On graded Brown--McCoy radicals of graded rings

TL;DR

This paper explores the properties and characterizations of graded Brown--McCoy radicals in graded rings, establishing relationships between different types of these radicals and the classical radical.

Contribution

It provides new characterizations of the graded Brown--McCoy radical and links the large graded Brown--McCoy radical to the classical radical in graded rings.

Findings

Multiple characterizations of the graded Brown--McCoy radical.

The large graded Brown--McCoy radical is the largest homogeneous ideal within the classical radical.

Established the relationship between graded and classical Brown--McCoy radicals.

Abstract

We investigate the graded Brown--McCoy and the classical Brown--McCoy radical of a graded ring, which is the direct sum of a family of its additive subgroups indexed by a nonempty set, under the assumption that the product of homogeneous elements is again homogeneous. There are two kinds of the graded Brown--McCoy radical, the graded Brown--McCoy and the large graded Brown--McCoy radical of a graded ring. Several characterizations of the graded Brown--McCoy radical are given, and it is proved that the large graded Brown--McCoy radical of a graded ring is the largest homogeneous ideal contained in the classical Brown--McCoy radical of that ring.

Equations

$$\xi\eta=\left\{\begin{array}[]{ll}0,&\textrm{if}\ A(\xi)A(\eta)=0;\\ \zeta,&\textrm{if}\ A(\xi)A(\eta)\subseteq A(\zeta)\\ \end{array}\right.$$

Full text

1 Introduction

Graded radicals and radicals of group-graded rings have been investigated by many authors. Particularly, the graded Brown–McCoy and the Brown–McCoy radical (classical) of group-graded rings have been studied recently, as well as in the recent decades, regarding many important open problems. For example, in [20], results related to the Brown–McCoy radical of a ring graded by the additive group of integers are obtained in order to examine open problems on the Brown–McCoy radical of a polynomial ring (see also [21]). These are, on the other hand, related to the famous Köthe’s Conjecture (consult also references of [20]). In [8], $G$-systems are introduced, where $G$ is a group, and their Brown–McCoy radicals are investigated, while in [4], the graded Brown–McCoy and the Brown–McCoy radical of $G$-graded rings are studied. For research related to some problems on monomial rings, see [12]. There is, however, a more general notion of a graded ring and the aim of this paper is to study the graded Brown–McCoy and the classical Brown–McCoy radical of such rings. Results dealing with the Jacobson radical of these rings can be found in [16]; see also [14] and [15].

Definition 1.1** ([16, 15]).**

Let $R$ be a ring, and $S$ a partial groupoid, that is, a set with a partial binary operation. Also, let $\{R_{s}\}_{s\in S}$ be a family of additive subgroups of $R.$ We say that $R=\bigoplus_{s\in S}R_{s}$ is $S$-graded and $R$ induces $S$ (or $R$ is an $S$-graded ring inducing $S$) if the following two conditions hold:

• $i)$

$R_{s}R_{t}\subseteq R_{st}$ whenever $st$ is defined;

• $ii)$

$R_{s}R_{t}\neq 0$ implies that the product $st$ is defined.

An $S$-graded ring inducing $S$ from above is also called a homogeneous sum [16, 15]. However, we will call it simply a graded ring in the sequel. If $R=\bigoplus_{s\in S}R_{s}$ is a graded ring, the set $A=\bigcup_{s\in S}R_{s}$ is called the homogeneous part of $R,$ and elements of $A$ are called homogeneous elements of $R.$ The previous definition applies to both associative and nonassociative rings, but all rings in this paper are assumed to be associative.

Clearly, every usual group (semigroup, groupoid) graded ring is graded in the above sense. However, there are many other important examples of such rings. For instance, let $A,$ $B$ be rings and $V,$ $W$ be an $(A,B)$-bimodule and a $(B,A)$-bimodule, respectively, and let the quadruple $(A,V,W,B)$ be a Morita context, that is, the set R=\left(\begin{array}[]{cc}A&V\\ W&B\\ \end{array}\right) of matrices forms a ring under matrix addition and multiplication (see, for instance, [7]). Then $R=\bigoplus_{(i,j)\in\{1,2\}^{2}}R_{(i,j)}$ is a graded ring in the above sense if $R_{(i,j)}$ denotes the set of matrices of $R$ whose all entries are zero except the $(i,j)$ entry which does not have to be zero. The following are also examples of graded rings in the above sense: a semidirect sum of rings (see [15]), particular case of which is the Dorroh extension of a ring, a ring which is the direct sum of its left ideals, a path algebra (see [15]).

A notion of a graded ring, equivalent to that in Definition 1.1, was studied in [6, 10, 18] from a different point of view. Namely, let $R$ be a graded ring and $A$ its homogeneous part. Consider $A$ with induced partial addition and induced multiplication from $R.$ Then $A$ is called an anneid [6, 10, 18]. Origins of this approach can be found in [17]. Anneid $A$ has a partial addition since the sum of nonzero elements $x,y\in A$ does not have to be a homogeneous element of $R.$ However, if $x+y$ belongs to $A,$ elements $x$ and $y$ are called addable and we write $x\#y$ [6, 10, 18]. Multiplication is defined everywhere, according to the very definition of a graded ring. As we will see in the next section, studying graded rings is equivalent to studying their corresponding anneids [6, 10, 18]. We will therefore, in this paper, conduct research on anneids, all of which are assumed to be associative.

The graded Jacobson radical of an anneid is thoroughly investigated in [9, 10]. Similar studies on $2$-graded rings can be found in [22]. Here, we use results of [9, 10] in order to introduce and investigate the graded Brown–McCoy radical of an anneid. Inspired by [9, 10], we give two notions of this radical, the graded Brown–McCoy and the large graded Brown–McCoy radical of an anneid. Several characterizations of the graded Brown–McCoy radical of an anneid are given, and as the main result we prove that the large graded Brown–McCoy radical of an anneid is the homogeneous part of the largest homogeneous ideal contained in the classical Brown–McCoy radical of the corresponding graded ring (Theorem 3.24).

Obtained results can be easily translated to the language of graded rings, and represent generalizations of results which hold for usual group-graded rings [8, 4].

2 Preliminaries

Here, we give some notions and properties related to graded rings described above. Everything stated in this section can be found in more detail in [18] and references therein (particularly [6, 10]) or in [19, 23].

