On the radicals of linear algebraic monoids

Abstract

In this paper we consider the Schwarz radical of linear algebraic semigroups as defined in semigroup theory. We give some new characterizations of the complete regularity, regularity and solvability of irreducible linear algebraic monoids in terms of Schwarz radical data. Moreover, we give a generalization about the results of the kernel to the results of completely regular $\mathscr {J}$-classes.

1 Introduction

Throughout this paper, $\mathbb {Z}^+$ will denote the set of all positive integers. Let S be a semigroup and $I\subseteq S$ a (two-sided) ideal of S. Let $\sqrt{I}$ denote the set of all elements of S which satisfy that some power of them belongs to I, i.e.,

$$\begin{aligned}\sqrt{I}=\{ a\in S\mid a^i \in I \text { for some } i\in \mathbb {Z}^+ \}.\end{aligned}$$

There are five concepts of radical of S with respect to I, called the Clifford radical, Luh radical, McCoy radical, Schwarz radical, and $\check{S}evrin$ radical (see Definition 3.1), which are natural extensions of the concepts of radical of a ring. Denote the Clifford radical, Luh radical, McCoy radical, Schwarz radical, and $\check{S}evrin$ radical of S with respect to I by $\mathcal {R}^*(I)$, $\mathcal {C}(I)$, $\mathcal {M}(I)$, $\mathcal {R}(I)$, $\mathcal {L}(I)$ respectively. If a semigroup S has a kernel $\ker (S)$ (the minimal ideal of S), then the Clifford radical, Luh radical, McCoy radical, Schwarz radical, and $\check{S}evrin$ radical of S with respect to $\ker (S)$ are simply called the Clifford radical, Luh radical, McCoy radical, Schwarz radical, and $\check{S}evrin$ radical of S. Bos$\check{a}$k [3] gives an example in abstract semigroups that all of the radicals mentioned above are distinct from one anther, and shows that for any semigroup S and any ideal I of S,

$$\begin{aligned}I\subseteq \mathcal {R}(I)\subseteq \mathcal {M}(I)\subseteq \mathcal {L}(I)\subseteq \mathcal {R}^*(I)\subseteq \sqrt{I}\subseteq \mathcal {C}(I)\subseteq S.\end{aligned}$$

Kme$\check{t}$ [14] prove that $\mathcal {R}^*(I)=\sqrt{I}=\mathcal {C}(I)$ if and only if $\sqrt{I}$ is an ideal of S. And S is a semilattice of archimedean semigroups (i.e., $b\in S^1aS^1 \Rightarrow b^i\in S^1a^2S^1$ for some $i\in \mathbb {Z}^+$, for all $a,b\in S$), if and only if for any ideal I of S, $\sqrt{I}$ is an ideal of S (see [5] and [15]).

Semigroups which satisfy that the set $\sqrt{I}$ is a semigroup for any ideal I, are characterized in [1] by Bogdanović and Ćirić.

In this paper, we consider the above radicals in linear algebraic semigroups. A linear (or affine) algebraic semigroup S over an algebraically closed field K is both an affine variety over K and a semigroup for which the product map $S\times S\rightarrow S$ is a morphism of varieties. Then S has a kernel $\ker (S)$ (see [22, Theorem 3.28]). Moreover, by [7, II 2.3.3] S is isomorphic to a (Zariski) closed subsemigroup of total n by n matrix monoid $M_n(K)$ for some $n\in \mathbb {Z}^+$, and is strongly $\pi$-regular (i.e., some power of each element of S lies in a subgroup of S) by a theorem of Clark (see [22, Theorem 3.18]). Then we have that the Clifford radical, McCoy radical, Schwarz radical, and $\check{S}evrin$ radical of S are coincide, i.e.,

$$\begin{aligned} \mathcal {R}(\ker (S))=\mathcal {M}(\ker (S))=\mathcal {L}(\ker (S)) =\mathcal {R}^*(\ker (S)). \end{aligned}$$

Hence we only need to consider the Schwarz radical of a linear algebraic semigroup. In particular, the radical-like property, introduced by J. Luh (see [17]), holds for the Schwarz (or Clifford) radical of a linear algebraic semigroup S, that is, the Rees factor semigroup $S/ \mathcal {R}(\ker (S))$ of S modulo $\mathcal {R}(\ker (S))$ has zero Schwarz (or Clifford) radical.

The following theorems play very important role to study the theory of linear algebraic monoids. Note that if a linear algebraic semigroup S has a zero element, the Schwarz radical $\mathcal {R}(\ker (S))$ of S is the maximum nilpotent ideal of S and $\sqrt{\ker (S)}$ is the set of all nilpotent elements of S.

Theorem 1.1

[21, Theorem 2.1] Let S be an irreducible linear algebraic semigroup with zero 0. Then the following conditions are equivalent:

• (i) S is completely regular;

• (ii) S has no non-zero nilpotent elements $(i.e.,~\sqrt{\ker (S)}=\{0\})$;

• (iii) S is a monoid and the unit group of S is a torus.

Theorem 1.2

[22, Theorem 7.3] Let M be an irreducible linear algebraic monoid with zero 0 and unit group G. Then the following conditions are equivalent:

• (i) G is reductive;

• (ii) M is regular;

• (iii) M has no non-zero nilpotent ideals $(i.e.,~\mathcal {R}(\ker (M))=\{0\})$.

Theorem 1.3

[19, Theorem 23] Let M be an irreducible linear algebraic monoid with zero 0 and unit group G. Then the following conditions are equivalent:

• (i) G is solvable;

• (ii) the nilpotents of M form an ideal of M $(i.e.,~\mathcal {R}(\ker (M))=\sqrt{\ker (M)})$;

• (iii) $J^2\subseteq J$ for all $J\in \mathcal {U}(M)$.

These facts imply that the structure of irreducible linear algebraic monoids with zero can be characterized in terms of Schwarz radical (or nilpotency) data. In general, for an irreducible linear algebraic monoid M, it need not have a zero element, that is, its kernel $\ker (M)$ is nontrivial. Brion shows in [4, Corollary 3.1.5] that for any irreducible non-affine algebraic monoid (that is, it is non-affine as a variety), its kernel must be nontrivial. The kernel of the linear algebraic monoid M carries a lot of structural information about M as well as of the unit group G which are well studied by Huang (cf. [9, 10, 12]).

The purpose of this paper is to study the structure of linear algebraic monoids in terms of Schwarz radical data. We give generalizations of the above results (Theorems 1.1, 1.2, 1.3) for an irreducible linear algebraic monoid (without zero). Namely, we prove the following theorem.

