Nil and Jacobson Radicals in Semigroup Graded Rings

Home , Jacobson radical, Jacobson ring, Nil ideal, Radical of a ring

Faculty of Science Departement of Mathematics

Nil and Jacobson radicals in semigroup graded rings

Master thesis submitted in partial fulﬁllment of the requirements for the degree of Master in Mathematics Carmen Mazijn

Promotor: Prof. Dr. E. Jespers

AUGUST 2015 Acknowledgements

When we started our last year of the Master in Mathematics at VUB, none of us knew how many hours we would spend on the reading, understanding and writing of our thesis. This final product as conclusion of the master was at that point only an idea. The subject was chosen, the first papers were read and the first words were written down. And more words were written, more books were consulted, more questions were asked to our promoters. Writing a Master thesis is a journey. Even though next week everyone will have handed in there thesis, we don’t yet understand clearly where this journey took us, for the future is unknown. First of all I would like to thank professor Eric Jespers, for giving me the chance to grow as mathematician in the past years. With every semester the interest in Algebra and accuracy as mathematician grew. Thank you for the guidance through all the books and papers to make this a consistent dissertation. Secondly I would like to thank all my classmates and compañerosde clase. For frowned faces when we didn’t get something in class, the laughter when we realized it was a ctually quite trivial or sometimes not even at all. For the late night calls and the interesting discussions. It was a pleasure. Tambiéngracias a todos que conoc´ı en Granada en mi añode Erasmus. Era mucho mas que un añode erasmus y sois mucho mas que mis amigos. Hasta pronto. Lastly, but probably mostly, I would like to thank my parents and little brother. For actually listening most of the times when I tried to explain something. For the constant support during these year. For making me come home now and then, as Bruges will always be one of my homes.

Carmen Mazijn Brussels, August 2015

i Samenvatting

In het kader van deze masterproef werken we in niet-commutatieve ring theorie, en meer speciﬁek in radicaal theorie van gegradeerde ringen. In het eerste hoofdstuk worden radicalen en gegradeerde ringen gedeﬁnieerd. Er worden respectievelijk voorbeelden gegeven van verscheidene radicalen zoals het nil, Jacobson, priem en Brown-McCoy radicaal alsook van gradaties op verschillende ringen. Het is nuttig om deze radicalen te bestuderen omdat het kan helpen bij het beantwoorden van het Koethe vermoeden. Het Koethe vermoden stelt dat een ring die geen niet nulle tweezijdige nil idealen heeft, ook geen niet nulle eenzijdige nil idealen heeft. Voor het Koethe vermoeden bestaan er equivalente stellingen die gebruik maken van het nil radicaal of het Jacobson radicaal. In het tweede hoofdstuk wordt een chronologisch overzicht gegeven van relevante stellingen en vragen in dit onderzoeksgebied. In het derde hoofdstuk wordt er bewezen dat het Jacobson en nil radicaal in een Z-gegradeerde ring homogeen zijn. Dit zijn respectievelijk resultaten van Bergman in 1975 en Smoktunowicz in 2006. Verder bewees Smoktunowicz dat de homogene deelringen van een Jacobson radicaal ring Jacobson radicaal zijn. Men kan zich afvragen voor welke semigroepgradaties de stellingen van Smok- tunowicz waar blijven. Dit is het uitgangspunt van hoofdstuk vier. Mazurek, Nielsen en Ziembowski gaven recent antwoord op deze vragen. Als de semigroep S abels, annuleerbaar en torsie-vrij is, dan is het nil radicaal van een S-gegradeerde ring homogeen. Verder is een nodige voorwaarde voor de semigroep S opdat de homogene deelringen van een Jacobson radicaal S-gegradeerde ring Jacobson radicaal zou zijn, dat \ T n = ∅ n≥1 voor alle eindig voortgebrachte deelsemigroepen T van S. Een voorbeeld zal tonen dat dit geen voldoende voorwaarde is. Als slot worden worden in het laatste hoofdstuk nog enkele open vragen opgelijst.

ii Contents

Acknowledgements i

Samenvatting ii

Introduction 1

1 Background 3 1.1 Rings and modules ...... 3 1.2 Introduction to radical theory ...... 4 1.2.1 Definitions ...... 4 1.2.2 Examples of radicals ...... 6 1.2.3 Examples of radical constructions ...... 9 1.2.4 The insignificance of the multiplicative unit in radicals . . . . 12 1.3 Graded rings ...... 13 1.3.1 Definitions ...... 13 1.3.2 Graded ideals ...... 15 1.3.3 Examples ...... 16 1.4 The radical of a graded ring ...... 18

2 History 20 2.1 A radical chain ...... 20 2.2 Historical overview ...... 21

3 Radicals of Z-graded rings 27 3.1 Introduction ...... 27 3.2 Amitsur’s result on the Jacobson radical of R[X] ...... 27 3.3 Bergman’s results ...... 31 3.3.1 Simple modules and ﬁnite extensions ...... 31 3.3.2 The Jacobson radical of Z-graded rings ...... 32 3.3.3 The Jacobson radical of rings graded by a torsion-free abelian group ...... 34

iii 3.4 Smoktunowicz’s results ...... 35 3.4.1 The prime radical of Z-graded rings ...... 35 3.4.2 The nil radical of Z-graded rings ...... 36 3.4.3 Jacobson radical rings that are Z-graded ...... 39 3.4.4 Graded-nil rings and the Brown-McCoy radical ...... 40

4 Radicals in semigroup graded rings 41 4.1 The nil radical in semigroup graded rings ...... 41 4.1.1 Ideal functions ...... 42 4.1.2 Cancellative semigroups ...... 43 4.1.3 Torsion-free semigroups ...... 44 4.1.4 Abelian, cancellative, torsion-free semigroups ...... 44 4.2 Homogeneous Jacobson radical subrings of Jacobson radical rings ...... 45 4.2.1 A power series construction ...... 46 4.2.2 A large class of examples ...... 47 4.2.3 More examples ...... 47

5 Open questions 49

Bibliography 50

iv Introduction

The scope of this thesis takes places in non-commutative ring theory. One of the important open problems is whether the Koethe conjecture is true or false.

Conjecture. A ring R with no nonzero nil (two-sided) ideals does not have a nonzero nil one-sided ideal.

Since 1930 mathematicians have tried to find a solution to this question up and till today without result. A proven method of tackling difficult questions is to find equivalent statements. As the Koethe conjecture talks about nil ideals, there exist equivalent statements concerning nil rings, the nilradical and also the Jacobson radical. An other tactic to handle problems is by splitting it up into smaller parts or dealing with specific but important cases. For example, consider rings with some inherent structure such as being a graded ring.

The main question of this thesis is to determine when radicals in graded rings are homogeneous. We focus on the nil and Jacobson radical as these are the most know and important ones, but we will also deal with examples of other radicals. The main emphasis is on Z-graded rings, but some results will be stated in a more general context as torsion-free abelian group graded and torsion-free cancellative abelian semigroup graded rings.

The first chapter starts with some background on the two main concepts mentioned in the title of this thesis, namely radicals and graded rings. First we give the definition of a radical class and consider the nil, Jacobson, prime and Brown-McCoy radical. Next we give the definition of graded rings and graded ideals. We consider examples of natural gradings but also of some more creative gradings. The second chapter gives an overview of the field of radicals since Koethe postu- lated his conjecture in 1930. Theorems, equivalent statements and more questions related to the Koethe conjecture are put in chronological order. In the third chapter the nil and Jacobson radical in Z-graded rings are studied. In 1965 Amitsur showed that the Jacobson radical of the polynomial ring R[X] is

1 homogeneous. Bergman showed in 1975 that the Jacobson radical of every ring graded by Z, and more general graded by a torsion-free abelian group, is homogeneous. It then took 30 years to be proven that the same holds for the nil radical. This was shown by Smoktunowicz in 2006. She also proved that for some Jacobson radical Z-graded rings each subring that is generated by homogeneous elements is Jacobson radical. We will give complete proves for these results. It is a natural question to investigate for which graded rings (by an arbitrary semigroup) the above results remain valid. In 2006 Smoktunowicz posted two concrete questions in this context. This is the topic of chapter four. First we will show a recent result of Mazurek, Nielsen and Ziembowski stating that the nil radical of a ring graded by a torsion-free, cancellative, abelian semigroup is homogeneous. They also investigated when Jacobson radical semigroup graded rings are such that all subrings are Jacobson radical. A suﬃcient but not necessary condition is that

\ T n = ∅ n≥1 for every finitely generated subsemigroup T . An example will show that this is not a necessary condition. Complete proofs will be given of the results. In the final chapter we finish by stating some open questions concerning related problems.

2 Chapter 1

Background

The scope of this work takes place in ring theory. After defining what is meant by rings and modules, we will define the two main concepts that are mentioned in the title, i.e. radicals and graded rings. We start with a short introduction to radical theory. We continue with defining and giving examples of graded rings. Afterwards we consider the definition of a graded radical.

1.1 Rings and modules

An associative ring (R, +, ·) is a set R with two binary operations, the addition + and the multiplication ·, such that (R, +, 0) is an abelian group with a neutral element 0, (R, ·) is a semigroup, such that the following conditions (the distributivity laws) are satisﬁed for all r, s, t ∈ R: r(s + t) = rs + rt and (s + t)r = sr + tr. Note that in this deﬁnition of a ring, and therefore in this work, an associative ring R not necessarily contains a multiplicative unit 1. As an example consider the ring 2Z. We get a non-commutative example if we consider the n-by-n matrices over 2Z. Another example is the ring consisting of the 3-by-3 matrices over a ring with all entries in the third row equal to 0. These examples also show that an ideal in a ring is itself a ring but it does not necessarily contain a multiplicative identity.

Recall that a left module M over a ring R is an abelian group M together with an opereration R × M → M, called the scalar multiplication, such that the following conditions are satisﬁed for all r, s ∈ R and x, y ∈ M: r(x + y) = rx + ry, (r + s)x = rx + sx, (rs)x = r(sx).

3 When R does not contain a multiplicative unit we also require that RM 6= 0. Simi- larly a right R-module is an abelian group with scalar multiplication M × R → M satisfying the dual right conditions of those mentioned above. A submodule of an R-module M is an additive subgroup N that is closed under the given scalar multiplication. The length of a module is the largest length of any of its strict chains of submodules. A left R-module M is called cyclic if it is generated by one element m ∈ M. If R contains a multiplicative unit this is Rm = {rm | r ∈ R}. When R does not contain a multiplicative unit Rm is not a module. A cyclic module is then the smallest module generated by 1 element, i.e. Zm + Rm. Similarly, one calls a right R-module N cyclic if N = yR or N = yZ + yR respectively for some y ∈ N. A module M over a ring R is called a simple (left or right) module if {0} and M are the only submodules. Equivalently, a module M is simple if and only if every cyclic submodule generated by a nonzero element of M equals M. Clearly every simple module is cyclic. A module is said to be semisimple if it is the direct sum of simple submodules. For more properties of (simple) modules we refer to [42].

1.2 Introduction to radical theory

From now on let R be an associative ring not necessarily containing a multiplicative unit. The content of this short introduction is based on [17].

1.2.1 Deﬁnitions One of the main problems in ring theory is to discover the structure of rings, which is not a trivial task. Wedderburn (1908) and Koethe (1930) considered a ”bad” property γ of rings and discarded the ideal γ(R) satisfying this bad property, called the γ-radical of the ring R such that the quotient ring R/γ(R) does not have this bad property anymore. It is clear that the larger the class of bad rings, the less there is left after discarding this class. We may now explain how this notion can be introduced formally.

Deﬁnition 1.2.1. A non-empty class γ of rings satisfying the following properties:

(a) the class γ is homomorphically closed, i.e. R ∈ γ and R Q imply that Q ∈ γ (where stands for a surjective homomorphism), P (b) for every ring R, the sum γ(R) = ICR,I∈γ I is in γ, (c) γ(R/γ(R)) = 0 for every ring R,

4 is called a radical class in the sense of Kurosh and Amitsur, or simply a radical. The ideal γ(R) is called the γ-radical of R. A ring R is called a γ-radical ring if R ∈ γ, that is, γ(R) = R. When γ(R/γ(R)) = 0 the ring R is called γ-semisimple. From the deﬁnition follows immediately next proposition. Property 1.2.2. If γ is a radical class, R ∈ γ a ring and I ⊆ γ(R) an ideal, then γ(R/I) = γ(R)/I. Proof. The canonical homomorphism R → R/I induces a homomorphism γ(R) → γ(R/I) with kernel γ(R) ∩ I = I. Hence, γ(R)/I ⊆ γ(R/I). converse, the canonical homomorphism R/I → R/γ(R) maps γ(R/I) onto zero, so γ(R/I) ⊆ γ(R)/I. Clearly (c) is a special case of this proposition. We will use this regularly in proofs without referring to it every time.

We say γ has respectively the inductive property or is closed under extensions if the following respective condition holds.

