On Homogeneous Co-maximal Graphs of Groupoid-Graded Rings

Abstract

Let R be a ring with unity which is graded by a cancellative partial groupoid (magma) S. A homogeneous element $0\ne x\in R$ is said to be locally right (left) invertible if there exist an idempotent element $e\in S$ and $x_r\in R$ ($x_l\in R$) such that $xx_r=1_e$ $(x_lx=1_e)$ where $1_e\ne 0$ is a unity of the ring $R_e.$ Element x is said to be locally two-sided invertible if it is both locally right and locally left invertible. The set of all locally invertible elements (left, right, two-sided) of R is denoted by $U_l(R).$ The homogeneous co-maximal graph $\Gamma ^h(R)$of R is defined as a graph whose vertex set consists of all homogeneous elements of R which do not belong to $U_l(R),$ and distinct vertices x and y are adjacent if and only if $xR+yR=R.$ If the edge set of $\Gamma ^h(R)$ is nonempty, then S (with zero) contains a single (nonzero) idempotent element. This condition characterizes the connectedness of $\Gamma ^h(R)\setminus \{0\}$ for a class of groupoid graded rings R which are graded semisimple, graded right Artinian, and which contain more than one maximal graded modular right ideal. If ${\mathbb {F}}_q$ is a finite field and $n\ge 2,$ then the full matrix ring $M_n({\mathbb {F}}_q)$ is naturally graded by a groupoid S with a single nonzero idempotent element. We obtain various parameters of $\Gamma ^h(M_n({\mathbb {F}}_q))\setminus \{0_{M_n({\mathbb {F}}_q)}\}.$ If R is S-graded, with the support equal to $S\setminus \{0\},$ and if $\Gamma ^h(R)\cong \Gamma ^h(M_n({\mathbb {F}}_q)),$ then we prove that R and $M_n({\mathbb {F}}_q)$ are graded isomorphic as S-graded rings.

1 Introduction

Assigning graphs to various algebraic structures is widely present in the literature. In particular, many useful properties of the Cayley graphs of groups (see for instance [7]) have motivated the study of the Cayley graphs of semigroups in general, and of some particular classes of semigroups (see [25, 27, 28, 31, 32] and references therein). Also, the power graphs of groups and semigroups have drawn a lot of attention, as it can be seen from [1]. For the more recent results on the Cayley graphs and the power graphs of semigroups, the reader is referred, for example, to the papers [11, 21, 22, 29, 30, 34, 45] and their references. There have also been various graphs assigned to a ring, and in particular, to a matrix ring, like for instance, the zero divisor graph, the annihilator graph, the ideal intersection graph, the ideal inclusion graph, the co-maximal graph, or generalized Cayley graphs, see for instance [3,4,5,6, 9, 12, 14, 40, 41, 43, 44] and references therein, as well as [36] for the Cayley graphs of ${\mathbb {Z}}$-graded rings. For graphs associated to lattices, see [39].

Throughout the article, by a graph we mean a simple undirected graph $G=(V,E)$ with the vertex set $V=V(G)$ and the edge set $E=E(G).$ If $V'\subseteq V,$ then by $G\setminus V'$ we denote a subgraph of G obtained from G by removing all of the vertices from $V'$ along with all of the edges incident to those vertices. For the standard notions of the graph theory, we refer the reader to for instance [43].

Let R be a commutative ring with unity. In [41], a graph, denoted by $\Gamma (R),$ is assigned to R,  with the vertex set R and distinct vertices x and y being adjacent if and only if $xR+yR=R.$ Many properties of the graph $\Gamma (R)$ have been investigated, like the chromatic number [41], the connectedness and the diameter of $\Gamma _2(R)\setminus J(R)$ [35], where $\Gamma _2(R)$ is a subgraph of $\Gamma (R)$ induced by non-unit elements of R,  and J(R) is the Jacobson radical of R. In [42], characterizations of the rings R for which $\Gamma _2(R)$ is a forest and for which $\Gamma _2(R)\setminus J(R)$ is Eulerian, are obtained. If R is a ring with unity, not necessarily commutative, then the co-maximal graph of R,  also denoted by $\Gamma (R),$ is defined in [43] as a graph whose vertex set consists of non-unit elements of R and distinct vertices x and y are adjacent if and only if $xR+yR=R.$

The aim of this article is to study a similar graph, which we call the homogeneous co-maximal graph, assigned to a ring with unity, not necessarily commutative, and graded in the following sense.

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,  called components. 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) [23, 24, 26] if the following two conditions hold:

• (i) $R_s R_t\subseteq R_{st}$ whenever st is defined;

• (ii) $R_s R_t\ne 0$ implies that the product st is defined.

The set $H_R=\bigcup _{s\in S}R_s$ is called the homogeneous part of R. Elements of $H_R$ are called homogeneous elements of R. The support $\mathrm {supp}(R)$ of R is defined as the set $\{s\in S\ |\ R_s\ne 0\}.$

After preliminaries on graded rings, in Sect. 3 we introduce the homogeneous co-maximal graph of an S-graded ring R inducing S,  with unity, denoted by $\Gamma ^h(R).$ Theorem 3.2 in [43] asserts that the graph $\Gamma (A)\setminus \{0\}$ of a semisimple right Artinian ring A with unity, and with more than one maximal right ideal, is connected. Here, under the assumption that S is cancellative, we find a necessary condition for the edge set of $\Gamma ^h(R)$ to be nonempty: S cannot have more than one idempotent element, unless S is with zero, in which case it contains two idempotent elements. It turns out that this condition characterizes the connectedness of $\Gamma ^h(R)\setminus \{0\}$ for a class of graded rings R with unity, which are graded semisimple, graded right Artinian, and which contain more than one maximal graded modular right ideal.

Inspired by the results obtained for the co-maximal graphs of matrix rings in [43], in Sect. 4, we turn our attention to matrix rings, graded in accordance with the findings from Sect. 3. Let ${\mathbb {F}}_q$ be a finite field with q elements, and $M_n({\mathbb {F}}_q)$ the ring of $n\times n$ matrices over ${\mathbb {F}}_q$ with respect to the usual matrix addition and multiplication, where $n\ge 2.$ Then $M_n({\mathbb {F}}_q)$ can naturally be regarded as an S-graded ring inducing S,  where S is a groupoid with a single nonzero idempotent element, and distinct from a group with zero. With respect to this grading, we obtain various parameters of $\Gamma ^h(M_n({\mathbb {F}}_q))\setminus \{0_{M_n({\mathbb {F}}_q)}\},$ where $0_{M_n({\mathbb {F}}_q)}$ denotes the zero matrix. If A is a ring with unity, it is known from [43] that $\Gamma (A)\cong \Gamma (M_n({\mathbb {F}}_q))$ implies that $A\cong M_n({\mathbb {F}}_q).$ Here we obtain that, if R is with unity, and graded by the same groupoid S as $M_n({\mathbb {F}}_q)$ is, such that $\mathrm {supp}(R)=S\setminus \{0\},$ and if $\Gamma ^h(R)\cong \Gamma ^h(M_n({\mathbb {F}}_q)),$ then R and $M_n({\mathbb {F}}_q)$ are graded isomorphic as S-graded rings.

