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
10.52 Length
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).
Lemma 10.52.2.slogan 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
Lemma 10.52.3.slogan 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
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
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
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
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
Let M be a B-module of finite length. Then
in particular \text{length}_ A(M) < \infty .
Proof. Choose a maximal chain
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
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
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 (2)
Comment #5519 by Alex on
Comment #5712 by Johan on