(b) If I1 ⊆ ... ⊆ Iλ ⊆ ... is an ascending chain of ideals of a ring R and if each Iλ is in γ, then ∪Iλ is in γ. (c) If I is an ideal of the ring R and if both I and R/I are in γ, then R itself is in γ. These conditions give us an easier way to discover if a class of rings is a radical class. Theorem 1.2.3. A class γ of rings is a radical class if and only if the following properties hold: (a) γ is homomorphically closed, (b) γ has the inductive property, (c) γ is closed under extensions. Proof. We prove the equivalence with the deﬁnition in two steps. Assuming conditions (a) and (b) we prove that condition (c) is equivalent with condition (c). Suppose that (c) holds and that I and R/I are both in γ, where R is a ring and I is an ideal of R. By (b) we have that I ⊆ γ(R) and R/I R/γ(R) ∼= γ(R)/I is a homomorphic image of R/I. Hence, by (a), R/γ(R) is in γ. By (b) and (c) on the other hand we have that

5 0 = γ(R/γ(R)) = R/γ(R). This shows that R = γ(R) and thus R is in γ. So we have shown (c). To prove the other implication, we suppose (c) holds. If we assume that γ(R/γ(R)) 6= 0 then γ(R/γ(R)) = K/γ(R) for some ideal K of R. Both γ(R) and K/γ(R) are in γ and so (c) tells us also that K is in γ. We now have that K ⊆ γ(R) and K/γ(R) = 0 what gives a contradiction. So condition (c) holds and we have proved the equivalence between (c) and (c). For the second part of this proof we assume the conditions (a) and (c) hold on a class γ of rings and show the equivalence between (b) and the inductive property (b). We suppose (b) holds. Put Q = ∪Iγ. To prove (b), let I1 ⊆ I2 ⊆ ... ⊆ Iλ ⊆ ... be an ascending chain of ideals of a ring R and assume each Iλ is in γ. Then by (b) each Iγ is contained in γ(Q) and therefore Q = γ(Q) is in γ and so (b) holds. For the converse, suppose (b) holds. Then, by Zorn’s lemma, we obtain a maximal γ-ideal Q in R. If K is any γ-ideal of R, then (Q + K)/K ∼= Q/(Q ∩ K) and this is in γ by (a). So both K and (Q + K)/K are in γ and by (c) also Q + K is in γ. Since Q is maximal with respect to this property, K must be in Q and thus γ(R) = Q which is in γ. We have proven that (b) holds.

Deﬁnition 1.2.4. A class of rings γ is called regular if for every ring R ∈ γ, every nonzero ideal of R has a nonzero homomorphic image in γ.

Deﬁnition 1.2.5. A class of rings γ is called hereditary if I C R ∈ γ implies I ∈ γ. We ﬁnish this section with a useful property. We will use this without referring to it every time.

Property 1.2.6. Let R be a ring and f any automorphism of R then f(γ(R)) = γ(R).

1.2.2 Examples of radicals There exist multiple speciﬁc radicals in a ring that help understand the structure of the considered ring. As is well known, a nilpotent element x of a ring is an element for which there exists a positive natural number n such that xn = 0. A ring is said to be nil if all its elements are nilpotent. A ring is said to be nilpotent of index n if Sn = 0 and Sn−1 6= 0. A locally nilpotent ring is a ring in which every ﬁnitely generated subring is nilpotent. Every nilpotent ring is locally nilpotent but the converse is not always true. Every locally nilpotent ring is nil.

The ﬁrst radical class of a ring R is the one discovered by Koethe in 1930, [17].

Deﬁnition 1.2.7. The nil radical class is the class consisting of all nil rings.

6 We check if this class meets the properties of the deﬁnition. For every epimor- phism ϕ : R → B and every element a ∈ R we have 0 = ϕ(0) = ϕ(an) = ϕ(a)n for some n. So B is nil and thus N is homomorphically closed. It has also the inductive property as ∪Iλ is in γ for each ascending chain of ideals. We check that N is closed under extensions. We take a ring R so that I and R/I are both in N . Then for every a ∈ R, a = a + I is nilpotent and thus for some n ≥ 1 is an ∈ I. But then an must also be nilpotent as I is a nil ring. So there exists a k ≥ 1 such that (an)k = ank = 0 what makes a nilpotent and so is R nil. The radical N (R) is called the nil radical of R. Clearly this is the largest nil ideal of R, also called the upper nil radical of R.

Our next example is the most known one.

Deﬁnition 1.2.8. In a ring R one can consider an other operation, called the circle operation and denoted ◦. It is deﬁned as follows

a ◦ b = a + b − ab for all a, b ∈ R. As we work in an associative ring this operation is also associative. Clearly (R, ◦) is a monoid with 0 as unit element. The Jacobson radical class J is the class of rings R such that (R, ◦) a group.

The class J is homomorphically closed and has the inductive property. To see that it is closed under extensions, let R be a ring and I an ideal of R such that I and R/I are in J . Let a ∈ R. Then a + I = a ∈ R/I, and there exists an x ∈ R/I such that x ◦ a = 0, that is x ◦ a ∈ I. Hence there exists an element y ∈ I such that y ◦ (x ◦ a) = (y ◦ x) ◦ a = 0. Thus R also is in the Jacobson radical class J . The Jacobson radical associated to this radical class is denoted as J (R).

Note that if R is a ring and a, b ∈ R then

(1 − a)(1 − b) = 1 − a − b + ab = 1 − (a + b − ab) or equivanlently a + b − ab = 1 − (1 − a)(1 − b).

The circle operation allows to deﬁne a new type of inverse in a ring.

Deﬁnition 1.2.9. Let R be a ring. An element x ∈ R is called left quasi-regular if it has an inverse y ∈ R for the circle operation, i.e. y ◦x = y +x−yx = 0. The element y ∈ R is called the left quasi-inverse. Similarly one deﬁnes the right quasi-inverse of a right quasi-regular element. An element is said to be quasi-regular if it is both left and right quasi-regular.

7 A ring R is called quasi-regular if all its elements are (left) quasi-regular. Every nilpotent element of R is quasi-regular. Indeed, suppose x ∈ R and xn = 0 for some n > 1. Take y = −x − x2 − ... − xn−1. Now y ◦ x = 0 = x ◦ y and therefore y is the quasi-inverse of x. With the concept of quasi-regular elements one can define the Jacobson radical class in an equivalent way. Definition 1.2.10. The Jacobson radical class J is the class of all quasi-regular rings. The Jacobson radical J (R) of a ring R is the sum of all quasi-regular ideals of R and also the sum of all quasi-regular left (right) ideals of R The ◦ operation was introduced by Jacobson. Some authors, like Smoktunowicz, use a different definition, i.e. a ∗ b = a + b + ab. One can see it is easy to change from one set-up to the other as (−a) ∗ (−b) = −(a ◦ b). Saying that a ∈ R has a (left) quasi-inverse for the circle operation is equivalent to say that −a has a (left) inverse for the star operation. We will always clearly indicate which set up we use in later sections.

There exist more equivalent definitions for the Jacobson radical. Depending on the context one can use the most suitable description. These notions are left-right symmetric. Recall that a left ideal I of a ring R is called modular if there exists an element e ∈ R such that a − ae ∈ I for every a ∈ R. A ring R is called a primitive ring if it has a faithful simple module M, that is a the annihilator of the simple module M is zero. An ideal P of R is called primitive if R/P is a primitive ring. Definition 1.2.11. The Jacobson radical J (R) of a ring R is the intersection of all maximal modular left ideals, as well as the intersection of all left primitive ideals of R. Definition 1.2.12. The Jacobson radical J (R) of a ring R is {r ∈ R | rM = 0 for every simple left R-module M}. The proofs of the equivalency between all these definitions can be found in [17].

Notice that for an ideal I C R we have that J (R) ∩ I ⊆ J (I).

We call a ring Jacobson radical or nil if the ring equals its respective Jacobson or nil radical. Note that this is not possible if R has a multiplicative unit 1 ∈ R and 1 6= 0. It is obvious that a subring of a nil ring is nil, but a subring of a Jacobson radical ring need not be Jacobson radical. Consider the power series ring with constant term zero R[[X]]X. This ring is equal to its Jacobson radical J (R[[X]]X). The

8 polynomial ring R[X]X is naturally embedded in R[[X]]X but its Jacobson radical J (R[X]X) is zero.

We continue with a third example of a radical class. Definition 1.2.13. The Levitzki radical class L is the class of all locally nilpotent rings. Clearly the class L is homomorphically closed and satisfies the inductive property. We now check if L is closed under extensions. Take a ring R and an ideal I of R such that I and R/I are both in L. Take S a subring of R generated by a finite set {r1, ..., rn} and consider the subring S of R/I generated by the finit set of cosets {r1 + I, ..., rn + I}. Then S is a subring of R/I and therefore nilpotent. This means there exists a k such that Sk = 0 or that Sk ⊆ I. Now Sk is finitely generated by all k products ai1 ...aik of k factors from {a1, ..., an}. Hence S is nilpotent as it is in I. So there exists an l such that (Sk)l = Skl = 0 for some 1 ≤ l. Hence also S is nilpotent and R is in L. We have proven that L is a radical class and thus L(R) is the largest locally nilpotent ideal of R.

1.2.3 Examples of radical constructions Clearly not every non-empty class of rings κ is a radical class. We continue with constructing two types of radicals, the lower and the upper radicals. As examples we consider the prime radical and the Brown-McCoy radical. The description of these radical constructions is from ([17], Section 2.2 Radical constructions). We can always construct the smallest radical class containing κ. One can do so by taking the intersection of all radical classes containing a class of rings κ. By Theorem 1.2.3, this is also a radical class. We denote this intersection by Lκ and is called the lower radical determined by κ. There exist other ways to conctruct Lκ. We give the Tangeman-Kreiling lower radical construction as example.

We start with a class of rings κ and define κ1 as the class of rings that are a homorphic image of a ring P ∈ κ. We call this the homomorphic closure of κ. We proceed inductively. If for all ordinals µ < λ the class κµ has been defined, then one defines κλ as the class of rings R for which there exists an ideal I C R such that I ∈ κλ−1 and R/I ∈ κλ−1, when λ − 1 exists. When λ is a limit ordinal, one defines κλ as the class of rings R such that R is the union of an ascending chain of ideals, each one in some κµ for µ < λ.

Theorem 1.2.14. The lower radical of κ equals the union of κλ over all ordinals λ, i.e. Lκ = ∪κλ.

9 Proof. We show first that κµ ⊆ κλ when µ < λ. It is easy to see that κ ⊆ κ1 and that κµ ⊆ κλ for any limit ordinal λ. To check that κµ ⊆ κµ+1 we note that the zero ring {0} is in κ1 and by induction in every κλ for λ ≥ 1. Then for every R ∈ κµ we take I = 0 ∈ κµ and R = R/I ∈ κµ. Thus R is in κµ+1 by definition. Next we want to prove that ∩κλ is a radical class. We start with the homomorphic closure property. Clearly κ1 is homomorphically closed. Assume that the same is true for κµ for µ < λ. Take R ∈ κλ and consider any image R/I of R. When λ is a limit ordinal, there exists a chain {Kι} of ideals of R such that ∪Kι = R and each Kι belongs to a class κµ for µ < λ. Then {(I + Kι)/I} is a chain of ideals of R/I such that R/I is the union of this chain. Since ∼ (I + Kι)/I = Kι/(I ∩ Kι), each of these ideals belongs to some κµ with µ < λ. This means that R/I ∈ κµ. If λ − 1 exists, then R contains an ideal J such that J and R/J are in κλ−1. By the hypothesis on κλ−1 we have ∼ (J + I)/I = J/(J ∩ I) ∈ κλ−1 and R/I ∼ (J+I)/I = R/(J + I) ∈ κλ−1, which gives us that R/I ∈ κλ. By transfinite induction every κλ is homomorphically closed and so is ∪κλ. Next we check the inductive property. Let I1 ⊂ ... ⊂ Iι ⊂ ... be a strictly ascending chain of ideals of a ring R such that each Iι is in ∪κλ. Since R is a set and the construction is monotonic, there exists an ordinal µ such that Iι ∈ κµ for each ι. If we now take any limit ordinal ν bigger than µ, we have ∪Iι ∈ κν ⊆ ∪κλ what proves the inductive property. Now it remains only to prove that ∪κλ is closed under extensions. Assume that both I and R/I are in κλ. By definition is R in κλ+1 and so R is in ∪κλ. This proves that ∪κλ is a radical class by Theorem 1.2.3. If γ is any radical class containing κ, then it must also contain κ1. If all κµ are contained in γ for µ < λ, then κλ must also be contained in γ by the properties (b) or (c) of the radical class. By transfinite induction we have that ∪κλ ⊆ γ, proving ∪κλ = Lκ. An example of a class of rings that is not a radical class is the class of all nilpotent rings. This class is homomorphically closed and closed under extensions but does not satisfy the inductive property. If we apply the lower radical construction on this class we get the prime radical class P. For the radical P(R) we use the following more usable description. First, reacll that a proper ideal P of a ring R is called prime if and only if for all a, b ∈ R, aRb ⊆ P implies a ∈ P or b ∈ P . A ring R is called prime if {0} is a prime ideal in R.

Deﬁnition 1.2.15. The prime radical P(R) of a ring R is the intersection of all prime ideals in R. This radical is also called the lower nil radical or the Baer radical.

10 Until now we have been studying the radical classes of rings with a ’bad’ property, like nil rings. However, there exist also classes of ’good’ rings, like division algebras, which we do not want to discard. We would like to ﬁnd a radical class τ that does not contain a ’good’ ring. In particular we want for the ’good’ class γ that τ ∩ γ = {0}. Next theorem gives some equivalences to see if a class γ is a radical clase. It will be used in the proof of the construction of the radical we are looking for. The proof of following theorem can be found in ([17], page 23).

Theorem 1.2.16. For any class γ the following conditions are equivalent

1. the class γ is a radical class,

2. (R1) if A ∈ γ then holds that for every A B 6= 0 ther is a C C B such that 0 6= C ∈ γ, (R2) if A is a ring in the universal class of rings A and for every A B 6= 0 there is a C C B such that 0 6= C ∈ γ, then A ∈ γ, 3. the class γ satisﬁes (R1), (b), and (c).

Theorem 1.2.17. If γ is a regular class of rings, then the class Uγ, of of rings R such that R has no nonzero homomorphic image in γ, is a radical class, γ ∩Uγ = {0} and Uγ is the largest radical such that the intersection with γ is zero.