Theorem 1.4

Let M be an irreducible linear algebraic monoid with unit group G. Then

• (i) (Theorem 4.1) M is completely regular if and only if $\sqrt{\ker (M)}=\ker (M)$;

• (ii) (Theorem 4.2) M is regular if and only if the Schwarz radical of M is a completely simple semigroup (i.e., $\mathcal {R}(\ker (M))=\ker (M))$;

• (iii) (Corollary 4.4) G is solvable if and only if $\sqrt{\ker (M)}$ forms an ideal of M (i.e., $\mathcal {R}(\ker (M))=\sqrt{\ker (M)})$ and a maximal subgroup of $\ker (M)$ is solvable.

Moreover, for any completely regular $\mathscr {J}$-class $J\in \mathcal {U}(M)$, we construct a submonoid $M_J$ of M with kernel J, defined by

$$\begin{aligned} M_J=\{a\in M\mid aJ\subseteq J\}. \end{aligned}$$

Then $M_J$ is a linear algebraic monoid with kernel J. Moreover, the unit group of $M_J$ is just the unit group of M. Hence we can generalize the known results about the kernel of a linear algebraic monoid (see [9, 10, 12]). For instance,

Theorem 1.5

(Corollary 5.5) Let M be an irreducible linear algebraic monoid with unit group G. Let $J\in \mathcal {U}(M)$ be completely regular, and $e\in E(J)$. Then

• (i) $\dim R(G)=\dim E(J) +\dim R(G_e) +\dim R(eGe)$;

• (ii) $\dim R_u(G) =\dim E(J)+ \dim R_u(G_e)+\dim R_u(eGe)$;

• (iii) G is reductive if and only if $G_e$ and J are both reductive groups.

The article is organized as follows. Section 2 is for notions and notations. In Sect. 3, we work with various properties of the Schwarz radical of algebraic semigroups. In Sect. 4, we give characterizations of the completely regularity, regularity and solvability of irreducible linear algebraic monoids in terms of Schwarz radical data. In Sect. 5, we generalize the results of Sect. 4 to the case in terms of completely regular $\mathscr {J}$-classes of irreducible linear algebraic monoids.

2 Preliminaries

We now assemble some notions and notations. $\mathbb {Z}^{+}$ will denote the set of all positive integers. If X is a set, then |X| denotes the cardinality of X. Let S be a semigroup. Let $S^1:=S\cup \{1\}$ be the natural monoid extension of S. The semigroup S is strongly $\pi$-regular (s$\pi$r) if for each $a\in S$, there exists $i\in \mathbb {Z}^+$ such that $a^i$ lies in a subgroup of S. If $a,b\in S$, then a|b (a divides b) if $xay=b$ for some $x,y\in S^1$. S is archimedean if for all $a,b\in S$, $a|b^i$ for some $i\in \mathbb {Z}^{+}$. Let $S_{\alpha }\ (\alpha \in \Omega )$ denote a partition of S into subsemigroups. Then S is a semilattice (union) of $S_{\alpha }\ (\alpha \in \Omega )$ if for all $\alpha ,\beta \in \Omega$, there exists $\gamma \in \Omega$ such that $S_{\alpha } S_{\beta } \cup S_{\beta } S_{\alpha } \subseteq S_{\gamma }$. According to [22, Theorem 1.15], S is a semilattice of archimedean semigroups if and only if for all $a,b\in S$, $a\mid b$ implies $a^2\mid b^i$ for some $i\in \mathbb {Z}^{+}$. Let E(S) denote the set of all idempotents of S. Let $e\in E(S)$. We denote by $J_e$, $L_e$, $R_e$ and $H_e$ the $\mathcal {J}$-, $\mathcal {L}$-, $\mathcal {R}$- and $\mathcal {H}$-classes of e in S under Green’s relations, respectively (see [22, Chap. 1]). Suppose S is an s$\pi$r-semigroup and J is a $\mathcal {J}$-class of S. Then J is regular if $E(J)\ne \varnothing$. Moreover, J is completely regular if J is regular and $J^2\subseteq J$. Let $\mathcal {U}(S)$ be the set of all regular $\mathcal {J}$-classes of S. For any $J_1,J_2\in \mathcal {U}(S)$, we denote $J_1\le J_2$ if $a_2\mid a_1$ for some (all) $a_i\in J_i$, $i=1,2$. We write $\mathcal {U}(S)$ for the partially ordered set $(\mathcal {U}(S),\le )$. Let $\varnothing \ne I \subseteq S$. Then I is a right ideal of S if $IS \subseteq I$; I is a left ideal of S if $SI\subseteq I$; I is an ideal of S if $S^1IS^1\subseteq I$. The minimum ideal of S, if it exists, is called the kernel of S, denoted by $\ker (S)$. A completely simple semigroup S is an s$\pi$r-semigroup with no ideals other than S.

Let K denote a fixed algebraically closed field. $M_n(K)$ will denote the algebra of all $n\times n$ matrices over K, and $GL_n(K)$ its unit group. Let S be a Zariski closed subsemigroup of $M_n(K)$. If $e\in E(S)$ and $a\in S$, then we let $\det _e(a)=det(eae+1-e)$. Thus $\det _e(a)\ne 0$ if and only if $eae\in H_e$ by [22, Remark 3.23].

Let M be an irreducible linear algebraic monoid over K with unit group G. M is regular (resp. completely regular) if it is so as a semigroup. We call M reductive (resp. semisimple, solvable, nilpotent, a d-monoid) if its unit group is reductive (resp. reductive with center 1-dimensional, solvable, nilpotent, a torus). We write R(G) (resp. $R_u(G)$) for the radical (resp. unipotent radical) of G. The rank of G, denoted $\text{ rank }(G)$, is referred to as the dimension of a maximal torus of G. Let W(G) denote the weyl group of G. Then by [13, Proposition 24.1A, Corollary 25.2C], W(G) is finite, and G is solvable if and only if $|W(G)|=1$. For a subset V of M, denote by $\overline{V}$ the Zariski closure of V in M. If N is a closed algebraic subsemigroup of M, let $N^c$ be the identity component of N. If $e\in E(M)$, then we denote $M_e=\{a\in M\mid ea=ae=e\}^c$, $G^r(e)=\{x\in G\mid xe=e\},\ G^l(e)=\{x\in G\mid ex=e\}$, and $\ G_e=(G^r(e)\cap G^l(e))^c$. For any subset X of M, $C_X^r(e)=\{a\in X\mid ae=eae\}$, $C_X^l(e)=\{a\in X\mid ea=eae\}$, $C_X(e)=\{a\in X\mid ae=ea\}$. Let T be a maximal torus of G. Then $\Lambda \subseteq E(\overline{T})$ is a cross-section lattice of M, if $|\Lambda \cap J|=1$ for all $J\in \mathcal {U}(M)$ and $J_e\ge J_f$ implies $e\ge f$ for all $e,f\in \Lambda$. If $J\in \mathcal {U}(M)$, then the width of J, $\omega (J)=|J\cap E(\overline{T})|$. If $e\in E(J)$, $\omega (e)=\omega (J)$. For completely regular $\mathcal {J}$-classes, we have the following characterizations.

