decomposes as a direct sum of modules having only one isomorphism class of simple module in their composition series. Is it worth including here?

Lemma ???4. Given ring $R$ and $R$ -module $M$ satisfying $\text{length} _ {R} (M) < \infty$ , there exists natural $k$ and (pairwise distinct) $k$ -tuple of maximal ideals $\left( \mathfrak{m} _ i \right) _ { i = 0} ^ {k-1}$ such that $M \cong \bigoplus _ { i = 0 } ^ {k-1} M _ { \mathfrak{m} _ i }.$

Here's one way to prove it, depending externally only on two basic facts (I don't know whether said basic facts are worth citing, and in the case that they are haven't checked where/if they occur in the Stacks Project; I've omitted their proofs in the interest of concision):

Basic fact ???0. Given ring $R$ , distinct maximal ideals $\mathfrak{m} _ 0 , \mathfrak{m} _ 1$ of $R$ , and naturals $d _ 0 , d _ 1$ , ${ \mathfrak{m} _ 0 } ^ { d _ 0 } + { \mathfrak{m} _ 1 } ^ { d _ 1 } = R.$

Basic fact ???1. Given ring $R$ , maximal ideals $\mathfrak{m}, \mathfrak{m}'$ of $R$ , and natural $d$ , $( R / \mathfrak{m} ^ d R ) _ {\mathfrak{m}' } \cong \begin{cases} &\text{if } \mathfrak{m} = \mathfrak{m}' &\text{ then } R / \mathfrak{m} ^ {d} R \newline &\text{if } \mathfrak{m} \neq \mathfrak{m}' &\text{ then } 0 \end{cases}.$

Lemma ???2: Given ring $R$ , $R$ -module $M$ , natural $k$ , pairwise distinct $k$ -tuple of maximal ideals $\left( \mathfrak{m} _ i \right) _ { i = 0} ^ {k-1}$ of $R$ , and $k$ -tuple of naturals $\left( d _ i \right) _ { i = 0 } ^ {k-1}$ satisfying $\left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) M = 0$ , $M \cong \bigoplus _ { i = 0 } ^ {k-1} M _ { \mathfrak{m} _ i }.$

Proof of Lemma ???2. Clearly $\left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) M = 0$ iff $M \simeq \left( R / \left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) R \right) \otimes _ R M$ . By Basic fact ???0, Lemma 00DT (lemma-chinese-remainder), and Basic fact ???1,

$M$

$\simeq \left( R / \left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) R \right) \otimes _ R M$

$\simeq \left( \bigoplus _ {i = 0} ^ {k-1} \left( R / \mathfrak{m} _ i ^ { d _ i } R \right) \right) \otimes _ R M$

$\simeq \left( \bigoplus _ {i = 0} ^ {k-1} \left( R / \left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) R \right) _ { \mathfrak{m} _ i } \right) \otimes _ R M$

$\simeq \bigoplus _ {i = 0} ^ {k-1} \left( \left( R / \left( \prod _ { i = 0 } ^ {k-1} \mathfrak{m} _ i ^ { d _ i } \right) R \right) \otimes _ R M \right) _ { \mathfrak{m} _ i }$

$\simeq \bigoplus _ {i = 0} ^ {k-1} M _ { \mathfrak{m} _ i }$

(I apologize for the bizarre formatting; I'm not sure whether align mode works in comments.) $\Box$

Lemma ???3: Given ring $R$ and $R$ -module $M$ satisfying $\text{length} _ {R} (M) < \infty$ , there exists natural $n$ and $n$ -tuple of maximal ideals $\left( \mathfrak{n} _ j \right) _ { j = 0 } ^ {n-1}$ of $R$ such that $\left( \prod _ { j = 0 } ^ {n-1} \mathfrak{n} _ j \right) M = 0.$

Proof of Lemma ???3. By hypothesis there exists a (finite) maximal filtration of $M$ , hence a natural $n$ , filtration $0 = M _ 0 \subset \dots \subset M _ n = M,$ $n$ -tuple of maximal ideals $\left( \mathfrak{n} _ j \right) _ { j = 0 } ^ {n-1}$ of $R$ , and $n$ -tuple of cokernel relationships $\left( M _ {j+1} / M _ j \simeq R / \mathfrak{n} _ j \right) _ { j = 0 } ^ {n-1}.$ In particular, given natural $j \in \{ 0 , \dots , n-1 \}$ , element $m \in M _ {j+1}$ , and element $f _ j \in \mathfrak{n} _ j$ it follows that $f _ j m \in M _ j$ . Hence (inductively from the right) for any element $m \in M$ and $n$ -tuple of elements $\left( f _ j \in \mathfrak{n} _ j \right) _ { j = 0 } ^ {n-1}$ we have $\left( \prod _ { j = 0 } ^ {n-1} f _ j \right) m = 0.$ The claim follows. $\Box$

Lemma ???4. Given ring $R$ and $R$ -module $M$ satisfying $\text{length} _ {R} (M) < \infty$ , there exists natural $k$ and (pairwise distinct) $k$ -tuple of maximal ideals $\left( \mathfrak{m} _ i \right) _ { i = 0} ^ {k-1}$ such that $M \cong \bigoplus _ { i = 0 } ^ {k-1} M _ { \mathfrak{m} _ i }.$

Proof of Lemma ???4. Immediate from Lemma ???2 and Lemma ???3. $\Box$

As mentioned above, Lemma ???4 is essentially a stronger form of Jordan-Holder for modules over commutative rings. It essentially says that modules of finite length over commutative rings are entirely max-local entities, carrying no nontrivial non-max-local information.

Note that current Lemma 00IW, lemma-length-infinite is just a weak form of (the contrapositive of) Lemma ???3 above. As the proof of Lemma ???3 is entirely self-contained, it should be feasible to replace the former with the latter.

We can also use Lemma ???4 to give an alternative proof of Lemma 00JA, lemma-product-local in 00J4, section-artinian as Lemma 00J8, lemma-artinian-radical-nilpotent amounts to saying that any Artinian ring $R$ is annihilated by a product of maximal ideals thereof.

(As an aside, the original motivation for Lemma ???4 is that it immediately implies that any collection of pairwise commuting matrices of (common) positive dimension over an algebraically closed field has a common eigenvector. But this elementary fact is almost surely outside the scope of the Stacks Project!)

If this proposal seem worth implementing in whole or in part at this time I'm happy to write up the corresponding patch and email it :)

10.52 Length

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