2 Preliminaries

2.1 Graded Rings

Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S. The degree $\deg (x)$ of a nonzero homogeneous element x of R is defined to be a unique $s\in S$ such that $x\in R_s.$ Let us define $0=\deg (0).$ Since the zero element of R can be regarded as a component of R,  without loss of generality, we may assume that $0\in S.$ We may moreover assume that $S\setminus \{0\}=\mathrm {supp}(R).$ Throughout the article, and without further notice, we make S a groupoid by putting $st=0$ for those pairs $(s,t)\in S\times S$ for which the product st is not originally defined (in which case $R_sR_t=0$) and $s0=0s=0$ for every $s\in S.$ We also write $S^*=S\setminus \{0\}.$ Hence $R_0=0,$ the zero subring of R,  and $R=\bigoplus _{s\in S}R_s=\bigoplus _{s\in S^*}R_s.$ If $x\in R,$ then x can be written uniquely as $\sum _{s\in S}x_s,$ where $x_s\in R_s$ is called the s-component of x. We also denote $x_s$ by $(x)_s.$

Note that for s,  t,  $u\in S^*,$ if $R_sR_tR_u\ne 0,$ then $(st)u=s(tu).$ In that case, as usual, we write this element as stu.

Throughout the article, a groupoid S with zero 0 is said to be cancellative if for s,  t,  $u\in S,$ each of the equalities $0\ne su=tu\in S$ or $0\ne us=ut\in S$ implies that $s=t.$ Also, the set of all idempotent elements of S is denoted by E(S). By $E(S)^*$ we denote the set $E(S)\setminus \{0\}.$

We note that the notions of an S-graded ring inducing S and of a graded ring studied in [15,16,17, 33] are equivalent.

Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S. A right (left, two-sided) ideal I of R is said to be homogeneous if $I=\bigoplus _{s\in S}R_s\cap I.$ Also recall that if I is a homogeneous ideal (two-sided) of R and $I_s=R_s\cap I,$ then $R/I=\bigoplus _{s\in S}R_s/I_s$ is an S-graded ring inducing S [17, 24, 33].

An S-graded ring inducing S is said to be graded right Artinian [15,16,17, 33] if it satisfies the descending chain condition on its homogeneous right ideals. If R is graded right Artinian, and if S is cancellative, it is easy to see that $R_e$ is right Artinian for every $e\in E(S)$ [17].

An S-graded ring inducing S is said to be a graded division ring [33] if its homogeneous part without the zero element forms a group with respect to the ring multiplication.

Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S,  and let $(M,+)=(\bigoplus _{d\in D}M_d,+)$ be a commutative group, where $\{M_d\}_{d\in D}$ is a family of subgroups of M. Then M is said to be a graded right R-module [17, 33] if for every $d\in D$ and every $s\in S$ there exists $d'\in D$ such that $M_dR_s\subseteq M_{d'}.$ If $0\ne x\in \bigcup _{d\in D}M_d,$ then $\deg (x)$ is a unique d for which $x\in M_d.$ Of course, every S-graded ring inducing S is a graded right module over itself.

Let R be an S-graded ring inducing S and let $M=\bigoplus _{d\in D}M_d$ and $M'=\bigoplus _{d'\in D'}M'_{d'}$ be graded right R-modules. Then an R-homomorphism $f:M\rightarrow M'$ is said to be homogeneous [15, 17, 33] if $f(\bigcup _{d\in D}M_d)\subseteq \bigcup _{d'\in D'}M'_{d'}$ and if for x,  $y\in \bigcup _{d\in D}M_d$ such that f(x),  $f(y)\ne 0,$ we have that $\deg (f(x))=\deg (f(y))$ implies that $\deg (x)=\deg (y).$ If, moreover, f is bijective, then M and $M'$ are said to be graded isomorphic.

2.2 The Graded Jacobson Radical

Throughout the article, the classical Jacobson radical of a ring A is denoted as usual by J(A).

Let R be an S-graded ring inducing S and let us assume that S is cancellative. A homogeneous right ideal I of R is said to be a graded modular right ideal [15, 17] if there exists a homogeneous element $u\in R$ such that $ux-x\in I$ for every homogeneous element $x\in R.$ The cancellativity of S gives that $\deg (u)$ is an idempotent element of S,  and that all such elements u are of the same degree, which is referred to as the degree of I.

The graded Jacobson radical $J^g(R)$of R [15, 17] is the intersection of all maximal graded modular right ideals of R. If $J^g(R)=0,$ we say that R is graded semisimple. For the study of other graded radicals of S-graded rings inducing S,  we refer the reader to [18,19,20, 23, 26] and references therein.

Let e be an idempotent element in $S^*.$ There exists a one-to-one correspondence between the set of all maximal graded modular right ideals of R of degree e and the set of all maximal modular right ideals of the ring $R_e,$ given by $I\mapsto I\cap R_e,$ see [15, 17]. As a corollary, one obtains the following results.

Theorem 1

( [15, 17]) Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S,  where S is cancellative. Then:

• (a) $J^g(R)=\bigoplus _{s\in S}I_s,$ where $I_s=\{x\in R_s\ |\ (\forall e\in E(S))\ xH_R\cap R_e\subseteq J(R_e)\}.$ In particular, $J^g(R)\cap R_e=J(R_e)$ for all $e\in E(S);$

• (b) $J^g(R)=0,$ that is, R is graded semisimple, if and only if the following two conditions are satisfied:

• (i) $J(R_e)=0$ for every $e\in E(S),$ that is, each ring component of R is semisimple;

• (ii) For every nonzero homogeneous element $x\in R$ there exists a homogeneous element $y\in R$ such that xy is a nonzero homogeneous element of an idempotent degree.

Throughout the article, by $\max _{rm}^e(R)$ we denote the set of all maximal graded modular right ideals of R of degree e,  and by $\max _r^h(R)$ we denote the set of all maximal homogeneous right ideals of R. Of course, if R is with a homogeneous unity, say of degree $e\in E(S)^*,$ then every maximal homogeneous right ideal of R is a maximal graded modular right ideal of R of degree e,  in which case $\max _{rm}^e(R)=\max _r^h(R).$

2.3 Graded Matrix Rings

Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S,  and let $n\ge 2$ be an integer. For each $s\in S,$ let $M_n(R_s)$ be the set of $n\times n$ matrices over the component $R_s.$ Then the full matrix ring $M_n(R)$ is S-graded with the components $(M_n(R))_s=M_n(R_s)$ $(s\in S).$ If R is trivially graded, then $M_n(R)$ can be graded by a Brandt semigroup, a rectangular band and by a group. One particular case of grading obtained from a Brandt semigroup grading will be discussed in Sect. 4. We refer the reader to [24] and references therein for more details on the other possible gradings.

If R is an S-graded ring inducing S,  then R is a direct product of graded rings if there exists a family of homogeneous ideals $\{I_\lambda \}_{\lambda \in \Lambda }$ of R such that $R=\prod _{\lambda \in \Lambda }I_\lambda .$ The following graded version of the Wedderburn–Artin Theorem holds true.

Theorem 2

( [16, 17]) Let R be an S-graded ring inducing S with a cancellative S. If R is graded semisimple and graded right Artinian, then there exist positive integers p and $n_1,$ $\dots ,$ $n_p$ such that R is graded isomorphic to a direct product $M_{n_1}(F_1)\times \dots \times M_{n_p}(F_p),$ where each $M_{n_i}(F_i)$ is a graded matrix ring over a graded division ring $F_i.$

3 Homogeneous Co-maximal Graphs of Groupoid-Graded Rings

Throughout this section, unless stated otherwise, $R=\bigoplus _{s\in S}R_s$ is an S-graded ring inducing S with unity 1,  and with S assumed to be cancellative. Then, as we know from [17], the set of all idempotent elements E(S) of S is finite, and the ring R is pseudo-unitary, that is:

• (i) For every $e\in E(S),$ the ring $R_e$ is a ring with unity $1_e;$

• (ii) For every $x\in H_R$ there exist e,  $f\in E(S)$ such that $1_ex=x=x1_f.$

Moreover, $1=\sum _{e\in E(S)}1_e.$

For a similar concept, in case ring is graded by an l.i.-semigroup and in case it is graded by a small category all of whose morphisms are invertible, we refer the reader to [2] and [10], respectively.