Theorem 2.1

[22, Remark 1.7(iii)] Let S be a linear algebraic semigroup, $J\in \mathcal {U}(S)$. Then the following conditions are equivalent:

• (i) J is completely regular;

• (ii) J is completely simple;

• (iii) $J^2\subseteq J$;

• (iv) $E(J)^2\subseteq J$.

Theorem 2.2

[22, Theorem 6.30, Corollary 6.34] Let M be an irreducible linear algebraic monoid with unit group G, $J\in \mathcal {U}(M)$ and $e\in E(J)$. Then the following conditions are equivalent:

• (i) J is completely regular;

• (ii) $E(J)\subseteq \overline{B}$ for some Borel subgroup B of G;

• (iii) $\omega (e)=1$;

• (iv) $e\in E(\overline{R(G)})$;

• (v) $G=C_G^l(e)C_G^r(e)$;

• (vi) eGe is the $\mathcal {H}$-class of e.

[22, 23] are our primary references for algebraic monoid theory, and [2, 13, 26] for algebraic group theory.

3 The radicals of linear algebraic semigroups

Let S be a linear algebraic semigroup and I a (two-sided) ideal of S. Let $\ker (S)$ be the kernel of S (the minimum ideal of S). The following concepts are used for abstract semigroups which are defined analogously to these concepts for rings.

An element a of S is termed a nilpotent element of S with respect to I if $a^i\in I$ for some $i\in \mathbb {Z}^+$. Let $\sqrt{I}$ denote the set of all nilpotent elements of S with respect to I, i.e.,

$$\begin{aligned} \sqrt{I}=\{ a\in S\mid a^i \in I\ for\ some\ i\in \mathbb {Z}^{+} \}. \end{aligned}$$

By [22, Corollary 3.30], the set $\sqrt{I}$ is closed in S. An ideal (left or right), or a subsemigroup A of S is nilpotent with respect to I if $A^i\subseteq I$ for some $i\in \mathbb {Z}^+$, and is nil with respect to I if every element of A is a nilpotent element of S with respect to I. For simplicity, an ideal A (left or right) is nilpotent (resp., nil) if A is nilpotent (resp., nil) with respect to $\ker (S)$. An ideal A of S is locally nilpotent with respect to I if every subsemigroup $S_1\subseteq S$, generated by a finite number of elements of A, is nilpotent with respect to I. An ideal P of S is called a prime ideal of S if $I_1I_2\subseteq P$ implies that $I_1\subseteq P$ or $I_2 \subseteq P$ where $I_1$ and $I_2$ are ideals of S. An ideal P of S is called a completely prime ideal of S if for any two elements $a, b\in S$, $ab\in I$ implies that $a\in I$ or $b\in I$. Evidently a completely prime ideal is a prime ideal. Now we define some radicals of S with respect to I.

Definition 3.1

Let S be a linear algebraic semigroup, and I an ideal of S.

• (i) The union $\mathcal {L}(I)$ of all locally nilpotent ideals of S with respect to I is called the $\check{S}evrin$ (or locally nilpotent) radical of S with respect to I.

• (ii) The union $\mathcal {R}^*(I)$ of all nil ideals of S with respect to I is called the Clifford (or nil) radical of S with respect to I [6].

• (iii) The union $\mathcal {R}(I)$ of all nilpotent ideals of S with respect to I is called the Schwarz (or nilpotent) radical of S with respect to I [24].

• (iv) The intersection $\mathcal {M}(I)$ of all prime ideals of S containing I is called the McCoy (or prime) radical of S with respect to I [17].

• (v) The intersection $\mathcal {C}(I)$ of all completely prime ideals of S containing I is called the Luh (or completely prime) radical of S with respect to I [17].

Obviously, every nilpotent (left or right) ideal of S with respect to I is nil with respect to I. By [22, Corollary 3.16], there exists $n\in \mathbb {Z}^+$ such that S is isomorphic to a (Zariski) closed subsemigroup of $M_n(K)$ , and thus for all $a\in S$, $a^n$ lies in a subgroup of S following [22, Theorem 3.18]. Hence

$$\begin{aligned} \sqrt{I} = \{ a\in S\mid a^n \in I\}. \end{aligned}$$

Then every nil (left or right) ideal of S with respect to I is nilpotent with respect to I. Therefore, an ideal of S is nil with respect to I if and only if it is nilpotent with respect to I. Thus $\mathcal {R}(I)=\mathcal {R}^*(I)$. It is known that (see [3])

$$\begin{aligned} I\subseteq \mathcal {R}(I)\subseteq \mathcal {M}(I)\subseteq \mathcal {L}(I)\subseteq \mathcal {R}^*(I)\subseteq \sqrt{I}\subseteq \mathcal {C}(I)\subseteq S. \end{aligned}$$

So the Clifford,  McCoy,  Sevrin ,  Schwarz radicals of S with respect to I coincide, that is,

$$\begin{aligned} \mathcal {R}(I)=\mathcal {M}(I)=\mathcal {L}(I)=\mathcal {R}^*(I). \end{aligned}$$

Hence we only consider the Schwarz radical of a linear algebraic semigroup S with respect to an ideal I.

Throughout this paper, we use the notation $\mathcal {R}(I)$ to denote the Schwarz radical of a linear algebraic semigroup S with respect to an ideal I of S. We write $\mathcal {R}_S(I)$, if we want to specify S. For simplicity, the Schwarz radical of S with respect to $\ker (S)$ is called the Schwarz radical of S, denoted by $\mathcal {R}\ker (S)$. Clearly, $\mathcal {R}(I)$ is the largest nilpotent (or nil) ideal of S with respect to I. Since every ideal of S contained in $\sqrt{I}$ is nil with respect to I, it is easy to see that $\mathcal {R}(I)$ is also the largest ideal of S contained in $\sqrt{I}$. The following lemma shows that $\mathcal {R}(I)$ is the largest nilpotent (or nil) left (or right) ideal of S with respect to I.

Lemma 3.2

Let S be a linear algebraic semigroup and I an ideal of S. Then $\mathcal {R}(I)$ contains every nilpotent left (or right) ideal of S with respect to I.

Proof

Let A be a nilpotent left ideal of S with respect to I. Then there exists some $i\in \mathbb {Z}^+$ such that $A^i\subseteq I$. So $(AS^1)^i\subseteq A^iS^1\subseteq IS^1\subseteq I$. Hence $AS^1$ is a nilpotent ideal of S with respect to I. Therefore $A\subseteq AS^1\subseteq \mathcal {R}(I)$. Similarly, we have that every nilpotent right ideal of S with respect to I is also contained in $\mathcal {R}(I)$. $\square$

By Lemma 3.2, we have the following lemma directly.

