THE LENGTH OF NOETHERIAN MODULES


 Christine Mathews
 2 years ago
 Views:
Transcription
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 γlength 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 γlength 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 γcritical 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 RNoeth the category of Noetherian left Rmodules. If A RNoeth, 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 79]. Since L (A) is a bounded Artinian partially ordered set, we can define the length, Krull dimension, Krull rank and γlength 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 RNoeth, 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 RNoeth 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 RNoeth 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 RNoeth 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 RNoeth. (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 RNoeth 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 RNoeth. (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 RNoeth. (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 RNoeth. 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 RNoeth, 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 RNoeth 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 RNoeth, 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 RNoeth is γcritical [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 γcritical if and only if L (A) is γcritical 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 RNoeth, 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 RNoeth. (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 RNoeth, 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 RNoeth, 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 RNoeth 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 RNoeth. (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
A KrullSchmidt Theorem for Noetherian Modules *
A KrullSchmidt Theorem for Noetherian Modules * Gary Brookfield Department of Mathematics, University of California, Riverside CA 925210135 Email: brookfield@math.ucr.edu We prove a version of the KrullSchmidt
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More informationTHE CLOSEDPOINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS. K. R. Goodearl and E. S. Letzter
THE CLOSEDPOINT ZARISKI TOPOLOGY FOR IRREDUCIBLE REPRESENTATIONS K. R. Goodearl and E. S. Letzter Abstract. In previous work, the second author introduced a topology, for spaces of irreducible representations,
More informationLecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).
Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an Alinear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More information4.3 Composition Series
4.3 Composition Series Let M be an Amodule. A series for M is a strictly decreasing sequence of submodules M = M 0 M 1... M n = {0} beginning with M and finishing with {0 }. The length of this series
More informationPiecewise Noetherian Rings
Northern Illinois University UNAM 25 May, 2017 Acknowledgments Results for commutative rings are from two joint papers with William D. Weakley,, Comm. Algebra (1984) and A note on prime ideals which test
More informationStructure of rings. Chapter Algebras
Chapter 5 Structure of rings 5.1 Algebras It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative ring. An Ralgebra is a ring A (unital as always) together
More informationCOMMUNICATIONS IN ALGEBRA, 15(3), (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY. John A. Beachy and William D.
COMMUNICATIONS IN ALGEBRA, 15(3), 471 478 (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY John A. Beachy and William D. Weakley Department of Mathematical Sciences Northern Illinois University DeKalb,
More informationCOHENMACAULAY RINGS SELECTED EXERCISES. 1. Problem 1.1.9
COHENMACAULAY RINGS SELECTED EXERCISES KELLER VANDEBOGERT 1. Problem 1.1.9 Proceed by induction, and suppose x R is a U and Nregular element for the base case. Suppose now that xm = 0 for some m M. We
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More informationSTRONGLY JÓNSSON AND STRONGLY HS MODULES
STRONGLY JÓNSSON AND STRONGLY HS MODULES GREG OMAN Abstract. Let R be a commutative ring with identity and let M be an infinite unitary Rmodule. Then M is a Jónsson module provided every proper Rsubmodule
More information4.2 Chain Conditions
4.2 Chain Conditions Imposing chain conditions on the or on the poset of submodules of a module, poset of ideals of a ring, makes a module or ring more tractable and facilitates the proofs of deep theorems.
More informationInfiniteDimensional Triangularization
InfiniteDimensional Triangularization Zachary Mesyan March 11, 2018 Abstract The goal of this paper is to generalize the theory of triangularizing matrices to linear transformations of an arbitrary vector
More informationHomework #05, due 2/17/10 = , , , , , Additional problems recommended for study: , , 10.2.
Homework #05, due 2/17/10 = 10.3.1, 10.3.3, 10.3.4, 10.3.5, 10.3.7, 10.3.15 Additional problems recommended for study: 10.2.1, 10.2.2, 10.2.3, 10.2.5, 10.2.6, 10.2.10, 10.2.11, 10.3.2, 10.3.9, 10.3.12,
More information2. Prime and Maximal Ideals
18 Andreas Gathmann 2. Prime and Maximal Ideals There are two special kinds of ideals that are of particular importance, both algebraically and geometrically: the socalled prime and maximal ideals. Let
More information2. Modules. Definition 2.1. Let R be a ring. Then a (unital) left Rmodule M is an additive abelian group together with an operation
2. Modules. The concept of a module includes: (1) any left ideal I of a ring R; (2) any abelian group (which will become a Zmodule); (3) an ndimensional Cvector space (which will become both a module
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationIntroduction to modules
Chapter 3 Introduction to modules 3.1 Modules, submodules and homomorphisms The problem of classifying all rings is much too general to ever hope for an answer. But one of the most important tools available
More information4.4 Noetherian Rings
4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)
More informationProceedings of the Twelfth Hudson Symposium, Lecture Notes in Math. No. 951, SpringerVerlag (1982), 4l 46.
