# THE LENGTH OF NOETHERIAN MODULES

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1 THE LENGTH OF NOETHERIAN MODULES GARY BROOKFIELD Abstract. We define an ordinal valued length for Noetherian modules which extends the usual definition of composition series length for finite length modules. Though originally defined by Gulliksen [1] in the 1970s, this extension has been seldom used in subsequent research. Despite this neglect, we will show that the ordinal valued length is a quite natural measure of the size of a Noetherian module, and has advantages over more familiar measures such as uniform dimension, Krull dimension, and reduced rank. We will also demonstrate how some familiar properties of left Noetherian rings can be proved efficiently using length and the arithmetic properties of ordinal numbers. 1. Introduction In the early 1970s, T. H. Gulliksen [1] showed how the definition of composition series length, defined only for finite length modules, could be extended to give an ordinal valued length for any Noetherian module. During the same time period, various people (e.g. [2], [3]) defined another ordinal valued measure for modules, the Krull dimension. This second line of research culminated in the article Krull Dimension by R. Gordon and J. C. Robson [4]. Since that time, the paper by Gulliksen has been rarely cited ([5], [6], [7], [8], [9]), whereas the Gordon and Robson article has been cited over 175 times. It is the main purpose of this paper to point out that, contrary to what one might expect from the above, for a Noetherian module B, its ordinal valued length, len B, is a more natural measure of its size than its Krull dimension, Kdim B. Moreover len B contains more information about the size of B. Length and Krull dimension are really measures of the size of the lattice L(B) of all submodules of B ordered by inclusion. We will write L (B) for the set of submodules of B ordered by reverse inclusion, that is, the dual of L(B). A module B is Noetherian if and only if L(B) is Noetherian if and only if L (B) is Artinian. 1

2 2 GARY BROOKFIELD Suppose B is a Noetherian uniserial module, meaning that L(B) is Noetherian and totally ordered. Then L (B) is a well ordered set with maximum element 0. Following the convention of counting the gaps rather than the modules in a finite chain, we define the length of B, len B, to be the ordinal represented by L (B) \ {0}. Using this definition and the arithmetic of ordinal numbers we can then prove various properties of uniserial modules. For a simple example, we notice that if 0 A B C 0 is exact, then len B = len C + len A. Consider the case when len B = len C. Here len B = len B + len A, and, since ordinal addition is cancellative on the left, we get len A = 0 and A = 0. Expressed differently, this says that a homomorphism ψ on B is injective if (and only if) len B = len ψ(b). As a special case, any surjective endomorphism of B is injective. The definition of length in this paper extends the above definition for uniserial modules to all Noetherian modules. It is natural because there is really only one possible way of making this extension: In short, for a Noetherian module B, we define len B = λ(0) where λ is the smallest possible strictly decreasing function from L(B) into the class of ordinal numbers. This is equivalent to Gulliksen s original definition. The function λ can also be defined inductively as follows: First set λ(b) = 0. Suppose, for an ordinal α, we have already identified those submodules B of B such that λ(b ) < α. Then λ(b ) = α if and only if B is maximal among those submodules of B on which λ has not yet been defined. Once again ordinal arithmetic comes into play. We will show (4.1) that if 0 A B C 0 is an exact sequence of Noetherian modules then len C + len A len B len C len A. Here is the natural sum on ordinals (2.7). We have already noted that if B is uniserial then len C + len A = len B. The other extreme case occurs if the sequence splits: If B = A C, then len B = len C len A. Since is a cancellative operation (2.8) on ordinals, we have immediately that A C = B C implies that len A = len B for Noetherian modules A, B and C. The relationship between len B and Kdim B is a simple one: If B is nonzero, then the length of B can be written uniquely in the long normal form len B = ω γ1 + ω γ2 + + ω γn where γ 1 γ 2 γ n are ordinals. Then Kdim B = γ 1. In fact, the (finite number of) possible values of len B for a submodule B B, are determined by len B. In particular, we have Kdim B { 1, γ 1, γ 2,..., γ n } (4.6). Thus len B contains a lot more information about B than Kdim B.

3 THE LENGTH OF NOETHERIAN MODULES 3 Much can be proved about Noetherian modules using only ordinal arithmetic and the above rule about short exact sequences. For more complicated theorems, the existence of the long normal form for ordinals permits proofs which are finite inductions on the number of terms in such a form. In the final section of this paper we demonstrate this technique in proving some familiar properties of left Noetherian rings. 2. The Length of a Partially Ordered Set As indicated in the introduction, both the length and Krull dimension of a Noetherian module B, are measures of the size of the lattice of submodules of B. Thus it is convenient to define these concepts first for lattices, or indeed for partially ordered sets. The key concept in this section is the length function on a partially ordered set L. This is a certain function from L into the class of ordinal numbers, Ord. We will find it convenient at first to define length functions, not just on partially ordered sets, but also on partially ordered classes such as Ord itself and Ord Ord. Thus we phrase our definitions below in this generality. Note that Ord only barely fails to be a set in the sense that, for any ordinal α, the class of elements of Ord which are less than α is a set. If L is a partially ordered class and x, y L, we will use the following notation: { x} = {z L z x} [x, y] = {z L x z y}. If ψ: K L is a function between partially ordered classes, then ψ is increasing if x y in K implies ψ(x) ψ(y) in L. ψ is an isomorphism (and K = L) if ψ is a bijection such that ψ and ψ 1 are increasing. Note that an increasing bijection may not be an isomorphism. ψ is strictly increasing if x < y in K implies ψ(x) < ψ(y) in L. ψ is exact if { ψ(x)} ψ({ x}) for all x K. One can easily check that if ψ 1 : L M and ψ 2 : K L are increasing (strictly increasing, exact) functions, then so is ψ 1 ψ 2 : K M. We will write L 1 L 2 for the Cartesian product of two partially ordered classes L 1 and L 2, with order given by (x 1, x 2 ) (y 1, y 2 ) (x 1 y 1 and x 2 y 2 ). Notice that { (x 1, x 2 )} = { x 1 } { x 2 }. It is easy to show that L 1 L 2 = L2 L 1 and (L 1 L 2 ) L 3 = L1 (L 2 L 3 ).