Proof. To prove that Uγ is a radical class we will prove conditions (R1) and (R2) from Theorem 1.2.16. The first condition will be proven by contradiction, this means we will prove that if A has a nonzero homomorphic image B such that B has no nonzero ideal in Uγ, then A/∈ Uγ. If we consider such B it can not be in Uγ and therefore B must have a homomorphic image D in γ. The composition of surjective homomorphisms A B D gives a nonzero homomorphic image of A in γ. This showes that A is not in γ which proves (R1). For the second condition (R2) we again use contradiction. Assume A is not in γ. By definition of Uγ this means that A has a nonzero homomorphic image D in γ. Since the class γ is regular, also all nonzero ideals of D have a nonzero homomorphic image in γ. This is a contradiction. By Theorem 1.2.16 the class Uγ is a radical class. We continue to prove that it is also the largest radical with intersection zero. Suppose that % is a radical and %∩γ = {0}. If % * Uγ, then there would exist a ring A ∈ % but A/∈ Uγ. By definition this gives us that A has a nonzero homomorphic image in γ, but this image also has to be in %, a contradiction. Hence % ⊆ Uγ is the largest radical such that γ ∩ Uγ = {0}. The construction of such a class is called the upper radical class construction and Uγ is then the upper radical of the class γ. We give an example.

11 Deﬁnition 1.2.18. Consider the class U of all nonzero simple rings with unit. The upper radical U(R) is called the Brown-McCoy radical. In a ring with an identity element, the Brown-McCoy radical is the intersection of all maximal two-sided ideals.

It turns out that U(R) is the intersection of all ideals I of R such that R/I is a nonzero simple ring with unit. A ring R is Brown-McCoy radical if it cannot be homomorphically mapped onto a simple ring with 1.

1.2.4 The insignificance of the multiplicative unit in radicals In all definitions in the previous section, except for the Brown-McCoy radical, there was no need for a multiplictive unit. But is this theory useless if one works in a ring that does not contain a multiplicative unit? Or what to do if one needs one for a specific and needed step in a proof? We will show that having a multiplicative unit will not change the results that will be proven in further chapters. Namely, the Jacobson and nil radical of R and R1 are the same, where R1 is obtained by adding a multiplicative unit to R as explained below.

To add a multiplicative unit to a ring R one can follow the next construction. De- fine R1 as the set Z×R with an addition defined componentwise and a multiplication defined by

(m, x)(n, y) = (mn, nx + my + xy), for all m, n ∈ Z and all x, y ∈ R. Cleary (R1, +, (0, 0)) is a group because (Z, +, 0) and (R, +, 0) are groups. We also have that (R1, ·, (1, 0)) is monoid with multiplicative unit (1, 0). Indeed, for all m ∈ Z, x ∈ R (m, x)(1, 0) = (m1, 1x1 + m0 + x0) = (m, x) = (1, 0)(m, x).

Lastly we check the distributivity laws. Take m, n, l ∈ Z and x, y, z ∈ R, then (m, x)((n, y) + (l, z)) = (m, x)(n + l, y + z) = (m(n + l), (n + l)x + m(y + z) + x(y + z)) = (mn + ml, nx + lx + my + mz + xy + xz) = (mn, nx, my, xy) + (ml, lx + mz + xz) = (m, x)(n, y) + (m, x)(l, z).

This proves that R1 indeed is a ring with R embedded naturally asa subring in R1.

We will now show that J (R) = J (R1). Recall that an element x ∈ R is right quasi-regular if there exist an y ∈ R such x ◦ y = 0. In unital rings this means that

12 1 − x is right invertible as (1 − x)(1 − y) = 1 − (x + y − xy) = 1. Therefore x ∈ J (R1) if and only if 1 − xy is right invertible for all y ∈ R1. We now determine the elements in J (R1). Take (m, x) ∈ J (R1) then (1, 0) − (m, x) must be right invertible, i.e. there exists (n, y) ∈ R1 such that

(1, 0) = (1 − m, x)(n, y) = ((1 − m)n, nx + (1 − m)x + xy).

Hence (1 − m)n = 1. As m, n ∈ Z this gives us 1 − m = 1 or 1 − m = −1 and therefore m = 0 or 2. But (1, 0) − (m, x)(−1) = (1, 0) + (−m, −x) also has to be right invertible. The same computation shows that 1 + m = 1 or 1 + m = −1 and therefore m = 0 or −2, so m = 0. Thus the elements in the Jacobson radical J (R1) are of the form (0, x). Hence J (R1) is an ideal in R. Now if (0, x) ∈ J (R1) then (1, 0) − (0, x) = (1, −x) is right invertible, that is there exist a (n, y) ∈ R1 so that (1, x)(n, y) = (1n, −nx + 1y − xy) = (1, 0). The ﬁrst term gives us that n = 1 and so −(x + (−y) − x(−y) = 0, that is x is quasi-regular. It follows that J (R1) ⊆ J (R). As J (R) is an ideal of R1 and all its elements are quasiregular, we obtain that J (R) ⊆ J (R1). So, indeed, J (R1) = J (R). Also for the nil radical we have N (R) = N (R1). For this note that if (m, x) ∈ R1 is nilpotent then m = 0, as the only nilpotent element in Z is 0. Thus, nil ideals of R1 are contained in R. Since ideals of R are also ideals of R1, it follows that the sum of nil ideals of R1 is the same as the sum of nil ideals of R. Therefore N (R) = N (R1).

Thus we can conclude that if it is necessary to have a multiplicative unit in the ring R, one can add one without causing any problems for the respective Jacobson and nil radical.

1.3 Graded rings

In this section we study graded rings. First we give a deﬁnition and some useful properties. Next we give the deﬁnition of graded ideals. Afterwards we consider examples of graded rings. The contents of this section is based on [43].

1.3.1 Deﬁnitions Deﬁnition 1.3.1. Let S be a semigroup. A ring R is said to be an S-graded ring if

M R = Rg, g∈S a direct sum of additive subgroups Rg, such that RgRh ⊆ Rgh. The direct sum de- composition is referred to as gradation or grading.

13 If the semigroup S is clear from the context then an S-graded ring is simply called a graded ring. A grading such that RgRh = Rgh, for all g, h ∈ S is called a strong grading. The additive subgroups Rg are called homogeneous components, the elements of Rg are called homogeneous elements of degree g and we denote by h(R) = ∪g∈SRg the set of all homogeneous elements. Every r ∈ R has a unique P expression r = g∈S rg, with rg ∈ Rg, for every g and rg = 0 for almost all g. The terms rg are called the homogeneous components of r. By supp(r) we denote the support of an element r, that is {s ∈ S | rs 6= 0}. A graded ring R is called graded- L L nil if all its homogeneous elements are nilpotent. Let R = g∈S Rg and Q = g∈S Qg be S-graded rings. A ring homomorphism f : R → Q is a graded ring homomorphism if f(Rg) ⊆ Qd. Deﬁnition 1.3.2. Let S be a semigroup. An element a ∈ S is left cancellative if ab = ac implies that b = c for all b, c ∈ S. If every element in S is left cancellative then S is called left cancellative. In a similar way one deﬁnes right cancellative. If a semigroup S both left and right cancellative then we call S cancellative.

Lemma 1.3.3. Assume R is an S-graded ring, where S is a cancellative monoid with identity e. If 1 ∈ R then 1 ∈ Re. P Proof. Let 1 = g∈S rg with rg ∈ Rg. For any sh ∈ Rh, h ∈ S we have that P sh = sh1 = g∈S shrg with shrg ∈ Rhg. Therefore and because of the cancellative condition, shrg = 0 when g 6= e and consequently srg = 0 for any s ∈ R. In particular for s = 1 we get that rg = 0 for any g 6= e. Hence 1 = re ∈ Re. We give two useful properties for graded rings from [34]. By R∗ we denote the multiplicative group consisting of the invertible elements of R. So,

R∗ = {r ∈ R | rr0 = 1 = r0r}.

Property 1.3.4. Let G be a group and R a G-graded ring with multiplicative unit 1.

1. The ring R is strongly graded if and only if 1 ∈ RgRg−1 for any g ∈ G. 2. If r ∈ h(R) and r ∈ R∗ then r0 ∈ h(R).

Proof. For the ﬁrst statement suppose that 1 ∈ RgRg−1 for any g ∈ G. Then, because 1 ∈ Re by Lemma 1.3.3 we have that for g, h ∈ G

Rgh = ReRgh = (RgRg−1 )Rgh = Rg(Rg−1 Rgh) ⊆ RgRh

14 and therefore Rgh = RgRh. Hence, R is strongly graded. Conversely, if R is strongly graded then, for g ∈ G, RgRg−1 = Re. As 1 ∈ Re by Lemma 1.3.3, we get that 1 ∈ RgRg−1 . −1 P −1 To prove the second statement, assume that r ∈ Rh. Write r = g∈G(r )g with −1 −1 P −1 −1 (r )g ∈ Rg. Then 1 = rr = g∈G r(r )g. Since 1 ∈ Re and r(r )g ∈ RhRg ⊆ −1 −1 −1 −1 Rhg, we have that r(r )g = 0 for g 6= h . Therefore r = (r )h−1 ∈ Rh−1 . L Let A be a subring of the S-graded ring R = g∈S Rg. We say that A is anS- graded subring of R if

P A = g∈S Ag, with Ag = A ∩ Rg. Let S be a semigroup and R = ⊕g∈SRg an S-graded ring. If X is a subset of S then we put

P RX = x∈X Rx.

If X is a subsemigroup of S then RX is an X-graded ring that is a graded subring of R. One can also consider RX as an S-graded ring by taking the homogeneous components of degree s ∈ S \ X equal to zero.

For modules we have a corresponding concept. An S-graded module over an S- graded ring R is an R-module M such that

M M = Mg and RgMh ⊆ Mgh g∈S with Mg an Rg-submodule. A submodule N of an S-graded module M is S-graded if N = ⊕g∈S(N ∩ Mg). A graded module M is called a simple-graded module of a graded R-module M is if M 6= {0} and {0} and M are the only graded submodules.

1.3.2 Graded ideals Now we will have a closer look at the (left, right) ideals of semigroup graded rings.

Lemma 1.3.5. Let S be a semigroup. For an ideal I of an S-graded ring R, the following three properties are equivalent.

(1) For every a ∈ I, all homogeneous components of a belong to I.

L (2) I = g∈S(I ∩ Rg). (3) I is generated as an (left, right) ideal by homogeneous elements.

15 Proof. (1) ⇒ (2) By (1) every element a of I has a unique expression of the form P a = g ag, a ﬁnite sum, with ag ∈ I ∩ Rg for every g ∈ S. So we have (2). (2) ⇒ (3) By (2) I is generated by ∪g∈S(I ∩ Rg) which is a set of homogeneous elements. (3) ⇒ (1) Let I C R be an ideal in R generated by homogeneous elements. If α ∈ I then we can write

n X α = βirgi γI i=1 P with βi, γi ∈ R, rgi ∈ I ∩ Rgi . As R is graded by S we also write βi = h∈S(βi)h, γi = P h0∈S(γi)h0 with (βi)h ∈ Rh and (γi)h0 ∈ Rh0 . Now we get X X X 0 α = (βi)hrgi (γi)h . i h h0

0 0 For each 1 ≤ i ≤ n the term (βi)hrgi (γi)h is part of I ∩ Rhgih . An other way to consider α is as a sum of homogeneous elements of I X X 0 α = (βi)hrgi (γi)h 0 s∈S s=hgih so the homogeneous compnents are X 0 αs = (βi)hrgi (γi)h ∈ I. 0 s=hgih

Deﬁnition 1.3.6. A graded ideal of a graded ring R is an ideal, satisfying any of the equivalent conditions of the previous lemma.

For an S-graded ring R and a homogeneous ideal I we have that R/I also is a graded ring with (R/I)g = Rg/Ig for g ∈ S, called the quotient gradation. A graded ideal I of R is a graded-maximal ideal if I 6= R and I is not contained in any other proper graded ideal of R.

1.3.3 Examples

Every ring can be graded by a monoid S in a trivial way by putting R = Re and Rh = 0 for all e 6= h ∈ S. We will now see some more interesting examples. Semigroup rings are a ﬁrst example of naturally graded rings. Recall that the semigroup ring R[S] of a semigroup S over a ring R is the free left R-module with basis the set S and with multiplication deﬁned by

(as)(br) = (ab)(st),

16 for a, b ∈ R, s, t ∈ S. Here ab is the multiplication in R and st is the multiplication in S. Now R[S] is S-graded with homogeneous components of degree s equal to Rs. For a general reference of semigroup rings we refer the reader to [35]. In case S is the free abelian monoid with basis X1, ..., Xn then R[S] = R[X1, ...Xn], the polynomial ring in n commuting variables. The following example shows that this ring can be considered with other gradations.

Example 1.3.7. The polynomial ring R = k[X1,X2, ..., Xr] over a ring k is a Z- graded with Rn consisting of the homogeneous polynomials of degree n, with n a non- negative integer, and Rn = 0 if n < 0. Homogeneous polynomials are polynomials whose terms all have the same total degree. In particular, we have that R0 = k and r X i1 i2 ir X Rn = { ki1,...,ir X1 X2 ...Xr | ij = n}. j=1 If R is G-graded and N C G is a normal subgroup of G, then R is G/N-graded with homogeneous components RgN . Example 1.3.8. Let S be the set of all nonzero homogeneous elements in a Z-graded integral commutative domain R. Then also the localization S−1R of R with respect to S is a Z-graded ring with homogeneous components of degree n ∈ Z equal to −1 −1 (S R)n = {r/s ∈ S R | s ∈ S ∩ Rr, r ∈ Rn+r}, where Rn = 0 for n < 0.