Theorem 2.1** ([6, 10, 18]).**

Anneid $A$ can be characterized by the following axioms:

• $a1)$

$A$* is a groupoid with respect to multiplication and possesses an element [math] for which we have $a0=0=0a$ for all $a\in A;$*

• $a2)$

[math]* is addable with every element of $A,$ and the relation of addibility is reflexive and almost transitive: $(\forall a,b,c\in A)\ a\#b\wedge b\#c\wedge b\neq 0\Rightarrow a\#c;$*

• $a3)$

for all $0\neq a\in A,$ the set $A(a)$ of all elements from $A,$ which are addable with $a,$ is a commutative group with respect to addition induced by that of $A$ and such a group is called the addibility group of $a;$

• $a4)$

for all $a,b,c\in A,$ $a\#b$ implies $ca\#cb,$ $ac\#bc,$ $c(a+b)=ca+cb,$ $(a+b)c=ac+bc.$

Clearly, [math] is the neutral element of all addibility groups. Also, two distinct addibility groups have only one common element, and it is $0.$ Let $\Delta^{*}$ denote the set of addibility groups of $A$ and let $\Delta=\Delta^{*}\cup 0.$ If $0\neq\delta\in\Delta,$ then let $A(\delta)$ denote the addibility group of $A$ that defines $\delta.$ Otherwise, that is, if $\delta=0,$ $A(\delta)=0.$ Then, if $\xi,\eta\in\Delta,$ we put

[TABLE]

(see, for instance, [6]). This multiplication on $\Delta$ is justified since it is known that if $x,y,a,b\in A,$ then $x\#y$ and $a\#b$ imply $xa\#yb$ (see, for instance, [18]). Now, to each anneid $A,$ we may associate a graded ring $\overline{A},$ called the linearization of $A$, in the following manner. We will follow the exposition from [10] (see also [6]). Define $\overline{A}$ to be the direct sum of $A(\delta),$ where $\delta$ runs through $\Delta^{*}.$ The multiplication of $\overline{A}$ is obtained by extending linearly the multiplication of $A.$ The set $\Delta^{*}$ plays the role of $S$ from Definition 1.1. Product of two elements of $\Delta^{*}$ is defined if and only if their product is distinct from [math] in $\Delta,$ in which case these products coincide. Then $\overline{A}=\bigoplus_{\delta\in\Delta^{*}}A(\delta)=\bigoplus_{\delta\in\Delta}A(\delta)$ is a graded ring whose homogeneous part is $A.$ If $0\neq a\in A,$ then the degree of $a$ [6, 10, 18], denoted by $\delta(a),$ is the element of $\Delta$ such that $a\in A(\delta(a)).$ It is assumed to be equal to [math] if $a=0.$ If $a\neq 0,$ then of course, $A(\delta(a))=A(a).$ Hence $\delta(ab)=\delta(a)\delta(b)$ if $ab\neq 0,$ and $\delta(a)(\delta(b)\delta(c))=(\delta(a)\delta(b))\delta(c)$ if $abc\neq 0$ [6, 10, 18]. If $\delta$ is an idempotent element of $\Delta^{*},$ that is, if $\delta\delta=\delta,$ then clearly, $A(\delta)$ is a ring with respect to operations induced from $A.$

Definition 2.2** ([6, 10, 18]).**

An anneid $A$ is said to be regular if each of the conditions $ac\#bc$ or $ca\#cb$ implies $a\#b$ for $ac\neq 0,$ $bc\neq 0,$ $ca\neq 0,$ $cb\neq 0,$ $a,b,c\in A.$

In the language of graded rings, regularity is equivalent to the notion of cancellativity [16, 15].

A nonempty subset $I$ of an anneid $A$ is called a right ideal of $A$ [6, 10, 18] if: $i)$ for all $a,b\in I,$ for which $a\#b,$ we have $a-b\in I;$ $ii)$ for all $a\in I$ and $x\in A,$ we have $ax\in I.$ A left ideal of an anneid $A$ is defined similarly, and a subset of $A$ which is a left and a right ideal is called an ideal or a two-sided ideal. If a nonempty subset $I$ of $A$ satisfies the first condition only, then $I$ is called a subanneid of $A.$

If $A$ and $A^{\prime}$ are two anneids, then the mapping $f:A\to A^{\prime}$ is called a homomorphism [6, 10, 18] if for all $a,b\in A:$

• $i)$

$a\#b\Rightarrow f(a)\#f(b)\wedge f(a+b)=f(a)+f(b);$

• $ii)$

$f(ab)=f(a)f(b);$

• $iii)$

$f(a)\#f(b)\wedge f(a),f(b)\neq 0\Rightarrow a\#b.$

Let $I$ be an ideal (left, right, two-sided) of an anneid $A$ and let us define $a\sim b$ if and only if either both $a$ and $b$ belong to $I$ or $a\#b$ and $a-b\in I.$ It is easy to verify that $\sim$ is an equivalence relation on $A.$ We also say that $a$ and $b$ are congruent modulo $I$ and write $a\sim b\mod I.$ Denote the set $A/\sim$ by $A/I.$ Clearly, $[a]_{\sim}=a+I=\{a+x\ |\ x\in I\wedge a\#x\}.$ Now, let $I$ be a two-sided ideal, and let $f:A\to A/I$ be the mapping defined by $f(a)=[a]_{\sim}.$ Put $[a]_{\sim}[b]_{\sim}:=f(ab),$ $a,b\in A.$ We say that $[a]_{\sim}$ and $[b]_{\sim}$ are addable and write $[a]_{\sim}\#[b]_{\sim}$ if $a\#b.$ If $[a]_{\sim}\#[b]_{\sim},$ we put $[a]_{\sim}+[b]_{\sim}:=f(a+b).$ Then $A/I$ becomes an anneid, called the factor anneid [6, 10, 18] (see also [13]). Also, $f$ is a homomorphism of anneids and if $A$ is regular, $A/I$ is regular too. Isomorphism theorems for anneids are analogous to those for ordinary rings.