4 4 GARY BROOKFIELD If ψ 1 : K 1 L 1 and ψ 2 : K 2 L 2 are maps between partially ordered classes, then we will write ψ 1 ψ 2 for the map from K 1 K 2 to L 1 L 2 given by (ψ 1 ψ 2 )(x 1, x 2 ) = (ψ 1 (x 1 ), ψ 2 (x 2 )). One can easily check that if ψ 1 and ψ 2 are increasing (strictly increasing, exact), then so is ψ 1 ψ 2. A partially ordered class L is Artinian (Noetherian) if every nonempty subclass has a minimal (maximal) element (equivalently, every strictly decreasing (increasing) sequence in L is finite.) If L 1 and L 2 are Artinian (Noetherian) partially ordered classes, then so is L 1 L 2. Other notation: N = {1, 2, 3,... } is the set of natural numbers, and Z + = {0, 1, 2, 3,... } is the set of nonnegative integers. The theorems in this paper depend heavily on the arithmetic of the ordinal numbers. For the details of ordinal arithmetic see W. Sierpinski, Cardinal and Ordinal Numbers [10] or M. D. Potter, Sets, An Introduction [11]. We collect here a few of those facts that are relevant: We will use lowercase Greek letters for ordinal numbers. The smallest infinite ordinal is written ω. Ordinal addition is associative but not commutative. For example, ω ω = ω. Ordinal addition is cancellative on the left: α + β = α + γ = β = γ. Also α + β α + γ = β γ. For a fixed ordinal α, the map from Ord to Ord given by β α+β is strictly increasing. If α β, then β α is the unique ordinal γ such that β = α + γ, hence β = α+(β α). For any α, β Ord, we have β = (α+β) α. For a fixed ordinal α, the map from {β Ord α β} to Ord given by β β α is strictly increasing. αn = α + α α (n times) when n N. Note: 3ω = ω ω3. The most important property of ordinal numbers is that any nonzero ordinal α can be expressed uniquely in long normal form α = ω γ1 + ω γ2 + + ω γn where γ 1 γ 2 γ n are ordinals. By collecting together terms which have identical exponents, this same form can be written α = ω γ1 n 1 + ω γ2 n ω γn n n where now γ 1 > γ 2 > > γ n and n 1, n 2,..., n n N. This we will call the short normal form for α. Certain parameters in these normal forms will have an important role in our discussion of Noetherian modules in later sections:

5 THE LENGTH OF NOETHERIAN MODULES 5 Definition 2.1. For a nonzero ordinal α = ω γ1 n 1 + ω γ2 n ω γn n n in short normal form we define the Krull dimension of α by Kdim α = γ 1 and the Krull rank of α by Krank α = n i=1 n i. For i = 1, 2,..., n, the number n i will be called the γ i -length of α, written len γi α. For an ordinal γ not in {γ 1, γ 2,..., γ n } we define len γ α = 0. By convention Kdim 0 = 1, Krank 0 = 0, and len γ 0 = 0. If α = ω γ1 + ω γ2 + + ω γn in long normal form, then Kdim α = γ 1 and Krank α = n. Most of the arithmetic properties of ordinals we will need are consequences of the fact that to add two ordinals in normal form one needs only the associativity of addition and the rule that ω γ + ω δ = ω δ if γ < δ. For example, (ω ω + ω 3 + ω2 + 1) + (ω 3 + ω) = ω ω + ω ω. Using this rule one can readily prove the following: Lemma 2.2. Suppose α, β, γ Ord with α > 0 and m, n Z +. (1) β + α α β + α = α Kdim β < Kdim α (2) α = ω γ for some γ Ord β + α = α for all β < α Kdim β < Kdim α for all β < α (3) β + ω γ n < ω γ m = β < ω γ (m n) The < symbols are necessary in 3: If β = γ = m = n = 1, then β + ω γ n ω γ m but β ω γ (m n). Now suppose we have a partially ordered class L. We would like to say something about the size of L by considering certain order preserving functions from L into Ord. Considering even the case when L is finite and totally ordered, it is easy to list some properties that such a function λ: L Ord should have. For example, we would certainly want λ to be strictly increasing. We would not want λ to skip any ordinals unnecessarily, that is, if α < λ(x) for some x L, then there ought to be some y < x such that λ(y) = α. Thus λ should be exact. The key result of this section is that these two conditions suffice to specify a unique order preserving function from L into Ord. Theorem 2.3. Let L be a partially ordered class. If there exists at least one strictly increasing function from L to Ord then L is Artinian and there exists a unique function λ L : L Ord satisfying the following equivalent conditions: (1) λ L is strictly increasing, and, if λ: L Ord is a strictly increasing function, then λ L (x) λ(x) for all x L. (2) For all x L and α Ord, λ L (x) = α if and only if x is minimal in K α = {y L λ L (y) α}.

6 6 GARY BROOKFIELD (3) λ L is strictly increasing and exact. Proof. Any strictly increasing function from L to Ord, maps infinite strictly decreasing sequences in L to infinite strictly decreasing sequences in Ord. Since no such sequences exist in Ord, there are no infinite strictly decreasing sequences in L either. Define λ L (x) = min{λ(x) λ: L Ord is strictly increasing} for all x L. Since we are assuming that at least one strictly increasing function exists, λ L is well defined by this equation. If x < y in L, then there is some strictly increasing function λ: L Ord such that λ(y) = λ L (y), so λ L (x) λ(x) < λ(y) = λ L (y). Thus λ L is itself strictly increasing and satisfies condition 1. The uniqueness of λ L is immediate from condition 1. We now show the equivalence of 1, 2 and 3: 1 2: Suppose λ L (x) = α. Then x K α. Since λ L is strictly increasing, for any y < x, we have λ L (y) < λ L (x) = α and so y / K α. Hence x is minimal in K α. Now suppose x is minimal in K α. In particular α λ L (x). Define λ: L Ord by { λ L (y) y x λ(y) = α y = x It is easy to check that λ is strictly increasing. From our hypothesis on λ L we get λ L (x) λ(x) = α, and so λ L (x) = α. 2 3: Suppose y < x in L and λ L (x) = α. Since x is minimal in K α we have y / K α and hence λ L (y) < α = λ L (x). Thus λ L is strictly increasing. For exactness we need to show { λ L (y)} λ L ({ y}) for all y L. Suppose then we have α λ L (y). Then y K α, and, since L is Artinian, there is some x y which is minimal in K α. By hypothesis, λ L (x) = α as desired. 3 1: Let λ: L Ord be strictly increasing. We will show that λ L (x) λ(x) for all x L. Suppose to the contrary that K = {x L λ(x) < λ L (x)} is nonempty. Let x be chosen in K so that λ(x) is minimum in λ(k). For any y < x we have λ(y) < λ(x), so y / K and λ L (y) λ(y) < λ(x) < λ L (x). Thus we have an ordinal α = λ(x) such that α < λ L (x) but there is no y < x with λ L (y) = α. This contradicts exactness. Definition 2.4. The function λ L : L Ord (when it exists) will be called the length function on L. If L has a maximum element, then we define the length of L by len L = λ L ( ). In addition, the Krull dimension,