Definition 1

Let x be a nonzero homogeneous element of R of degree s. We say that x is locally right invertible if there exist $e\in E(S)^*$ and $x^r\in R$ such that $xx^r=1_e.$ Element $x^r$ is called a right inverse of x. Analogously, we say that x is locally left invertible if there exist $f\in E(S)^*$ and $x^l\in R$ such that $x^lx=1_f.$ Element $x^l$ is called a left inverse of x. We say that x is locally invertible or locally two-sided invertible if it is both locally right and locally left invertible. The set of all locally right, locally left and locally two-sided invertible elements of R is denoted by $U_l(R).$

Of course, if $|E(S)^*|=1,$ then a locally right invertible (locally left invertible, locally invertible) element is a right invertible (left invertible, invertible) homogeneous element in the classical sense and vice-versa.

By the following proposition, the inverses of elements from $U_l(R)$ may be assumed to be homogeneous.

Proposition 1

Let x be a nonzero homogeneous element of R of degree s. Then x is a locally right (locally left) invertible element if and only if there exist $e\in E(S)^*$ $(f\in E(S)^*)$ and a homogeneous element $x^r$ $(x^l)$ of R of degree $s^{-1}\in S$ such that $xx^r=1_e$ $(x^lx=1_f).$ Moreover, $s^{-1}$ is a unique element of S such that $ss^{-1}=e,$ $es=s,$ and $s^{-1}e=s^{-1}$ $(s^{-1}s=f,$ $sf=s,$ and $fs^{-1}=s^{-1})$. If x is locally invertible, then $\deg (x^r)=\deg (x^l).$

Proof

If there exist $e\in E(S)^*$ and a homogeneous element $x^r\in R$ such that $xx^r=1_e,$ then x is locally right invertible by the very definition. So, let x be a locally right invertible element. By the definition of a locally right invertible element, there exist $e\in E(S)^*$ and $x^r\in R$ such that $xx^r=1_e.$ Let $x^r=\sum _{t\in S}(x^r)_t$ be a unique homogeneous decomposition of $x^r.$ Then, since S is cancellative, there exists a unique $t\in S$ such that $x(x^r)_t=1_e.$ Let $s^{-1}=t.$ Then, $ss^{-1}=e.$ Since R is pseudo-unitary, there exist $e', e''\in E(S)^*$ such that $1_{e'}x=x$ and $(x^r)_t1_{e''}=(x^r)_t.$ Moreover, for all distinct g,  $h\in E(S),$ we have that $1_{g}1_{h}=0.$ Since $1_ex(x^r)_t=x(x^r)_t1_e=x(x^r)_t=1_e\ne 0,$ it follows that $e'=e=e''.$ Therefore, $es=s,$ and $s^{-1}e=s^{-1}.$

The statement regarding a locally left invertible element can be proved analogously.

Now, let x be a locally invertible element. According to what we have just proved, there exist e,  $f\in E(S)^*$ and $x^r,$ $x^l\in H_R,$ such that $xx^r=1_e$ and $x^lx=1_f.$ Then, $s\deg (x^r)=e,$ $\deg (x^l)s=f,$ $es=sf=s,$ $f\deg (x^l)=\deg (x^l),$ and $\deg (x^r)e=\deg (x^r).$ Since $es=s=sf,$ we have that $1_ex=x=x1_f.$ Hence, we obtain that $xx^rx=1_ex=x\ne 0$ and $xx^lx=x1_f=x\ne 0.$ It follows that $s\deg (x^r)s=s$ and $s\deg (x^l)s=s.$ Therefore, since S is cancellative, $\deg (x^r)=\deg (x^l).$ $\square$

The following lemma represents a graded version of Remark 1 from [4], and it holds for rings which are not necessarily with unity. By a graded domain, we mean a graded ring without nontrivial homogeneous zero divisors (left, right or two-sided).

Lemma 1

Let $R=\bigoplus _{s\in S}R_s$ be an S-graded ring inducing S,  not necessarily with unity, and let S be cancellative. If the set of all homogeneous zero divisors $D^h(R)$ of R is finite, then $H_R$ is either finite or R is a graded domain.

Proof

Let us assume that R is not a graded domain. Then $D^h(R)^*=D^h(R)\setminus \{0\}$ is nonempty, and by assumption, $|D^h(R)|<\infty .$ Let $x\in D^h(R)^*$ be a right zero divisor. Since S is cancellative, $(0:x)_r=\{a\in R\ |\ xa=0\}$ is a homogeneous right ideal of R. Clearly, $H_{(0:x)_r}=\bigcup _{s\in S}(0:x)_r\cap R_s\subseteq D^h(R),$ and so, $H_{(0:x)_r}$ is finite. Now, xR is a homogeneous right ideal of R and $H_{xR}=\bigcup _{s\in S}xR\cap R_s\subseteq D^h(R).$ Therefore, $H_{xR}$ is also finite. Since $(0:x)_r$ is a homogeneous right ideal, $R/(0:x)_r$ is an S-graded right R-module. It is known from [17], and easy to verify, that the mapping $f:R\rightarrow xR$ defined by $f(a)=xa$ $(a\in R)$ is a surjective homogeneous R-homomorphism, and therefore, $R/(0:x)_r$ and xR are graded isomorphic as graded right R-modules (see also for instance [18]). It is also easy to verify that the relation $\sim$ on $H_R,$ defined by $a\sim b$ if and only if either both a and b belong to $H_{(0:x)_r}$ or $\deg (a)=\deg (b)$ and $a-b\in H_{(0:x)_r},$ is an equivalence relation on $H_R$ (see for instance [33]). However, this implies that $|H_R|=|H_{(0:x)_r}||H_{xR}|\le |D^h(R)|^2,$ which completes the proof. $\square$

Definition 2

The homogeneous co-maximal graph of R,  denoted by $\Gamma ^h(R),$ is a graph whose vertex set is $H_R\setminus U_l(R)$ and vertices x and y are adjacent if and only if $xR+yR=R.$

Remark 1

It is clear that 0 is an isolated vertex in $\Gamma ^h(R).$ Hence, $\Gamma ^h(R)$ is a disconnected graph. It is also easy to see that the notion of $\Gamma ^h(R)$ cannot be extended to graphs which contain loops. Namely, if $x\ne 0$ is adjacent to x,  then $xR=R.$ Since R is pseudo-unitary, there exists $e\in E(S)^*$ such that $1_ex=x.$ Now, $xR=R$ implies that there exists $y\in R$ such that $xy=1_e.$ Hence, x is locally right invertible, a contradiction.

Lemma 2

Let s,  $t\in S^*$ be such that $st=e\in E(S)^*.$ If $u\in S^*$ and $f\in E(S)^*$ are such that $su=f,$ then $u=t$ and $f=e.$

Proof

Since $st=e\in E(S)^*,$ there exist $x\in R_s$ and $y\in R_t$ such that $xy\ne 0$ and $xy\in R_e.$ Now, $xy=1_exy\ne 0.$ Therefore, $st=(es)t.$ Since S is cancellative, this implies that $es=s.$ Similarly, we obtain that $fs=s.$ Since $es=s$ and $fs=s,$ the cancellativity of S implies that $f=e.$ Now, since $st=e$ and $su=f=e,$ it follows that $u=t.$ $\square$

Lemma 3

If the edge set of $\Gamma ^h(R)$ is nonempty, then $|E(S)^*|=1.$

Proof

Let x and y be adjacent vertices of $\Gamma ^h(R),$ and let $\deg (x)=s$ and $\deg (y)=t.$ Then $xR+yR=R.$ Therefore, there exist a,  $b\in R$ such that $xa+yb=1.$ Moreover, we know that $1=\sum _{e\in E(S)}1_e.$ We have the following cases.