3 The graded Brown–McCoy radical

Let $A$ be an anneid. We will fix some notation. If $I$ is a right ideal of $A,$ then $\check{I}$ denotes the largest ideal of $A$ contained in $I.$ The class of simple regular anneids with unity, that is, of regular anneids with unity whose only ideals are [math] and the anneid itself, will be denoted by $\mathcal{M}.$ Also, if $a$ is an element of $A,$ $\langle a\rangle$ denotes the ideal of $A$ generated by $a,$ that is, the set comprised of all elements of the form $na+\sum_{\textrm{finite}}ax_{i}+\sum_{\textrm{finite}}y_{i}a+\sum_{\textrm{finite}}x^{\prime}_{i}ay^{\prime}_{i},$ where $n$ is an integer, $x_{i},y_{i},x^{\prime}_{i},y^{\prime}_{i}\in A,$ and where all the summands are mutually addable [6, 10, 18].

Following [1], it makes sense to define the graded Brown–McCoy radical $G(A)$ of an anneid $A$ to be the intersection of all maximal right ideals $I$ of $A$ such that $A/\check{I}\in\mathcal{M}.$

Remark 3.1*.*

The linearization $\overline{G(A)}=\bigoplus_{\delta\in\Delta^{*}}G(A)\cap A(\delta)$ of $G(A)$ is a homogeneous (graded) ideal of a graded ring $\overline{A}.$ This graded ideal will be referred to as a graded Brown–McCoy radical of a graded ring. We could have simply called $G(A)$ the Brown–McCoy radical of an anneid $A,$ just as it was done in [9, 10] in the case of the graded Jacobson radical of an anneid, which was called just radical in accordance with [11]. We added the adjective ‘graded’ in order to avoid confusion with the classical Brown–McCoy radical of a ring while discussing relations between these notions.

A right ideal $I$ of an anneid $A$ is called a modular right ideal if there exists $e\in A$ such that $ea\sim a\mod I$ for all $a\in A$ [9]. If $I$ is a proper modular right ideal, then the degree of $e$ is an idempotent element of $\Delta^{*}$ [9, 10]. This notion served in a description of the graded Jacobson radical of a regular anneid in [9, 10] by analogy with classical rings. In order to do the same with the graded Brown–McCoy radical of a regular anneid, we introduce the following natural notion.

Definition 3.2**.**

An ideal $I$ of an anneid $A$ is called modular if there exists an element $e\in A$ such that for all $a\in A,$ $ea\sim a\mod I$ and $ae\sim a\mod I.$ Element $e$ is called a unity modulo $I.$ We also say that an ideal $I$ is modular with respect to $e.$

Remark 3.3*.*

If $I$ is a proper modular ideal of an anneid $A$ with respect to $e,$ then $\delta(e),$ the degree of $e,$ is an idempotent element of $\Delta^{*},$ that is, $\delta(e)\delta(e)=\delta(e).$ We proceed as in [10]. Indeed, since $e$ is a unity modulo $I,$ for all $a\in A,$ either $ea,$ $a\in I$ or $ea\#a$ and $ea-a\in I,$ and, either $ae,$ $a\in I$ or $ae\#a$ and $ae-a\in I.$ Since $I$ is proper, $e\notin I.$ Also, $e^{2}\#e$ and $e^{2}-e\in I.$ Since $e\notin I,$ we have that $e^{2}\neq 0.$ This and the fact that $e^{2}$ and $e$ belong to the same addibility group, that is, $e^{2}\#e,$ imply the claim.

The proof of the following lemma is included for the sake of completeness.

Lemma 3.4**.**

An ideal $I$ of an anneid $A$ is a maximal modular ideal if and only if $A/I$ is a simple anneid with unity.

Proof.

Let $I$ be a maximal modular ideal and $e$ a unity modulo $I.$ If $a+I$ is an arbitrary element of $A/I,$ then, since $e$ is a unity modulo $I,$ $ea+I=a+I=ae+I.$ Hence $(e+I)(a+I)=(a+I)(e+I)=a+I,$ which proves that $e+I$ is a unity of $A/I.$ Clearly, $A/I$ is a simple anneid, since $I$ is a maximal ideal.

Conversely, if $A/I$ is a simple anneid with unity $e+I,$ then, for all $a\in A,$ we have $(e+I)(a+I)=(a+I)(e+I)=a+I,$ that is, $ea+I=a+I=ae+I.$ Hence $ea\sim a\mod I$ and $ae\sim a\mod I.$ Therefore $I$ is a modular ideal. Also, since $A/I$ is simple, $I$ is a maximal ideal. ∎

Corollary 3.5**.**

If $A$ is a regular anneid, then $G(A)$ coincides with the intersection of all maximal right ideals $I$ of $A$ such that the ideals $\check{I}$ are modular.

Proof.

If $I$ is a maximal right ideal of $A$ such that $\check{I}$ is modular, then $A/\check{I}$ is a simple anneid with unity. Moreover, $A/\check{I}$ is a regular anneid since $A$ is regular. Conversely, if $I$ is a maximal right ideal of $A$ such that $A/\check{I}$ is a regular simple anneid with unity, then $\check{I}$ is a modular ideal of $A.$ ∎

Remark 3.6*.*

If $I$ is a maximal right ideal such that $\check{I}$ is modular with respect to $e,$ then $e$ is also a unity modulo $I,$ that is, $ea\sim a\mod I$ and $ae\sim a\mod I$ for all $a\in A.$

It is sometimes more pleasable to describe radicals of rings by their modules. It is shown in [2] that any special radical of an associative ring may be defined by some class of modules. For analogous results on special radicals of group-graded rings, see [3]. Here, we aim to describe the graded Brown–McCoy radical of an anneid $A$ by the means of right $A$-moduloids, and so in the next paragraph we recall the notion of a moduloid.