Proceedings of the Twelfth Hudson Symposium, Lecture Notes in Math. No. 951, SpringerVerlag (1982), 4l 46. MAXIMAL TORSION RADICALS OVER RINGS WITH FINITE REDUCED RANK John A. Beachy Northern Illinois
More informationA MODELTHEORETIC PROOF OF HILBERT S NULLSTELLENSATZ
A MODELTHEORETIC PROOF OF HILBERT S NULLSTELLENSATZ NICOLAS FORD Abstract. The goal of this paper is to present a proof of the Nullstellensatz using tools from a branch of logic called model theory. In
More informationInjective Modules and Matlis Duality
Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective Rmodules. The following
More informationSolutions to Homework 1. All rings are commutative with identity!
Solutions to Homework 1. All rings are commutative with identity! (1) [4pts] Let R be a finite ring. Show that R = NZD(R). Proof. Let a NZD(R) and t a : R R the map defined by t a (r) = ar for all r R.
More informationTopics in Module Theory
Chapter 7 Topics in Module Theory This chapter will be concerned with collecting a number of results and constructions concerning modules over (primarily) noncommutative rings that will be needed to study
More informationJónsson posets and unary Jónsson algebras
Jónsson posets and unary Jónsson algebras Keith A. Kearnes and Greg Oman Abstract. We show that if P is an infinite poset whose proper order ideals have cardinality strictly less than P, and κ is a cardinal
More informationRing Theory Problems. A σ
Ring Theory Problems 1. Given the commutative diagram α A σ B β A σ B show that α: ker σ ker σ and that β : coker σ coker σ. Here coker σ = B/σ(A). 2. Let K be a field, let V be an infinite dimensional
More informationChapter 1 : The language of mathematics.
MAT 200, Logic, Language and Proof, Fall 2015 Summary Chapter 1 : The language of mathematics. Definition. A proposition is a sentence which is either true or false. Truth table for the connective or :
More informationADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS
ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material.
More information11. Finitelygenerated modules
11. Finitelygenerated modules 11.1 Free modules 11.2 Finitelygenerated modules over domains 11.3 PIDs are UFDs 11.4 Structure theorem, again 11.5 Recovering the earlier structure theorem 11.6 Submodules
More informationRings and Fields Theorems
Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a nonempty set. Let + and (multiplication)
More informationON STRONGLY PRIME IDEALS AND STRONGLY ZERODIMENSIONAL RINGS. Christian Gottlieb
ON STRONGLY PRIME IDEALS AND STRONGLY ZERODIMENSIONAL RINGS Christian Gottlieb Department of Mathematics, University of Stockholm SE106 91 Stockholm, Sweden gottlieb@math.su.se Abstract A prime ideal
More informationINJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA
INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to
More informationModules Over Principal Ideal Domains
Modules Over Principal Ideal Domains Brian Whetter April 24, 2014 This work is licensed under the Creative Commons AttributionNonCommercialShareAlike 4.0 International License. To view a copy of this
More informationand this makes M into an Rmodule by (1.2). 2
1. Modules Definition 1.1. Let R be a commutative ring. A module over R is set M together with a binary operation, denoted +, which makes M into an abelian group, with 0 as the identity element, together
More informationMODULES OVER A PID. induces an isomorphism
MODULES OVER A PID A module over a PID is an abelian group that also carries multiplication by a particularly convenient ring of scalars. Indeed, when the scalar ring is the integers, the module is precisely
More informationA NOTE ON SIMPLE DOMAINS OF GK DIMENSION TWO
A NOTE ON SIMPLE DOMAINS OF GK DIMENSION TWO JASON P. BELL Abstract. Let k be a field. We show that a finitely generated simple Goldie kalgebra of quadratic growth is noetherian and has Krull dimension
More informationCriteria for existence of semigroup homomorphisms and projective rank functions. George M. Bergman
Criteria for existence of semigroup homomorphisms and projective rank functions George M. Bergman Suppose A, S, and T are semigroups, e: A S and f: A T semigroup homomorphisms, and X a generating set for
More informationFormal power series rings, inverse limits, and Iadic completions of rings
Formal power series rings, inverse limits, and Iadic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely
More informationII. Products of Groups
II. Products of Groups HongJian Lai October 2002 1. Direct Products (1.1) The direct product (also refereed as complete direct sum) of a collection of groups G i, i I consists of the Cartesian product
More informationAzumaya Algebras. Dennis Presotto. November 4, Introduction: Central Simple Algebras
Azumaya Algebras Dennis Presotto November 4, 2015 1 Introduction: Central Simple Algebras Azumaya algebras are introduced as generalized or global versions of central simple algebras. So the first part
More informationADVANCED CALCULUS  MTH433 LECTURE 4  FINITE AND INFINITE SETS
ADVANCED CALCULUS  MTH433 LECTURE 4  FINITE AND INFINITE SETS 1. Cardinal number of a set The cardinal number (or simply cardinal) of a set is a generalization of the concept of the number of elements
More informationALGEBRA HW 4. M 0 is an exact sequence of Rmodules, then M is Noetherian if and only if M and M are.