Case 1 $\deg (x)=\deg (y)=s.$

Subcase 1a. $s\notin E(S).$ Since S is cancellative and $\deg (x)=\deg (y),$ by Lemma 2 we obtain that there exists a unique $e\in E(S)^*$ such that $1=(xa)_e+(yb)_e.$ Note that both $(xa)_e$ and $(yb)_e$ are nonzero. Namely, suppose for instance that $(yb)_e=0.$ Then $(xa)_e=1,$ which implies that $E(S)^*=\{e\}$ and x is a locally right invertible element, a contradiction. Therefore, $1=u+v,$ where $u=(xa)_e,$ $v=(yb)_e,$ and u,  $v\ne 0.$ It follows that $1=1_e,$ and so, $|E(S)^*|=1.$

Subcase 1b. $s\in E(S)^*.$ Since R is pseudo-unitary, $st=e\in E(S)^*$ implies that $t=e.$ However, in a pseudo-unitary ring, $ef=0$ for all distinct e,  $f\in E(S).$ Hence, we obtain that $1=(xa)_s+(yb)_s.$ Like in the previous subcase, we conclude that $1=1_s.$ Therefore, again $|E(S)^*|=1.$

Case 2 $\deg (x)\ne \deg (y).$

Subcase 2a. s,  $t\notin E(S).$ By Lemma 2, there exist idempotent elements e,  $f\in S^*$ such that $1=(xa)_e+(yb)_f.$ Since x,  $y\notin U_l(R),$ we have that $(xa)_e,$ $(yb)_f\ne 0.$ If $e\ne f,$ this implies that $(xa)_e=1_e$ and $(yb)_f=1_f.$ Therefore, x and y are locally right invertible elements, a contradiction. Hence, $e=f,$ and so, $1=1_e.$ Thus, $|E(S)^*|=1.$

Subcase 2b. Either s or t belongs to $E(S)^*.$ Assume for instance that $s\in E(S)^*$ and $t\notin E(S).$ By reasoning like in the Subcase 1b, we conclude that the only possible nonzero component of xa of an idempotent degree is $(xa)_s.$ Moreover, by Lemma 2, there exists a unique $e\in E(S)^*$ such that $1=(xa)_s+(yb)_e.$ Like in the previous subcase, we must have that $(xa)_s,$ $(yb)_e\ne 0$ and that $s=e.$ Hence, $1=1_e,$ and so, $|E(S)^*|=1.$

Subcase 2c. s,  $t\in E(S)^*.$ This case cannot occur. Namely, proceeding as in the previous cases, we obtain that $1=(xa)_s+(yb)_t$ and $(xa)_s,$ $(yb)_t\ne 0.$ Since $s\ne t,$ it follows that $(xa)_s=1_s$ and $(yb)_t=1_t.$ However, this is impossible, since x and y are not locally right invertible elements of R. $\square$

We finish this section with a result that characterizes the connectedness of the graph $\Gamma ^h(R)\setminus J^g(R).$ We remove the vertices that come from $J^g(R),$ since every vertex of $\Gamma ^h(R)$ that belongs to $J^g(R)$ is isolated. Indeed, let $x\in J^g(R).$ Since 0 is an isolated vertex in $\Gamma ^h(R),$ let us assume that $x\ne 0.$ If y is a homogeneous element of R adjacent to x,  then, by Lemma 3, there exists a single nonzero idempotent element $e\in S.$ Since $xR+yR=R,$ there exist homogeneous elements a,  $b\in R$ such that xa,  $yb\in R_e,$ and $xa+yb=1_e.$ However, $x\in J^g(R).$ Therefore, we obtain that $xa\in J^g(R)\cap R_e,$ and so, $xa\in J(R_e)$ by Theorem 1a). Hence, $yb=1_e-xa$ is a unit in $R_e,$ a contradiction. Therefore, x is an isolated vertex in $\Gamma ^h(R).$ Also, if R contains a unique maximal graded modular right ideal M,  it coincides with $J^g(R),$ and $\Gamma ^h(R)$ is totally disconnected. Hence, taking into account Lemma 3, if $\Gamma ^h(R)\setminus J^g(R)$ is connected, R must have at least two maximal graded modular right ideals of the same degree.

Theorem 3

Let R be a graded semisimple and a graded right Artinian ring. Also, let R be such that $|\max _{rm}^e(R)|\ge 2$ for every $e\in E(S)^*.$ Then $\Gamma ^h(R)\setminus \{0\}$ is a connected graph if and only if $|E(S)^*|=1.$ Moreover, in that case, the diameter of $\Gamma ^h(R)\setminus \{0\}$ is at most 5.

Proof

$(\Rightarrow )$ Let $\Gamma ^h(R)\setminus \{0\}$ be connected. Since there are at least two maximal graded modular right ideals of R of degree e,  for every $e\in E(S)^*,$ all of them are distinct from 0. Having in mind that there is a one-to-one correspondence between the maximal graded modular right ideals of R of degree e and the maximal modular right ideals of $R_e,$ for every $e\in E(S),$ none of the maximal graded modular right ideals of R contains a locally right invertible element. So, there must be at least two distinct vertices in $\Gamma ^h(R)\setminus \{0\}.$ Since $\Gamma ^h(R)\setminus \{0\}$ is connected, its edge set is nonempty. Hence, $|E(S)^*|=1$ by Lemma 3.

$(\Leftarrow )$ Let $E(S)^*=\{e\}.$ Let us observe a subgraph $\Gamma '$ of $\Gamma ^h(R)\setminus \{0\}$ that has only the elements from $R_e$ as vertices. We claim that $\Gamma '=\Gamma (R_e){\setminus }\{0\}.$ Let x,  $y\in \Gamma '$ be adjacent in $\Gamma ^h(R)\setminus \{0\}.$ Then $xR+yR=R.$ Now, $xR_e+yR_e\subseteq R_e.$ On the other hand, R is with unity $1=1_e\in R_e\subseteq R=xR+yR.$ Hence, there exist a,  $b\in R$ such that $1_e=xa+yb.$ Since x,  $y\in R_e,$ we get that $1_e=xa_e+yb_e\in xR_e+yR_e,$ where $a_e$ and $b_e$ are the e-components of a and b,  respectively. Therefore, for any $z\in R_e,$ we have that $z=1_ez\in (xR_e+yR_e)z\subseteq xR_e+yR_e.$ Thus, $xR_e+yR_e=R_e,$ that is, x and y are adjacent as vertices of $\Gamma (R_e)\setminus \{0\}.$ It follows that $\Gamma '\subseteq \Gamma (R_e)\setminus \{0\}.$ Now, let x,  $y\in \Gamma (R_e)\setminus \{0\}$ be adjacent. Then $xR_e+yR_e=R_e.$ Hence, for some a,  $b\in R_e,$ we have that $1_e=xa+yb.$ However, $1=1_e.$ Therefore, for every $z\in R,$ we get that $z=1_ez=(xa+yb)z\subseteq xR+yR,$ and so, $R\subseteq xR+yR.$ Therefore, $xR+yR=R,$ that is, x and y are adjacent as vertices of $\Gamma ^h(R)\setminus \{0\}.$ Since x,  $y\in R_e,$ they are adjacent as vertices of $\Gamma '.$ Hence, indeed, $\Gamma '=\Gamma (R_e)\setminus \{0\}.$ Now, since R is graded right Artinian, and since S is cancellative, $R_e$ is right Artinian. Moreover, there exists a one-to-one correspondence between the set of all maximal graded modular right ideals of R of degree e and maximal modular right ideals of $R_e.$ Since R is with unity $1_e,$ every maximal homogeneous right ideal of R is a maximal graded modular right ideal of R of degree e. Hence, by the hypotheses, $R_e$ contains at least two maximal right ideals. Also, since R is graded semisimple, the ring $R_e$ is semisimple by Theorem 1b). Therefore, $\Gamma (R_e)\setminus \{0\}=\Gamma '$ is a connected graph according to Theorem 3.2 in [43]. By the same theorem, the diameter of $\Gamma '$ is at most 3. Now, let $x\in \Gamma ^h(R)\setminus \{0\}$ be such that $\deg (x)=s\ne e.$ Since R is graded semisimple, by Theorem 1b), there exists a homogeneous element $a\in R$ such that $0\ne xa\in R_e.$ Since x is not locally right invertible in R,  nor is xa right invertible in $R_e.$ However, $\Gamma '$ is connected. Therefore, for every $w\in \Gamma (R_e)\setminus \{0\},$ distinct from xa,  there exists a path between xa and w,  of length at most 3. Since $R_e$ contains at least two maximal right ideals, it follows that there exists $y\in \Gamma (R_e)\setminus \{0\},$ distinct from xa,  such that $xaR_e+yR_e=R_e.$ Thus, $1=1_e=xau+yv,$ for some u,  $v\in R_e.$ Hence, $1\in xR+yR,$ and therefore, $xR+yR=R.$ In other words, x and y are adjacent in $\Gamma ^h(R)\setminus \{0\}.$ Therefore, for every nonzero $w\in R_e,$ there exists a path in $\Gamma ^h(R)\setminus \{0\}$ between x and w of length at most 4. Since x was chosen arbitrarily, $\Gamma ^h(R)\setminus \{0\}$ is a connected graph with the diameter at most 5. $\square$