If a group is the direct sum of a family of its subgroups, indexed by a nonempty set, then such a group is called a graded group [18]. Let $\overline{M}=\bigoplus_{d\in D}\overline{M}_{d}$ be a commutative graded group and $\overline{A}=\bigoplus_{\delta\in\Delta}\overline{A}_{\delta}$ a graded ring, where $D$ and $\Delta$ are nonempty sets. Moreover, suppose that $\overline{M}$ is a right $\overline{A}$-module. A right $\overline{A}$-module $\overline{M}$ is called graded [6, 10, 18] if for all $d\in D$ and $\delta\in\Delta$ there exists $t\in D$ such that $\overline{M}_{d}\overline{A}_{\delta}\subseteq\overline{M}_{t}.$ A right $A$-moduloid [6, 10, 18] $M$ is just the homogeneous part of $\overline{M},$ that is, $M=\bigcup_{d\in D}\overline{M}_{d},$ with induced partial addition from $\overline{M}$ and induced outer operation $M\times A\to M$ from $\overline{M}\times\overline{A}\to\overline{M},$ where $A$ is the corresponding anneid of $\overline{A}.$ Similarly to the case of anneids, to an $A$-moduloid $M$ one can associate a graded $\overline{A}$-module $\overline{M},$ called the linearization of an $A$-moduloid $M,$ where $\overline{A}$ is the linearization of an anneid $A$ (see, for instance, [19]).

A right $A$-moduloid $M$ is said to be regular [6, 10, 18] if $xa\#xb,$ $xa,xb\neq 0,$ implies $a\#b,$ where $x\in M,$ $a,b\in A.$

If $M$ and $M^{\prime}$ are two right $A$-moduloids, then the mapping $f:M\to M^{\prime}$ is called a homomorphism [6, 10, 18] if for all $x,y\in M,$ $a\in A:$ $i)$ $x\#y\Rightarrow f(x)\#f(y)\wedge f(x+y)=f(x)+f(y);$ $ii)$ $f(xa)=f(x)a;$ $iii)$ $f(x)\#f(y)\wedge f(x),f(y)\neq 0\Rightarrow x\#y.$

Submoduloids are defined as usual (see [6]). A right $A$-moduloid $M$ is said to be irreducible [10] if $MA\neq 0$ and only submoduloids of $M$ are [math] and $M.$ If $x$ is an element of a right $A$-moduloid, then $(0:x)$ denotes the set $\{a\in A\ |\ xa=0\},$ which is a right ideal of an anneid $A$ [10]. Correspondingly, the annihilator of a right $A$-moduloid $M,$ that is, the set $\{a\in A\ |\ Ma=0\},$ will be denoted by $(0:M)_{A}$ or just by $(0:M)$ (see [10]). If $A$ is an anneid and $I$ a right ideal of $A,$ then $A/I$ is a right $A$-moduloid with respect to $(a+I)+(b+I)=(a+b)+I$ if $a\#b,$ and $(a+I)c=ac+I$ for $a+I,b+I\in A/I,$ $c\in A$ (see [10]). Also, $f:A\to A/I,$ defined by $f(a)=a+I$ for $a\in A,$ is a homomorphism of right $A$-moduloids [10].

The notion of a simplicity is commonly used to mean that an algebraic structure in hand has only trivial substructures. Inspired by the notion from [3] for usual group-graded modules, we give simplicity a different meaning in the following definition.

Definition 3.7**.**

An irreducible right $A$-moduloid $M$ over an anneid $A$ is called simple if $MA\neq 0$ and if for all ideals $I$ of $A$ for which $MI\neq 0,$ there exists $b\in\overline{I}=\bigoplus_{\delta\in\Delta^{*}}A(\delta)\cap I$ such that $xb=x$ for all $x\in M.$

We will soon see, after proving two lemmas, that the graded Brown–McCoy radical of a regular anneid $A$ coincides with the intersection of annihilators of all regular simple right $A$-moduloids. Therefore, inspired by the notion of the large graded Jacobson radical of an anneid from [9, 10], it makes sense to introduce the following notion.

Definition 3.8**.**

The intersection of annihilators of all simple right $A$-moduloids, which are not necessarily regular, is called the large graded Brown–McCoy radical of an anneid $A$ and is denoted by $G_{l}(A).$

Lemma 3.9**.**

Let $M$ be a regular simple right $A$-moduloid. Then, if $I$ is an ideal of $A$ such that $MI\neq 0,$ then there exists $b\in I$ such that $xb=x$ for all $x\in M,$ and, moreover, the degree of $b$ is an idempotent element of $\Delta^{*}.$

Proof.

Let $M$ be a regular simple right $A$-moduloid, and $I$ an ideal of $A$ such that $MI\neq 0.$ By definition, there exists $b\in\overline{I}$ such that $xb=x$ for all $x\in M.$ Then we have $b=b_{1}+\dots+b_{n}$ for some natural number $n,$ and where $b_{i}\in I,$ $i\in\{1,\dots,n\}.$ For an arbitrary $x\in M,$ $x=xb=x(b_{1}+\dots+b_{n})=xb_{1}+\dots+xb_{n}.$ This means that $xb_{i}$ are mutually addable, and since $M$ is regular, $b_{i}$ are mutually addable. Therefore $b=b_{1}+\dots+b_{n}\in I.$ Let us now prove that the degree of $b$ is an idempotent element of $\Delta^{*}.$ Indeed, since $xb=x$ for all $x\in M,$ we have $xb^{2}=xb\neq 0,$ for $x\neq 0.$ Now, the regularity of $M$ implies $b^{2}\#b,$ and the claim follows. ∎

Lemma 3.10**.**

If $M$ is a regular simple right $A$-moduloid, then $M\cong A/I,$ where $I$ is a maximal right ideal of $A$ such that $\check{I}$ is a modular ideal of $A.$ Conversely, if $I$ is a maximal right ideal of $A$ such that $\check{I}$ is a modular ideal of $A,$ then $A/I$ is a simple right $A$-moduloid.

Proof.