ALGEBRA HW 4 CLAY SHONKWILER (a): Show that if 0 M f M g M 0 is an exact sequence of Rmodules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationREPRESENTATION THEORY, LECTURE 0. BASICS
REPRESENTATION THEORY, LECTURE 0. BASICS IVAN LOSEV Introduction The aim of this lecture is to recall some standard basic things about the representation theory of finite dimensional algebras and finite
More informationarxiv:math/ v1 [math.gm] 21 Jan 2005
arxiv:math/0501341v1 [math.gm] 21 Jan 2005 SUBLATTICES OF LATTICES OF ORDERCONVEX SETS, I. THE MAIN REPRESENTATION THEOREM MARINA SEMENOVA AND FRIEDRICH WEHRUNG Abstract. For a partially ordered set P,
More informationNOTES FOR COMMUTATIVE ALGEBRA M5P55
NOTES FOR COMMUTATIVE ALGEBRA M5P55 AMBRUS PÁL 1. Rings and ideals Definition 1.1. A quintuple (A, +,, 0, 1) is a commutative ring with identity, if A is a set, equipped with two binary operations; addition
More information11 Annihilators. Suppose that R, S, and T are rings, that R P S, S Q T, and R U T are bimodules, and finally, that
11 Annihilators. In this Section we take a brief look at the important notion of annihilators. Although we shall use these in only very limited contexts, we will give a fairly general initial treatment,
More information2.3 The KrullSchmidt Theorem
2.3. THE KRULLSCHMIDT THEOREM 41 2.3 The KrullSchmidt Theorem Every finitely generated abelian group is a direct sum of finitely many indecomposable abelian groups Z and Z p n. We will study a large
More informationMath 210B. Artin Rees and completions
Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an Amodule. In class we defined the Iadic completion of M to be M = lim M/I n M. We will soon show
More informationAlgebra Homework, Edition 2 9 September 2010
Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.
More informationReview of Linear Algebra
Review of Linear Algebra Throughout these notes, F denotes a field (often called the scalars in this context). 1 Definition of a vector space Definition 1.1. A F vector space or simply a vector space
More informationCHAPTER 8: EXPLORING R
CHAPTER 8: EXPLORING R LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN In the previous chapter we discussed the need for a complete ordered field. The field Q is not complete, so we constructed
More informationGEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS
GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions
More information32 Divisibility Theory in Integral Domains
3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible
More informationUniversal localization at semiprime Goldie ideals
at semiprime Goldie ideals Northern Illinois University UNAM 23/05/2017 Background If R is a commutative Noetherian ring and P R is a prime ideal, then a ring R P and ring homomorphism λ : R R P can be
More informationNOTES ON SPLITTING FIELDS
NOTES ON SPLITTING FIELDS CİHAN BAHRAN I will try to define the notion of a splitting field of an algebra over a field using my words, to understand it better. The sources I use are Peter Webb s and T.Y
More informationSkew Polynomial Rings
Skew Polynomial Rings NIU November 14, 2018 Bibliography Beachy, Introductory Lectures on Rings and Modules, Cambridge Univ. Press, 1999 Goodearl and Warfield, An Introduction to Noncommutative Noetherian
More informationNotes on ordinals and cardinals
Notes on ordinals and cardinals Reed Solomon 1 Background Terminology We will use the following notation for the common number systems: N = {0, 1, 2,...} = the natural numbers Z = {..., 2, 1, 0, 1, 2,...}
More informationMath 210B: Algebra, Homework 1
Math 210B: Algebra, Homework 1 Ian Coley January 15, 201 Problem 1. Show that over any field there exist infinitely many nonassociate irreducible polynomials. Recall that by Homework 9, Exercise 8 of
More informationLIVIA HUMMEL AND THOMAS MARLEY
THE AUSLANDERBRIDGER FORMULA AND THE GORENSTEIN PROPERTY FOR COHERENT RINGS LIVIA HUMMEL AND THOMAS MARLEY Abstract. The concept of Gorenstein dimension, defined by Auslander and Bridger for finitely
More informationPrimal, completely irreducible, and primary meet decompositions in modules
Bull. Math. Soc. Sci. Math. Roumanie Tome 54(102) No. 4, 2011, 297 311 Primal, completely irreducible, and primary meet decompositions in modules by Toma Albu and Patrick F. Smith Abstract This paper was
More informationThis is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:
Math 6130 Notes. Fall 2002. 7. Basic Maps. Recall from 3 that a regular map of affine varieties is the same as a homomorphism of coordinate rings (going the other way). Here, we look at how algebraic properties
More informationCRing Project, Chapter 5
Contents 5 Noetherian rings and modules 3 1 Basics........................................... 3 1.1 The noetherian condition............................ 3 1.2 Stability properties................................