4 Homogeneous Co-maximal Graphs of Groupoid-Graded Matrix Rings

Let ${\mathbb {F}}_q$ be a finite field with q elements, with unity 1,  and let $R=M_n({\mathbb {F}}_q)$ be the ring of $n\times n$ matrices over ${\mathbb {F}}_q$ under the usual matrix addition and multiplication, where n is a positive integer.

We know that R can be graded by a Brandt semigroup, a rectangular band and by a group, see [24]. In accordance with Lemma 3, there is only interest in gradings whose grading sets (with zero) contain a single (nonzero) idempotent element. The case of a trivial grading is covered by the results obtained in [43]. So, there are two interesting options left: a nontrivial group grading and a nontrivial grading obtained from a Brandt semigroup, distinct from a group with zero, with a single nonzero idempotent element.

R can be naturally graded by a groupoid with a single nonzero idempotent element, and which is distinct from a group with zero. In this article, we consider this grading only, which moreover “encodes” the good group gradings (see [13, 24, 37]). Indeed, let $R_e$ be the set $D_n({\mathbb {F}}_q)$ of all diagonal matrices of R. For $i\ne j,$ let $R_{(i,j)}$ be the set of all matrices of R with entries from ${\mathbb {F}}_q$ at the (i, j) position and zeroes elsewhere. Define

$$\begin{aligned}S:=\{(i,j)\ |\ i,j\in \{1,\dots ,n\},\ i\ne j\}\cup \{0,e\}\end{aligned}$$

and put $R_0$ to be the zero matrix. Then $R=\bigoplus _{s\in S}R_s$ is an S-graded ring inducing S,  and S is a groupoid with respect to the induced operation $(i,j)(k,l)=\delta _{jk}(i,l),$ for $i\ne l,$ and $(i,j)(k,i)=\delta _{jk}e,$ where $\delta _{jk}$ is the Kronecker delta, 0 is the zero element of S,  and $e^2=e$ is such that $e(i,j)=(i,j)e=(i,j)$ for all (i, j). Moreover, R is a pseudo-unitary S-graded ring inducing S with a homogeneous unity $1_e=1_{M_n({\mathbb {F}}_q)},$ where $1_{M_n({\mathbb {F}}_q)}$ is the identity matrix of $M_n({\mathbb {F}}_q).$ Hence, $\max _{rm}^e(R)=\max _r^h(R).$ Note that $R=\bigoplus _{s\in S^*}R_s,$ and $S^*=S\setminus \{0\}$ is a partial groupoid. Of course, if $n=1,$ then $R=R_e={\mathbb {F}}_q$ is trivially graded.

Throughout this section, for any full matrix ring $M_n(F)$ over a finite field F with unity 1,  by $e_{(i,j)}$ we denote the matrix having 1 at the (i, j) position and zeroes elsewhere. The zero matrix of $M_n(F)$ is denoted by $0_{M_n(F)},$ and the subset of all diagonal matrices of $M_n(F)$ is denoted by $D_n(F).$ Also, when we say that a matrix ring over a finite field is S-graded, we mean that it is graded in the above described way.

If A is a ring with unity, the set of all maximal right ideals of A is denoted by $\max _r(A).$

Lemma 4

Let ${\mathbb {F}}_q$ be a finite field with q elements, and let us observe $R=M_n({\mathbb {F}}_q)$ as an S-graded ring, where n is a positive integer. Then the following statements hold:

• (i) $|\max _r^h(R)|=n;$

• (ii) Every maximal homogeneous right ideal of R is generated by a homogeneous matrix of rank $n-1,$ and the number of such matrices is ${\left\{ \begin{array}{ll} 2(q-1) &{} \mathrm {if}\quad n=2;\\ (q-1)^{n-1} &{} \mathrm {if}\quad n\ne 2. \end{array}\right. }$

Proof

• (i) Since R is a ring with a homogeneous unity, there exists a one-to-one correspondence between the maximal homogeneous right ideals of R and the maximal right ideals of $R_e.$ Now, $R_e$ is a subset of R which consists of all diagonal matrices of R. Hence, $|\max _r(R_e)|=n,$ which proves the claim.

• (ii) Let I be a maximal homogeneous right ideal of R. We know that I is principal, observed as a right ideal of R,  and every matrix of I of rank $n-1$ can serve as a generator of I (see for instance Lemma 4.2 in [43]).

Case 1 $n=1.$ This is a trivial case when $R={\mathbb {F}}_q$ and $\{0\}$ is the only maximal right ideal of R. Hence, the number of homogeneous generators of rank 0 is $1=(q-1)^{n-1}=(q-1)^0,$ namely, the zero element of ${\mathbb {F}}_q.$

Case 2 $n=2.$ There are two maximal homogeneous right ideals of R,  namely, $\left( \begin{array}{cc} 0 &{} 0 \\ {\mathbb {F}}_q &{} {\mathbb {F}}_q \end{array}\right)$ and $\left( \begin{array}{cc} {\mathbb {F}}_q &{} {\mathbb {F}}_q \\ 0 &{} 0 \end{array}\right) .$ Let, for instance, $I=\left( \begin{array}{cc} 0 &{} 0 \\ {\mathbb {F}}_q &{} {\mathbb {F}}_q \end{array}\right) .$ Then the homogeneous matrices of rank $n-1=1$ that generate I are the matrices $\left( \begin{array}{cc} 0 &{} 0 \\ a_{(2,1)} &{} 0 \end{array}\right) ,$ $\left( \begin{array}{cc} 0 &{} 0 \\ 0 &{} a_{(2,2)} \end{array}\right) ,$ where $a_{(2,1)},$ $a_{(2,2)}$ are the arbitrary nonzero elements of ${\mathbb {F}}_q.$ Hence, I has $2(q-1)$ homogeneous generators of rank 1. Of course, the number of generators is the same for the other ideal too.