Let $M$ be a regular simple right $A$-moduloid. Then, since $MA\neq 0,$ and since $M$ is regular, there exist $0\neq x\in M$ and $a\in A$ such that $xa\neq 0.$ Therefore $M\langle a\rangle\neq 0.$ Since $M$ is simple and regular, Lemma 3.9 implies that there exists $a^{\prime}\in\langle a\rangle$ such that $ma^{\prime}=m$ for all $m\in M.$ Let $x\neq 0$ be an element of $M.$ Since $M$ is irreducible, we know from [10] that $M=xA,$ by analogy with the case of classical modules [11]. On the other hand, from [10] we know that $xA\cong A/(0:x)$ by the first isomorphism theorem for moduloids (for the theorem, see [6, 10]). Therefore, according to [10], $(0:x)$ is a maximal modular right ideal of $A,$ that is, there exists $e\in A$ such that $ea\sim a\mod(0:x)$ for all $a\in A.$ Denote $(0:x)$ by $I.$ It suffices now to prove that $\check{I}$ is a modular ideal of $A.$ Since $I$ is a modular right ideal of $A,$ we have that $\check{I}=\{a\in A\ |\ Aa\subseteq I\},$ by analogy with classical rings (see, for instance, [5]). We already know that there exists $a^{\prime}\in A$ such that $ma^{\prime}=m$ for every $m\in M,$ and particularly, $xa^{\prime}=x.$ Also, for every $b\in A,$ and every $c\in A,$ we have $x(cba^{\prime})=(xcb)a^{\prime}=xcb.$ Then, if $x(cba^{\prime})=xcb\neq 0,$ the regularity of $M$ implies $cba^{\prime}\#cb$ and $ba^{\prime}\#b.$ Hence $cba^{\prime}-cb\in I$ and $ba^{\prime}-b\in\check{I}.$ If $x(cba^{\prime})=xcb=0,$ then, of course, both $cba^{\prime}$ and $cb$ belong to $I,$ meaning that $ba^{\prime},$ $b\in\check{I}.$ Also, for every $b\in A,$ and every $c\in A,$ $x(ca^{\prime}b)=(xca^{\prime})b=xcb,$ which again, by the regularity of $M,$ implies $ca^{\prime}b-cb\in I$ if $x(ca^{\prime}b)=xcb\neq 0,$ and $ca^{\prime}b,cb\in I$ if $x(ca^{\prime}b)=xcb=0.$ Therefore $\check{I}$ is a modular ideal of $A.$

Conversely, if $I$ is a maximal right ideal of $A$ such that $\check{I}$ is modular, it is clear that $M=A/I$ is an irreducible right $A$-moduloid. Let us recall, if $\check{I}$ is a modular ideal with respect to $e,$ then $e$ is a unity modulo $I.$ If $K$ is an ideal of $A$ such that $MK\neq 0,$ then $MK=M.$ Hence there exists $f\in K$ such that $e+I=f+I,$ and so for all $a+I\in M,$ $af+I=a+I.$ Therefore $M$ is a simple right $A$-moduloid. ∎

Corollary 3.11**.**

The graded Brown–McCoy radical of a regular anneid $A$ coincides with the intersection of annihilators of all regular simple right $A$-moduloids.

Remark 3.12*.*

If $A$ is a regular anneid, notice that we have $G_{l}(A)\subseteq G(A).$

Let us now deal with the elementwise characterization of the graded Brown–McCoy radical of a regular anneid, inspired by a similar study on the graded Jacobson radical of a regular anneid from [9, 10].

The following lemma follows the proof of the analogous lemma for modular right ideals of regular anneids from [10].

Lemma 3.13**.**

If $I$ is a proper modular ideal of a regular anneid $A,$ then all unities modulo $I$ have the same degree. Moreover, this degree is an idempotent element of $\Delta^{*}.$

Proof.

Let $f:A\to A/I$ be the canonical mapping, and $e,$ $e^{\prime}$ be two unities modulo $I.$ Then for all $a\in A,$ $f(a)=f(ea)=f(e^{\prime}a)=f(ae)=f(ae^{\prime}).$ Hence in particular, $\bar{0}\neq f(ea)\#f(e^{\prime}a)\neq\bar{0}$ if $a\notin I.$ Since $f$ is a homomorphism, it follows that $ea\#e^{\prime}a.$ Now, the regularity of $A$ implies $e\#e^{\prime}.$ Therefore, according to Remark 3.3, $e$ and $e^{\prime}$ have the same degree, which is an idempotent element of $\Delta^{*}.$ ∎

Definition 3.14**.**

The degree of all unities modulo a proper modular ideal $I$ of a regular anneid is called the degree of $I.$

It is stated in [9] and proved in [10] that there exists a one-to-one correspondence between the maximal modular right ideals of a regular anneid $A$ of degree $\epsilon$ and the maximal modular right ideals of $A(\epsilon).$ Here, we prove the same for maximal modular ideals if we additionally assume that $\Delta$ is such that the product of nonidempotent elements cannot be a nonzero idempotent. We assume the same in Theorem 3.17, Theorem 3.18, Corollary 3.19 and Theorem 3.20.

Theorem 3.15**.**

Let $A$ be a regular anneid and $\epsilon$ an idempotent element of $\Delta^{*}.$ If $I$ is a maximal right ideal of $A$ such that $\check{I}$ is a modular ideal of $A$ of degree $\epsilon,$ then $I_{\epsilon}=I\cap A(\epsilon)$ is a maximal right ideal of $A(\epsilon)$ such that $\check{I_{\epsilon}}$ is a modular ideal of $A(\epsilon).$ If $I$ is a maximal right ideal of $A(\epsilon)$ such that $\check{I}$ is modular, then $K=\{x\in A\ |\ xA\cap A(\epsilon)\subseteq I\}$ is a maximal right ideal of $A$ such that $\check{K}$ is a modular ideal of $A$ of degree $\epsilon.$

Proof.

Let $I$ be a maximal right ideal of $A(\epsilon)$ such that $\check{I}$ is a modular ideal of $A(\epsilon)$ with respect to $e.$ Since $A$ is regular, according to [9, 10] we have that $K=\{x\in A\ |\ xA\cap A(\epsilon)\subseteq I\}$ is a maximal modular right ideal of $A$ with respect to $e,$ and $K\cap A(\epsilon)=I.$ It is clear that $\check{K}\cap A(\epsilon)\subseteq\check{I}.$ Let $x\in\check{I},$ and let $a\in A.$ If $b\in A$ is such that $0\neq axb\in A(\epsilon),$ then, by our assumption on $\Delta,$ we have that both $a$ and $b$ belong to $A(\epsilon).$ Since $x\in\check{I},$ it follows that $axb\in\check{I}.$ Therefore $\check{I}\subseteq\check{K}\cap A(\epsilon).$ Thus $\check{K}\cap A(\epsilon)=\check{I},$ and so $\check{K}\neq 0.$ It follows that $\check{K}$ is a maximal ideal of $A.$ Let $a\in A.$ If $a\in\check{K},$ then $ea,$ $ae\in\check{K}.$ If $a\notin\check{K},$ then there exist $x,$ $y\in A$ such that $xay\in A(\epsilon)$ but $xay\notin I.$ In particular, $xay\neq 0.$ Since $A$ is regular, our assumption on $\Delta$ implies that $\delta(x)=\delta(y)=\delta(a)=\epsilon.$ By assumption, $\check{I}$ is a modular ideal of $A(\epsilon)$ with respect to $e.$ Therefore $ea\#a,$ $ae\#a,$ and $ea-a,$ $ae-a\in\check{I}\subseteq\check{K}.$ Since $\check{K}$ is a maximal ideal of $A,$ it follows that $A/\check{K}$ is a regular simple anneid with unity $e+\check{K}.$ Therefore $\check{K}$ is modular.