Lemma 3.3

Let S be a closed subsemigroup of $M_n(K)$, I an ideal of S. Then

$$\begin{aligned} \mathcal {R}(I)&=\{a\in S\mid (xay)^n\in I \text { for all } x,y\in S^1\}\\&=\{a\in S\mid (ay)^n\in I \text { for all } y\in S^1\}\\&=\{a\in S\mid (xa)^n\in I \text { for all } x\in S^1\}\\ \end{aligned}$$

Proposition 3.4

Let M be a linear algebraic monoid with unit group G and let I be an ideal of M. Then

$$\begin{aligned} \mathcal {R}(I)&=\{a\in M\mid GaG \subseteq \sqrt{I} \}\\&=\{a\in M\mid aG \subseteq \sqrt{I} \}\\&=\{a\in M\mid Ga \subseteq \sqrt{I} \}.\\ \end{aligned}$$

Proof

Without loss of generality, we may assume that M is a closed submonoid of $M_n(K)$ for some $n\in \mathbb {Z}^+$. By Lemma 3.3, we have

$$\begin{aligned} \mathcal {R}(I)=\{a\in M\mid MaM \subseteq \sqrt{I} \}=\{a\in M\mid aM \subseteq \sqrt{I} \}=\{a\in M\mid Ma \subseteq \sqrt{I} \}. \end{aligned}$$

If $a\in \mathcal {R}(I)$, then $GaG\subseteq MaM\subseteq \sqrt{I}$. If $b\in M$ with $GbG\subseteq \sqrt{I}$, since $\sqrt{I}$ is closed in M, we have

$$\begin{aligned} MbM\subseteq \overline{GbG}=\overline{MbM}\subseteq \sqrt{I}, \end{aligned}$$

which implies $b\in \mathcal {R}(I)$. Therefore,

$$\begin{aligned} \mathcal {R}(I)=\{a\in M\mid GaG \subseteq \sqrt{I} \}. \end{aligned}$$

Similarly, we can get

$$\begin{aligned} \mathcal {R}(I) =\{a\in M\mid aG \subseteq \sqrt{I} \}=\{a\in M\mid Ga \subseteq \sqrt{I} \}. \end{aligned}$$

$\square$

Proposition 3.5

Let S be a linear algebraic semigroup, I an ideal of S and $e\in E(S)$. Then $\mathcal {R}(I)$ is closed in S.

In particular,

$$\begin{aligned} \mathcal {R}(SeS)=\bigcap _{f\in E(S)\backslash E(SeS)}I(f), \end{aligned}$$

where $I(f)=\{a\in S \mid a\not \mid f\}$.

Proof

Consider the product $\mu :S\times S \rightarrow S$ by $\mu (a,b)=ab$ for any $a,b\in S$. As $\mu (S\times \mathcal {R}(I) )=S\mathcal {R}(I)\subseteq \mathcal {R}(I)$, we deduce that

$$\begin{aligned} \mu (S\times \overline{\mathcal {R}(I)})=\mu (\overline{S\times \mathcal {R}(I)})\subseteq \overline{\mu (S\times \mathcal {R}(I))}\subseteq \overline{\mathcal {R}(I)}. \end{aligned}$$

Similarly, we can get $\mu (\overline{\mathcal {R}(I)}\times S)\subseteq \overline{\mathcal {R}(I)}$. So $S\overline{\mathcal {R}(I)}S \subseteq \overline{\mathcal {R}(I)}$, which implies that $\overline{\mathcal {R}(I)}$ is an ideal of S. Since $\sqrt{I}$ is closed in S, $\overline{\mathcal {R}(I)}\subseteq \sqrt{I}$. By the maximality of $\mathcal {R}(I)$, we get $\overline{\mathcal {R}(I)}=\mathcal {R}(I)$, thus $\mathcal {R}(I)$ is closed in S.

Let $a\in \mathcal {R}(SeS)$ and $f \in E(S)\backslash E(SeS)$. If $a \mid f$, then $f\in SaS \subseteq \mathcal {R}(SeS)$, thus $f\in E(\mathcal {R}(SeS))=E(SeS)$, a contradiction. So $a\not \mid f$, and thus $a\in I(f)$. Hence

$$\begin{aligned} \mathcal {R}(SeS)\subseteq \bigcap _{f\in E(S)\setminus E(SeS)}I(f). \end{aligned}$$

On the other hand, suppose $a\in S$ and $a \not \mid f$ for every $f \in E(S)\backslash E(SeS)$. We claim that $a\in \mathcal {R}(SeS)$. In fact, if not, there exists $y\in S$, such that $(ay)^n \notin SeS$ for some $n\in \mathbb {Z}^+$ by Lemma 3.3. Thus there exits $f\in E(S)\backslash E(SeS)$ such that $(ay)^n \in H_f$. Then $a\mid f$, a contradiction. So $a\in \mathcal {R}(SeS)$. Therefore,

$$\begin{aligned} \mathcal {R}(SeS)=\bigcap _{f\in E(S)\backslash E(SeS)} I(f). \end{aligned}$$

$\square$

For a linear algebraic semigroup S with kernel $\ker (S)$, recall that an idempotent $e\in E(S)$ is called primitive if $E(eSe)\setminus E(\ker (S))=\{e\}$, that is, for any $f\in E(S)$, $e\gneq f$ implies $f\in E(\ker (S))$. The following corollary gives a characterization of the Schwarz radical of an irreducible linear algebraic monoid in terms of primitive idempotents as follows.

Corollary 3.6

Let M be an irreducible linear algebraic monoid with unit group G. Let T be a maximal torus of G, $\Lambda \subseteq E(\overline{T})$ a cross-section lattice of M. Then

$$\begin{aligned} \mathcal {R}\ker (M)= \bigcap _{e\in \Lambda _0} I(e), \end{aligned}$$

where $I(e)=\{a\in M \mid a\not \mid e\}$ and $\Lambda _0=\{e\in \Lambda \mid e \text { is primitive}\}$.