7 THE LENGTH OF NOETHERIAN MODULES 7 Krull rank and γ-length of L are defined by Kdim L = Kdim(len L), Krank L = Krank(len L) and len γ L = len γ (len L). Theorem 2.3(2) suggests that we could define λ L inductively using the relationship λ L (x) = α if and only if x is minimal in K α = {y L λ L (y) α}. Notice in particular that, from this definition, λ L (x) = 0 if and only if x is minimal in L. Certainly, if L has a length function then, from 2.3, this definition produces the length function of L. If L is not known to have a length function then this definition will produce the length function of some (possibly empty) subclass of L. Theorem 2.5. Let L be an Artinian partially ordered class. If { x} is a set for all x L, then L has a length function. Proof. Let λ L be defined inductively as above. Suppose x L is minimal among elements for which λ L is undefined. Then for every z < x, λ L (z) is defined. Let α = sup{λ L (z) + 1 z < x}. This is well defined since any subset of Ord has a supremum [12, Section 20]. It is then easy to show that x is minimal in K α = {y L λ L (y) α} and so λ L (x) = α. This contradicts our assumption that λ L (x) is undefined. Consequently, λ L is defined on all of L, and then by 2.3, λ L is the length function of L. From the proof of this theorem we notice that for any x L, λ L (x) = sup{λ L (z) + 1 z < x}. This equation could also be used inductively to define λ L. In fact, this formula, written in terms of the lattice of submodules of a Noetherian module, is part of Gulliksen s original definition of the length of a Noetherian module. Lemma 2.6. Let L and K be partially ordered classes with length functions and maximum elements. (1) For all x L, len{ x} = λ L (x). (2) For all x y L, len{ x} + len[x, y] len{ y}. (3) For any ordinal α len L, there is some x L such that λ L (x) = α. (4) If λ : L K is a strictly increasing function, then len L len K. Proof. (1) It is easy to see that the restriction of λ L to { x} is strictly increasing and exact, so by 2.3, λ L = λ { x} on { x}. In particular, λ L (x) = λ { x} (x) = len{ x}.

8 8 GARY BROOKFIELD (2) Define λ: [x, y] Ord by λ(z) = λ L (z) λ L (x). The function λ is strictly increasing, so len[x, y] λ(y) = λ L (y) λ L (x) = len{ y} len{ x}. Hence len{ x} + len[x, y] len{ y}. (3) This follows immediately from the exactness of λ L. (4) The function λ K λ: L Ord is strictly increasing, so, from 2.3(1), λ L (x) λ K (λ(x)) for all x L. In particular, len L = λ L ( ) λ K (λ( )) λ K ( ) = len K. The identity map on Ord is strictly increasing and exact, so we have λ Ord (α) = α for all ordinals α. In addition, len{ α} = α and len[α, β] = β α for all ordinals α β, as can be easily checked. Now consider Ord Ord. This partially ordered class is Artinian and for any (α, β) Ord Ord we have that { (α, β)} = { α} { β} is a set. Thus Ord Ord has a length function λ Ord Ord which we can use to define a new operation on ordinals: Definition 2.7. Define the natural sum of ordinals α and β by α β = λ Ord Ord (α, β). Note that, from 2.6(1), α β = len{ (α, β)} = len({ α} { β}). The natural sum of ordinals was originally defined by G. Hessenberg [13, pages ] as in Definition 2.11 (see also [10, page 363]). In 2.12, we will show that these two definitions for the natural sum are equivalent. Lemma 2.8. The operation is commutative, associative and, for all α, β, γ Ord, (1) α β = α γ = β = γ (2) α β α γ = β γ Proof. Since the Cartesian product operation on partially ordered classes is commutative and associative, is a commutative and associative operation on ordinals. The cancellation properties follow easily from the fact that, for a fixed α Ord, the map β (α, β) λ Ord Ord (α, β) = α β is strictly increasing and hence injective. Notice that, being commutative, is cancellative on both sides, unlike ordinary ordinal addition. Note also that for any α Ord we have 0 α = α 0 = len({ α} {0}) = len{ α} = α. The importance of this operation is already apparent in the following easy theorem.

9 THE LENGTH OF NOETHERIAN MODULES 9 Theorem 2.9. Let K and L be partially ordered classes with length functions. Then λ K L (x, y) = λ K (x) λ L (y) for all (x, y) K L. In particular, len(k L) = len K len L if L and K have maximum elements. Proof. The map λ: K L Ord defined by λ(x, y) = λ K (x) λ L (y) is the composition of the strictly increasing exact functions λ K λ L and λ Ord Ord, and so is itself strictly increasing and exact. Hence, from 2.3, we get λ = λ K L. It turns out that the natural sum of two ordinals can be calculated very easily from their normal forms. To show this we need the following lemma. Lemma Let α, β, α 1, β 1,..., α n, β n Ord. (1) α + β α β (2) (α 1 β 1 )+ +(α n β n ) (α 1 +α 2 + +α n ) (β 1 +β 2 + +β n ) (3) α 1 + β α n + β n (α 1 + α α n ) (β 1 + β β n ) Proof. For claim 2, set x 0 = (0, 0), x 1 = (α 1, β 1 ), x 2 = (α 1 + α 2, β 1 + β 2 ),..., x n = (α 1 + α α n, β 1 + β β n ) in Ord Ord. Then for i = 1, 2,..., n we have [x i 1, x i ] = { α i } { β i } so len[x i 1, x i ] = α i β i. Applying 2.6(2) inductively to the sequence x 0 x 1... x n yields len[x 0, x 1 ] + len[x 1, x 2 ] len[x n 1, x n ] len[x 0, x n ]. Rewriting this in terms of α i and β i yields 2. Claim 1 is a special case of 2: α+β = (α 0)+(0 β) (α+0) (0+β) = α β. Then, since α i + β i α i β i, for i = 1, 2,..., n, the inequality 3 is immediate from 2. Consider now the natural sum of two ordinals which are given in short normal form, for example, α = ω ω + ω 3 + ω2 + 1 and β = ω 3 + ω. Using 2.10(3), we can interleave the terms of these two normal forms in various ways and add them to get lower bounds for α β. There is a unique way of doing this so that no terms are lost, namely: Write down the terms gathered from both the short normal forms in decreasing order and then add. In the example, we have the six terms ω ω, ω 3, ω 3, ω2, ω, 1 so, from the lemma, ω ω + ω 3 + ω 3 + ω2 + ω + 1 = ω ω + ω ω3 + 1 α β. We will show that this method actually gives us the natural sum of α and β, not just a lower bound for it, but first we need to formalize this construction:

10 10 GARY BROOKFIELD Definition Let α and β be nonzero ordinals. With suitable relabeling, the short normal forms for these ordinals can be written using the same strictly decreasing set of exponents γ 1 > γ 2 > > γ n : α = ω γ1 m 1 + ω γ2 m ω γn m n where n i, m i Z +, that is, we allow m i, n i to be zero. Now we define the operation by β = ω γ1 n 1 + ω γ2 n ω γn n n α β = ω γ1 (m 1 + n 1 ) + ω γ2 (m 2 + n 2 ) + + ω γn (m n + n n ). This is a well defined operation because of the uniqueness of the normal forms for ordinals. In addition, we define 0 α = α 0 = α, and 0 0 = 0. Theorem The operations and are identical. Proof. From the discussion following 2.10 we have α β α β for any ordinals α and β. To show the opposite inequality we need only show that : Ord Ord Ord is strictly increasing: Since is commutative and increasing, it suffices to show only that, for all α, β Ord, we have α (β + 1) > α β. But this follows easily from the definition of, in fact, α (β + 1) = (α β) + 1. It is apparent from this theorem and 2.11 that, if α and β are finite ordinals, then α β = α + β and hence both ordinal addition and coincide with the usual addition of integers. Notice also that Krank(α β) = Krank α + Krank β for any ordinals α, β Ord, so that, if K and L are partially ordered classes with length functions and maximum elements, then Krank(L K) = Krank L+Krank K. 3. The Length of a Bounded Artinian Modular Lattice We will now specialize to the case of length functions on bounded Artinian modular lattices. In this section we will no longer need to consider proper classes, and so we define a lattice to be a partially ordered set L such that every pair of elements, x, y L, has a supremum, x y, and an infimum, x y. A bounded lattice is a lattice which has a maximum element and a minimum element. A lattice L is modular if (x 1 x 2 = (x 1 y) x 2 = x 1 (y x 2 ) ) for all x 1, x 2, y L.

11 THE LENGTH OF NOETHERIAN MODULES 11 Lemma 3.1. Let L be a modular lattice and x, y L. (1) The map λ: L L L L given by (x, y) (x y, x y), is strictly increasing. (2) The maps φ: [x, y x] [y x, y] given by z z y, and ψ: [y x, y] [x, y x] given by z z x are inverse isomorphisms. Proof. (1) Suppose that (x 1, y 1 ) (x 2, y 2 ) with λ(x 1, y 1 ) = λ(x 2, y 2 ). Then x 1 x 2, y 1 y 2, x 1 y 1 = x 2 y 2 and x 1 y 1 = x 2 y 2. Using modularity we get x 2 = (x 2 y 2 ) x 2 = (x 1 y 1 ) x 2 = x 1 (y 1 x 2 ) x 1 (y 2 x 2 ) = x 1 (y 1 x 1 ) = x 1. Hence x 1 = x 2, and by symmetry, y 1 = y 2. Thus (x 1, y 1 ) = (x 2, y 2 ). Now suppose (x 1, y 1 ) < (x 2, y 2 ). Since λ is an increasing function, we have λ(x 1, y 1 ) λ(x 2, y 2 ). From the above argument, λ(x 1, y 1 ) = λ(x 2, y 2 ) is impossible, and so we must have λ(x 1, y 1 ) < λ(x 2, y 2 ). (2) [14, Theorem 13, page 13] The functions ψ and φ are clearly increasing. If z [x, y x], then ψ(φ(z)) = (z y) x = (y x) z = z. The second equality comes from applying the modularity of L to the inequality x z. Thus ψ φ is the identity map on [x, y x], and similarly φ ψ is the identity map on [y x, y]. From 2.5 we know that a bounded Artinian modular lattice L has a length function. The main property of the length function in this circumstance is that if x L then len[, x] len[x, ] is an upper bound for len L. Without the hypothesis that L is a modular lattice, we know only a lower bound, namely len[, x] + len[x, ]. Theorem lattice L Let x and y be elements of a bounded Artinian modular (1) len[, x] + len[x, ] len L len[, x] len[x, ] (2) len[, x y] + len[, x y] len[, x] len[, y] len[, x y] len[, x y] (3) len[x y, ] + len[x y, ] len[x, ] len[y, ] len[x y, ] len[x y, ] Proof. (1) The first inequality is directly from 2.6(2). To prove the second inequality, consider the restriction of the map λ from 3.1(1) to the domain L {x}. This map is strictly increasing and its image

12 12 GARY BROOKFIELD is contained in [, x] [x, ]. From 2.6(4) we get len L = len(l {x}) len([, x] [x, ]) = len[, x] len[x, ]. (2) To prove the first inequality we apply 1 to the lattices [, x y] and [, y]. This yields len[, x y] len[, x] len[x, x y] and len[, x y] + len[x y, y] len[, y] respectively. From 3.1(2) we also have len[x, x y] = len[x y, y]. Hence len[, x y] + len[, x y] len[, x y] + (len[, x] len[x, x y]) = len[, x y] + (len[, x] len[x y, y]) len[, x] (len[, x y] + len[x y, y]) len[, x] len[, y] We have also used the fact that α 1 + (α 2 β 2 ) (α 1 + α 2 ) β 2 which follows from 2.10(2). To prove the second inequality, consider the restriction of the map λ from 3.1(1) to the domain [, x] [, y]. This map is strictly increasing and its image is contained in [, x y] [, x y], and so, from 2.6(4) we get len[, x] len[, y] len[, x y] len[, x y]. (3) Proof is similar to that of 2. Note that, if len L is finite, then ordinal addition and coincide and all the inequalities in this theorem become equalities. One important corollary of this theorem follows from the observation that for any nonzero ordinals α and β, α+β and α β have the same leading term in their short normal forms. Further, this leading term depends only on the leading terms of α and β. For example, if α = ω ω + ω 3 + ω2 + 1 and β = ω 3 + ω, then α + β = ω ω + ω ω and α β = ω ω + ω ω3 + 1, both having the leading term ω ω. Formulating this observation in terms of Krull dimension and γ-length we have, for example, from 3.2(1) Lemma 3.3. Let L be a bounded Artinian modular lattice, γ = Kdim L and x L. (1) Kdim L = max{kdim[, x], Kdim[x, ]} (2) len γ L = len γ [, x] + len γ [x, ] We leave it to the reader to write down the corresponding lemma derived from 3.2(2 and 3). Theorem 3.2 directs our attention to pairs of ordinals α and β such that α + β = α β. Some easy ordinal arithmetic shows when this happens:

13 THE LENGTH OF NOETHERIAN MODULES 13 Lemma 3.4. Suppose α + β = α β = ω γ1 + ω γ2 + + ω γn in long normal form. Then α = 0 or β = 0, or there is some i {1, 2,..., n 1} such that α = ω γ1 + ω γ2 + + ω γi and β = ω γi+1 + ω γi ω γn. Lemma 3.5. Let L be a bounded Artinian modular lattice. Suppose we have α, β Ord such that α + β = α β. (1) If len L = α + β, then there is some x L such that len[, x] = α and len[x, ] = β. (2) If there is x L such that len[, x] = α and len[x, ] = β, then len L = α + β. Proof. (1) From 2.6(3), there is some x L such that len[, x] = α. From 3.2(1), α + len[x, ] α + β = α β α len[x, ]. Cancellation in the first inequality gives len[x, ] β. Cancellation in the second inequality gives β len[x, ]. Hence len[x, ] = β (2) This follows directly from 3.2(1). For example, if len L = ω ω + ω , then the previous two lemmas guarantee the existence of an x L such that len[x, ] is any of the following ordinals: 0, 1, ω 3 + 1, ω , ω ω + ω Lemma 3.6. Let L be a bounded Artinian modular lattice with len L = 0 (that is, L is nontrivial). Then the following are equivalent: (1) len L = ω γ for some γ Ord (2) len[x, ] = len L for all x < (3) Kdim[, x] < Kdim L for all x < Proof. Put α = len L. Since, by 2.6(3), for any β < α there is some x in L such that len[, x] = β, the claim is an easy consequence of 3.2(1), 3.3 and 2.2(2). Definition 3.7. A nontrivial bounded Artinian modular lattice L is critical if it satisfies any of the conditions of the previous lemma. More specifically, we will say L is γ-critical if len L = ω γ. A critical series for a bounded Artinian modular lattice L, is a sequence = z 0 < z 1 < < z n = in L such that [z i 1, z i ] is γ i -critical for all i, and γ 1 γ 2 γ n. If len L = ω γ1 + ω γ2 + + ω γn in long normal form, then from 3.4 and 3.5, there is an element z L such that len[z, ] = ω γn and len[, z] = ω γ1 +ω γ2 + +ω γn 1. In particular, [z, ] is γ n -critical. A simple induction

14 14 GARY BROOKFIELD then shows that any nontrivial bounded Artinian modular lattice has a critical series:

15 THE LENGTH OF NOETHERIAN MODULES 15 Lemma 3.8. Let L be a bounded Artinian modular lattice. Then the following are equivalent (1) len L = ω γ1 + ω γ2 + + ω γn in long normal form. (2) L has a critical series = z 0 < z 1 < < z n = with [z i 1, z i ] γ i -critical for i = 1, 2,..., n. Given a bounded Artinian modular lattice L with a critical series, the next theorem shows how to construct a critical series for the sublattice [x, ] for any element x L. Theorem 3.9. Let L be a bounded Artinian modular lattice with the critical series = z 0 < z 1 < < z n = with [z i 1, z i ] γ i -critical for i = 1, 2,..., n. Let x L and set x i = x z i for i = 0, 1, 2,..., n. Then for i = 1, 2,..., n, len[x i 1, x i ] is either zero or ω γi. Further, the sequence x = x 0 x 2 x n =, after removal of duplicate entries, is a critical series for [x, ]. Proof. From 3.1(2) we get [x i 1, z i x i 1 ] = [z i x i 1, z i ] for i = 1, 2,..., n. Since z i x i 1 = z i (z i 1 x) = z i x = x i, and, using the modularity of the lattice, z i x i 1 = z i (z i 1 x) = z i 1 (z i x), we get [x i 1, x i ] = [z i 1 (z i x), z i ]. We also have z i 1 z i 1 (z i x) z i, and so [x i 1, x i ] is isomorphic to a final segment of [z i 1, z i ]. Because [z i 1, z i ] is γ i -critical, 3.6(2) applies and either x i 1 = x i or len[x i 1, x i ] = ω γi. The claim that x = x 0 x 2 x n =, after removal of duplicate entries, is a critical series for [x, ] is then clear. From this theorem we see that the factors in a critical series for [x, ] have lengths which are among the lengths of the factors in a critical series for L. Combining this with 3.8 we get Corollary Let L be a bounded Artinian modular lattice with len L = ω γ1 n 1 + ω γ2 n ω γn n n in short normal form. Then for x L, len[x, ] = ω γ1 m 1 + ω γ2 m ω γn m n for some m i Z + such that m i n i for all i. In particular, (1) Krank[x, ] Krank L with equality if and only if len[x, ] = len L. (2) Kdim[x, ] { 1, γ 1, γ 2,..., γ n }. Continuing with the example len L = ω ω + ω , if x L, then len[x, ] is one of the following ordinals: 0, 1, ω 3, ω 3 + 1, ω 3 2, ω , ω ω, ω ω + 1, ω ω + ω 3, ω ω + ω 3 + 1, ω ω + ω 3 2, ω ω + ω

16 16 GARY BROOKFIELD This result is to be contrasted with the possible values of len[, x] which, by 2.6(3), include all ordinals less than ω ω + ω The mere fact that len[x, ] can take on only a finite number of different values is significant. As an application of this we prove a simple property of complemented lattices: A bounded lattice L is complemented if for every x L there is some y L, called a complement of x, such that x y = and x y =. Corollary [15, 0.4] Any complemented Artinian modular lattice has finite length. Proof. Let L be such a lattice. Then for any α len L, there is some x L such that len[, x] = α. From 3.1(2) we have len[, x] = len[y, ] where y is a complement of x. But by 3.10, there are only a finite number of possible values for len[y, ]. Thus len L is also finite. 4. The Length of a Noetherian Module In this section we apply the results of our study of length functions to Noetherian modules. Throughout this section, R will be a fixed ring and R-Noeth the category of Noetherian left R-modules. If A R-Noeth, we will write L(A) for the set of submodules of A ordered by set inclusion, and L (A) for the set of submodules of A ordered by reverse set inclusion, that is, the dual of L(A). Since A is Noetherian, L (A) is Artinian. In addition, both L(A) and L (A) are bounded modular lattices. In particular, in L (A), we have A 1 A 2 = A 1 + A 2 and A 1 A 2 = A 1 A 2 for all A 1, A 2 in L (A). For the details of these claims about L(A), see L. Rowen, Ring Theory, Volume 1, [16, pages 7-9]. Since L (A) is a bounded Artinian partially ordered set, we can define the length, Krull dimension, Krull rank and γ-length of A by len A = len L (A) Krank A = Krank L (A) Kdim A = Kdim L (A) len γ A = len γ L (A). If A is a finite length module, then len A is finite and has the usual meaning as the length of a composition series for A. We will show in 4.10 that the definition of Krull dimension here coincides with the usual one for Noetherian modules. Let 0 A B C 0 be an exact sequence in R-Noeth, and A the image of A in B. Then L (C) = [B, A ] L (B), and L (A) = L (A ) = [A, 0] L (B). So, from 3.2(1) and 3.3, we get the main result of this section:

