The Stacks project

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10.51 Length

Definition 10.51.1. 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 10.51.2. 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 10.51.3. 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 10.51.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $M$ be an $R$-module.

  1. If $M$ is a finite module and $\mathfrak m^ n M \not= 0$ for all $n\geq 0$, then $\text{length}_ R(M) = \infty $.

  2. 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 $f_1f_2 \ldots f_ n x \not= 0$. By Nakayama's Lemma 10.19.1 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. This can also be seen directly. 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, i.e., (1) holds. Combine (1) and Lemma 10.51.2 to see (2). $\square$

Lemma 10.51.5. 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 10.51.6. 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 10.51.5. 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 10.51.7. 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 10.51.8. 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 10.51.6 applies. We conclude by additivity, see Lemma 10.51.3. $\square$

Definition 10.51.9. 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 10.51.10. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is simple,

  2. $\text{length}_ R(M) = 1$, and

  3. $M \cong R/\mathfrak m$ for some maximal ideal $\mathfrak m \subset R$.

Proof. Let $\mathfrak m$ be a maximal ideal of $R$. By Lemma 10.51.6 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 10.51.11. 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

  1. $n = \text{length}_ R(M)$,

  2. each $M_ i/M_{i-1}$ is simple,

  3. each $M_ i/M_{i-1}$ is of the form $R/\mathfrak m_ i$ for some maximal ideal $\mathfrak m_ i$,

  4. 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 10.51.10 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 10.51.3, 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 10.51.12. 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

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

by $B$-submodules as in Lemma 10.51.11. Then each quotient $M_ i/M_{i - 1}$ is isomorphic to $\kappa (\mathfrak m_{j(i)})$ for some $j(i) \in \{ 1, \ldots , n\} $. Moreover $\text{length}_ A(\kappa (\mathfrak m_ i)) = [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m)]$ by Lemma 10.51.6. The lemma follows by additivity of lengths (Lemma 10.51.3). $\square$

Lemma 10.51.13. 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 10.38.17. 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 10.51.14. 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 10.51.13 applied to the ring map $B \to C$ and the $B$-module $M = B/\mathfrak m_ A B$ $\square$


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