In a matrix ring Mn(k) over a ring k with unit 1 we denote by eij the matrix with 1 in the (i, j)-entry and zeros in all other entries.

Example 1.3.9. Let Zn be the cyclic group of order n. The matrix ring Mn(k) over a ring k is Zn-graded. Indeed, write Mn(k) = V0 ⊕ ... ⊕ Vn−1, where Vi is the k-span of the matrix units eab with b − a = i (mod n). Clearly one has that ViVj ⊆ Vi+j, where one takes i + j modulo n. For n = 3 the three homogeneous components are       a11 0 0 0 a12 0 0 0 a13       V0 =  0 a22 0  ,V1 =  0 0 a23 and V2 = a12 0 0  . 0 0 a33 a21 0 0 0 a32 0 There exist other ways to give a grading on a matrix ring. We give two examples of a Z2-grading, the checkerboard grading and a block grading.

Example 1.3.10. [42] Let Z2 be the cyclic group of order 2. Let k be a ring and write Mn(k) = V0 ⊕ V1 where V0, respectively V1, is the span of the matrix units eab with a + b even, respectively odd. For n = 3 the two homogeneous components are     a11 0 a13 0 a12 0     V0 =  0 a22 0  and V1 = a21 0 a23 . a31 0 a33 0 a32 0

17 Example 1.3.11. [34] Let Z2 be the cyclic group of order 2. Let k be a ring and P P write Mn(k) = V0 ⊕ V1, where V0 = 1≤i≤j≤n−1 keij + kenn and V1 = 1≤i≤n−1 kein + P 1≤j≤n−1 kenj. For n = 3 the two homogeneous components are     a11 a12 0 0 0 a13     V0 = a21 a22 0  and V1 =  0 0 a23 . 0 0 a33 a31 a32 0

We notice that the three examples of graded matrix rings are all strongly graded.

Deﬁnition 1.3.12. Let G be a group. A G-graded ring R = ⊕g∈GRg is called a ∗ crossed product if for every g ∈ G there exists ug ∈ R ∩ Rg. Note that it follows that −1 Rg = Rgug ug = Reug and similarly Rg = ugRe. So each ug induces via conjugation an automorphism, say σg, on Re. Furthermore uguh = γ(g, h)ugh for some γ(g, h) ∈ ∗ Re. Hence, the multiplication in R is determined by the multiplication in Re, the ∗ maps σg and γ : G × G → Re. Examples of crossed products can be obtained as follows. Let R be a ring and N a normal sugroup of a group G. Then the group ring R[G] is a crossed product

R[G] = R[N] ∗ [G/N], with homogeneous components of R[G] of degree gN ∈ G/N. As last example we consider the grading of a cyclotomic ﬁeld.

Example 1.3.13. Let p be prime and ζp a primitive p-th rooth of unit in C. The 2 p−1 cyclotomic ﬁeld Q[ζp] = Q ⊕ Qζp ⊕ Qζp ⊕ ... ⊕ Qζp is a Zp-graded ring with homo- i geneous components of degree i ∈ Zp the set Qζp.

1.4 The radical of a graded ring

In this work we will always work with the Jacobson radical as defined in Definition 1.2.8 but for the sake of completeness we mention that there are graded versions of the Jacobson radical. However, as Bergman mentioned in his paper [9], there are differences between the Jacobson radical in the graded sense and the Jacobson radical in the classical version. Bergman used the following definition for the graded Jacobson radical, denoted as JG(R), with G the group by which the ring R is graded.

Deﬁnition 1.4.1. [9] Let G be a group and R a G-graded ring. Then by JG(R) we will denote the ideal which may be characterized equivalently as follows:

1. The set of elements of R annihilating all simple G-graded R-modules.

18 2. The largest homogeneous ideal J ⊆ R such that 1+x is a unit for all x ∈ J ∩Re with e ∈ G the identity element of G.

3. The largest homogeneous ideal J ⊆ R such that J ∩ Re ⊆ J (Re).

P 4. If H is any subgroup of G, and RH denotes the H-graded subring h∈H Rh ⊆ R, then JG(R) can also be described inductively as the largest homogeneous ideal J ⊆ R such that J ∩ RH ⊆ JH (RH ). Passman proved in his book [37] following theorem, that shows the connection between the two notions of the Jacobson radical.

Theorem 1.4.2. Let R be a G-graded ring with G a ﬁnite group.

1. JG(R) is the largest graded ideal contained in J (S).

|G| −1 2. J (R) ⊆ JG(R) ⊆ J (R). Furthermore JG(R) = J (R) if |G| ∈ R.

|G| 3. JG(R) ⊆ RJ (R)R ⊆ JG(R).

Bergman stated in his paper [9] that if G = Z then J (R) ⊆ JG(R). He also stated that if G is ﬁnite it appeared likely for the other inclusion to hold, JG(R) ⊆ J (R). He proved it is true for G ﬁnite solvable.

19 Chapter 2

History

In this chapter we consider a radical chain and give examples of strict inclusions. Afterwards we give a historical overview of the ﬁeld of radicals in graded rings.

2.1 A radical chain

So far we have given five examples of radicals. For a given ring R these radicals are related as follows: P(R) ⊆ L(R) ⊆ N (R) ⊆ J (R) ⊆ U(R). We give some examples that show that these inclusions can be strict. We start with the inclusion between the Jacobson and nil radical. For each local domain R the Jacobson radical J (R) is equal to the unique maximal ideal but the nil radical will be N (R) = {0} as there are no nonzero nilpotent elements. The examples for the other inclusions are less trivial. Baer [4] showed in 1943 that there exist an example that makes the first inclusion strict. For a more recnt work see ([17], page 188). Theorem 2.1.1. There exists a ring R which is locally nilpotent as well as prime, so that P(R) = 0 while L(R) = R. An example that makes the second inclusion strict is due to Golod and Shafarevich [20]. The construction can also be found in ([17], page 190). Theorem 2.1.2. There exists a finitely generated ring R that is in N but is not in L. For some classes of rings one can show that the inclusions are equalalities. For example, we give a result due to Levitzki [30]. Recall that a ring R is Noetherian if every strictly ascending chain of left ideals ends in a finite number of steps. One says that the ring satisfies the ascending chain condition (a.a.c.) on left ideals. We need following lemma. For the complete proof see [30].

20 Lemma 2.1.3. Let R be a noetherian ring. If R has a nonzero nil one-sided ideal, it has a nonzero nilpotent ideal. Theorem 2.1.4. If R is a noetherian ring, then N (R) is nilpotent. Proof. Let R be a noetherian ring. The set of nilpotent ideals of R has a maximal element N. As the sum of nilpotent ideals is nilpotent, N contains all nilpotent ideals. Clearly R/N is noetherian. So, if N (R/N) 6= 0, then by previous lemma we k have that R/N has a nilpotent ideal I/N with I C R. This means that I ⊆ N for some integer k > 1. Hence, I is a nilpotent ideal of R. It follows that I = N and that I/N = 0, a contradiction. We conclude that N (R) ⊆ N. The converse inclusion is clear. Corollary 2.1.5. If R is a noetherian ring, then

P(R) = L(R) = N (R)

2.2 Historical overview

We mention some of the main contributions and questions in the ﬁeld of radicals of polynomial rings and graded rings since 1930. We do this in chronological order. Unless stated otherwise, we use the notation from the previous chapter. For undeﬁned terminology we refer to [37]. 1930 Koethe [23]

• (KC): A ring R with no nil two-sided ideals does not have a nonzero nil one-sided ideal. • Theorem: (KC) holds if and only if every left nil ideal of a ring R is contained in N (R). • A radical γ(R) is called left strong if I is a left ideal of R such that γ(I) = I implies that I is in the radical of R. Question: Is the nil radical N (R) left strong?

1956 Amitsur [2]

∗ • Theorem : J (R[X]) = N[X] with N = J (R[X]) ∩ R C R a nil ideal. • Question: Is J (R[X] = N (R)[X]? • Question: Is J (R[X]) = N (R[X])? • Question: Is N (R[X]) = N (R)[X]?

1972 Krempa [24, 25]

21 • Theorem: For a radical γ holds that γ(R[X]) = (γ(R[X]) ∩ R)[X] for all R if and only if γ(R[X]) ∩ R = 0 implies γ(R[X]) = 0 for all R. • Theorem: (KC) holds if and only if for each ring R holds that J (R[X]) = N (R)[X].

• Theorem: (KC) holds if and only if for each nil ring R holds that M2(R) is nil. • Theorem: (KC) holds if and only if N is a nil ring than N[X] is Jacobson radical. 1973 Lenagan [29] • Theorem: (KC) holds for rings with the right Krull dimension. 1974 Razmyslov [41]

• Let k be a field of characteristic zero. Let X = {X1,X2, ...} be an countable infinite set. A ring R is called a polynomial identity (PI) ring if it satisfies a nontrivial polynomial identity f = 0, i.e. f = f(X1, ..., Xn) is a nonzero element of K < X > such that f(r1, ..., rn) = 0 for all r1, ..., rn ∈ R. Theorem: (KC) holds for (PI) rings. 1975 Bergman [9]

• Theorem∗: If R is a Z-graded ring then J (R) is homogeneous. 1978 Krempa and Sierpinski [26]

−1 −1 • Theorem: If R is a ring then J (R[X,X ]) = I[X,X ] with I C R such that J (R[X]) = I[X]. 1980 Bedi and Ram [5] • Theorem: If σ is an endomorphism of a ring R then J (R[X, σ]) = (J (R)∩ I) + I[X, σ]X, where I = {r ∈ R | rX ∈ J (R[X, σ])}. • Theorem: If σ is an automorphism of a ring R then J (R[X,X−1, σ]) = K[X,X−1, σ], where K is an σ-invariant ideal of R such that for all r ∈ K hold rX ∈ J (R[X,X−1, σ]). • Question: Are the ideals I and K in the previous theorems equal? 1983 Fisher and Krempa [16] • Theorem: Let G be the group of automorphisms on the ring R. (KC) holds if and only if RG = {r ∈ R | rg = r for all g ∈ G} is nil implies R is nil.

22 1987 Levitzki [33, 27]

• A ring R satisfies a.c.c⊕ if R does not have an infinite independent set of left ideals. A ring is called Goldie if R satisfies a.c.c on left annihilators and a.c.c.⊕. Theorem: (KC) holds for Goldie rings. • Theorem: Let R be a right noetherian ring. Every nil one-sided ideal in R is nilpotent. This means that (KC) holds for right noetherian rings.

1988 Amitsur [27]

• Theorem: (KC) holds for an algebra R over a ﬁeld k such that k is an uncountable ﬁeld or such that dimk(R) < |k|. 1989 Ferrero and Puczylowski [14]

• Theorem: (KC) is true if and only if every ring which is a sum of a nilpotent subring and a nil subring is nil.

1992 Kelarev [22]

• Theorem: Let S be an abelian semigroup. The Jacobson radical is S- graded if and only if S is embeddable in a torsion-free abelian group.

1993 Puczylowski [39]

• Question: Let R be a nil ring. Is R[X] Brown-McCoy radical? • Question: Let R be a nil ring. Let X be a set of cardinality ≥ 2 a) Is the polynomial ring R[X] in commuting indeteminates from X Brown-McCoy radical? b) Is the polynomial ring R < X > in non-commuting indeterminates from X Brown-McCoy radical? • Question: Let R be a nil Z-graded ring. Is N (R) homogeneous? 1998 Smoktunowicz and Puczylowski [38]

• A nonzero ring R is called prime if for any two elements a, b ∈ R, arb = 0 for all r ∈ R implies that either a = 0 or b = 0. Theorem: For a given ring R, R[X] is Brown-McCoy radical if and only if R cannot be homomorphically mapped onto a prime ring R0 then Z(R0)∩I 6= 0 0 for every 0 6= I C R . • Theorem: U(R[X]) = P(R)[X] for all rings R.

23 Beidar and Fong [6]

• Theorem: (KC) holds for monomial algebras.

2000 Smoktunowicz [44]

• Theorem: There exists a nil algebra R over a countable ﬁeld such that the polynomial ring R[X] is not nil.

2001 Smoktunowicz and Puczylowski [51]

• Theorem: There exists a polynomial ring that is Jacobson radical and not nil.

Beider, Fong and Puczylowski [7]

• A ring is called Behrens radical if it cannot be homomorphically mapped onto a ring with a nonzero idempotent. A ring R is Behrens radical if and only if every left ideal of R is Brown-McCoy radical. Theorem: If R is a nil ring, then R[X] is Behrens radical.

2002 Ferrero and Wisbanner [15]

• Theorem: If X is inﬁnite then for any (not necessarily nil) ring R, R[X] is Brown-McCoy radical if and only if R < X > is Brown-McCoy radical.

Beidar, Puczylowski and Wiegandt [8]

• A ring R is called uniformly stronly prime if there exists a ﬁnite subset F such that aF b = 0 implies that a = 0 or b = 0 for all a, b ∈ R. A ring R is called Von Neuman regular if for every a ∈ R there exists an x ∈ R such that a = axa. Theorem: If R is nil then R[X] is in the upper radical class of the uniformly strongly prime rings as well as in that of Von Neuman regular rings.

2003 Ferrero and Wisbauer [15]

• Let Z be the class of prime rings R with large center, i.e., I ∩ Z(R) 6= 0 for every 0 6= I C R. Deﬁne V(R) = ∩{I C R | R/I ∈ Z}. Question: Is V(R[X]) = V(R)[X] for all rings R? • Question: Let R be an algebra of positive character. Is the polynomial ring R[X] in commuting indeteminates from X Brown-McCoy radical?

Tumurbat and Wiegandt [53]

24 • A (radical) class C of rings is said to be polynomially extensible, if R ∈ C implies R[X] ∈ C. Theorem: (KC) holds if and only if the nil radical class N is polynomial extensible to the Jacobson radical class J .