Proof

First, we claim that if $e,f\in E(M)$ with $e\le f$, then $I(e) \subseteq I(f)$. In fact, for any $a\in I(e)$, if $a\notin I(f)$, then $a\mid f$. Since $f\mid e$, we have $a\mid e$, a contradiction. Let $\Omega =\{e\in E(M)\mid e \text { is primitive}\}$. By [22, Corollary 6.9, Theorem 6.20], for any $e\in E(M)\backslash E(\ker (M))$, there exists $e^{'}\in \Omega$ such that $e\ge e^{'}$. Hence, by Proposition 3.5,

$$\begin{aligned} \mathcal {R}\ker (M)=\bigcap _{e\in E(M)\backslash E(\ker (M))}I(e)=\bigcap _{e\in \Omega } I(e). \end{aligned}$$

Next, we claim that if $e,f\in E(M)$ with $J_e=J_f$, then $I(e)=I(f)$. Since $J_e=J_f$, by [22, Corollary 6.8], there exists $x\in G$ such that $x^{-1}ex=f$. Let $a\in I(e)$. If $a\mid f$, then $f\in MaM$. Thus $e=xfx^{-1} \in MaM$, which implies $a\mid e$, a contradiction. So $I(e) \subseteq I(f)$. Similarly, we get $I(f) \subseteq I(e)$.Therefore, $I(e)=I(f)$. According to [22, Corollary 6.10], $E(M)=\bigcup _{x\in G}x^{-1}E(\overline{T})x$. So

$$\begin{aligned} \mathcal {R}\ker (M)=\bigcap _{e\in \Omega } I(e)=\bigcap _{e\in \Lambda _0} I(e), \end{aligned}$$

where $\Lambda _0=\{e\in \Lambda \mid e \text { is primitive}\}$. $\square$

Remark 3.7

• (i) For an irreducible linear algebraic monoid M with unit group G, if $e\in E(M)$ such that the $\mathscr {J}$-class $J_e$ of M is completely regular, then by the proof of Proposition 5.8,

  $$\begin{aligned} I(e)=\{a\in M\mid a\not \mid e\}=\{a\in M\mid det_e(a)=0\}, \end{aligned}$$

  and I(e) is the union of some irreducible components of the non-units of M. In particular, if M is solvable with zero, then for any $e\in E(M)$, the $\mathscr {J}$-class $J_e$ of M is completely regular by [22, Corollary 6.32]. Hence following Corollary 3.6, we obtain that

  $$\begin{aligned} \mathcal {R}\ker (M)=\{a\in M\mid det_{e_i}(a)=0,\ 1\le i\le n\}, \end{aligned}$$

  Where $\{e_i\mid 1\le i\le n\}$ are the set of all primitive idempotents of $E(\overline{T})$ for a maximal torus T of G.

  For example, let M be the set of all upper triangular matrices in $M_3(K)$, i.e.,

  $$\begin{aligned} M=\begin{pmatrix}K&{} \quad K&{} \quad K\\ 0&{} \quad K&{} \quad K\\ 0&{} \quad 0&{} \quad K \end{pmatrix}:=\left\{ \begin{pmatrix}a_1&{} \quad b_1&{} \quad c\\ 0&{} \quad a_2&{} \quad b_2 \\ 0&{} \quad 0&{} \quad a_3\end{pmatrix}\mid a_i,b_j,c\in K, 1\le i\le 3,1\le j\le 2\right\} . \end{aligned}$$

  Then $\ker (M)=\{0\}$, and the ${ Schwarz}$ radical of M, $\mathcal {R}\ker (M)\!=\!\begin{pmatrix}0&{} \quad K&{} \quad K\\ 0&{} \quad 0&{} \quad K\\ 0&{} \quad 0&{} \quad 0\end{pmatrix}.$ Choose

  $$\begin{aligned} e_1=\begin{pmatrix}1&{} \quad 0&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix},\ e_2=\begin{pmatrix}0&{} \quad 0&{} \quad 0\\ 0&{} \quad 1&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix},\ e_3=\begin{pmatrix}0&{} \quad 0&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\\ 0&{} \quad 0&{} \quad 1\end{pmatrix}. \end{aligned}$$

  Thus

  $$\begin{aligned} I(e_1)=\begin{pmatrix}0&{} \quad K&{} \quad K\\ 0&{} \quad K&{} \quad K\\ 0&{} \quad 0&{} \quad K\end{pmatrix},\ I(e_2)=\begin{pmatrix}K&{} \quad K&{} \quad K\\ 0&{} \quad 0&{} \quad K\\ 0&{} \quad 0&{} \quad K\end{pmatrix},\ I(e_3)=\begin{pmatrix}K&{} \quad K&{} \quad K\\ 0&{} \quad K&{} \quad K\\ 0&{} \quad 0&{} \quad 0\end{pmatrix}. \end{aligned}$$

  So

  $$\begin{aligned} \mathcal {R}\ker (M)=\bigcap _{1\le i\le 3}I(e_i). \end{aligned}$$

  Note that the $e_i$s are primitive idempotents of M, and the $I(e_i)$s are exactly the irreducible components of the set of all non-units of M.

• (ii) Even thought the Schwarz radical of any linear algebraic semigroup is closed which is showed in Proposition 3.5, it may not be irreducible. Let

  $$\begin{aligned} M=\left\{ a\otimes b\mid a,b\in \begin{pmatrix}K&{}K\\ 0&{}K\end{pmatrix}\right\} ,\ E_{12}=\begin{pmatrix}0&{} \quad 1\\ 0&{} \quad 0\end{pmatrix}. \end{aligned}$$

  Then the Schwarz radical of M, $\mathcal {R}\ker (M)=S_1\cup S_2$, where

  $$\begin{aligned} S_1= & {} \left\{ a\otimes E_{12}\mid a\in \begin{pmatrix}K&{} \quad K\\ 0&{} \quad K\end{pmatrix}\right\} ,\\ S_2= & {} \left\{ E_{12}\otimes b\mid b\in \begin{pmatrix}K&{} \quad K\\ 0&{} \quad K\end{pmatrix}\right\} . \end{aligned}$$

  So $\mathcal {R}\ker (M)$ is reducible.

Now we discuss the properties of the minimal non-nilpotent ideals of linear algebraic semigroups determined by primitive idempotents. We have the following proposition which is similar to the result of [16, Theorem 1] in compact semigroups with zero.

Proposition 3.8

Let S be a linear algebraic semigroup, $e\in E(S)$. Then the following are equivalent:

• (i) e is a primitive idempotent of S;

• (ii) the set $eSe\setminus \sqrt{\ker (S)}$ is a group;

• (iii) Se is a minimal non-nilpotent left ideal of S;

• (iv) eS is a minimal non-nilpotent right ideal of S;

• (v) SeS is a minimal non-nilpotent ideal of S;

• (vi) each idempotent in $SeS\setminus \ker (S)$ is a primitive idempotent of S.

Proof

• (i) $\implies$ (ii): Assume that e is a primitive idempotent of S. Then we have $H_e\subseteq eSe\setminus \sqrt{\ker (S)}$. Let $a\in eSe\setminus \sqrt{\ker (S)}$. Then $a^m\in H_f$ for some $f\in E(eSe)$ and some $m\in {\mathbb {Z}}^+$. Since $a\notin \sqrt{\ker (S)}$, we have $f\notin E(\ker (S))$, which implies $f=e$. Hence $a^m\in H_e$, and then $a\in H_e$. Therefore, the set $eSe\setminus \sqrt{\ker (S)}=H_e$ is the maximal subgroup of S containing e.