17 THE LENGTH OF NOETHERIAN MODULES 17 Theorem 4.1. [1, 2.1] Let 0 A B C 0 be an exact sequence in R-Noeth and γ = Kdim B. (1) len C + len A len B len C len A (2) Kdim B = max{kdim A, Kdim C} (3) len γ B = len γ A + len γ C Of course, if A, B, C are finite length modules, then len A, len B and len C are finite ordinals; ordinal addition, and the usual addition of natural numbers coincide; and the inequalities in this theorem become equalities. Corollary 4.2. If A, B R-Noeth and φ: A B, then φ is injective if and only if len A = len φ(a). Proof. We have the short exact sequence 0 ker φ A φ(a) 0. So from 4.1, len φ(a) + len(ker φ) len A. If len A = len φ(a), then we can cancel from this inequality to get len(ker φ) = 0 and hence ker φ = 0. The converse implication is clear since if φ is injective, then φ(a) = A. A simple special case of this corollary is the well known property of A R-Noeth that if φ End A is surjective, then φ is injective. From 3.2(2,3) we get the following. Lemma 4.3. Let A 1, A 2 A R-Noeth. (1) len(a/(a 1 + A 2 )) + len(a/(a 1 A 2 )) len(a/a 1 ) len(a/a 2 ) len(a/(a 1 + A 2 )) len(a/(a 1 A 2 )) (2) Kdim(A/(A 1 A 2 )) = max{kdim(a/a 1 ), Kdim(A/A 2 )} (3) len(a 1 + A 2 ) + len(a 1 A 2 ) len A 1 len A 2 len(a 1 + A 2 ) len(a 1 A 2 ) (4) Kdim(A 1 + A 2 ) = max{kdim A 1, Kdim A 2 } Claim 3 of this lemma can also be obtained by applying 4.1(1) to the usual exact sequence 0 A 1 A 2 A 1 A 2 A 1 + A 2 0. Claims 2 and 4 are easy consequences of 1 and 3 obtained as in 3.3. Lemma 4.4. Let A, B, C R-Noeth and n N. (1) len(a B) = len A len B (2) A C = B C = len A = len B (3) A n = B n = len A = len B Proof. (1) The special case of 4.3(3) when A 1 A 2 = 0. (2) From A C = B C and 1 we have len A len C = len B len C. Since is cancellative (2.8) it follows that len A = len B. (3) Follows, as for 2, from the cancellativity of.

18 18 GARY BROOKFIELD We will write udim A for the uniform (or Goldie) dimension of a module A [17, ]. Krull rank has many properties in common with uniform dimension. Lemma 4.5. Let A, B R-Noeth. (1) If A B, then Krank A Krank B with equality if and only if len A = len B. (2) Krank(A B) = Krank A + Krank B (3) udim A Krank A Proof. (1) Immediate from (2) From 4.4(1) and the fact that Krank(α β) = Krank α + Krank β for any ordinals α, β Ord. (3) Any nonzero module has nonzero Krull rank, so if A contains a direct sum of udim A nonzero submodules, then using 1 and 2, we must have udim A Krank A. Applying 2.6(3), 3.4, 3.5 and 3.10 to Noetherian modules we get the following. Theorem 4.6. Let A R-Noeth. (1) For every ordinal α len A there exists a submodule A A such that len(a/a ) = α. (2) Suppose len A = ω γ1 +ω γ2 + +ω γn in long normal form. Then for ordinals α = ω γ1 +ω γ2 + +ω γi and β = ω γi+1 +ω γi+2 + +ω γn, for some i {0, 1, 2,..., n 1} there exists a submodule A A such that len(a/a ) = α and len A = β. (3) Suppose len A = ω γ1 n 1 + ω γ2 n ω γn n n in short normal form. Then for any submodule A A we have len A = ω γ1 m 1 + ω γ2 m ω γn m n for some m i Z + such that m i n i for all i. In particular, Kdim A { 1, γ 1, γ 2,..., γ n }. For example, suppose that len A = ω ω +ω Then 4.6(2) guarantees the existence of submodules of A with lengths 0, 1, ω 3 + 1, ω , ω ω + ω , and 4.6(3) says that the length of any submodule of A is one of the following ordinals: 0, 1, ω 3, ω 3 + 1, ω 3 2, ω , ω ω, ω ω + 1, ω ω + ω 3, ω ω + ω 3 + 1, ω ω + ω 3 2, ω ω + ω As we have seen already in 3.11, the fact that there are only a finite number of possible values for len A when A A is already useful.

19 THE LENGTH OF NOETHERIAN MODULES 19 Corollary 4.7. Let A, B R-Noeth. Then the partially ordered set K = {ker φ φ : A B} L (A) has finite length. Proof. Consider the restriction of λ L (A) to K. This map is, of course, strictly increasing, so any chain in K maps injectively into λ L (A)(K). Given a homomorphism φ : A B, we have A/ ker φ = im φ B and so λ L (A)(ker φ) = len(a/ ker φ) = len(im φ). Thus λ L (A)(K) is contained in the set of the lengths of all submodules of B. By 4.6(3), this set is finite, so there is a finite bound on the length of chains in λ L (A)(K). This same bound then limits the length of chains in K. As a special case of this corollary we have that, if R is a left Noetherian ring and B R-Noeth, then the set of annihilators of elements of B has finite length. Finally in this section we need to show that, for Noetherian modules, Krull dimension as we have defined it coincides with the usual definition. Definition 4.8. The Krull dimension (in the sense of Gordon and Robson) [4], [18, Chapter 13], of a module A, which we will denote by Kdim A { 1} Ord, is defined inductively as follows: Kdim A = 1 if and only if A = 0. Let γ Ord and assume that we have defined which modules have Kdim equal to δ for every δ < γ. Then Kdim A = γ if and only if (a) A does not have Kdim less than γ, and (b) for every countable decreasing chain A 1 A 2... of submodules of A, Kdim (A i /A i+1 ) < γ for all but finitely many indices. This definition does not provide a Kdim for all modules. However, any Noetherian module has a Kdim. See [18, 13.3]. Lemma 4.9. Let A R-Noeth with Kdim A = γ 1. (1) For any ordinal δ < γ, there is an infinite sequence A 1 A 2... of submodules of A such that Kdim(A i 1 /A i ) = δ for all i. (2) If A 1 A 2... is an infinite sequence of submodules of A, then Kdim(A i /A i+1 ) < γ for all but a finite number of indices. Proof. (1) Set A 1 = A. Since Kdim A 1 = γ, we have len A 1 ω γ > ω δ. By 4.6(1), there is some A 2 A 1 such that len(a 1 /A 2 ) = ω δ. We have Kdim(A 1 /A 2 ) = δ and Kdim A 1 = max{kdim A 2, Kdim(A 1 /A 2 )}, so Kdim A 2 = γ, and we can repeat the process to get A 3, A 4... as required. (2) Since the sequence of ordinals len A 1 len A 2... is decreasing, there is some n N such that len A i+1 = len A n for all i n. If i