More informationCHAPTER 0 PRELIMINARY MATERIAL. Paul Vojta. University of California, Berkeley. 18 February 1998
CHAPTER 0 PRELIMINARY MATERIAL Paul Vojta University of California, Berkeley 18 February 1998 This chapter gives some preliminary material on number theory and algebraic geometry. Section 1 gives basic
More informationGroups of Prime Power Order with Derived Subgroup of Prime Order
Journal of Algebra 219, 625 657 (1999) Article ID jabr.1998.7909, available online at http://www.idealibrary.com on Groups of Prime Power Order with Derived Subgroup of Prime Order Simon R. Blackburn*
More informationPrimary Decomposition and Associated Primes
Chapter 1 Primary Decomposition and Associated Primes 1.1 Primary Submodules and Ideals 1.1.1 Definitions and Comments If N is a submodule of the Rmodule M, and a R, let λ a : M/N M/N be multiplication
More informationMath 762 Spring h Y (Z 1 ) (1) h X (Z 2 ) h X (Z 1 ) Φ Z 1. h Y (Z 2 )
Math 762 Spring 2016 Homework 3 Drew Armstrong Problem 1. Yoneda s Lemma. We have seen that the bifunctor Hom C (, ) : C C Set is analogous to a bilinear form on a Kvector space, : V V K. Recall that
More informationTROPICAL SCHEME THEORY
TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),
More informationSchemes via Noncommutative Localisation
Schemes via Noncommutative Localisation Daniel Murfet September 18, 2005 In this note we give an exposition of the wellknown results of Gabriel, which show how to define affine schemes in terms of the
More informationREPRESENTATION THEORY WEEK 9
REPRESENTATION THEORY WEEK 9 1. JordanHölder theorem and indecomposable modules Let M be a module satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence
More informationA NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS
A NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS RAVI JAGADEESAN AND AARON LANDESMAN Abstract. We give a new proof of Serre s result that a Noetherian local ring is regular if
More informationHomological Methods in Commutative Algebra
Homological Methods in Commutative Algebra Olivier Haution LudwigMaximiliansUniversität München Sommersemester 2017 1 Contents Chapter 1. Associated primes 3 1. Support of a module 3 2. Associated primes
More information5 Set Operations, Functions, and Counting
5 Set Operations, Functions, and Counting Let N denote the positive integers, N 0 := N {0} be the nonnegative integers and Z = N 0 ( N) the positive and negative integers including 0, Q the rational numbers,
More information5 Dedekind extensions
18.785 Number theory I Fall 2016 Lecture #5 09/22/2016 5 Dedekind extensions In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also
More informationA GENERAL THEORY OF ZERODIVISOR GRAPHS OVER A COMMUTATIVE RING. David F. Anderson and Elizabeth F. Lewis
International Electronic Journal of Algebra Volume 20 (2016) 111135 A GENERAL HEORY OF ZERODIVISOR GRAPHS OVER A COMMUAIVE RING David F. Anderson and Elizabeth F. Lewis Received: 28 April 2016 Communicated
More informationMath 412, Introduction to abstract algebra. Overview of algebra.