Case 3 $n>2.$ In this case, the only homogeneous matrices of R of rank $n-1$ are matrices that come from $R_e.$ For $i\in \{1,\dots ,n\},$ they are of the form

$$\begin{aligned} \left( \begin{array}{cccccccc} a_{(1,1)} &{} \dots &{} 0 &{} 0 &{} 0 &{} \dots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} \dots &{} a_{(i-1,i-1)} &{} 0 &{} 0 &{} \dots &{} 0 \\ 0 &{} \dots &{} 0 &{} 0 &{} 0 &{} \dots &{} 0 \\ 0 &{} \dots &{} 0 &{} 0 &{} a_{(i+1,i+1)} &{} \dots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} \dots &{} 0 &{} 0 &{} 0 &{} \dots &{} a_{(n,n)} \end{array}\right) , \end{aligned}$$

(1)

where $a_{(k,k)}\in {\mathbb {F}}_q^*,$ $k=1,\dots ,i-1,i+1,\dots ,n.$ Hence, there are $(q-1)^{n-1}$ such matrices. $\square$

In what follows, if $G=(V,E)$ is a graph, the degree of $v\in V(G)$ is denoted by $\deg _G(v).$ As usual, the minimal degree and the maximal degree among the vertices of G are denoted by $\delta (G)$ and $\Delta (G),$ respectively. Recall that G is said to be regular of degree r(G) if $r(G)=\delta (G)=\Delta (G).$ The connectivity number, the diameter, and the chromatic number of G are denoted by $\kappa (G),$ d(G),  and $\chi (G),$ respectively. By $\omega (G)$, we denote the least upper bound of the cardinal numbers of all the cliques in G.

Theorem 4

Let ${\mathbb {F}}_q$ be a finite field with q elements, and let us observe $R=M_n({\mathbb {F}}_q)$ as an S-graded ring, where $n\ge 2.$ Also, let $G=\Gamma ^h(R)\setminus \{0_{M_n({\mathbb {F}}_q)}\}.$ Then the following statements hold:

• (i) $|V(G)|=q^n-1+n(n-1)(q-1)-(q-1)^n;$

• (ii) G is connected and $d(G)={\left\{ \begin{array}{ll}2 &{} \mathrm {if}\quad n=2,\\ 3 &{} \mathrm {if}\quad n>2;\end{array}\right. }$

• (iii) If $n\ge 3,$ then $\delta (G)=(q-1)^{n-1}$ and $\Delta (G)=n(q-1)+\sum _{i=1}^{n-2}\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (q-1)^{i+1};$

• (iv) G is regular if and only if $n=2.$ In that case, $r(G)=2(q-1);$

• (v) $\kappa (G)=\delta (G)$ and $\omega (G)=\chi (G)=|\max ^h_r(R)|;$

• (vi) $|\{a\in G\ |\ \deg _G(a)=\delta (G)\}|=n^2(q-1).$

Proof

• (i) It is clear that $|R_e|=q^n.$ Also, $|R_{(i,j)}|=q$ for all $(i,j)\in S.$ On the other hand, the number of invertible homogeneous elements of R is equal to the number of invertible elements of $R_e.$ Hence, there are $(q-1)^n$ such elements. Since $|\{(i,j)\in S\}|=n(n-1),$ and since the zero matrix of R is not a vertex in G,  we therefore obtain that $|V(G)|=q^n-1+n(n-1)(q-1)-(q-1)^n.$

• (ii) According to Lemma 4, the number of maximal homogeneous right ideals of R is n,  and $n\ge 2$ by the hypothesis. Theorem 1b) implies that R is graded semisimple. Moreover, R is right Artinian. Hence, R is graded right Artinian. So, G is connected by Theorem 3. By the same theorem, $d(G)\le 5.$

Let us assume first that $n=2.$ Then each homogeneous matrix of R of rank 1 generates a maximal homogeneous right ideal of R. Let $a\in R_e$ be a homogeneous matrix of rank 1 with zero at the (2, 2) position. Then a is adjacent to every nonzero matrix from $R_{(2,1)}$ and to every nonzero matrix from $R_e$ which has zero at the (1, 1) position. The same what is said for a can be said for a nonzero matrix $b\in R_{(1,2)}.$ Hence, a and b are connected by a path of length 2. We analogously conclude that two matrices of rank 1 from the second row, and from distinct columns, are connected by a path of length 2. Hence, if $n=2,$ we have that $d(G)=2.$

Let now $n>2.$ Take non-invertible a,  $b\in R^*$ such that $aR+bR\ne R.$ Note that all of the homogeneous elements of R,  which are not of degree e,  are matrices of rank 1.

Case 1 Both a and b are of rank 1.

Subcase 1a. a,  $b\in R_e.$ Let us assume first that a and b have a nonzero entry at the same position, say (i, i). Let $c\in R_e$ be of rank $n-1,$ which is of the form (1). Then c is adjacent to both a and b. So, a and b are connected through c. Now, let us assume that a and b have nonzero entries at different positions. Then, there exist $c\in R_e$ and $d\in R_e,$ both of rank $n-1,$ such that $aR+cR=R$ and $bR+dR=R.$ Clearly, $c\ne d,$ and c and d are adjacent. Hence, in that case, a and b are connected by a path $a-c-d-b.$

Subcase 1b. $a\in R_e$ and $b\notin R_e$ or $a\notin R_e$ and $b\in R_e.$ For instance, let $a\in R_e$ and $b\notin R_e.$ Assume that the nonzero entry of a is at the (i, i) position. Now, let us suppose first that $b\in R_{(i,j)},$ for some $(i,j)\in S.$ Let $c\in R_e$ be a matrix of the form (1). Then c is adjacent to both a and b. So, a and b are connected by a path $a-c-b.$ Let us now assume that $b\in R_{(k,j)},$ for some $(k,j)\in S,$ where $k\ne i.$ Again, take $c\in R_e$ of the form (1). Then a and c are adjacent. However, there exists $d\in R_e$ of rank $n-1$ with the zero entry at the (k, k) position. Then b and d are adjacent. Since $k\ne i,$ we have that c and d are adjacent. Therefore, a and b are connected by a path $a-c-d-b.$

Subcase 1c. a,  $b\notin R_e.$ Assume that $a\in R_{(i,j)}$ and $b\in R_{(k,l)},$ for some (i, j),  $(k,l)\in S.$ By reasoning similarly to the previous subcases, if $i=k,$ we obtain that a and b are connected by a path of length 2,  and, if $i\ne k,$ we obtain that a and b are connected by a path of length 3.

Case 2 One of the matrices is of rank 1 and the other has the rank at least 2 and at most $n-1.$ For instance, let the rank of a be 1. Then $b\in R_e.$

Subcase 2a. $a\in R_e$ with the nonzero (i, i) entry. Let $c\in R_e$ be a matrix of the form (1). Then a and c are adjacent. Of course, if $b=c,$ then a and b are adjacent. So, let $b\ne c.$ Assume that b has a zero entry at the (i, i) position. Let $d\in R_e$ be of rank $n-1$ whose zero entry is not at the (i, i) position. Then b and d are adjacent, and so are c and d. Hence, a and b are connected by a path $a-c-d-b.$ Assume now that b has a nonzero entry at the (i, i) position. Then b and c are adjacent. So, a and b are connected by a path $a-c-b.$

Subcase 2b. $a\notin R_e.$ Let $a\in R_{(i,j)},$ for some $(i,j)\in S.$ We may proceed as in the previous case and conclude that either a and b are adjacent or they are connected by a path of length 2 or they are connected by a path of length 3.