20 20 GARY BROOKFIELD n, then, from 4.1(1), len(a i /A i+1 ) + len A i+1 len A i = len A i+1, and hence, from 2.2(1), Kdim(A i /A i+1 ) < Kdim A i+1 Kdim A = γ. Theorem For all A R-Noeth, Kdim A = Kdim A. Proof. Suppose the claim is not true. Let A be a counterexample of smallest possible Kdim. Set γ = Kdim A. Then γ > 1 and for any module B with Kdim B < γ we have Kdim B = Kdim B. From 4.9(1), we see that, for any δ < γ, the module A fails part (b) of the definition of having Kdim equal to δ. Thus A does not have Kdim less than γ, and A satisfies part (a) of the definition of having Kdim A = γ. Also, by 4.9(2), A satisfies part (b) of this definition. Thus Kdim A = γ, and A is not a counterexample. Let γ Ord. Then a module A R-Noeth is γ-critical [18, page 227] if Kdim A = γ and Kdim(A/A ) < γ for all nonzero submodules A A. From 3.6 and 3.7 is is clear that a Noetherian module A is γ-critical if and only if L (A) is γ-critical if and only if len A = ω γ. More generally A 0 is critical if and only if len A = len A for all nonzero submodules A A. Notice in particular that from 4.6(2), any nonzero Noetherian module contains a critical submodule. Specifically, if len A = ω γ1 + ω γ2 + + ω γn in long normal form, then A has a submodule of length ω γn. A critical series [18, page 229] for a module A is a submodule series 0 = A n < A n 1 <... < A 0 = A such that A i 1 /A i is a γ i -critical module for i = 1, 2,..., n and γ 1 γ 2... γ n. Comparison with 3.7, shows that a submodule series 0 = A n < A n 1 <... < A 0 = A is a critical series if and only if = A = A 0 < A 1 <... < A n = 0 = is a critical series in L (A). From 3.8 we see that an equivalent definition of the length of a nonzero Noetherian module A is len A = ω γ1 + ω γ2 + + ω γn where γ 1 γ 2... γ n are the Krull dimensions of the factors in a critical series for A. Of course, for this to be a useful definition, it is necessary to establish first the existence of critical series and then the uniqueness of the Krull dimensions of the factors of such series. See [18, 13.9]. 5. Some Applications In this section we demonstrate the use of length in proving some familiar properties of left Noetherian rings. The important thing to notice in this section is that no further use is made of the ascending chain condition. The main theorems are proved by finite induction on the Krull rank of

21 THE LENGTH OF NOETHERIAN MODULES 21 some module involved. The lemmas require only ordinal arithmetic, 4.1 and 4.4 in their proofs. Many of the results in this section can be seen as providing conditions under which an ideal or module contains a big cyclic submodule. Here big has the following technical meaning: Definition 5.1. Given a module A R-Noeth, any submodule A A such that len A = len A is said to be big in A. This situation is denoted A A. Of course, if A is a finite length module and A A, then A = A. Other basic properties of this relationship are collected in the next lemma: Lemma 5.2. Let A, A, A, B, B R-Noeth. (1) If A A A, then A A if and only if A A and A A. (2) ψ: A B and B B = ψ 1 (B ) A (3) A, A A = A A A (4) 0 A A = Kdim A/A < Kdim A (5) A A = A is essential in A Proof. (1) Immediate from the definition. (2) We have B B + ψ(a) B, and so len B = len(ψ(a) + B ) = len B. Now consider the exact sequence 0 ψ 1 (B ) σ A B τ ψ(a) + B 0 where σ(a) = (a, ψ(a)) and τ(a, b) = ψ(a) b for all a A and b B. Using 4.1(1) and 4.4(1), we get len A len B = len(a B ) len(ψ 1 (B )) len(ψ(a) + B ) = len(ψ 1 (B )) len B. Cancellation from this inequality yields len A len(ψ 1 (B )). Since ψ 1 (B ) A, the opposite inequality is, of course, true and we have len A = len(ψ 1 (B )). (3) Apply 2 to the inclusion map ψ: A A. (4) From 4.1 we get len(a/a ) + len A len A, and then 2.2(1) implies Kdim A/A < Kdim A. (5) If A B A, then len A len B = len(a B) len A. Since len A = len A, we can cancel from this inequality to get len B = 0, that is, B = 0. Essential submodules are not necessarily big any finite length module which has a proper essential submodule serves as an example.

22 22 GARY BROOKFIELD For the remainder of this section, we will suppose that R is a left Noetherian ring. If A R-Noeth, then A is finitely generated and so, from 4.1(2), Kdim A Kdim R. For any a A we will write φ a : R Ra for the homomorphism defined by r ra. Of course, since R R-Noeth, we can apply the results in the previous section to the exact sequence 0 ann a R φa Ra 0. By definition, any nonzero submodule of a critical module is big. This fact has some easy consequences for critical left ideals: Lemma 5.3. Let I, J be critical left ideals in a left Noetherian ring R. (1) If IJ 0 and len I len J, then len I = len J and there is some x J such that φ x is injective on I, I = Ix Rx J and I ann x R. (2) If I 2 0 then there is some x I such that φ x is injective on I, I = Ix Rx I and I ann x R. (3) If I is nil, then I 2 = 0. Proof. (1) Let x J be chosen so that 0 Ix J. Since J is critical we have len J = len Ix = len Rx. Since Ix = φ x (I) is an image of I, we also have len Ix len I and so len Ix = len I = len Rx = len J. From 4.2 we have that φ x is injective on I so I = Ix, ann x I = 0 and ann x I R. From the exact sequence 0 ann x R Rx 0 we get len R (len Rx) (len ann x) = len(i ann x) and so I ann x R. (2) The special case of 1 when I = J. (3) Suppose, contrary to the claim, that I 2 0. Then, from 2, there is some x I such that φ x is an injective map from I to I. But this is impossible since x n = 0 for some n N, and hence φ n x = 0. If I is a critical left ideal in a semiprime left Noetherian ring R, then I 2 0 and so from 5.3(2), there is some x I such that I = Ix Rx I and I ann x R. In the next theorem we extend this result to all left ideals of R. Theorem 5.4. Let R be a semiprime left Noetherian ring, and I a left ideal such that len I = ω γ1 + ω γ ω γn in long normal form. Then there are x 1, x 2,..., x n I such that (1) Rx 1 Rx 2... Rx n I (2) len Rx i = ω γi for i = 1, 2,..., n. (3) x i x j = 0 whenever i < j with i, j = 1, 2,..., n. Setting x = x 1 + x x n we also have 4. φ x is injective on I