Math 412, Introduction to abstract algebra. Overview of algebra. A study of algebraic objects and functions between them; an algebraic object is typically a set with one or more operations which satisfies
More information11. Dimension. 96 Andreas Gathmann
96 Andreas Gathmann 11. Dimension We have already met several situations in this course in which it seemed to be desirable to have a notion of dimension (of a variety, or more generally of a ring): for
More informationCongruence Boolean Lifting Property
Congruence Boolean Lifting Property George GEORGESCU and Claudia MUREŞAN University of Bucharest Faculty of Mathematics and Computer Science Academiei 14, RO 010014, Bucharest, Romania Emails: georgescu.capreni@yahoo.com;
More informationRIGHTLEFT SYMMETRY OF RIGHT NONSINGULAR RIGHT MAXMIN CS PRIME RINGS
Communications in Algebra, 34: 3883 3889, 2006 Copyright Taylor & Francis Group, LLC ISSN: 00927872 print/15324125 online DOI: 10.1080/00927870600862714 RIGHTLEFT SYMMETRY OF RIGHT NONSINGULAR RIGHT
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #7 09/26/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #7 09/26/2013 In Lecture 6 we proved (most of) Ostrowski s theorem for number fields, and we saw the product formula for absolute values on
More informationCOUNTING SUBMODULES OF A MODULE OVER A NOETHERIAN COMMUTATIVE RING
COUNTING SUBMODULES OF A MODULE OVER A NOETHERIAN COMMUTATIVE RING YVES CORNULIER Abstract. We count the number of submodules of an arbitrary module over a countable noetherian commutative ring. We give,
More informationIntroduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013
18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local
More informationUniversal Localization of Piecewise Noetherian Rings
Universal Localization of Piecewise Noetherian Rings Northern Illinois University UCCS November 29, 2017 Acknowledgments For commutative rings, results on piecewise Noetherian rings are from two joint
More informationALGEBRA EXERCISES, PhD EXAMINATION LEVEL
ALGEBRA EXERCISES, PhD EXAMINATION LEVEL 1. Suppose that G is a finite group. (a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer. (b) Use (a)
More informationShort notes on Axioms of set theory, Well orderings and Ordinal Numbers
Short notes on Axioms of set theory, Well orderings and Ordinal Numbers August 29, 2013 1 Logic and Notation Any formula in Mathematics can be stated using the symbols and the variables,,,, =, (, ) v j
More informationBENJAMIN LEVINE. 2. Principal Ideal Domains We will first investigate the properties of principal ideal domains and unique factorization domains.
FINITELY GENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN BENJAMIN LEVINE Abstract. We will explore classification theory concerning the structure theorem for finitely generated modules over a principal
More informationThe category of linear modular lattices
Bull. Math. Soc. Sci. Math. Roumanie Tome 56(104) No. 1, 2013, 33 46 The category of linear modular lattices by Toma Albu and Mihai Iosif Dedicated to the memory of Nicolae Popescu (19372010) on the occasion
More informationHomework #2 Solutions Due: September 5, for all n N n 3 = n2 (n + 1) 2 4
Do the following exercises from the text: Chapter (Section 3):, 1, 17(a)(b), 3 Prove that 1 3 + 3 + + n 3 n (n + 1) for all n N Proof The proof is by induction on n For n N, let S(n) be the statement
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More informationExercises for Unit VI (Infinite constructions in set theory)
Exercises for Unit VI (Infinite constructions in set theory) VI.1 : Indexed families and set theoretic operations (Halmos, 4, 8 9; Lipschutz, 5.3 5.4) Lipschutz : 5.3 5.6, 5.29 5.32, 9.14 1. Generalize
More informationNOTES ON WELL ORDERING AND ORDINAL NUMBERS. 1. Logic and Notation Any formula in Mathematics can be stated using the symbols
NOTES ON WELL ORDERING AND ORDINAL NUMBERS TH. SCHLUMPRECHT 1. Logic and Notation Any formula in Mathematics can be stated using the symbols,,,, =, (, ) and the variables v j : where j is a natural number.
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More information(Rgs) Rings Math 683L (Summer 2003)
(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that
More informationMAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9
MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended
More informationMATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA
MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA These are notes for our first unit on the algebraic side of homological algebra. While this is the last topic (Chap XX) in the book, it makes sense to
More informationNoncommutative Generalizations of Theorems of Cohen and Kaplansky
Algebr Represent Theor (2012) 15:933 975 DOI 10.1007/s1046801192737 Noncommutative Generalizations of Theorems of Cohen and Kaplansky Manuel L. Reyes Received: 12 July 2010 / Accepted: 3 January 2011
More informationMath 145. Codimension
Math 145. Codimension 1. Main result and some interesting examples In class we have seen that the dimension theory of an affine variety (irreducible!) is linked to the structure of the function field in
More information