Smoktunowicz [45]

• Theorem: If R[X] is Jacobson radical then R[X,Y ] is Brown-McCoy radical.

2005 Smoktunowicz [46]

• Theorem: (KC) holds if and only if for all nil rings R the polynomial ring R[X] is not left or right primitive.

2006 Beidar [40]

• Question: Does there exist a prime ring R with trivial centre such that the central closure of R is a simple ring with unit? • Question: Does there exist a prime nil ring R such that the central closure of R is a simple ring with unit?

2008 Chebotar, Ke, Lee, Puczylowski [11]

• Theorem: If R is a nil algebra over a ﬁeld of positive character then V(R[X]) = R[X] and R[X,Y ] is Brown-McCoy radical. • Question: What happens when charR = 0 in the previous theorem?

Smoktunowicz [48, 49]

• Theorem: There exists a positively graded ring, which is graded-nil and Jacobson radical. • Theorem: If R is positively graded, graded-nil and I is a primitive ideal of R, then I is homogeneous. • Theorem: If R is a nil ring and P is a primitive ideal of R[X] (thus under the assumption that the Koethe problem has a negative solution), then P = I[X] for an ideal I of R.

2013 Smoktunowicz [50]

• Theorem∗: If R is Z-graded then N (R) is homogeneous. • Theorem∗: If R is positively graded, then every homogeneous subring of a Jacobson radical ring is Jacobson radical.

25 • Theorem∗: Every Z-graded ring which is graded-nil, is Brown-McCoy radical.

2014 Mazurek, Nielsen and Ziembowski [32]

• Theorem∗: Let S be an abelian semigroup. The nilradical N (R) of a ring R is homogeneous whenever R is an S-graded ring if and only if S is cancellative and torsion-free. ∗ n • Theorem : Let S be a semigroup such that ∩n≥1T = ∅ for every ﬁnitely generated subsemigroup T ⊆ S. If A is a homogeneous subring of an S-graded ring R, then A ∩ J (R) ⊆ J (A). In particular, when R is a Jacobson radical, then so is A ⊆ R.

One can see that some of the questions posed over the years have found a (partial) solution. Although there are many equivalent propositions to the Koethe conjecture, it is still an open problem. The main goal of this thesis is to prove the theorems indicated with ∗.

26 Chapter 3

Radicals of Z-graded rings

3.1 Introduction

Of all gradings on rings, Z-gradings are historically the ﬁrst. It was shown by Bergman in 1975 that the Jacobson radical of a Z-graded ring is homogeneous but only in 2006 Smoktunowicz showed that the same is true for nil radicals. First we consider as an example the Jacobson radical in polynomial rings.

3.2 Amitsur’s result on the Jacobson radical of R[X]

In this section we will give a discription of J (R[X]). This result is due to Amitsur [2]. Throughout the section R is a ring. By Fp we denote the ﬁnite ﬁeld with p elements. The following lemma is crucial in determining J (R[X]).

Lemma 3.2.1. Let N = J (R[X]) ∩ R. If J (R[X]) 6= 0 then N 6= 0.

We mention following theorem by Amitsur [1] without proof. It will be used in the proof of Lemma 3.2.1.

Theorem 3.2.2. If the kernel of the homomorphic mapping σ of R onto σ(R) is contained in the Jacobson radical J (R), then J (σ(R)) = σ(J (R)).

Proof. By Lemma 1.2.6 the Jacobson radical J = J (R[X]) is invariant under automorphisms of R[X], in particular invariant under the automorphism deﬁned by X 7→ X + 1. If the Lemma were not to be true, then J 6= 0 but J ∩ R = N = 0. Take f(x) a nonzero polynomial of minimal degree in J. Note that because of the assumption, the degree is at least 1. Consider the automorphism deﬁned by X 7→ X + 1.

27 Then we also have f(x + 1) ∈ J. Hence f0(x) = f(x + 1) − f(x) ∈ J but as the degree of f0(x) is smaller than the degree of f(x), the minimality of f(x) implies that f0(x) = 0 and so that f(x + 1) = f(x). For a prime p denote by Rp the set of all elements r ∈ R that have additive order a divisor of p, i.e. pr = r+...+r = 0. Clearly Rp is an ideal of R and Rp[X] is an ideal of R[X]. We may assmume that f(x) ∈ Rp[X]. n n−1 Indeed, let f(x) = anx + an−1x + ... + a0 with ai ∈ R. Recall that the degree of n−1 f(x) is n ≥ 1. Since f(x + 1) − f(x) = nanx + ...+ = 0 it follows that nan = 0. Let m be the minimal interger such that man = 0 and let a prime p dividing m. Thus we have that (m/p)an 6= 0 and that (m/p)an ∈ Rp. The highest degree coefficient of (m/p)f(x) belongs to Rp and (m/p)f(x) ∈ J. As deg((m/p)f(x)) = deg(f(x)) we may replace f(x) by (m/p)f(x). But now also pf(x) ∈ J and it has a smaller degree n n than f(x) because p(m/p)anx = 0x = 0. Hence the minimality of f(x) implies pf(x) = 0. This means that f(x) ∈ Rp[X]. The next step is to show that if a polynomial g(x) ∈ Rp[X] satisfies g(x+1) = g(x) p p then g(x) is a polynomial h(x − x) in x − x with coefficients in Rp. This will be shown by induction on the degree of g(x). If g(x) has degree zero then there is nothing to prove. Let g(x) be of degree k < p. Since g(x) = g(x + 1), it follows that g(x) = g(x + ν) for all integers ν. We can write

k k g(x + ν) = g(ν) + xg1(ν) + ... + x gk(ν) = b0 + xb1 + ... + x bk = g(x) (3.1) with gi(ν), bi ∈ Rp for all i. The equality yields that g(ν) = b0 for all integers ν. Clearly, Rp is an algebra over the ﬁnite ﬁeld Fp of p elements. This gives us that g(ν) − b0 = 0 for the p elements ν in Fp. But as the degree of g(x) − b0 is smaller than p it follows that g(x) − b0 = 0. Indeed, write the left equality of Equation 3.1 as

n g(x) − b0 = xb1 + ... + x bn n−1 = x(b1 + ... + x bn). For all 1 ≤ ν ≤ p − 1 this yields zero, n−1 g(ν) − b0 = ν(b1 + ... + ν bn) = 0.

In order to know the solutions of this equation in bi with 1 ≤ i ≤ n < p, we consider the system of equations:

n−1 b1 + b2 + ... + 1 bn = 0 n−1 b1 + 2b2 + ... + 2 bn = 0 ... = 0 n−1 b1 + (p − 1)b2 + ... + (p − 1) bn = 0.

28 We also can write this in matrix form

 0 1 n−1      1 1 ... 1 b1 0  20 21 ... 2n−1  b   0     2       =   .  ......  ... ... 0 1 n−1 (p − 1) (p − 1) ... (p − 1) bn 0

The square matrix is a Vandermonde matrix. As all integers are diﬀerent, it has a T T nonzero determinant. There exists a solution for (b1, ..., bn) , namely (0, ..., 0) . For 1 ≤ i ≤ p − 1 we have bi = 0 and it follows that g(x) = b0 ∈ Rp. This proves the statement for polynomials with degree smaller than p. Now take g(x) of arbitrary degree. Then we can write g(x) = h(x)(xp − x) + k(x) with the degree of k(x) < p. We also have g(x + 1) = h(x + 1)(xp − x) + k(x + 1) and since g(x + 1) = g(x) we get

g(x + 1) − g(x) = 0 ⇔ (h(x + 1) − h(x))(xp − x) = k(x + 1) − k(x).

At the left hand side we have degree ≥ p, if not zero, and at the right hand side we have degree less than p. Hence, h(x + 1) = h(x) and k(x + 1) = k(x). As the degree of k(x) is lower than p we have that k(x) = k0 ∈ Rp and by induction follows p p p that h(x) = h0(x − x). Therefore g(x) = h0(x − x)(x − x) + k0 is a polynomial in p Rp[X − X]. p Next we prove that if a polynomial h(x − x) belongs to J (Rp[X]), then it also p belongs to J (Rp[X − X]). Let

p p k(x) ∈ h(x − x)Rp[X − X].

As (x + 1)p − (x + 1) = xp + 1 − x − 1 = xp − x, p we get that k(x) = k(x + 1). As we assume that h(x − x) ∈ J (Rp[X]), also 0 0 k(x) ∈ J (Rp[X]). Hence its quasi-inverse k (x) is unique. Write k (x + 1) for the unique quasi-inverse of k(x+1). Now k(x) = k(x+1) implies k0(x) = k0(x+1). From 0 p the previous remark it follows that k (x) ∈ Rp[X − x] and therefore the right ideal p p p h(x − x)Rp[X − X] is quasi-regular in Rp[X − X]. Thus

p p h(x − x) ∈ J (Rp[X − X]).

Since the nonzero polynomial of minimal degree f(x) ∈ J ∩ Rp[X] and Rp[X] is an ideal in R[X], we have that f(x) ∈ J (Rp[X]) = Rp[X] ∩ J . As it was shown p that f(x) = f(x + 1) we have that f(x) ∈ Rp[X − X]. By the previous remarks p p p this gives us that f(x) = g(x − x) and g(x − x) ∈ J (Rp[X − X]). The mapping

29 p p h(x) 7→ h(x − x) determines an automorphism between Rp[X] and Rp[X − X]. p It follows, by Theorem 3.2.2, that J (Rp[X]) is the image of J (Rp[X − X]). As p p g(x − x) = f(x) ∈ J (Rp[X − X]) we get that

g(x) ∈ J (Rp[X]) = J ∩ Rp[X] ⊆ J . As the degree of g(x) is lower than the degree of f(x), we get a contradiction with the minimality of f(x). Lemma 3.2.3. The Jacobson radical of the polynomial ring R[X] is J (R[X]) = N[X], where N = J (R[X]) ∩ R. Proof. Write J = J (R[X]). Since N ⊆ J it follows that N[X]R[X] ⊆ NR[X] ⊆ JR[X] ⊆ J and therefore N[X] ⊆ J. Consider the natural homomorphism R[X] → R[X]/N[X] with kernel N[X] ⊆ J. By Remark 1.2.2 it follows that J (R[X]/N[X]) = J/N[X]. Put R = R/N. Then R[X] ∼= R[X]/N[X]. Now J (R[X]) ∩ R = J (R[X]/N[X]) ∩ (R/N) = (J (R[X])/N[X]) ∩ (R/N). Let f(x) + N[X] ∈ (J (R[X])/N[X]) ∩ (R/N). Let f ∈ J (R[X]) such that f + N[X] = r + N[X] with r ∈ R, then f − r ∈ N[X]. n n Write f = r0 + r1X + ... + rnX , ri ∈ R. Then (r0 − r) + r1X + ... + rnX ∈ N[X] n with r1, ..., rn, r0 − r ∈ N[X]. As N ⊆ J it follows that r0 = f − (r1X + ... + rnX ) ∈ J ∩ R = N. So r0 ∈ N and thus f ∈ N[X]. Hence f + N[X] = 0 + N[X]. Thus J (R[X]) ∩ R = {0}. By the previous Lemma {0} = J (R[X]) = J/N[X] and therefore J = N[X]. Lemma 3.2.4. Let N = J (R[X]) ∩ R, then N is a nil ideal in R. Proof. Cleary N is an ideal in R. Take r ∈ N ⊆ J = J (R[X]), then rrx = r2x ∈ J. Hence there exists q(x) ∈ R[X] so that r2x so q(x) + r2x + q(x)r2x = 0, or q(x) = −r2x − q(x)r2x. Substitute in the right hand side q(x) by the expression −r2x−q(x)r2x. On the right side one can subsitute q(x) by the whole expression on the right side of this equality. Repeating this process we get q(x) = −r2x + (r2x)3 + ... + (−1)n(r2x)n + (−1)n+1(r2x)n+1 + (−1)n+1q(x)(r2x)n+1. Take n greater than the degree of the polynomial q(x). Comparing both sides of the equation the coeﬃcient of xn is r2n = 0. This shows that N is a nil ideal. We conclude by putting these three lemmas together. Theorem 3.2.5. J (R[X]) = N[X], where N = J (R[X]) ∩ R is a nil ideal.

30 3.3 Bergman’s results

In this section we prove Bergman’s result on the Jacobson radical of a Z-graded ring and more general of rings graded by torsion-free abelian groups. First we need some preparation. The results in this section are taken from [9].

3.3.1 Simple modules and ﬁnite extensions We follow the notation of Bergman. Recall that a ring S is said to be a ﬁnite normalizing extension of a subring R if

S = Ra1 + ... + Ran, and rai = air = 0 for all ai, r ∈ R. Lemma 3.3.1. Let S be a ﬁnite normalizing extension of R. Let M be a simple left S-module. Then M is semisimple of ﬁnite length as a left R-module, and all its simple R-submodules are isomorphic.

Proof. For every nonzero x ∈ M we can write M = Sx = Ra1x + ... + Ranx. Hence, M is a ﬁnitely generated R-module and therefore has a maximal R-submodule N ⊆ M. Because each ai commutes with all elements of R, multiplication with ai : M → M : y 7→ aiy is an R-module endomorphism of M. Denote

Ni = {y ∈ M | aiy ∈ N} ⊆ M.

These are the inverse images of the maximal R-submodule N. Hence they are either a maximal sumodule of M or N = M. Let L = N1 ∩ ... ∩ Nn, and consider the quotient module M/L. This is an R-module of finith length. We note that X X X SL = RaiL ⊆ RaiNi ⊆ RN = N 6= M, and, as M is a simple S-module, L has to be 0. With both L and M/L having finith length, also M has finith length. We can then choose a simple R-submodule M0 ⊆ M. We have X X X M = SM0 = RaiM0 = aiRM0 = aiM0 a finite sum of homomorphic images of the simple module M0 as desired. For the next lemmas we will make use of tensor products. For a detailled construction of tensor products we refer to [42].