• (ii) $\implies$ (iii): Assume (ii) and that $I_l$ is a left non-nilpotent ideal of S contained in Se. We claim that there exists an idempotent $f\in E(I_l)\setminus E(\ker (S))$. Since S is a linear algebraic semigroup, $I_l$ is a non-nil ideal of S. Thus there exists $x\in I_l\setminus \sqrt{\ker (S)}$. But by [22, Theorem 3.18], there exist $f\in E(S)$ and $m\in \mathbb {Z}^+$ such that $x^m\in H_f\cap I_l$. So there exists $y\in S$ such that $yx^m=f\in I_l$, since $I_l$ is a left ideal of S. Since $x\notin \sqrt{\ker (S)}$ and $x^m\in H_f$, we have $f\notin E(\ker (S))$. Now $f\in Se$, implying $f=fe$. Then $ef=efef\in E(eSe)$. Since $fef=f\notin E(\ker (S))$, $ef\notin E(\ker (S))$. Since $eSe\setminus \sqrt{\ker (S)}$ is a group and $ef\in E(eSe\setminus \sqrt{\ker (S)})$, $e=ef\in I_l$. Hence $I_l=Se$.

• (iii) $\implies$ (v): Suppose I is a non-nilpotent ideal of S contained in SeS. Then I is a non-nil ideal of S, as S is a linear algebraic semigroup. Thus there exist $x\in I\setminus \sqrt{\ker (S)}$ and $m\in \mathbb {Z}^+$ such that $x^m\in H_f\cap I$ for some $f\in E(S)\setminus E(\ker (S))$. So there exists some $y\in S$ such that $f=yx^m\in I\subseteq SeS$. Hence there exist a, $b\in S$ such that $f=aeb$. Let $g=bfae\in I$. Then $g^2=bfaebfae =bfae=g\in E(Se)$. Since $f\notin E(\ker (S))$ and $f=aegb$, $g\notin E(\ker (S))$. By the minimality of Se, we have $Se=Sg$. Thus $SeS=SgS\subseteq I$, implying $I=SeS$

• (v) $\implies$ (i): Assume (v). If e were not primitive, there would exist $f\in E(eSe)\setminus E(\ker (S))$ such that $e\ne f$. Thus SfS in a non-nilpotent ideal of S contained in SeS. By the minimality of SeS, $SfS=SeS$. Hence $f\in eSe\cap J_e$. By [22, Theorem 1.4(iii)], $eSe\cap J_e=H_e$. Then $f=e$, a contradiction. Therefore, e is a primitive idempotent of S;

The result (vi) $\iff$ (i) follows from (v) $\iff$ (i). And by symmetry, we finish the proof. $\square$

In a similar way in [6, Theorem 1.1] and by Proposition 3.8, we have the following proposition.

Proposition 3.9

Let S be a linear algebraic semigroup and let I be a minimal non-nilpotent ideal of S. Then any proper ideal of I is nilpotent. In particular, there exists a primitive idempotent $e\in E(S)$ such that $I=SeS$.

Proof

Note that the kernel of I is equal to $\ker (S)$ by [25]. Let A be a proper ideal of I. Suppose (by way of contradiction) A is not a nilpotent ideal of I with respect to $\ker (S)$. Then IAI is a two-sided ideal of S contained in I, and $IAI\subseteq A \subsetneq I$. By the minimality of I, IAI is a nilpotent ideal of S, and thus $IAI\subseteq \mathcal {R}\ker (S)$. Then $(IA)^2=IAIA\subseteq \mathcal {R}\ker (S)A\subseteq \mathcal {R}\ker (S)$. So IA is a nilpotent left ideal of S. Similarly, AI is a nilpotent right ideal of S.

Now, SAS is a two-sided ideal of S contained in I. By the minimality of I, we have that SAS is a nilpotent ideal of S or $SAS=I$. In either event, since IA is nilpotent, we have that $(SA)^2=SASA$ is nilpotent. Hence SA is a nilpotent left ideal of S. Therefore, $SA\subseteq \mathcal {R}\ker (S)$. Thus there exists some $m\in \mathbb {Z}^+$ such that $(SA)^m\subseteq \ker (S)$. So $A^{2m}=(A\cdot A)^m \subseteq (S\cdot A)^m \subseteq \ker (S)$. Hence A is a nilpotent ideal of I, a contradiction. Therefore, we have that any proper ideal of I is nilpotent.

Since I is non-nilpotent, I is non-nil, and thus there exists $a\in I$ such that $a\notin \sqrt{\ker (S)}$. By [22, Theorem 3.18], there exist some $n\in \mathbb {Z}^+$ and $e\in E(S)$ such that $a^n\in H_e$. Clearly, $e\notin E(\ker (S))$. Hence there exists $b\in S$ such that $e=a^{n}b$, which implies $e\in E(I)\setminus E(\ker (S))$. So SeS is a non-nilpotent ideal of S contained in I, which implies $SeS=I$. By Proposition 3.8, e is a primitive idempotent of S. $\square$

4 The structure of linear algebraic monoids in terms of Schwarz radical data

For an irreducible linear monoid M, it is easy to see that

$$\begin{aligned} \ker (M)\subseteq \mathcal {R}\ker (M) \subseteq \sqrt{\ker (M)}. \end{aligned}$$

In the Sect. 4.1, we give a characterization of the condition that $\ker (M)=\mathcal {R}\ker (M)$. And in the Sect. 4.2, we give a characterization of the condition that $\mathcal {R}\ker (M)=\sqrt{\ker (M)}$.

4.1 Completely regularity and regularity conditions

Theorem 4.1

Let M be an irreducible linear algebraic monoid. Then the following are equivalent:

• (i) M is completely regular;

• (ii) $\sqrt{\ker (M)}=\ker (M)$;

• (iii) $\{a \in M\mid a^i=f$ for some $i\in \mathbb {Z}^+\}\subseteq fGf$ for every $f \in E(\ker (M))$.

Proof

$\mathrm (i)\implies \mathrm (ii)$. Suppose M is completely regular and $a\in \sqrt{\ker (M)}$. Then there exists $f\in E(\ker (M))$ such that $a\in H_f$. Hence $a\in \ker (M)$ as $H_f\subseteq \ker (M)$. So $\sqrt{\ker (M)}=\ker (M)$.

$\mathrm (ii)\implies \mathrm (iii)$. Assume that $\sqrt{\ker (M)}=\ker (M)$. Let $f\in E(\ker (M)$. If $a^i=f$ for some $i\in \mathbb {Z}^+$, then $a\in \sqrt{\ker (M)}$, and thus $a\in \ker (M)$. So there exists $e \in E(M)$ such that $a\in H_e$. Then $a^i=f \in H_e$ which implies $e=f$. Therefore, $a\in H_f$. By Theorem 2.2, we have that $H_f=fGf$. Therefore, $a\in fGf$.