Conversely, if $I$ is a maximal right ideal of $A$ such that $\check{I}$ is modular, then clearly, $I_{\epsilon}=I\cap A(\epsilon)$ is a modular right ideal of $A(\epsilon)$ such that $\check{I_{\epsilon}}$ is modular. Also, according to [9, 10], $I_{\epsilon}$ is maximal. ∎

Definition 3.16**.**

An element $x$ of an anneid $A$ is said to be $G$-regular if there exists no proper ideal $I$ of $A$ such that $x$ is a unity modulo $I.$ An anneid (ideal) is called $G$-regular if every of its elements is $G$-regular.

The following theorem is an analogue of the characterization of quasi-regular elements of a regular anneid from [9, 10] with the same proving technique.

Theorem 3.17**.**

Let $A$ be a regular anneid. An element $x\in A$ is $G$-regular if and only if one of the following two conditions is satisfied:

• $i)$

$\delta(x)$* is not an idempotent element of $\Delta^{*};$*

• $ii)$

$\epsilon=\delta(x)$* is an idempotent element of $\Delta^{*}$ and $x$ is a $G$-regular element of the ring $A(\epsilon).$*

Proof.

According to Remark 3.3, every unity modulo a proper modular ideal has the degree which is an idempotent element of $\Delta^{*}.$ Hence, if $x\in A$ satisfies $i),$ then $x$ is $G$-regular. If $x\in A$ is such that $\epsilon=\delta(x)$ is an idempotent element of $\Delta^{*},$ we claim that $x$ is $G$-regular if and only if $x$ is $G$-regular in $A(\epsilon).$ It suffices to prove that $x\in A$ is a unity modulo a maximal modular ideal of $A$ if and only if it is a unity modulo a maximal modular ideal of $A(\epsilon).$ However, this follows from Theorem 3.15. ∎

Theorem 3.18**.**

Let $A$ be a regular anneid. Then $G(A)=A$ if and only if $A$ is a $G$-regular anneid.

Proof.

Let $G(A)=A.$ According to Remark 3.6, no element of $A$ can be a unity modulo a proper ideal, whence every element of $A$ is a $G$-regular element.

Conversely, suppose that every element of a regular anneid $A$ is $G$-regular. Take any right ideal $I$ of $A$ such that $I\neq A.$ Suppose that $A/\check{I}$ has a unity $a+\check{I},$ $a\notin\check{I}.$ Then $a^{2}+\check{I}=a+\check{I}.$ This and the fact that $a\notin\check{I}$ imply $0\neq a^{2}\#a.$ Hence $\epsilon=\delta(a)$ is an idempotent element of $\Delta^{*}.$ According to the previous theorem, $a$ is a $G$-regular element of $A(\epsilon),$ that is, $a$ belongs to $G(a):=\{ax-x+\sum_{i=1}^{n}(y_{i}az_{i}-y_{i}z_{i})\ |\ x,y_{i},z_{i}\in A(\epsilon),\ n\ \textrm{a natural number}\}$ (see, for instance, [7]). However, for any element $ax-x+\sum_{i=1}^{n}(y_{i}az_{i}-y_{i}z_{i})$ from $G(a),$ we have $(a+\check{I})(x+\check{I})-(x+\check{I})+\sum_{i=1}^{n}((y_{i}+\check{I})(a+\check{I})(z_{i}+\check{I})-(y_{i}+\check{I})(z_{i}+\check{I}))=0+\check{I},$ since $a+\check{I}$ is by assumption a unity of $A/\check{I}.$ Hence $ax-x+\sum_{i=1}^{n}(y_{i}az_{i}-y_{i}z_{i})\in\check{I}.$ Therefore $G(a)\subseteq\check{I},$ and particularly, $a\in\check{I},$ a contradiction. We proved that there is no proper right ideal $I$ of $A$ such that $A/\check{I}$ is a simple regular anneid with unity, and so $A$ is a graded Brown–McCoy radical anneid, that is, $G(A)=A.$ ∎

Corollary 3.19**.**

Let $A$ be a regular anneid. Then $G(A)$ is a $G$-regular ideal that contains all $G$-regular ideals of $A.$

If $A$ is an anneid and if $\epsilon$ is an idempotent element of $\Delta^{*},$ then we will denote the classical Brown–McCoy radical of the ring $A(\epsilon)$ by $G(A(\epsilon)).$ The proof of the following theorem follows Halberstadt’s proof of the equality $J(A(\epsilon))=J(A)\cap A(\epsilon),$ where $J(A(\epsilon))$ denotes the classical Jacobson radical of the ring $A(\epsilon),$ and $J(A)$ the graded Jacobson radical of a regular anneid $A$ [10]. We include it for completeness.

Theorem 3.20**.**

If $A$ is a regular anneid, then for every idempotent element $\epsilon$ of $\Delta^{*},$ we have $G(A(\epsilon))=G(A)\cap A(\epsilon).$

Proof.