Lemma 3.3.2. Let M be a simple R-module. Then S ⊗R M is semisimple and of ﬁnite length as an R-module, and it is a direct sum of copies of M.

31 Proof. For each ai, the tensor product ai ⊗ M is a homomorphic image of M. These span S ⊗R M as an R-module. Lemma 3.3.3. Let S be a ﬁnite normalizing extension of R, then J (S)∩R ⊇ J (R), with equality if S ⊗ M is nonzero for every simple R-module.

Proof. Take r ∈ R \J (S). Then there exists a simple S-module M that is not annihilated by r. By Lemma 3.3.1, we know that M is semisimple as an R-module. Hence M has a simple R-submodule, say M0, that is not annihilated by r. This shows that r∈ / J (R) and proves the inclusion. We will now prove equality assuming that S ⊗ M 6= 0 for every simple R-module. Take r ∈ R \J (R). We can again ﬁnd a simple R-module that is not annihilated by r and we form S ⊗R M. As all simple modules are generated by one element, S ⊗R M is a nonzero cyclic S-module. Hence we can ﬁnd a simple quotient module 0 0 M = (S ⊗R M)/N. From the previous lemma we know that M , as R-module, is a direct sum of copies of M. This shows us that M 0 is not annihilated by r and hence r∈ / J (S). In the case that R contains a multiplicative unit the inclusion of previous lemma is also an equility when R is rationally closed in S. Recall that a ring R is rationally closed in the ring S when x ∈ R, x−1 ∈ S implies x−1 ∈ R. In this case we use the desciption of the Jacobson radical that says that x ∈ J (R) if and only if 1 − RxR consists of invertible elements in R. Since we have here that r∈ / J (R) there is an element x ∈ RrR such that 1 − x is not invertible in R. Hence 1 − x is not invertible in S, hence x∈ / J (S).

3.3.2 The Jacobson radical of Z-graded rings We now have all the necessary results to prove the main results of Bergman.

Theorem 3.3.4. Let R = ⊕i∈Zn Ri be a Zn-graded ring. If a0 + ... + an−1 ∈ J (R) with ai ∈ Ri, then nai ∈ J (R) for each 0 ≤ i < n.

Proof. With ζ a primitive complexe nth root of unit. Let Z[ζ] be the extension of Z in the complex numbers. Consider the ring S = R ⊗Z Z[ζ]. Clearly , R is in a natural way, a subring of S, with S = X Rζi 0≤i≤n−1 is a normalizing extension of R. Note that Z[ζ] is a ﬁnitely generated free Z-module,a dn thus a direct sum of copies of Z. Hence, as R-module, S is a direct sum of copies of R. So, if M is a

32 simple R-module with

S ⊗ M ∼= R ⊗ M ⊕ ... ⊕ R ⊗ M ∼= M ⊕ ... ⊕ M 6= 0.

Therefore, by Lemma 3.3.3, J (S)∩R = J (R). In particular, a0 +...+an−1 ∈ J (S) ⊆ J (R). We grade S by Zn with Si = Ri ⊗ Z[ζ] as homogeneous components of degree i ∈ Zn. Deﬁne the automorphism X X i σ : S → S : si 7→ ζ si for si ∈ Si, i ∈ Zn. With σ(J (S)) = J (S) we get that for any i, j ∈ Zn, the j −ij j elements σ (a0 + ... + an−1) and also ζ σ (a0 + ... + an−1) are in J (S). We calculate P −ij −j j ζ σ (a0 + ... + an−1) for all i ∈ Zn. Fix i = 0, then

X −j σ (a0 + ... + an−1) j n−1 = (a0 + ... + an−1) + (a0 + ζa1 + ... + ζ an−1) + ... n−1 n−1 n−1 + (a0 + ζ a1 + ... + (ζ ) an−1) 2 n−1 = na0 + (1 + ζ + ζ + ... + ζ )a1 + ... n−1 n−1 n−1 + (1 + ζ + ... + (ζ ) )an−1

= na0 + 0a1 + ... + 0an−1 = na0. For i = 1, we have X −j −j ζ σ (a0 + ... + an−1) j −1 n−1 = (a0 + ... + an−1) + ζ (a0 + ζa1 + ... + ζ an−1) + ... −(n−1) n−1 n−1 n−1 + ζ (a0 + ζ a1 + ... + (ζ ) an−1) −1 −2 −(n−1) = (1 + ζ + ζ + ... + ζ )a0 + (1 + 1 + ... + 1)a1 + ... n−1 n−1 n−1 + (1 + ζ + ... + (ζ ) )an−1

= 0a0 + na1 + ... + 0an−1 = na1.

P −ij −j For each i ∈ Zn we get j ζ σ (a0 + ... + an−1) = nai ∈ J (S) ∩ R = J (R), as claimed. Now the next result is just a corollary.

Theorem 3.3.5. Let R be a Z-graded ring. Then J (R) is a homogeneous ideal.

Proof. Let ar + ... + as ∈ J (R), where r ≤ s ∈ Z and ai ∈ Ri. The Z-grading on R induces a Zn-grading for each positive integer n. If we take n > r − s, the Zn-homogeneous components of our element will be precisely the ai’s (and 0). Hence by Theorem 3.3.4, nai ∈ J (R) for each i. But similarly we have (n + 1)ai ∈ J (R), so (n + 1)ai − nai = ai ∈ J (R).

33 3.3.3 The Jacobson radical of rings graded by a torsion-free abelian group Theorem 3.3.5 can be extended to all rings generated by an arbitrary torsion-free abelian group. To prove this we will ﬁrst need some lemmas. We ﬁx a torsion-free P abelian group G and a G-graded ring R = g∈G Rg. Observe that if f :(G, .) → (Z, +) is a surjective group homomorphism then R also is Z-graded. Indeed, M X M R = ( Rg) = Rz z∈Z f(g)=z z∈Z g∈G P with Rz = f(g)=z Rg. Further, g∈G

X X RzRz0 = ( Rg)( Rh) f(g)=z f(h)=z0 g∈G h∈G X = RgRh. f(g)=z f(h)=z0 g,h∈G 0 As f is a group homomorfism f(gh) = f(g) + f(h) = z + z . So RzRz0 ⊆ Rz+z0 and thus R indeed is Z-graded. As example we consider the abelian group G = Z×..×Z, a direct product of n copies of Z. The projections pi : G → Z :(a1, ..., ai, ..., an) 7→ ai, for 1 ≤ i ≤ n, are group homomorphisms. Hence, by the above R admits natural Z- gradings. For example, the first projection p1 defines a Z-grading on R as R = ⊕z∈ZRz P with Rz = a2,...,an∈Z R(z,a2,...,an). Corollary 3.3.6. If G = Z × .. × Z, a finitely generated torsion-free abelian group, and R is G-graded, then J (R) is homogeneous. ∼ Proof. Because of the assumptions, G = Z × ... × Z, and thus it admits Z-gradings via the natural projections pi : G → Z. By Theorem 3.3.5, we know that if X r = r(a1,...,an) ∈ J (R) a1,...,an∈Z then, for every 1 ≤ i ≤ n, the component of degree ai for this grading X r(a1,...,ai−1,ai,ai+1,...,an) ∈ J (R). a1,...,ai−1,ai+1,...,an∈Z

By repeatingly taking an aj ﬁxed, it follows that for all (a1, ..., an) ∈ Z × .. × Z we have that r(a1,...,an) ∈ J (R). Hence, every homogeneous component of r ∈ J (R) also belongs to J (R).

34 Lemma 3.3.7. Let R be a G-graded ring and r ∈ J (R). If H is a subgroup of G and supp(r) ⊆ H, then r ∈ J (RH ).

∗ Proof. Let r ∈ J (R). For all a, b ∈ RH we have 1 + arb ∈ R ∩ RH . It follows that ∗ 1 + arb ∈ RH . This means that r ∈ J (RH ).

Lemma 3.3.8. Let R be G-graded and r ∈ R. If r ∈ J (RH ) for all H ≤ G with H ﬁnitely generated and supp(r) ⊆ H then r ∈ J (R). Proof. Assume r satisﬁes the stated conditions. Let a, b ∈ R and let

< supp(a), supp(b), H >= H1,

a ﬁnitely generated torision-free abelian subgroup of G. By assumptions r ∈ J (RH1 ). ∗ ∗ Thus 1 + arb ∈ RH1 ⊆ R . It follows that r is quasiregular and r ∈ J (R) Corollary 3.3.9. Let G be a torsion-free abelian group, R a G-graded ring. Then J (R) is homogeneous.

Proof. Let r ∈ J (R). By Lemma 3.3.7, r ∈ J (RH ) for all ﬁnitely generated torision- free abelian subgroups of G with supp(r) ⊆ H. By Corollary 3.3.6, J (RH ) is H- homogeneous. This means that every homogeneous component of r is contained in J (RH ) for each such H. By Lemma 3.3.8, these homogeneous components are also in J (R).

3.4 Smoktunowicz’s results

In this section we show that the nil radical of a Z-graded ring R is homogeneous. Next it will be shown that homogeneous subrings of a Jacobson radical ring R are Jacobson radical. It also is observed that a ring graded by the additive semigroup of positive integers that is graded-nil, is Brown-McCoy radical. The results and proves in this section are based on [50].

3.4.1 The prime radical of Z-graded rings Before we study the nil radical of a Z-graded ring, we consider the prime radical of a Z-graded ring. These results are due to Jespers [21]. A set X admits a total order (or linear order) ≤ if for all a, b, c ∈ X, ≤ is antisymmetrical, i.e. if a ≤ b and b ≤ a then a = b, transitive, i.e. if a ≤ b and b ≤ c then a ≤ c, and total, i.e. either a ≤ b or b ≤ a. Deﬁnition 3.4.1. Let S be a semigroup. If for any two non-empty ﬁnite subsets X,Y of S, one of which is not a singleton, there exists at least one element that can be uniquely written as xy, with x ∈ X and y ∈ Y , then S is called an u.p.-semigroup.

35 Clearly, Z is an example of an u.p.-semigroup. More general, any linearly ordered group is an u.p.-semigroup. Lemma 3.4.2. Let S be an u.p.-semigroup. If P is a nonzero prime ideal of an S-graded ring R and R is not prime then P contains a nonzero homogeneous ideal.

Proof. Let a = a + ... + a and b = b 0 + ... + b 0 be nonzero elements of s1 sn s1 sm R such that aRb = 0, with n, m minimal. As S is an u.p.-semigroup the set 0 0 0 {s1s, ..., sns}.{s1, ..., sm} contains a uniquely presented element sissj, for every s ∈ S. Thus for any r ∈ R , a rb 0 = 0. Certainly arb 0 Rb = 0 and since arb 0 = s si sj sj sj a rb 0 +...+a rb 0 has fewer terms than a, then a rb 0 = 0 for 1 ≤ k ≤ n. Similarly s1 sj sn sj sk sj we obtain from the equality aRa rb = 0 that a rb 0 = 0 for all 1 ≤ k ≤ n, 1 ≤ l ≤ m. sk sk sl This holds for any s ∈ S and r ∈ R so a Rb 0 = 0. The fact that P is prime implies s s1 s1 that a ∈ P or b 0 ∈ P , so P contains a nonzero homogeneous ideal. s1 s1 Theorem 3.4.3 (Jespers,[21]). Let S be an u.p.-semigroup. Any minimal prime ideal of R is homogeneous. Proof. Let P be a minimal prime ideal of the ring R. Let I be the largest homogeneous ideal of R contained in P . If P 6= I then I is not prime. The previous lemma applied to P/I and R/I leads to a contradiction. A direct consequence of this theorem is the following corollary. Corollary 3.4.4 (Jespers,[21]). Let S be an u.p.-semigroup. The prime radical of an S-graded ring R is homogeneous.

3.4.2 The nil radical of Z-graded rings Recall that a ring R is semiprime if it is a subdirect product of prime rings.

Lemma 3.4.5. Let R = ⊕i∈ZRi be a Z-graded ring and N = N (R) its nil radical. If N contains no nonzero homogeneous elements then N = 0.

Proof. We will prove the following statement for every k, n ∈ Z, n ≥ 0 : if a = ak + ... + ak+n ∈ N, with ai ∈ Ri, then ak = ak+1 = ... = ak+n = 0. We do this by induction on n. If n = 0 we get that a = ak ∈ N. Since, by assumption, N does not contain nonzero homogeneous elements and ak is homogeneous we have that a = ak = 0. Let now n > 0, and suppose that the result is true for all elements a ∈ N with less than n nonzero homogeneous terms. The homogeneous prime radical in a Z- graded ring is contained in N. Therefore the prime radical of R is zero because by assumption N does not contain nonzero homogeneous elements. It follows that R is a semiprime ring.

