The Length of a Module

Part IV Local Rings

12.1 The Length of a Module

The goal of the following definition is to measure the size of a module M over a ringR. In the theory of vector spaces, size is measured by the number of vectors in a basis. But in general, modules have no basis (if they do, they are called free). In view of this, we resort to the idea of considering chains of submodules.

Definition 12.1. Let M be a module over a ring R. The length of M, written as length(M), is the supremum of the lengthsn of chains

177

M₀$M₁$· · ·$M_n of submodules M_i⊆M. Solength(M)∈N0∪ {∞}.

Example 12.2. (1) For m a positive integer, the Z-moduleM =Z/(m) has length equal to the number of prime factors (with multiplicities) ofm.

(2) Zhas infinite length as a module over itself.

(3) IfR=K is a field andV a vector space, then length(V) = dimK(V).

(4) An R-moduleM is simple (i.e., M 6= {0} and there exist no non-zero, proper submodule) if and only if length(M) = 1. Moreover, M ={0} if

and only if length(M) = 0. /

Recall the definition of a maximal chain on page116. In particular, a finite chain M₀ $ M₁ $· · · $M_n of submodules of M is maximal if no further submodule can be added into the chain by insertion or by appending at either end. This means that M0 ={0}, Mn =M, and all Mi/M_i−1 (i= 1, . . . , n) are simple modules. A finite maximal chain of submodules ofM is also called a composition seriesofM.

Theorem 12.3(Basic facts about length). Let M be a module over a ring R.

(a) If M has a finite, maximal chain M0 $M1$· · ·$Mn of submodules, thenlength(M) =n. So in particular, all maximal chains have the same length.

(b) M has finite length if and only if it is Artinian and Noetherian. In par- ticular, R has finite length as a module over itself if and only if it is Artinian.

(c) LetN ⊆M be a submodule. Then

length(M) = length(N) + length(M/N).

Proof. (a) We use induction on n. If n = 0, then M = {0} and so length(M) = 0. Therefore we may assume n > 0. Let N $ M be a proper submodule, and setNi :=N∩Mi. TheNi need not be distinct, but the setC :={N0, . . . , Nn} is a chain of submodules ofN. We have N0 = {0} and Nn = N. Moreover, for 1 ≤ i ≤ n, the natural map Ni →Mi/Mi−1 induces an isomorphism from Ni/Ni−1 to a submodule ofMi/Mi−1, soNi/Ni−1is either simple or zero. ThereforeCis a maximal chain of submodules ofN. Clearly length(C)≤n. By way of contradic- tion, assume that length(C) =n. We will show by induction onithat this impliesN_i =M_i for all i. This is true fori = 0. For i > 0,N_i/N_i−1 is non-zero by assumption, so the mapN_i→M_i/M_i−1is surjective. Using induction, we concludeM_i⊆N_i+M_i−1=N_i+N_i−1=N_i⊆M_i, proving our claim. In particular, we obtainN =N_n=M_n =M, a contradiction.

So length(C)< n. By induction onn, this implies length(N)< n. Since this holds for all proper submodulesN, we conclude length(M)≤n. On

12.1 The Length of a Module 179 the other hand, we are given a chain of submodules ofM of lengthn, so part (a) follows.

(b) If is clear from the definition that a module of finite length has to be Artinian and Noetherian. Conversely, ifM is Noetherian andM 6={0}, there exists a maximal proper submodule M₁ $ M. If M₁ 6= {0}, we can continue and findM2$M1 with no submodules in between, and so on. IfM is Artinian, this process stops, so we end up with a finite chain {0} =Mk $M_k−1$· · · $M1 $M0 :=M. This chain is maximal by construction, so length(M) =k by part (a). The last statement follows since every Artinian ring is Noetherian by Theorem2.8.

(c) If length(N) =∞or length(M/N) =∞, then by part (b) at least one of these modules fails to be Noetherian or Artinian, so by Proposition 2.4 the same is true forM, and length(M) =∞, too. So we may assume that N and M/N have finite length. Taking maximal chains of submodules of N and of M/N, lifting the latter into M, and putting these chains together yields a maximal chain of submodules of M. So (c) follows by

part (a). ut

By applying Theorem 12.3(c) several times, we see that if {0} = M₀ ⊆ M₁⊆ · · · ⊆M_k =M is a chain of submodules, then

length(M) =

k

X

i=1

length (Mi/M_i−1). (12.1) A special case of Theorem12.3(c) says that a direct sumM⊕N ofR-modules has length(M ⊕N) = length(M) + length(N). With Theorem12.3(b), this shows that the free moduleM =Rⁿover an Artinian ringRhas finite length.

Since any finitely generated module is a factor module of some Rⁿ, another application of Theorem12.3(c) shows that a finitely generated module over an Artinian ring has finite length. Exercise12.2deals with a further consequence of Theorem12.3(c) on exact sequences of modules.

An important example where Theorem 12.3(b) applies is the following.

If R is a Noetherian local ring with maximal ideal m and q ⊆ R is an ideal with √

q=m, thenR/q has dimension 0. Therefore it is Artinian and Noetherian (as a ring and therefore also as an R-module) by Theorem2.8, so length (R/q)<∞. In particular, this applies to q = m^d, a power of the maximal ideal. The following result is about the lengths of the modulesR/m^d. Lemma 12.4. LetR be a Noetherian local ring with maximal idealm. Then there exists a polynomial p∈Q[x]of degreen= dim(R)such that

length R/m^d

≤p(d) for all d∈N0.

Proof. The main idea is to use a system of parametersa1, . . . , an∈m, whose existence is guaranteed by Corollary 7.9. Setting q:= (a₁, . . . , a_n), we have q^d ⊆ m^d for all d ∈ N0, so there is an epimorphism R/q^d → R/m^d. By

Theorem12.3(c), this implies length R/m^d

≤length R/q^d

. (12.2)

Consider the chain

{0}=q^d/q^d⊆q^d−1/q^d⊆ · · · ⊆q²/q^d⊆q/q^d⊆R/q^d,

which has factors qⁱ/qⁱ⁺¹. For every i, qⁱ is generated (as anR-module) by the monomials of degree i in a1, . . . , an. By Remark 11.5, there are ki :=

i+n n

− ⁱ⁻¹⁺ⁿ_n

such monomials. Soqⁱ/qⁱ⁺¹can be generated byki elements as an R/q-module, giving an epimorphism (R/q)^ki →qⁱ/qⁱ⁺¹. Using (12.2), (12.1) and Theorem12.3(c), we conclude that

length R/m^d

≤length R/q^d

=

d−1

X

i=0

length qⁱ/qⁱ⁺¹

≤

d−1

X

i=0

length

(R/q)^ki

=

d−1

X

i=0

ki·length (R/q) =

d−1 +n n

·length (R/q). By Corollary 7.9 we have √

q =m, so length (R/q) <∞ by the discussion preceding the lemma. Therefore the above inequality yields the desired upper

bound. ut

In Chapter 11, we have found that the Hilbert function of an ideal is essentially a polynomial function. Somewhat similarly, Lemma 12.4 relates the functiond7→length R/m^d

to a polynomial. Is there a connection? The answer is yes. In fact, we will interpret the functiond7→length R/m^d+1

as the Hilbert function of the associated graded ring, to be defined in the next section.