# 4. CLASSES OF RINGS 4.1. Classes of Rings class operator A-closed Example 1: product Example 2:

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1 4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets in an uncontrolled way. For example the Russell Paradox comes from considering the set S = {x x x} and asking whether S S. Constructions such as the set of all sets are just not on, and if we were pursuing axiomatic set theory we d have to lay down certain axioms to restrict which properties are allowed to give rise to a set. But, we re doing ring theory and we want to turn adjectives that describe properties of rings into nouns. We use the word class in place of set to avoid these logical problems. Now it s not the fact that we re using a different word that avoids logical difficulties. It s the fact that, while sets are allowed to be elements of other sets, classes are not. So we ll talk about the class of all rings, the class of all nilpotent rings etc. The only assumption that we make is that if a certain ring belongs to a class, all isomorphic copies belong to it as well. In other words, in this context, we consider all rings isomorphic to a single one as being identical. We re not concerned with the nature of the underlying set, but rather the structure of the ring. As well as having a rich supply of classes to play with we want to consider certain recipes for obtaining a new class from a given one. We define these as class operators. Essentially they re functions on the class of all classes of all rings, but as we ve said, classes are not allowed to be considered as elements of other classes. So a class operator is not to be considered as a set of ordered pairs, or even a class of ordered pairs it s simply a rule that associates with each class X a certain class fx. We shall write operators on the left but avoid the usual f(x) notation as a way of reminding ourselves that these operators are not exactly functions in the usual sense in set theory. A class operator A is a rule that creates a ring class AX from a class X. If a ring R belongs to a class X we say that R is an X-ring. If AX X (subclass being defined in exactly the same way as subset) we say that X is A-closed. Example 1: We define the class operator S by stating that a ring is in SX if it is a subring of an X-ring. If N denotes the class of all nil rings, N is S-closed is just shorthand for the statement every subring of a nil ring is a nil ring. Another construct that produces new classes out of existing ones is the product of classes. We define the product of class operators by defining (AB)X = A(BX). Example 2: If QX is defined as the class of all quotients of X-rings then (SQ)X would become the class of all rings that are subrings of quotients of X-rings, while (QS)X would be all quotients of subrings of X-rings. Multiplication of class operators is clearly associative and, as is typical, when we define any sort of associative product, we can define powers. We define A 0 X = X; A 2 X = A(AX) and so on. Moreover we define A X to be the union of the A n X. 19

2 Examples 3: Here are a few classes of rings: F = class of finite rings; C = class of commutative rings; T = class of torsion rings (every element having finite additive order); D = class of all integral domains. Here are a few class operators: QX = quotients of X-rings; X = rings that are the sum of an ideal and a subring, each of which is an X-ring; UX = unions of ascending chains of X ideals of some ring; PX = rings R with an ideal I such that both I and R/I are X -rings. D isn t closed under P, Q or. Is it U-closed? C is closed under Q, U. But it is not closed under P (or as we say, not poly-closed ). For example if R = * * 0 *, meaning the set of all matrices of the form a b 0 d and I = 0 * 0 0, then I is an ideal of R. But, although both I and R/I are commutative rings, R is not. 0 * * 0 0 * Nor is C -closed. For example and 0 0 * are commutative ideals yet their sum, * * * 0 * * is not commutative. 0 0 * F is closed under Q,, P, but not U. Consider the infinite restricted direct sum of countably many finite rings. T is closed under Q,, U and P. The first three of these closure properties are obvious. We can show that T is poly-closed as follows. Suppose that R/I and I are torsion rings and r R. Then nr I for some n > 0. Hence (mn)r = m(nr) = 0 for some m > 0. Theorem 1: Closure under both P and Q implies -closure. Proof: Suppose I, J X with J a 2-sided ideal of R. Then (I + J)/J I/(I J) X and hence I + J X Zorn s Lemma The rings we re interested in are infinite and infinite sets can have some strange properties. Given a certain property we want to find the largest ideal that has that property. But such a largest might not exist. One reason why a set fails to have a largest is because there may be more than one possible candidate. For example we might ask which proper subspace of R 3 is largest. There are plenty of 2-dimensional subspaces but none contains any of the others. The other, and more subtle reason why a set may not have a largest is that there may be larger and larger elements, but no largest. For example there s no largest real number x for which x 2 < 2. We shall focus on the second problem. This will take us into the rarefied world of Axiomatic Set Theory where we discuss Zorn s Lemma. It isn t our role here to give a complete account of the necessary set theory, but, before we start using Zorn s Lemma, we need some discussion as to its logical status. 20