36 For each homogeneous r ∈ R, ak+nra − arak+n ∈ N as a ∈ N. It follows that 0 0 0 0 ak+ak+1+...+ak+n−1 ∈ N where ai = an+krai−airan+k. By the inductive assumption, 0 0 0 0 ak = ak+1 = ... = ak+n−1 = 0. It follows that ai = an+krai − airan+k = 0, for all r ∈ R. To get the desired result it suﬃces to prove that ak+n = 0 because then, by the inductive assumption, ak = ak+1 = ... = ak+n−1 = 0. So a = ak + ... + ak+n−1 = 0. Suppose on the contrary that ak+n 6= 0. As ak+n is homogeneous, it follows that P ak+n ∈/ N. That means that there are pi, qi ∈ R, such that the element i pian+kqi P d is not nilpotent. Notice that a ∈ N implies ( i piaqi) = 0, for some d. Note that P d ( i pian+kqi) 6= 0. It follows from an+kra = aran+k that X X X pian+kqi RaR = piaqi Ran+kR ⊆ piaqi R. i i i Applying this argument again, we get

X 2 X 2 pian+kqi RaRRaR ⊆ piaqi R. i i Continuing this way, we get

X d d X d pian+kqi RaR ⊆ piaqi R = 0. i i So X d d X d d pian+kqi RaR ⊆ pian+kqi RaR = 0. i i Therefore, X d pian+kqi RaR = 0 i and thus also X d pian+kqi Ra = 0 i This follows because in a semiprime ring R, if a 6= 0, then (aR)m 6= 0 for all m. We P d will now show that ( i pian+kqi) Ran+k = 0. We write

X d pian+kqi = et + ... + ej, i with each ei ∈ Ri for some t, j ∈ Z, t ≤ j and ej 6= 0. Recall that a = ak + ...ak+n. Let r be a homogeneous element of R. Now et + ...ej r ak + ... + ak+n = 0

37 implies ejrak+n = 0 (by comparing elements of a given degree). Moreover, for each v < j, X evran+k = − eiran+k+v−i. i>v j By repeatedly applying this last equation to the elements from (et + ...ej)(Ran+k) we get j et + ... + ej Ran+k ⊆ ejR. Therefore, j+1 et + ... + ej Ran+k ⊆ ejRan+k.

As ejRan+k = 0, we get that j+1 et + ... + ej Ran+k = 0.

j+1 j+1 Observe now that ((et + ... + ej)(Ran+kR)) ⊆ (et + ... + ej)(Ran+kR) ⊆ (et + j+1 j+1 ... + ej)(Ran+k) R = 0. Because R is semiprime ((et + ... + ej)(Ran+kR) = 0 implies et + ... + ej Ran+k = 0. P d P d Because ( i pian+kqi) = et + ... + ej we have ( i pian+kqi) Ran+k = 0, and therefore P d ( i pian+kqi) Ran+kR = 0, as claimed. P To ﬁnd the proof, notice that i pian+kqi ⊆ Ran+kR. Hence, X d X d pian+kqi R pian+kqi = 0. i i

P d Since R is semiprime, we get i pian+kqi = 0, a contradiction. We will now prove the main result by Smoktunowicz. Theorem 3.4.6. The nil radical of a Z-graded ring is homogeneous. Proof. Let M be the sum of all nil ideals in R which are generated by homogeneous P elements. Let (a) denote the ideal generated by the element a. Hence M = a∈W (a), where W is the set of all homogeneous elements of R which are in the nil radical of R. Clearly M is a nil ideal by deﬁnition, and M is homogeneous. Consider the ring R/M and let a ∈ R/M be a homogeneous element which is in the nil radical of R/M. We need to prove that a = 0. Since a is homogeneous, a = r + M for some homogeneous element r ∈ R. Since a is in the nil radical of R/M and M is nil, it follows that r is in the nil radical of R. Hence r ∈ M, and so a = r + M = M. It follows that the nil radical of R/M does not contain nonzero homogeneous elements. So, by the previous lemma it is zero. It follows that the nil radical of R equals M.

38 Corollary 3.4.7. If R is a ring graded by a torsion-free abelian group G then its nil radical is homogeneous.

Proof. This is proved in a similar manner as in Corrolary 3.3.6, Lemma 3.3.7, Lemma 3.3.8 and Corrolary 3.3.9.

3.4.3 Jacobson radical rings that are Z-graded In this section we prove the second result of Smoktunowicz (see [50]). In the prove one makes use of an embedding a positive graded ring into a ’power series’ type of ring. This idea generalizes an idea of Krempa to study the Jacobson radical of a polynomial ring by embedding it into the power series ring.

Property 3.4.8. Let R be a ring graded by the semigroup of strict positive integers. If R is a Jacobson radical ring then every subring of R which is generated by homogeneous elements, is a Jacobson radical ring.

P∞ Proof. Consider the set [R], the elements of which are formal sums r = i=1 ri, with ri ∈ Ri for all i ≥ 1. By abuse of terminology we call the ri the homogeneous components of r. This becomes a ring for the component wise addition and multiplication deﬁned as follows ∞ ∞ ∞ X X X X ri si = rksl . i=1 i=1 i=1 k+l=i k,l Obviously R can be considered in a natural way as a subring of [R]. Clearly, if a ∈ [R] 0 P∞ i i then also a = i=1(−1) a ∈ [R] and

∞ ∞ a + a0 + aa0 = a + X(−1)iai + X(−1)iai+1 = 0. i=1 i=1 Hence a0 is the quasi-inverse of a in [R]. So [R] = J ([R]). Note that if a ∈ R = J (R) then its quasi-inverse in R and [R] must coincide, as the quasi-inverse is uniquely determined by a. Hence, if a ∈ R then almost all ’homogeneous’ components of P∞ i i i=1(−1) a are zero. Let S be a subring of R generated by homogeneous elements. Let a ∈ S. Write a = a1 + a2 + ... + an, with ai ∈ Ri. As S is homogeneous, ai ∈ S for all i ≤ n. 0 Pk 0 Let a = i=1 ri ∈ R be the quasi-inverse of a, with ri ∈ Ri. By the above, a = P∞ i i Pk 0 i=1(−1) a = i=1 ri. It follows that all ri ∈ S because all aj ∈ S. Hence, a ∈ S. So S is a Jacobson radical ring.

39 3.4.4 Graded-nil rings and the Brown-McCoy radical Let R be a ring graded by the positive integers. In [50] Smoktunowicz also proved that if R is a graded-nil ring then R is a Brown-McCoy radical. The proof makes use of the following result, also by Smoktunowicz.

∞ Property 3.4.9. [48] Let R = ⊕i=1Ri be a graded-nil ring and let I be a primitive ideal in R. Then I is homogeneous.

Property 3.4.10. Let R be a ring graded by the additive semigroup of positive integers. If R is graded-nil then R is Brown-McCoy radical.

Proof. Suppose that R is not a Brown-McCoy radical ring. Then, R/I is a simple ring with an identity element, for some proper ideal I in R. Clearly I is a maximal ideal and since R/I has an identity element it can be shown that it is also a primitive ideal. To see that we proceed as follows: let e in R be such that e + I is the identity element of the ring R/I. Then ea − a ∈ I for all a ∈ R, hence I is a modular right ideal in R. Notice that e can not be in I as otherwise I = R. By Zorn’s Lemma, there exists a maximal right ideal Q in R which contains I and does not contain e. Then Q is a maximal modular right ideal in R because for all a ∈ R we have ea − a ∈ I ⊆ Q. On the other hand, we have that I the largest two-sided ideal contained in Q since R/I is simple. We have proven that I is a primitive ideal in R. By Proposition 3.4.9, we know that primitive ideals in graded-nil rings are homogeneous. It follows that R/I is graded. However, a ring graded by the additive group of positive integers cannot have an identity element, as for large n, the lowest degree term of en is greater than the highest degree term of e. So, en 6= e for all n ≥ 1.

40 Chapter 4

Radicals in semigroup graded rings

The results stated in the previous chapter lead to obvious and natural questions. These were posed by Smoktunowicz in [50].

Question 4.0.11. For which semigroups S is the nil radical of a s-graded ring homogeneous?

Question 4.0.12. For which semigroups S are homogeneous subrings of S-graded Jacobson radical rings also Jacobson radical rings?

These questions were solved to some extend by R. Mazurek, P.P. Nielsen and M. Ziembowski in [32]. This will be the premise for this chapter. First we have a look at the nil radical in a semigroup graded ring to answer the ﬁrst question. Next we have a look at the homogeneous subrings of an Jacobson radical S-graded ring to answer the second question. Recall that for a semigroup S one denotes by S1 the smallest monoid containing S, i.e.  S if S is a monoid, S1 = S ∪ {1} otherwise, where in the later 1 is a ’formal’ identity added to S.

4.1 The nil radical in semigroup graded rings

To solve Question 4.0.11 we will see that S has to be cancellative and torsion-free. In the abelian case the converse implication also holds. We start with a useful concept, ideal functions. The deﬁnitions and theorems in this section are due to Mazurek, Nielsen and Ziembowski, [32].

41 4.1.1 Ideal functions Deﬁnition 4.1.1. An ideal function F is a ’function’ that assigns to each ring R a two-sided ideal F(R) C R All radicals are ideal functions but there are more examples.

Example 4.1.2. By W(R) we denote the Wedderburn radical, i.e. the sum of all nilpotent ideals in R. This is not a real radical as W(R/W(R)) might not be zero. The Wedderbrun radical is containted in the prime radical and clearly the Wedderburn radical is an ideal funtion.

Example 4.1.3. For the ring R, put

Ll(R) = {x ∈ R | xR = 0}.

It is called the left annihilator of R. The right annihilator Lr of R is deﬁned similarly. Note that in a ring with an unit these two ideals are the zero ideal. Put L(R) = Ll(R) + Lr(R) C R. It is called the annihilator of R. Deﬁnition 4.1.4. Let C be an isomorphically closed class of rings, and let F and B be ideal functions. One says that the ideal function F is B-C-sandwiched if for any ring R we have B(R) ⊆ F(R) ⊆ I for any ideal I C R such that R/I ∈ C. Example 4.1.5. We consider a new radical. Consider the class D of all division rings. Clearly this is isomorphically closed class. The Thierren radical is the upper radical F(R) = ∩ICR,R/I∈DI. Obviously, an ideal function F is W-D-sandwiched if and only if W(R) ⊆ F(R) ⊆ F(R) for each ring R. The prime radical, the nil radical, the Jacobson radical, the Brown-McCoy radical all are examples of W-D-sandwiched ideal functions. The function which assigns the sum of all nil left ideals of a ring R also is a W-D-sandwiched ideal function. A last example of a W-D-sandwiched ideal function is given by the upper radical T determinded by the class ε of matrix rings over division rings, that is T (R) = ∩ICR,R/I∈εI. Remark 4.1.6. Consider an ideal I and an isomorphically closed class of rings C. We suppose that F is a B-C-sandwiched ideal function and that R is a subring of a semigroup ring R0[S], with R0 ∈ C. We denote ω : R0[S] → R0 for the augmentation P P map, that is ω( s∈S rss) = s∈S rs. Assume ω(R) = R0. Then F(R) ⊆ ICR ⊆ R0[S] ∼ for an ideal I such that R/I ∈ C. As R0[S]/ker(ω|R) = R0 ∈ C it follows that F(R) ⊆ ker(ω|R). So either F(R) = 0 or F(R) is not a homogeneous ideal in R0[S] (since we know that ker(ω|R) does not contain nonzero homogeneous elements).

42 4.1.2 Cancellative semigroups In a paper of Clase and Kelarev in 1994 ([12]) it was proven that for the Jacobson radical of a semigroup graded ring to be homogeneous the semigroup has to be cancellative. Mazurek, Nielse and Ziembowski use some of the arguments from that paper to prove that the same condition is needed for the nil radical to be homogeneous. We first recall the concept of a reduced ring. Definition 4.1.7. A ring R is called reduced if it does not contain nonzero nilpotent elements. Theorem 4.1.8. Let S be a semigroup and let F be a L-C-sandwiched ideal function, where C is isomorphically closed class of rings containing a non-reduced ring R0. If for every S-graded ring R the ideal F(R) is homogeneous, then the semigroup S is right cancellative. Proof. Assume S is not right cancellative. Then, there exist elements u, v, w ∈ S with u 6= v but so that uw = vw. We will try to find a ring R which is S-graded, but such that F(R) is not a homogeneous ideal. Let A = R0[S], the semigroup ring of S over R. As R0 is not reduced, we can choose an element x ∈ R0 such that x2 = 0 but x 6= 0. Put I = wS1, the smallest right ideal of S containing w and let J = xZ + xR0, the right ideal of R0 generated by x. The subring R consisting of P the elements s∈S rss with rs ∈ J for any s ∈ S \ I is the S-graded ring R we were looking for. Put d = xu − xv ∈ R. Now dx = 0 and thus dJ = 0. Also dt = 0 for 1 t ∈ ωS because uw = vw. Therefore dR = 0. This implies that d ∈ Ll(R) ⊆ F(R) and d 6= 0. The remark shows us that F(R) is not homogeneous. Corollary 4.1.9. Let S be a semigroup and let ϑ be an ideal function such that, for every S-graded ring R, the ideal ϑ(R) is homogeneous.

(1) If for every ring R we have Ll ⊆ ϑ(R) ⊆ T (R), then S is right cancellative.

(2) If for every ring R we have Lr ⊆ ϑ(R) ⊆ T (R), then S is left cancellative. (3) If for every ring R we have L ⊆ ϑ(R) ⊆ T (R), then S is cancellative.

Proof. We prove (1). We know that M2(D) contains always a nonzero nilpotent element whenever D is a division ring. We can take C to be the collection ξ of matrix rings over division rings and then apply the previous theorem. The other two parts of this corollary are proven in a similar way. We conclude this section with the two main corollaries. Corollary 4.1.10. [12] If S is a semigroup for which the Jacobson radical J (R) is homogeneous whenever a ring R is S-graded, then S is cancellative. Corollary 4.1.11. If S is a semigroup for which the nil radical N is homogeneous whenever a ring R is S-graded, then S is cancellative.