$\mathrm (iii)\implies \mathrm (i)$. Suppose (iii) holds. For any $f\in E(\ker (M)$, if a is a nilpotent element of $M_f$, then $a^i=f$ for some $i\in \mathbb {Z}^+$. Thus $a\in fGf\cap M_f=\{f\}$, then $a=f$. This shows that $M_f$ has no non-zero nilpotent elements, hence by [22, Theorem 5.12], $M_f$ is completely regular. Following [22, Theorem 7.4], we obtain that M is regular. Moreover, by[18, Theorem 3.7], $E(M_f)=\{e\in E(M)\mid e\ge f\}$ is finite. Therefore, M is completely regular following [20, Theorem 3.10]. $\square$

Theorem 4.2

Let M be an irreducible linear algebraic monoid. Then M is regular if and only if the Schwarz radical of M is completely simple $(i.e.,\ \mathcal {R}\ker (M)=\ker (M))$.

Proof

By [22, Theorem 3.15], M is isomorphic to a closed submonoid of $M_n(K)$ for some $n\in \mathbb {Z}^+$, thus $\sqrt{\ker (M)} = \{ a\in M\mid a^n \in \ker (M) \}$. Suppose M is regular. For any $a \in \mathcal {R}\ker (M)$, there exists $e\in E(M)$ such that $MaM=MeM$. So

$$\begin{aligned} e\in MeM=MaM \subseteq \mathcal {R}\ker (M), \end{aligned}$$

which implies $e \in \mathcal {R}\ker (M)$. Thus $e=e^n\in \ker (M)$. Therefore,

$$\begin{aligned} MaM =MeM=\ker (M), \end{aligned}$$

and then $a\in \ker (M)$. This prove that $\mathcal {R}\ker (M)=\ker (M)$.

Suppose $\mathcal {R}\ker (M)=\ker (M)$. Let G be the unit group of M. If M is not regular, then $\overline{R(G)}$ is not completely regular following [22, Theorem 7.4]. Hence, from Theorem 4.1, there exist $a \in \overline{R(G)}$ and $f\in E(\ker (\overline{R(G)})$, such that $a^n=f$ and $a\notin f\overline{R(G)}f$.

Let $x\in M$. Then $x\in \overline{B}$ for some Borel subgroup B of G. Now $a\in \overline{R(G)}\subseteq \overline{B}$. From [22, Corollaries 1.16, 1.17, 3.20], we have that for all $b_1,b_2 \in \overline{B}$, $b_1\mid b_2$ implies $b_{1}^{n}\mid b_{2}^{n}$. So $a\mid ax$ implies $a^n\mid (ax)^n$, i.e. $f\mid (ax)^n$. But $(ax)^n\mid f$ following $f\in E(\ker (\overline{R(G)})$ and $E(\ker (\overline{R(G)})\subseteq E(\ker (M))$. So $f\mathscr {J} (ax)^n$. Thus $(ax)^n\in \ker (M)$, $ax\in \sqrt{\ker (M)}$. This proves that $aM \subseteq \sqrt{\ker (M)}$. Thus by Lemma 3.3, $a\in \mathcal {R}\ker (M) =\ker (M)$. Since $a^n=f$, we have $f\in aM, fM\subseteq aM$. But by [9, Lemma 3(2)], aM is a minimal right ideal of M since $a\in \ker (M)$. So $fM=aM$, $a\in fM$. Similarly, $a\in Mf$. Therefore, $a\in fMf$, and thus $a\in f\overline{R(G)}f$, a contradiction. So $\overline{R(G)}$ is completely regular, and hence M is regular.

Finally, by Proposition 3.5, $\mathcal {R}\ker (M)$ is closed. Therefore, $\mathcal {R}\ker (M)$ is a linear algebraic semigroup which implies that $\mathcal {R}\ker (M)$ is an $s\pi r$-semigroup. Note that the kernel of $\mathcal {R}\ker (M)$ is just $\ker (M)$. Hence we have that $\mathcal {R}\ker (M)=\ker (M)$ if and only if $\mathcal {R}\ker (M)$ is completely simple. $\square$

4.2 The solvability condition

Theorem 4.3

Let M be an irreducible linear monoid with unit group G and $e\in E(\ker (M))$. Then the following are equivalent:

• (i) $G_e$ is solvable;

• (ii) $|W(G)|=|W(H_e)|$;

• (iii) M is a semilattice of archimedean semigroups;

• (iv) $\sqrt{\ker (M)}$ forms an ideal of M (i.e., $\ \mathcal {R}\ker (M)=\sqrt{\ker (M)})$;

• (v) $\mathcal {U}(M)$ is a relatively complement lattice.

Proof

That $\mathrm (i)\iff \mathrm (iii) \iff \mathrm (v)$ follows from [20, Theorem 2.15]. Let T be a maximal torus of G. By [22, Proposition 6.25] and [22, Theorem 6.16], we know that $|W(G)|=\omega (e)|W(C_G(e)|$ where $\omega (e)=|J_e\cap E(\overline{T})|$, and $|W(C_G(e))|=|W(G_e)|\cdot |W(H_e)|$. Since $e\in E(\ker (M))$, we have $\omega (e)=1$, thus $|W(G)|=|W(C_G(e)|=|W(G_e)|\cdot |W(H_e)|$. So $\mathrm (i)\iff \mathrm (ii)$.

By [22, Theorem 3.15], M is isomorphic to a closed submonoid of $M_n(K)$ for some $n\in \mathbb {Z}^+$, thus

$$\begin{aligned} \sqrt{\ker (M)} = \left\{ a\in M\mid a^n \in \ker (M)\right\} . \end{aligned}$$

If M is a semilattice of archimedean semigroups, by [22, Theorem 1.15], for every $a,b\in M$, $a\mid b$ implies $a^2\mid b^i$ for some $i\in \mathbb {Z}^{+}$. Let $a\in \sqrt{\ker (M)}$, $b\in M$. since $a\mid ab$, we have $a^n\mid (ab)^n$. Moreover, $(ab)^n\mid a^n$ following $a^n \in \ker (M)$. So $a^n \mathscr {J} (ab)^n$, $ab\in \sqrt{\ker (M)}$. Similarly, $ba\in \sqrt{\ker (M)}$. Hence $MaM \in \sqrt{\ker (M)}$. This proves that $\sqrt{\ker (M)}$ is an ideal of M, that is $\mathcal {R}\ker (M)=\sqrt{\ker (M)}$. Thus $\mathrm (iii)\implies \mathrm (iv)$.