Ideal $G(A)\cap A(\epsilon)$ is a $G$-regular ideal of $A(\epsilon),$ according to Theorem 3.17 and the previous corollary. Hence, according to [7, Corollary 4.8.3], $G(A)\cap A(\epsilon)$ is contained in $G(A(\epsilon)).$

Conversely, let $x\in G(A(\epsilon))$ and suppose that $x\notin G(A).$ Then, according to Theorem 3.15 and Corollary 3.5, there exists $y\in A$ such that $\epsilon^{\prime}=\delta(xy)$ is an idempotent element of $\Delta^{*},$ but $xy\notin G(A(\epsilon^{\prime})).$ Clearly, $\epsilon\neq\epsilon^{\prime}.$ Since $\epsilon\delta(y)=\epsilon^{\prime}$ and $A$ is a regular anneid, $\epsilon,\epsilon^{\prime}$ and $\delta(y)$ are mutually distinct. Notice that in any ring $R,$ the set $I=\{r\in R\ |\ RrR=0\}$ is an ideal of $R$ and $I^{3}=0,$ which implies that $I$ is contained in the Jacobson radical of $R,$ and hence, $I$ is contained in the Brown–McCoy radical of $R.$ In our case, $xy\notin G(A(\epsilon^{\prime})),$ and so there exists $z\in A(\epsilon^{\prime})$ such that $zxy\neq 0.$ This implies $\epsilon^{\prime}(\epsilon\delta(y))=(\epsilon^{\prime}\epsilon)\delta(y),$ and so $(\epsilon^{\prime}\epsilon)\delta(y)=\epsilon^{\prime 2}=\epsilon^{\prime}=\epsilon\delta(y)\neq 0.$ Therefore, since $A$ is regular, $\epsilon^{\prime}\epsilon=\epsilon=\epsilon^{2}\neq 0.$ Again, by the regularity of $A,$ we have $\epsilon^{\prime}=\epsilon,$ a contradiction. ∎

Remark 3.21*.*

One cannot discard the assumption made on $\Delta.$ Namely, according to Example in [8], page 352, there exists a simple $C_{2}$-graded ring $R$ without unity such that $R_{e}$ is not a Brown–McCoy radical ring. Since a simple graded ring is graded simple, it follows that $A=R_{e}\cup R_{g}$ is a graded Brown–McCoy radical anneid. Therefore, $G(R_{e})\subsetneqq G(A)\cap R_{e}.$ (Here, $A(e)=R_{e}.$)

However, the first statement of Theorem 3.15 is true in general, and implies the following result.

Theorem 3.22**.**

Let $A$ be a regular anneid and let $\epsilon$ be a nonzero idempotent element of $\Delta.$ Then $G(A(\epsilon))\subseteq G(A)\cap A(\epsilon).$

Proof.

Let $I$ be a maximal right ideal of $A$ such that $\check{I}$ is a modular ideal of $A$ of degree $\epsilon.$ Also, let $e$ be a unity modulo $\check{I}.$ Then $I$ is a maximal modular right ideal of $A$ with respect to $e.$ Therefore for every $x\in A$ we either have that $ex,$ $x\in I$ or $ex\#x$ and $ex-x\in I.$ Let $x\in A(\phi),$ where $\phi$ is a nonzero idempotent element of $\Delta$ distinct from $\epsilon.$ Then, since $A$ is regular, we must have that $x\in I.$ Hence $I\cap A(\phi)=A(\phi).$ On the other hand, according to the first statement of Theorem 3.15, we have that $I_{\epsilon}=I\cap A(\epsilon)$ is a maximal right ideal of $A(\epsilon)$ such that $\check{I_{\epsilon}}$ is a modular ideal of $A(\epsilon)$ with respect to $e.$ Therefore, $G(A)\cap A(\epsilon)=\bigcap_{\check{I}\in\mathcal{M}}I\cap A(\epsilon)$ equals the intersection of maximal right ideals $I$ of $A,$ such that $\check{I}$ are modular ideals of $A$ of degree $\epsilon,$ with $A(\epsilon).$ Hence $G(A(\epsilon))\subseteq G(A)\cap A(\epsilon).$ ∎

In case $R=\bigoplus_{\delta\in\Delta}R_{\delta}$ is a strongly graded ring (see [15]) with unity, where $\Delta$ is a finite group with identity $\epsilon,$ it is known from [12] that $G(R_{\epsilon})=G(R)\cap R_{\epsilon},$ where $G$ denotes the classical Brown–McCoy radical of a ring. In the proof of this result, the fact that unity element of $R$ belongs to $R_{\epsilon}$ is essential.

According to [10], if $A$ is a regular anneid, and $\overline{A}$ its linearization, then $\overline{A}$ is a graded ring with unity $1$ if and only if the following hold:

• $i)$

For every nonzero idempotent element $\epsilon\in\Delta,$ the ring $A(\epsilon)$ is with unity $1_{\epsilon};$

• $ii)$

For every $x\in A$ there exist idempotent elements $\epsilon,$ $\phi\in\Delta^{*}$ such that $1_{\epsilon}x=x=x1_{\phi};$

• $iii)$

$\Delta^{*}$ contains only a finite number of idempotent elements.

Moreover, $1=\sum_{\epsilon}1_{\epsilon}.$ Therefore, if $A$ is a regular anneid with unity $1,$ then $\Delta^{*}$ contains exactly one idempotent element $\epsilon$ and $1\in A(\epsilon).$

If $A$ is a regular anneid, by $J(A)$ we denote the graded Jacobson radical [9] of $A.$

Theorem 3.23**.**

Let $A$ be a regular anneid with unity such that $\overline{A}$ is strongly graded. Also, let us assume that $J(A)\cap A(\delta)$ is properly contained in $A(\delta)$ for every $\delta\in\Delta^{*}.$ If $\epsilon$ is the idempotent element of $\Delta^{*},$ then $G(A(\epsilon))=G(A)\cap A(\epsilon).$

Proof.