23 5. I = Ix Rx I 6. I ann x R THE LENGTH OF NOETHERIAN MODULES 23

24 24 GARY BROOKFIELD Proof. Proof by induction on n = Krank I. From 4.6(2), the left ideal I contains a critical left ideal I n of length ω γn. Since R is semiprime, In 2 0, and from 5.3(2) there is some x n I n such that len I n x n = len Ix n = len Rx n = ω γn and ann x n Rx n = 0. Let I = ann x n I. Then I Rx n I, so that len I ω γn len I. From the obvious short exact sequence 0 I I Ix n 0 we get len I len I len Ix n = len I ω γn. Thus, in fact, len I = len I ω γn. Canceling ω γn from this equation we get len I = ω γ1 + ω γ ω γn 1, and so, Krank I = n 1 < Krank I = n. By induction there are x 1, x 2,..., x n 1 I satisfying the above conditions with respect to I. We claim that x 1, x 2,..., x n satisfy these conditions with respect to I: By induction we have I ann(x 1 + x x n 1 ) = 0. We also have R(x 1 + x x n 1 ) Rx n I Rx n = 0, from which it follows that ann x = ann(x 1 +x x n 1 ) ann x n. A simple calculation then yields I ann x = 0. Claims 4, 5 and 6 follow from this as in the proof of 5.3(1). The remaining claims are easy to check. Corollary 5.5. Let R be a semiprime left Noetherian ring, I R a left ideal and r R. (1) ann r R r = 0 (2) ann r = 0 Rr R r is regular. (3) I is essential in R I R I contains a regular element. (4) If I is nil, then I = 0. (5) udim I = Krank I. Proof. (1) Applying 5.4(6) to the left ideal Rr, we see that there is some s R such that Rr ann sr = 0. But ann sr = φ 1 s (ann r), so from 5.2(2), we have ann sr R, and, in particular, ann sr is essential in R. Thus Rr = 0 and r = 0. (2) Suppose Rr R. Then by 4.2, the homomorphism φ r is injective and hence ann r = 0. Further, if rs = 0 for some s R, then Rr ann s, and so ann s R and then, by 1, s = 0. Thus r is regular. The remaining claims are easy. (3) If I is essential, then, from 5.4(6), I contains an element x such that ann x = 0. From 2, x is regular. The remaining claims are easy. (4) From 5.4(4), there is some x I such that φ x is injective on I. Since x n = 0 for some n N, we have φ n x = 0 and hence I = 0. For an alternative proof which avoids 5.4, notice first that if I is critical, then from 5.3(3), I 2 = 0, and then, because R is semiprime,

25 THE LENGTH OF NOETHERIAN MODULES 25 I = 0. Thus R has no critical nil left ideals. The claim then follows from the fact that any nonzero nil left ideal must contain a critical nil left ideal. (5) From 5.4(1), I contains a direct sum of Krank I nonzero submodules, and so Krank I udim I. The opposite inequality is 4.5(3). Notice also that, from 5, I is uniform if and only if it is critical. We now specialize to left Noetherian prime rings. Theorem 5.6. If R is a left Noetherian prime ring, then len R R = ω γ n where γ = Kdim R R and n = udim R R. Further, for A R-Noeth and m N we have (1) ω γ m len A if and only if A has a submodule isomorphic to a direct sum of m critical left ideals. (2) (len R R)m len A if and only if A has a submodule isomorphic to R m. Proof. First we notice that for any two critical left ideals I, J R we have IJ 0 and JI 0 and so from 5.3(1), len I = len J, I has a submodule isomorphic to J, and vice versa. In this situation I and J are said to be subisomorphic [17, 3.3.4]. If len R R = ω γ1 + ω γ ω γn in long normal form, then from 5.4(1), there are critical left ideals of length ω γ1, ω γ2,..., ω γn. From above, we must have have γ 1 = γ 2 =... = γ n and so we can write len R R = ω γ n as required. This means, in particular, that any critical left ideal of R has length ω γ. (1) Proof by induction on m, the case m = 0 being trivial. Suppose 0 < m and ω γ m len A. Then by 4.6(1) there is some submodule A A such that len A/A = ω γ. Using 4.1(1), we have ω γ m len A len(a/a ) len A = ω γ len A, so by cancellation ω γ (m 1) len A. By induction, A contains a submodule isomorphic to a direct sum of m 1 critical left ideals. Let a A \ A. Then len(ra + A )/A = ω γ. From the exact sequence 0 ann(a + A ) R (Ra + A )/A 0 and 4.1(1) we get len((ra+a )/A)+len(ann(a+A )) len R, that is, ω γ + len(ann(a + A )) ω γ n. Cancellation from this inequality gives len(ann(a + A )) ω γ (n 1) < len R. From 5.5(3), ann(a + A ) is not essential in R, and there is a critical left ideal I of R such that I ann(a + A ) = 0. The map

26 26 GARY BROOKFIELD φ a+a : I (Ia + A )/A is then an isomorphism. In particular φ a+a is injective on I, so for any u I, ua A implies u = 0. Thus Ia A = 0, and Ia = Ia + A /A = I is critical. Since Ia A = 0, A contains a direct sum of m critical modules. (2) In view of 1, to prove 2, it suffices to show that any (external) direct sum of n critical left ideals, contains a submodule isomorphic to R. Since critical left ideals are pairwise subisomorphic, it suffices to show this for any particular direct sum of n critical left ideals. Now, from 5.4, there are critical left ideals I 1, I 2,..., I n such that I 1 I 2... I n R, and x I 1 I 2... I n such that Rx = R. Thus this particular direct sum contains a submodule isomorphic to R as required. We will next show that, for a Noetherian module A over a left Noetherian prime ring, len A encodes the reduced rank ρ(a) [16, 3.5.4] [19, 6.3], and also whether or not A is torsion or torsion free. By definition, an element a A is torsion if ann a is essential in R. From 5.5(3), a is torsion if and only if ann a R. Applying 4.1 and 2.2 to the exact sequence 0 ann a R Ra 0 we get a is torsion if and only if Kdim Ra < Kdim R, if and only if len Ra < ω γ, if and only if len γ Ra = 0, where γ = Kdim R. The module A is torsion if all its elements are torsion, and is torsion free if 0 is the only torsion element. Corollary 5.7. Let R be a left Noetherian prime ring, γ = Kdim R R, and A R-Noeth. (1) A is torsion if and only if Kdim A < γ. (2) A is torsion free if and only if len A = ω γ k for some k Z +. (3) ρ(a) = len γ A. Proof. (1) If Kdim A < γ, then for any a A, we have Kdim Ra < γ and hence, from the preceding discussion, a is torsion. If A is torsion with generators a 1, a 2,..., a n, then Kdim Ra i < γ for i = 1, 2,..., n. From 4.3(4), Kdim A = max i {Kdim Ra i } < γ. (2) Follows from 1, since, Kdim A Kdim R = γ and then, by 4.6, A has no nonzero submodule with Krull dimension less than γ if and only if len A = ω γ k for some k Z +. (3) Set k = len γ A. Since Kdim A γ we have ω γ k len A < ω γ (k + 1). From 5.6(1), A contains a submodule A isomorphic to a direct sum of k critical (and hence uniform) left ideals of R. Since uniform left ideals have reduced rank 1 [19, 6.11(f)] we have ρ(a ) = k. From 4.1(1), we have len A/A +len A len A and consequently len A/A + ω γ k len A < ω γ (k + 1). From 2.2(3), len A/A < ω γ

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