Therefore, taking into account all of the cases, if $n>2,$ we have that $d(G)=3.$

• (iii) Every nonzero homogeneous element x of R,  which is of rank 1,  is adjacent to y only if yR is a maximal homogeneous right ideal of R. Therefore, by Lemma 4, if $x\in R_{(i,j)}$ or if $x\in R_e,$ but with a nonzero entry at the (i, i) position, then $y\in R_e$ is of the form (1), and there are $(q-1)^{n-1}$ such elements. It follows that $\delta (G)=(q-1)^{n-1}.$ Now, let x be a homogeneous element of R that generates a maximal right ideal of R. Then x is of the form (1), for some $i\in \{1,\dots ,n\}.$ Hence, as a vertex of G,  it is adjacent to every nonzero matrix from $R_{(i,j)}$ for all $j=1,\dots ,i-1,i+1,\dots ,n.$ There are $(n-1)(q-1)$ such matrices. Element x is also adjacent to every matrix of rank 1 that belongs to $R_e,$ but has the only nonzero entry at the (i, i) position. Of course, there are $q-1$ such matrices. Moreover, x is adjacent to every homogeneous element $y\in R_e$ of rank greater than 1 and less than n,  which has a nonzero entry at the (i, i) position. There are $\sum _{i=1}^{n-2}\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (q-1)^{i+1}$ such elements. Hence, $\Delta (G)=n(q-1)+\sum _{i=1}^{n-2}\left( {\begin{array}{c}n-1\\ i\end{array}}\right) (q-1)^{i+1}.$

• (iv) Let $n=2.$ As we have already pointed out in the proof of ii),  every nonzero matrix from $e_{(1,1)}Re_{(1,1)}$ is adjacent to every nonzero matrix from both $e_{(2,1)}Re_{(1,1)}$ and $e_{(2,2)}Re_{(2,2)}.$ The same holds true for every nonzero matrix from $e_{(1,2)}Re_{(2,2)}.$ Symmetrically, every nonzero matrix from $e_{(2,1)}Re_{(1,1)}$ and every nonzero matrix from $e_{(2,2)}Re_{(2,2)}$ is adjacent to every nonzero matrix from both $e_{(1,1)}Re_{(1,1)}$ and $e_{(1,2)}Re_{(2,2)}.$ Hence, $\delta (G)=\Delta (G)=2(q-1).$ If $n>2,$ then, clearly by iii), 

  $$\begin{aligned} \delta (G)=(q-1)^{n-1}=\deg _G(e_{(1,1)})<\deg _G(e_{(1,1)}+e_{(2,2)}) \le \Delta (G). \end{aligned}$$

  So, indeed, G is regular if and only if $n=2.$

• (v) These statements are proved analogously to the way the corresponding statements for the non-graded case are proved (cf. statements v) and vii) of Theorem 4.3 in [43]). For the readers’ convenience, we provide a proof for the assertion concerning the chromatic number. We know that $\omega (G)\le \chi (G).$ According to Lemma 4, the number of all maximal homogeneous right ideals of R is n. Let $\{M_1,\dots ,M_n\}$ be the set of all homogeneous maximal right ideals of R. Put $X_1=M_1$ and $X_i=M_i{\setminus } M_i\cap (M_1\cup \dots \cup M_{i-1})$ for every $i=2,\dots ,n.$ Then each $X_i$ is an independent set, $G=\bigcup _{i=1}^nX_i,$ and $X_i\cap X_j=\emptyset$ for all $i\ne j.$ It follows that $\chi (G)\le n.$ Let $x_i\in G$ be a homogeneous generator of $M_i.$ By Lemma 4, the set $\{x_1,\dots ,x_n\}$ forms a clique of order n. Hence, $n\le \omega (G),$ which concludes the proof.

• (vi) In order to find the cardinality of $\{a\in G\ |\ \deg _G(a)=\delta (G)\},$ it is enough to count the number of homogeneous elements of R of rank 1. The number of matrices from $R_e$ of rank 1 is obviously $n(q-1).$ Also, for every $(i,j)\in S,$ we have that $|R_{(i,j)}\setminus \{0_{M_n({\mathbb {F}}_q)}\}|=q-1.$ Therefore,

  $$\begin{aligned} |\{a\in G\ |\ \deg _G(a)=\delta (G)\}|=n(q-1)+n(n-1)(q-1)=n^2(q-1), \end{aligned}$$

  as asserted.

$\square$

Let A and B be arbitrary rings, and let $\Gamma (A)$ and $\Gamma (B)$ be the co-maximal graphs of A and B,  respectively. Example 4.4 in [43] shows that the implication $\Gamma (A)\cong \Gamma (B)\Rightarrow A\cong B$ does not hold true in general. Namely, if $A=M_n({\mathbb {Z}}_4)$ and $B=M_n({\mathbb {Z}}_2[x]/(x^2)),$ then $\Gamma (A)\cong \Gamma (B)$ but A is not isomorphic to B. This particular example also shows that in the case of two S-graded rings inducing S,  say A and B,  the graph isomorphism of graphs $\Gamma ^h(A)$ and $\Gamma ^h(B)$ does not in general imply that A and B are graded isomorphic as S-graded rings, by observing $M_n({\mathbb {Z}}_4)$ and $M_n({\mathbb {Z}}_2[x]/(x^2))$ as trivially graded rings. However, by Theorem 4.5 in [43], if R is a ring with unity and $n\ge 2,$ then $\Gamma (R)\cong \Gamma (M_n({\mathbb {F}}_q))$ implies that $R\cong M_n({\mathbb {F}}_q).$ Here, we obtain an S-graded version of this result.

Let us recall a few notions first. If $G=(V,E)$ is a graph and $V'\subset V,$ then $\langle V'\rangle$ denotes the induced subgraph, whose vertex set is $V'$ and with vertices being adjacent if and only if they are adjacent in G. Let $G_1=(V_1,E_1)$ and $G_2=(V_2,E_2)$ be such that $V_1\cap V_2=\emptyset$ and $E_1\cap E_2=\emptyset .$ The join of $G_1$ and $G_2$ is a graph $G_1\cup G_2$ with the edges that join $V_1$ and $V_2$ being added.

Theorem 5

Let ${\mathbb {F}}_q$ be a finite field with q elements, and let us observe $M_n({\mathbb {F}}_q)$ as an S-graded ring, where $n\ge 2.$ Also, let R be an S-graded ring with unity, and $\mathrm {supp}(R)=S^*.$ If $G'=\Gamma ^h(R)\cong \Gamma ^h(M_n({\mathbb {F}}_q))=G,$ then R and $M_n({\mathbb {F}}_q)$ are graded isomorphic as S-graded rings.