3 A partially-ordered set is a set X, together with a relation, that we shall denote by (though keep in mind that it may not bear any relation to any ordering we ve ever met). But, like ordinary less-than-or-equals we do insist that it have the following three properties: (i) x x for all x X (the reflexive property); (ii) if x y and y x then x = y (the anti-symmetric property); (iii) if x y and y z then x x (the transitive property). An example of such an ordering is that of subset of. Now notice that subset of fails to have a fourth property that we expect of ordinary : (iv) for all x, y X either x y or y x. We can easily have subsets where neither is a subset of the other. This fourth property is not required for a set to be a partial ordering. But if it does hold we say that the set is totally ordered, or simply that the set is a chain. By analogy with ordinary of numbers we define, < and >. If X is a partially ordered set, and S X then b X is an upper bound for S if x b for all x S. There is no need for the upper bound to be in the set, but if it is it is called a greatest element. A set may not have a greatest element, but it may have maximal elements. These are elements where nothing in the subset is bigger. A maximal element of S is thus an element m S such that if x m for some x S then x = m. Finally we can combine a couple of these together and talk of a least upper bound. Similarly we define lower bound, least element, minimal element and greatest lower bound. Example 4: Consider the set {A, B, C, D, E, F, G} where the ordering is described by the following diagram (x y if and only if you can get from x to y in the diagram by a path consisting of one or more edges, each traversed in an upwards direction). F G E C D Let S = {B, C, D}. Then E and G are upper bounds for S, with E the least upper bound. But S has no largest element, though C, D are maximal elements. S has a least element, B and this is the only minimal element. A, B are both lower bounds, but B is the greatest lower bound. In finite partially ordered sets there will always be at least one maximal element. If x is not maximal there s an element y with x y. If y is not maximal then there is an element z with y z, and so on. With finitely many elements we must eventually reach a maximal element. On the other hand, with infinite partially ordered sets, there may be no maximal element. For example the set of real numbers is partially ordered in the usual way and the subset {x 0 x 2 < 2} has no maximal element. But it does have a least upper bound, 2. If we restricted ourselves to just rational numbers there will not even be a least upper bound. B A 21

4 ZORN S LEMMA: If every chain in the non-empty partially ordered set X has an upper bound then X has a maximal element. Proof : A pseudo proof of this goes along the following lines. Take an element x 1 X. If x 1 isn t maximal there s an element x 2 that s bigger than x 1. If that isn t maximal there s an x 3 bigger than x 2, and so on. So if X doesn t have a maximal element there must be a chain x 1 < x 2 < in X. By our assumption, every chain in X has an upper bound, and so there exists y 1 that is bigger than all of the x i. By the same argument we can produce a chain of y i s so that x 1 < x 2 < < y 1 < y 2 < So far so good. The pseudo nature of the pseudo proof comes when we say surely the process must eventually terminate. We haven t called Zorn s Lemma a theorem because it isn t one. We can prove that there can be no proof of it. But that doesn t mean that it s false, because we can prove there cannot exist a counter-example. This is an example of an undecidable proposition in set theory. We set up a collection of axioms that attempt to avoid a contradiction like the Russell Paradox. Now if you were to set up a random collection of axioms it s possible that they would be self-contradictory. The simplest way to show that they re consistent is to give a concrete example that satisfies them all. The group axioms are consistent because we can give examples of groups. But when it comes to the axioms of set theory, how are we going to produce an example of a universe of sets satisfying the axioms without making use of sets in order to construct such a model? So it would seem that we ll probably never be able to prove that any set of axioms for set theory is consistent. What has been proved is that if the axioms of set theory are consistent, and therefore that there is a model that satisfies them, then it is possible to construct within it another model that satisfies all the set theory axioms together with Zorn s Lemma. This proves that Zorn s Lemma is consistent with the other axioms of set theory. If ever a contradiction is reached after using Zorn s Lemma, the problem would lie with the basic set theory axioms and not with Zorn s Lemma. On the other hand, again assuming that a model for set theory exists, we can construct within it a different model that satisfies the basic axioms of set theory but in which Zorn s Lemma is false! This proves that Zorn s Lemma is independent from the other axioms of set theory. It can never be proved from them, otherwise such a model couldn t exist. So Zorn s Lemma isn t a theorem it isn t even a lemma it is in fact an optional extra axiom that we are logically free to accept or reject. There are other axioms that are equivalent to Zorn s Lemma. The most famous of these is the Axiom of Choice. At least this one goes by the correct name of axiom. Axiom of Choice: If X is a set of non-empty sets, and Y is the set of all the elements of the elements of X, there is a function f:x Y such that f(x) x. Example 5: Let X 1 = {1, 2, 3}, X 2 = {1, 5, 9, 10}, X 3 = the set of prime numbers. Let X = {X 1, X 2, X 3 }. Then Y consists of all the prime numbers, together with 1, 9 and 10. An example of a function described in the Axiom of Choice is: f(x 1 ) = 2, f(x 2 ) = 10, f(x 3 ) = 17. In other words, the function f is a choice function it chooses a specific element from each of the non-empty sets. It may well seem to you obvious that the Axiom of Choice is obvious. If we have a bunch of non-empty sets surely we can choose one element from each! Certainly if the sets 22