Since $A(\delta)\nsubseteq J(A)$ for every $\delta\in\Delta^{*},$ it follows by [9, 10] that for every $\delta\in\Delta^{*}$ there exists a unique $\delta^{-1}\in\Delta^{*}$ such that $\delta\delta^{-1}=\epsilon.$ Since $\epsilon\delta=\delta\epsilon=\delta$ for every $\delta\in\Delta,$ and since $A$ is regular, it follows easily that $\delta\delta^{-1}=\delta^{-1}\delta=\epsilon.$ Now, as in the case of group-graded rings, if $I$ is an ideal of $A,$ then $I_{\epsilon}=I\cap A(\epsilon)$ is an ideal of $A(\epsilon)$ and $A(\delta)I_{\epsilon}A(\delta^{-1})=I_{\epsilon}$ for every $\delta\in\Delta^{*}.$ We refer to such ideals of $A(\epsilon)$ as $\Delta^{*}$-invariant. Also, it can be proved that $I$ is an ideal of $A$ if and only if $I=AI_{\epsilon}=I_{\epsilon}A.$ If $I$ is an ideal of $A(\epsilon),$ then the mapping $I\mapsto IA$ defines a one-to-one correspondence between the maximal $\Delta^{*}$-invariant ideals of $A(\epsilon)$ and the maximal ideals of $A.$ Since $A$ is with unity, if $I$ is a maximal ideal of $A(\epsilon),$ then $A(\delta)IA(\delta^{-1})$ is a maximal ideal of $A(\epsilon)$ for every $\delta\in\Delta^{*},$ and $\bigcap_{\delta\in\Delta^{*}}A(\delta)IA(\delta^{-1})$ is a maximal $\Delta^{*}$-invariant ideal of $A(\epsilon),$ just as in the proof of Proposition 2 in [12]. It follows that $G(A(\epsilon))$ equals the intersection of all maximal $\Delta^{*}$-invariant ideals of $A(\epsilon).$ Therefore $G(A(\epsilon))=G(A)\cap A(\epsilon).$ ∎

Next we prove that the homogeneous part of the largest homogeneous ideal contained in the classical Brown–McCoy radical of a graded ring coincides with the large graded Brown–McCoy radical of the corresponding anneid. For a similar result on the Jacobson radical, see [9, 10].

Theorem 3.24**.**

Let $A$ be an anneid, $\overline{A}$ its linearization, and $G(\overline{A})$ the classical Brown–McCoy radical of the ring $\overline{A}.$ Then $G_{l}(A)=G(\overline{A})\cap A.$

Proof.

Let $\overline{M}$ be a simple right $\overline{A}$-module. Then $\overline{M}$ may be viewed as a simple right $A$-moduloid. Indeed, $\overline{M}$ is obviously a right $A$-moduloid and $\overline{M}A\neq 0.$ Also, let $I$ be an ideal of $A$ such that $\overline{M}I\neq 0.$ Suppose that for all $a\in\overline{I}$ there exists $y\in\overline{M}$ such that $ya\neq y.$ On the other hand, $\overline{M}\,\overline{I}\neq 0$ and so there exists $b\in\overline{I}$ such that $xb=x$ for every $x\in\overline{M}.$ In particular, $yb=y,$ a contradiction. Hence $\overline{M}$ is a simple right $A$-moduloid. Also, it is obvious that $(0:\overline{M})_{A}\subseteq(0:\overline{M})_{\overline{A}},$ that is, the annihilator of a right $A$-moduloid $\overline{M}$ is contained in the annihilator of a right $\overline{A}$-module $\overline{M}.$ Since the classical Brown–McCoy radical of a ring equals the intersection of annihilators of all simple right modules over that ring [2], we have $G_{l}(A)\subseteq G(\overline{A}).$ Therefore $G_{l}(A)\subseteq G(\overline{A})\cap A.$

Conversely, let $a\in G(\overline{A})\cap A$ and let $M$ be an arbitrary simple right $A$-moduloid. We claim that $a\in(0:M).$ Suppose $a\notin(0:M).$ Then there exists $x\in M$ such that $xa\neq 0.$ This implies $M\langle a\rangle\neq 0.$ Since $M$ is simple, there exists $b\in\overline{\langle a\rangle}$ such that $mb=m$ for all $m\in M.$ Particularly, $xb=x.$ For all $\bar{a}\in\overline{A},$ we have $x\bar{a}=xb\bar{a},$ which implies $\bar{a}-b\bar{a}\in(0:x).$ Since $b\in\overline{\langle a\rangle},$ and $a\in G(\overline{A}),$ we have $b\in G(\overline{A}).$ Therefore there exist $w,$ $y_{i}$ $z_{i}\in\overline{A}$ and a natural number $n$ such that $b=bw-w+\sum_{i=1}^{n}y_{i}(bz_{i}-z_{i})$ (see, for instance, [7, Theorem 4.8.2]). However, as we have seen, for all $\bar{a}\in\overline{A},$ we have $\bar{a}-b\bar{a}\in(0:x),$ that is, $b\bar{a}-\bar{a}\in(0:x).$ Particularly, $bw-w,$ $bz_{i}-z_{i}\in(0:x),$ and therefore $b\in(0:x),$ which implies $0=xb=x,$ a contradiction. Hence $a\in(0:M),$ and so $G(\overline{A})\cap A\subseteq G_{l}(A).$ Therefore $G_{l}(A)=G(\overline{A})\cap A.$ ∎

Remark 3.25*.*

Let $A$ be a regular anneid. Since $G_{l}(A)\subseteq G(A),$ the previous theorem implies that $\bigoplus_{\delta\in\Delta^{*}}G(\overline{A})\cap A(\delta)\subseteq\overline{G(A)},$ that is, the largest homogeneous ideal contained in the classical Brown–McCoy radical of the ring $\overline{A}$ is contained in the graded Brown–McCoy radical of $\overline{A}$ (contrast with [4, Proposition 3.5]).

If $R=\bigoplus_{s\in S}R_{s}$ is a group-graded ring, and if $e$ is the neutral element of $S,$ then we know from [8] that the Brown–McCoy radical of $R_{e}$ is contained in the Brown–McCoy radical of $R.$ Here we have the following result.

Theorem 3.26**.**

Let $A$ be a regular anneid, $\overline{A}$ its linearization, and $\epsilon$ an idempotent element of $\Delta^{*}.$ If $G(A)=G_{l}(A),$ then the classical Brown–McCoy radical $G(A(\epsilon))$ of $A(\epsilon)$ is contained in the classical Brown–McCoy radical $G(\overline{A})$ of $\overline{A}.$

Proof.

Since by assumption, $G(A)=G_{l}(A),$ and since by Theorem 3.22, $G(A(\epsilon))\subseteq G(A)\cap A(\epsilon),$ we have $G(A(\epsilon))\subseteq G_{l}(A).$ On the other hand, Theorem 3.24 tells us that $G_{l}(A)=G(\overline{A})\cap A,$ and so $G(A(\epsilon))\subseteq G(\overline{A}).$ ∎