Proof

Since R is S-graded and R is with unity 1,  let us note that 1 is a homogeneous element $1_e.$ So, $\max ^e_{rm}(R)=\max ^h_r(R).$ Every homogeneous zero divisor of R is a vertex in $G'.$ On the other hand, the number of homogeneous elements of $M_n({\mathbb {F}}_q)$ is finite. Therefore, since $G\cong G',$ the number of homogeneous zero divisors of R is finite. Hence, by Lemma 1, it follows that the homogeneous part of R is finite. In particular, R is graded right Artinian. If $J^g(R)\ne 0,$ then $G'$ contains more than one isolated vertex, according to the discussion preceding Theorem 3. This implies that the same holds true for G,  which is a contradiction by Theorem 4ii). Therefore, R is graded semisimple. Then, by Theorem 2, there exist positive integers p and $n_1,$ $\dots ,$ $n_p$ such that R is graded isomorphic to a direct product $M_{n_1}(F_1)\times \dots \times M_{n_p}(F_p),$ where each $M_{n_i}(F_i)$ is a graded matrix ring over a graded division ring $F_i.$ Each $M_{n_i}(F_i)$ is a homogeneous ideal of R,  that is, an S-graded ring. Hence, each $M_{n_i}(F_i)$ is an $S_i$-graded ring inducing $S_i,$ where $S_i$ is a subgroupoid of S. On the other hand, we know, and it is easy to check, that a graded division ring is graded by a group, and, that its ring component is a division ring (see for instance [33]). Now, $\{e\}$ is the only nonzero subgroup of S. Hence, each $F_i$ is trivially graded, that is, each $F_i$ is a division ring. Moreover, $H_R$ is finite, which implies that each $F_i$ is a finite field, say $q_i=|F_i|.$ So, each $M_{n_i}(F_i)$ is an $S_i$-graded ring $M_{n_i}({\mathbb {F}}_{q_i}).$ Let $\phi$ be a graph isomorphism from G onto $G'.$ Note that $\phi$ is also a graph isomorphism between the graphs $G\setminus \{0_{M_n({\mathbb {F}}_q)}\}$ and $G'\setminus \{0_R\}.$ For every i,  denote by $M_i$ the set of all maximal right ideals of R of the form

$$\begin{aligned}M_{n_1}(F_1)\times \dots \times M_{n_{i-1}}(F_{i-1})\times M\times M_{n_{i+1}}(F_{i+1})\times \dots \times M_{n_p}(F_p),\end{aligned}$$

where M is a maximal homogeneous right ideal of $M_{n_i}(F_i).$ Then $\max ^h_r(R)=\bigcup _{i=1}^pM_i.$ Since $\omega (G)=\omega (G'),$ by Theorem 4v), we get $|\max ^h_r(M_n({\mathbb {F}}_q))|=|\max ^h_r(R)|.$ By Lemma 4, we have that $|\max ^h_r(M_n({\mathbb {F}}_q))|=n.$ There exists a one-to-one correspondence between the set of all maximal homogeneous right ideals of R and the set of all maximal right ideals of $R_e.$ So, $|\max ^h_r(R)|=|\max _r(R_e)|.$ On the other hand, $R_e\cong D_{n_1}(F_1)\times \dots \times D_{n_p}(F_p).$ Therefore, by Lemma 4, we obtain that $|\max ^h_r(R)|=\sum _{i=1}^pn_i.$ Hence,

$$\begin{aligned} n=\sum _{i=1}^pn_i. \end{aligned}$$

(2)

Let u and $u_i$ be the number of elements that can be generators in any maximal homogeneous right ideal of R and $M_i,$ respectively. Then, by reasoning as in the proof of Theorem 4.5 in [43], we get from (2) that $u=u_i.$ Namely, since

$$\begin{aligned}\langle \{a\in V(G)\ |\ \deg _G(a)=\Delta (G)\}\rangle \cong \langle \{a\in V(G')\ |\ \deg _{G'}(a)=\Delta (G')\}\rangle ,\end{aligned}$$

we have that $\langle \{a\in V(G)\ |\ \deg _G(a)=\Delta (G)\}\rangle$ is the join of n copies of the complement graph of $K_u,$ and $\langle \{a\in V(G')\ |\ \deg _{G'}(a)=\Delta (G')\}\rangle$ is the join of $\sum _{i=1}^pn_i$ copies of the complement graph of $K_{u_i}.$ (Here, as usual, $K_m$ denotes the complete graph with m vertices.) According to Lemma 4, note that for every i we have that

$$\begin{aligned}u={\left\{ \begin{array}{ll} 2(q-1) &{} \mathrm {if}\quad n=2;\\ (q-1)^{n-1} &{} \mathrm {if}\quad n>2, \end{array}\right. }\quad \mathrm {and}\quad u_i={\left\{ \begin{array}{ll} 2(q_i-1) &{} \mathrm {if}\quad n_i=2;\\ (q_i-1)^{n_i-1} &{} \mathrm {if}\quad n_i\ne 2. \end{array}\right. }\end{aligned}$$

So, if $n=2,$ then $u>1.$ Since $u=u_i$ for every i,  it follows that $n_i\ne 1$ for every i. Therefore, by (2), we get that $p=1$ and $n_1=2.$ Thus, $2(q-1)=2(q_1-1),$ which implies that $q=q_1.$ Hence, R and $M_2({\mathbb {F}}_q)$ are graded isomorphic as S-graded rings. So, from now on, we assume that $n>2.$ Then, since $\mathrm {supp}(R)=S^*,$ we conclude from (2) that $n_i>2$ for every i.

Case i $q=2.$ Then $u_i=(q-1)^{n-1}=1$ for every i. Since $n_i>2,$ it follows that $(q_i-1)^{n_i-1}=1$ for every i. Hence, $q_i=2$ for every i. By Theorem 4vi),  we have that $n^2=\sum _{i=1}^pn_i^2.$ However, this, together with (2), implies that $p=1$ and $n_1=n.$ Therefore, R and $M_n({\mathbb {F}}_2)$ are graded isomorphic as S-graded rings.

Case ii $q>2.$ Then, $u_i=u=(q-1)^{n-1}>1$ for every i. Now, let M be a maximal homogeneous right ideal of $M_n({\mathbb {F}}_q),$ and a a generator of M. It follows that $\deg _{G'}(\phi (a))=\Delta (G)=\Delta (G')$ and $\phi (a)$ generates a maximal homogeneous right ideal $M'$ of R. Analogously to the proof of Theorem 4.5 in [43], we conclude that $\phi$ induces a one-to-one correspondence between the set of all maximal homogeneous right ideals of $M_n({\mathbb {F}}_q)$ and those of R. Namely, M is a unique maximal independent set in G that contains a,  and therefore, $M'$ is a unique maximal independent set in $G'$ that contains $\phi (a).$ Hence, $\phi (M)$ is a maximal homogeneous right ideal of R. Let

$$\begin{aligned}B_i=0_{M_{n_1}(F_1)}\times \dots \times 0_{M_{n_{i-1}}(F_{i-1})} \times e_{(1,1)}\times 0_{M_{n_{i+1}}(F_{i+1})}\times \dots \times 0_{M_{n_p}(F_p)}.\end{aligned}$$

Then $\deg _G(e_{(1,1)})=\delta (G)=\delta (G')=\deg _{G'}(B_i)$ for every i. Since $n>2$ and $n_i>2$ for every i,  we get that

$$\begin{aligned} \deg _G(e_{(1,1)})=(q-1)^{n-1}\quad \mathrm {and}\quad \deg _{G'}(B_i)=\frac{\prod _{j=1}^p(q_j-1)^{n_j}}{q_i-1}. \end{aligned}$$

(3)

Now, $\deg _G(e_{(1,1)})=\deg _{G'}(B_i)$ for every i,  so (3) implies that $q_1=\dots =q_p.$ Since $u_i=(q_i-1)^{n_i-1}=(q-1)^{n-1}>1$ for every i,  and since $q_1=\dots =q_p,$ we get that $n_1=\dots =n_p.$ Therefore, from (2) we obtain that $n=pn_1.$ This, together with $\deg _G(e_{(1,1)})=\deg _{G'}(B_i)$ and (3), implies that

$$\begin{aligned}(q-1)^{n-1}=(q_1-1)^{pn_1-1}=(q_1-1)^{n-1}.\end{aligned}$$

Hence, $q=q_1.$ By Theorem 4vi), we have that $n^2(q-1)=pn_1^2(q_1-1).$ Since $n=pn_1$ and $q=q_1,$ it follows that $p=1.$ Hence, $n=n_1.$ Thus, R and $M_n({\mathbb {F}}_q)$ are graded isomorphic as S-graded rings, which completes the proof. $\square$