6 4.4. Nil and Jacobson Radicals Theorem 6: The class, N, of nil rings is a radical class. Proof: N is clearly Q-closed. If I 1 I 2 is an ascending chain of nil ideals then I r is clearly nil, for if x I r, x I r for some r. Since I r is nil, x n = 0 for some n. Finally we must show that N is poly-closed. Suppose I is an ideal of R and both I and R/I are nil rings. Let x R. Since R/I is nil, (r + I) n = r n + I = I, for some n. Thus r n I. Since I is nil, (r n ) m = 0 for some m. The class of nilpotent rings is probably a more useful class than the class of nil rings, but unfortunately it isn t a radical class. Theorem 7: The class of nilpotent rings is closed under P, Q and, but not under U. Proof: Let R = x a x x R where a x a y = a x+y if x + y < 1 0 if x + y 1 0<x<1 This is a non-nilpotent ring. For each positive integer n, let R n = are nilpotent ideals, and R is the union of them all. 1/2 n <x<1 x a x x R. These Example 6: R = {all n n upper triangular matrices over a field F}. The nil-radical R N. consists of those matrices with 0 s on the diagonal and R/R N. F F (n copies). Theorem 8: If R is a commutative ring the set of all nilpotent elements is an ideal and is thus the nil radical of R. Proof: Let a, b be elements of the commutative ring R. If a m = b n = 0 then (a + b) m+n1 = 0 and (ab) min(m,n) = 0. Commutativity is needed since in the ring of 22 matrices and are nilpotent yet their sum and product aren t. Let J be the class of rings in which every element is right quasi-regular. Example 7: J contains, the set of all rational numbers of the form even odd. Theorem 9: J is a radical class. Proof: P-closure follows from the associativity of. Theorem 10: N is a proper subclass of J. Proof: If x is nilpotent, y = x + x 2 x 3... exists as a finite sum and x y = 0. So N is a subclass of J. Since J N, it is a proper subclass. Theorem 11 (Machke s Theorem): Let G = {g 1,..., g n } be a finite group and suppose that char F does not divide G. Then the group ring FG is nil semi-simple. Proof: Suppose 0 S is a nil ideal and let 0 x = a g g S. Since S is an ideal we may suppose, without loss of generality, that a 1 = 1. Let :FG End F FG be the ring homomorphism defined by x(y) = xy. Since x is nilpotent, so is x and hence tr(x) = 0. Now tr(g) = 0 for g 1 and tr(1) = n. Hence n1 = 0 so the characteristic of F divides n. 24

8 Theorem 19: Suppose (M, N) is any partition of the class of simple rings. If X is any radical class with N = S X and M = S X then N X M +. Proof: H n QN H n QX X for all n. If R/I M X X = {0}. Theorem 20: If (M, N) is any partition of S then N M +. Proof: Zp is the Prüfer p-group, being the abelian group generated by a 0, a 1, and subject to the relations pa 0 = 0, pa n+1 = a n, for all n 0. It is an infinite p-group where all non-zero quotients are isomorphic to itself. If H < G let n = max[a n H]. Then H = a n and G/H [a n+1, a n+2, pa n+1 = 0, pa m+1 = a m ] Zp. Z is the infinite cyclic group and every non-zero subgroup is isomorphic to itself. Denote Zero(G) by G. Case I: M contains a zero ring: Z p M for some prime p. Let n = min[ zero ring on a 0 p-group H n N]. Since all 0 quotients of a 0 p-group are 0 p-groups, n = 0. But Z p N. Hence H N = H QN = N contains no 0 zero ring on a p-group. Hence Zp N. If there exists a non-zero quotient Zp/I Zp M, we get a contradiction. Hence Zp NM NIM = M +. So Zp M + N. Case II: M contains no zero rings: Let n = min[z H n N]. Since all non-zero ideals of Z are isomorphic to Z, n = 0. But Z N. Hence Z N. Every quotient of Z is a zero ring and so is not in M. Hence Z NM NIM = M + and so Z M + N. 26

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