43 4.1.3 Torsion-free semigroups Now we will see that the second necessary condition, for the nil radical to be homogeneous, is for the grading semigroup to be torsion-free. Definition 4.1.12. A semigroup S is called torsion-free if sn = tn for commuting elements s, t ∈ S and some positive integer n imply s = t Mazurek, Nielsen and Ziembowski use this definition which is weaker than what other authors define as torsion-free where s and t do not necessarily commute. They prefer this one as it agrees with the definition of a torsion-free group. Recall that a ring R is said to be of characteristic p if pR = 0. Property 4.1.13. Let S be a semigroup. Let C be an isomorphically closed class of rings such that for every prime p the class C contains a nonzero commutative ring of characteristic p. Consider F, a W-C-sandwiched ideal function. If for every S-graded ring R the ideal F(R) is homogeneous, then the semigroup S is torsion-free. Proof. Assume that for some n ≥ 1 there exist commuting elements s, t ∈ S such that sn = tn. Denote by T the subsemigroup of S generated by s and t. We want to show that s = t. It suffies to consider the case when n = p, with p prime. By assumption there exists a nonzero commutative ring Rp ∈ C of characteristic p. Let p 0 6= r ∈ Rp. Clearly, (rs − rt) = 0 in Rp[T ], and because Rp[T ] is commutative, d = rs − rt generates a nilpotent ideal. Therefore d is in W(Rp[T ]) ⊆ F(Rp[T ]). By assumption the latter is homogeneous since Rp[T ] has an S-grading. Hence by Remark 4.1.6, F(Rp[T ]) = 0. So t = s, as desired. Corollary 4.1.14. Let S be a semigroup and let F be an ideal function such that for every ring R we have W(R) ⊆ F(R) ⊆ F(R), where W(R) is the Wedderbrun ideal function and F(R) is the ideal function determined by the class of matrix rings over division rings. If for every S-graded ring R the ideal F(R) is homogeneous, then the semigroup S is torsion-free. Proof. Apply Property 4.1.13 with C = D the class of division rings. This class contains for every prime p, the commutative field Fp of characteristic p.

4.1.4 Abelian, cancellative, torsion-free semigroups We recall one more theorem before classifying the abelian semigroups for which the nilradical is homogeneous in every S-graded ring. The following is wel known and classic. For a proof we refer to [19]. Theorem 4.1.15. The abelian semigroup S admits a total order compatible with the semigroup operation if and only if S is cancellative and torsion-free.

44 Theorem 4.1.16. Let S be an abelian semigroup. The nilradical N (R) of a ring R is homogeneous whenever R is an S-graded ring if and only if S is cancellative and torsion-free.

Proof. ⇒ The necessity of the condition follows from Corollary 4.1.11 and and Corol- lary 4.1.14. ⇐: To prove the suﬃciency, assume S is an abelian, cancellative and torsion- L free semigroup. Let R = s∈S Rs be an S-graded ring and assume that N (R) is not homogeneous. Let I C R be the largest homogeneous ideal contained in N (R). Clearly, N (R/I) = N (R)/I. So N (R/I) does not contain nonzero homogeneous elements. By assumption there exists an element a = as1 .as2 + ... + ask ∈ N (R), with s1, .., sk ∈ S distinct elements, asi ∈ Rsi \N (R) for each 1 ≤ i and such that k ≥ 2 is minimal. For any homogeneous element rs ∈ Rs we have that arsask −ask rsa ∈ N (R) and it has fewer homogeneous components. Hence, from the minimality of k, we get arsask = ask rsa. This holds for all rs so we also have that arask = ask ra for all r ∈ R.

Since the homogeneous components ask are not in the nil radical N (R), there exist Pn elements pi, qi ∈ R, for 1 ≤ i ≤ n and some integer n ≥ 1, such that α = i=1 piask qi is not nilpotent. On the other hand we have that a ∈ N (R) and therefore there exists Pn m some m ≥ 1 such that ( i=1 piaqi) = 0. We obtain

n n m m m m X m m X m m α RaR ⊆ α RaR = piask qi RaR ⊆ piaqi Rask R = 0. i=1 i=1 So, if P is a minimal prime ideal of R then αm ∈ P or am ∈ P . By Theorem 4.1.15 we know that S is a totally ordered semigroup and therefore a u.p.-semigroup. By previous corollary, it follows that P is homogeneous and thus that if a ∈ P then m ask ∈ P . Consequently α ∈ P is in all minimal prime ideals of R. This means that αm belongs to the prime radical of R and therefore must be nilpotent. Hence α is nilpotent, what gives a contradiction.

4.2 Homogeneous Jacobson radical subrings of Jacobson radical rings

In this section we will focus on the second question posed by Smoktunowicz. The semigroups complying for this question will turn out to be the semigroups for which n all ﬁnitely generated subsemigroups T satisfy ∩n≥1T = ∅. Other examples will show that this is not a necessary condition. An immediate restricting on the semigroup S is that S can not contain idempotents. Indeed, assume an indempotent s0 ∈ S and let R be a Jacobson radical ring containing a subring R0 that is not Jacobson radical. Grade R by setting Rs = 0 for

45 s 6= s0, and Rs0 = R. Clearly, R0 is a homogeneous subring of R but not Jacobson radical.

4.2.1 A power series construction For rings graded by the semigroup of positive integers Smoktunowicz relied on the construction of a ring of ”formal series”. This can be generalized as follows. Let S be a semigroup and R an S-graded ring. Put

X P = {r = rs | supp(r) is contained in a ﬁnitely generated subsemigroup of S}, s∈S P where s∈S rs is considered as a formal set, with rs ∈ Rs. The set P is an abelian group for the component-wise addition, and clearly R ⊆ P . One would like P to be a ring for the multiplication deﬁned by

0 X X 0 X X 0 rr = rs rt = rsrt , (4.1) s∈S t∈S u∈S st=u s,t

0 0 with r, r ∈ P . If r, r ∈ P , then there exist finitely generated subsemigroups Sr,Sr0 ⊆ 0 S such that supp(r) ⊆ Sr and supp(r ) ⊆ Sr0 . Let U =< Sr,Sr0 >⊆ S, a finitely generated subsemigroup of S. On the right hand of Equation (4.1), the nonzero ”homogeneous components” are those such that u ∈ U and therefore supp(rr0) ⊆ U. P 0 To make sure that for each u ∈ U the summation s,t:st=u rsrt makes sense, one demands that Xu = {(s, t) ∈ Sr × Sr0 | st = u} is finite for each u ∈ U. The following lemma shows that it suffies to assume that

\ T n = ∅ n≥1 for every ﬁnitely generated subsemigroup T ⊆ S.

n Lemma 4.2.1. Let T be a finitely generated semigroup such that ∩n≥1T = ∅. For any u ∈ T the set Xu = {(t1, t2) ∈ St1 × St2 | t1t2 = u} is finite. Proof. Let u ∈ T . Because of the assumption, there exists a maximal positive integer m such that u ∈ T m. As T is finitely generated, we can write T =< C > for some m+k finite set C. The maximality of m implies that u∈ / ∪1≤kC . Therefore if t1, t2 ∈ T k and t1t2 = u, then t1, t2 ∈/ C for k > m. Thus, t1 and t2 belong to the finite set 2 m C ∩ C ∩ ... ∩ C . Therefore Xu is finite as well. With these restrictions P is a ring that contains R naturally as a subring.

46 4.2.2 A large class of examples Now we can describe a large class of semigroups for which Question 4.0.12 can be answered.

n Theorem 4.2.2. Let S be a semigroup such that ∩n≥1T = ∅ for every ﬁnitely generated subsemigroup T ⊆ S. If A is a homogeneous subring of an S-graded ring R, then A ∩ J (R) ⊆ J (A). In particular, when R is a Jacobson radical, then so is A ⊆ R.

Proof. Suppose R is an S-graded ring and deﬁne the ring P as above. Notice that P is 0 P∞ i a Jacobson radical as each element p ∈ P has a unique quasi-inverse p = i=1(−p) ∈ P , this means that p + p0 + pp0 = p + p0 + p0p = 0.

Let A be a homogeneous subring of R and let a ∈ A ∩ J (R). Write a = as1 + as2 +...+ask , with asi ∈ Rsi for 1 ≤ i ≤ k. Each asi also is in A as A is homogeneous. 0 Because a ∈ J (R), we can write a = rt1 + rt2 + ... + rtl ∈ R with rtj ∈ Rtj for the 00 P∞ i quasi-inverse of a in R. Let a = i=1(−a) be the quasi-inverse of a in P . As quasi- inverses are unique, we must have that a0 = a00. This means that each homogeneous component in a0 comes from the corresponding homogeneous component in a00. The homogeneous components of a00 belong to the subring generated by the homogeneous components of a. Therefore a0 ∈ A and J (R) ∩ A ⊆ J (A).

4.2.3 More examples The condition on the semigroup S in Theorem 4.2.2 is not the most general one. In this section we will give a construction of a new semigroup which yields a positive answer to Question 4.0.12.

Lemma 4.2.3. Let R,R1 and R2 be rings such that R = R1 ⊕ R2 is a direct sum as abelian groups and R2 C R. Let K be a hereditary radical, then R ∈ K if and only if R1,R2 ∈ K.

Proof. ⇒: As the ring R is in a hereditary radical K, we have that R2 ∈ K. Fur- ∼ thermore, as R1 = R/R2 is a homomorphic image of a ring in K it follows that R1 ∈ K. ⇐: Assume R1,R2 ∈ K. We have that R2 ⊆ K. So, R/K(R) is both a quotient of R1 and therefore in K, and K-semisimple. Hence, R = K(R). Property 4.2.4. Let S be a semigroup which is the disjoint union of two subsemi- P groups S1 and S2, with S2 an ideal in S. Let R = s∈S Rs be a Jacobson radical P P S-graded ring. Write R1 = s ∈ S1Rs and R2 = s∈S2 Rs. Every homogeneous subring of R is Jacobson radical if and only if every homogeneous subring of R1 and R2 is Jacobson radical.

47 Proof. ⇒: Let Ai be a homogeneous subring of Ri, and clearly also of R. Thus, Ai is Jacobson radical. ⇐: Let A be a homogeneous subring of R. Write A = A1 ⊕ A2 for homogeneous subrings Ai ⊆ Ri, i = 1, 2. By assumption, A1 and A2 are Jacobson radical. Clearly A2 is an ideal of A. By the previous lemma, A is a Jacobson radical ring. Now we will give an example of a ﬁnitly generated, cancellative semigroup S for T n which n≥1 T 6= ∅ but every homogeneous subring of the Jacobson radical S-graded ring R is Jacobson radical.

Example 4.2.5. Define S as the semigroup generated by the three letters t1, t2, t3 T n with relation t2t1t3 = t1. Clearly n≥1 S 6= ∅ and S is cancellative. The degree of a word in S is defined as the total number of times t1 occurs. The relation preserves the number of times of times t1 appears in a word so it is well-defined. Define S2 = n ∩n≥1S and S1 = S\S2. Clearly S2 is an ideal of S which consists of the monomials of positive degree, and S1 =< t2, t3 > is a subsemigroup which consists of the monomials of degree zeo. Let R be a Jacobson radical ring graded by S. Let R1 and R2 be the subrings graded by S1 and S2 respectively. By Lemma 4.2.3, R1 and R2 are Jacobson radical rings. n For i = 1, 2 we have that ∩n≥1Si = ∅. Indeed, for i = 1 S1 is a free semigroup. For i = 2 one can consider the degrees. By Theorem 4.2.2 we have that the homogeneous subrings of Ri, i = 1, 2 are Jacobson radical. By applying Proposition 4.2.4, R has the stated property.

48 Chapter 5

Open questions

Let S be a semigroup and R be an S-graded ring. For the nil and Jacobson radical it is determined which are the necessary and suﬃcient conditions on S such that respective N (R) and J (R) are homogeneous. Question 4.0.12 is not completely answered yet. It still has to be determined what the necessary conditions on a semigroup S are such that if an S-graded ring R is Jacobson radical then all its homogeneous subrings are Jacobson radical. We capitulate here some related open questions.

Question 5.0.6. Is a primitive ideal in a Z-graded ring which is graded-nil, homogeneous?

Question 5.0.7. Is a polynomial ring over a graded Jacobson ring is Brown-McCoy radical?

Question 5.0.8. Let R be a nil ring. Is R[X,X−1] a Jacobson radial ring?

Question 5.0.9. Does there exist a ring R such that R[X] is Jacobson radical but R[X,X−1] is not Jacobson radical?

Question 5.0.10. Does there exist simple nil algebras over uncountable ﬁelds?

Question 5.0.11. If R is nil, is R[X] Behrens radical?

Note that it might be possible that one of these questions is equivalent to the Koethe conjecture.

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53 Index

R∗, 14 modular ideal, 8 γ-semisimple, 5 module, 3 supp(r), 14 nil radical, 7 annihilator, 42 nil radical class, 6 nil ring, 8 Behrens radical, 24 nilpotent, 6 Brown-McCoy radical, 12 prime ideal, 10 cancellative, 14 prime radical, 10 crossed product, 18 primitive ideal, 8 graded ideal, 16 primitive ring, 8 graded Jacobson radical, 18 quasi-inverse, 7 graded module, 15 quasi-regular, 7 graded ring, 13 quotient gradation, 16 graded ring homomorphism, 14 graded submodule, 15 radical class, 4 graded subring, 15 reduced, 43 graded-maximal, 16 regular class of rings, 6 graded-nil, 14 ring, 3 hereditary class of rings, 6 sandwiched ideal function, 42 homogeneous component, 14 semigroup ring, 16 homogeneous element, 14 simple-graded module, 15 homomorphic closure, 9 strong grading, 14 submodule, 4 ideal function, 42 Thierren radical, 42 Jacobson radical, 7 torsion-free, 44 Jacobson radical class, 7 total order, 35 Jacobson radical ring, 8 u.p.-semigroup, 35 Levitzki radical, 9 Levitzki radical class, 9 Wedderburn radical, 42