Now we clam that $\mathrm (iv)\implies \mathrm (iii)$. Suppose $\mathcal {R}\ker (M)=\sqrt{\ker (M)}$. Let $J\in \mathcal {U}(M)$ such that J covers $\ker (M)$. By [19, Theorem 23, Remark 24], we just need to show that $\omega (J)=|J\cap E(\overline{T})|=1$. Suppose $\omega (J)>1$, then there exist $e,f\in J\cap E(\overline{T})$, such that $e\ne f$. By [22, corollary 6.8], there exists $x\in G$ such that $f=xex^{-1}$. Since $e,f\in E(\overline{T})$, $ef=fe\in E(M)$. So $ef=fe\le e,f$. But $ef\ne e,\ ef\ne f$. Thus $ef=fe\in E(\ker (M))$ as J covers $\ker (M)$. Then $exex^{-1}=xex^{-1}e \in E(\ker (M))$. So $exe,ex^{-1}e \in \ker (M)$. Hence

$$\begin{aligned} (ex)(ex)=(exe)x \in \ker (M),\ (x^{-1}e)(x^{-1}e)=x^{-1}(ex^{-1}e) \in \ker (M). \end{aligned}$$

Therefore, $ex,x^{-1}e\in \sqrt{\ker (M)}$. Since $\sqrt{\ker (M)}$ is an ideal of M, $e=(ex)(x^{-1}e)\in \sqrt{\ker (M)}$, which implies that $e\in E(\ker (M))$, a contradiction. This shows that $\omega (J)=1$. Therefore, M is a semilattice of archimedean semigroups. $\square$

Corollary 4.4

Let M be an irreducible monoid with unit group G. Then G is solvable if and only if $\sqrt{\ker (M)}$ forms an ideal of M and a maximal subgroup of $\ker (M)$ is solvable.

Proof

It is known that G is solvable if and only if $|W(G)|=1$. Let $e\in E(\ker (M))$. From the proof of Theorem 4.3, we have that $|W(G)|=|W(G_e)|\cdot |W(H_e)|$, and $|W(G_e)|=1$ if and only if $\sqrt{\ker (M)}$ forms an ideal of M. So G is solvable if and only if $\sqrt{\ker (M)}$ forms an ideal of M and the subgroup $H_e$ is solvable. $\square$

5 On completely regular $\mathcal {J}$-classes

It is known that the kernel of a linear algebraic semigroup S is a completely regular $\mathcal {J}$-class of S, and every completely regular $\mathcal {J}$-class of S is completely simple (as a semigroup). In this section, we generalize the results about the kernel to completely regular $\mathcal {J}$-classes of linear algebraic semigroups.

The following construction plays an important role in this section.

Proposition 5.1

Let S be a linear algebraic semigroup. Let $J\in \mathcal {U}(S)$ be completely regular and let $S_J=\{a\in S\mid aJ\subseteq J\}$. Then $S_J$ is a linear algebraic semigroup with kernel J. If S is irreducible, $S_J$ is irreducible.

Proof

By [22, Corollary 3.16], we can assume that S is a closed subsemigroup of some $M_n(K)$. For any $a,b\in S_J$, $abJ\subseteq aJ\subseteq J$, so $ab\in S_J$, which implies $S_J$ is a semigroup. Since J is completely regular, $J^2\subseteq J$, thus $J\subseteq S_J$. Let $e\in E(J)$. Now we claim that $S_J=\{a\in S\mid det_e(a)\ne 0\}$. Let $a\in S$ such that $eae\in J$. Then $e\mid ea \mid eae \mid e$, thus $ea\in J$. Then for any $b\in J$, we have $eab\in J$. So $ab\mid eab\mid b\mid ab$, thus $ab\in J$ which implies $aJ\subseteq J$. On the other hand, if $b\in S$ with $bJ\subseteq J$, then $ebe\in J$ since $J^2\subseteq J$. Hence we get $eae\in J$ if and only if $aJ\subseteq J$, i.e., $S_J=\{a\in S\mid det_e(a)\ne 0\}$. Similarly, we have $eae\in J$ if and only if $Ja\subseteq J$. But by [22, Remark 3.23], we obtain that $eae\in J$ if and only if $\det _e(a)\ne 0$. So

$$\begin{aligned} S_J=\{a\in S\mid Ja\subseteq J\}=\{a\in S\mid det_e(a)\ne 0\}. \end{aligned}$$

Note that J is an ideal of $S_J$ and J is completely simple, thus J is the kernel of S. Since $S_J=\{a\in S\mid det_e(a)\ne 0\}$ which is open in S, $S_J$ is a linear algebraic semigroup with kernel J, and if S is irreducible, $S_J$ is irreducible. $\square$

Throughout this section, we use the notation $S_J$ defined by

$$\begin{aligned} S_J=\{a\in S\mid aJ\subseteq J\}, \end{aligned}$$

for a linear algebraic semigroup S and a completely regular $\mathcal {J}$-class J of S. Let $e\in E(S_J)$. By [22, Remark 1.3(iii)], the $\mathcal {H}$-class of e in S is equal to the $\mathcal {H}$-class of e in $S_J$. Moreover, if $f\in E(S)$ with $e\mathcal {L} f$ in S, then $ef=e,\ fe=f$. Thus $f\in S_J$ and $e\mathcal {L} f$ in $S_J$. Therefore, we have that the two sets of idempotents of the $\mathcal {L}$-class of e in S and in $S_J$ coincide. Similarly, the two sets of idempotents of the $\mathcal {R}$-class of e in S and in $S_J$ coincide. Then applying the results of the structure of the kernel of S described in [9] for $S_J$, it is easy to give the structure of completely regular $\mathcal {J}$-classes of S by the following two corollaries directly.

Corollary 5.2

Let S be a linear algebraic semigroup. Let $J\in \mathcal {U}(S)$ be completely regular. Then

• (i)
  $$\begin{aligned} J=\bigcup _{h\in E(J)}H_h; \end{aligned}$$

• (ii) for any $e, f \in E(J)$, $H_e$ is isomorphic to $H_f$ as an algebraic group under the morphism given by $x\mapsto gxf$ for some $g\in E(J)$.

Corollary 5.3

Let S be a linear algebraic semigroup. Let $J\in \mathcal {U}(S)$ be completely regular, and $e\in E(J)$. Then

• (i) $J=E(L_e)H_eE(R_e)$;

• (ii) under the Rees construction $J=E(L_e)\times H_e \times E(R_e)$,

  $$\begin{aligned} E(J)=\left\{ (f, (gf)^{-1}, g)\mid f\in E(L_e),\ g\in E(R_e)\right\} , \end{aligned}$$

  which is isomorphic to $E(L_e)\times E(R_e)$ as an algebraic variety.

Theorem 5.4

Let M be an irreducible linear algebraic monoid with unit group G, $J\in \mathcal {U}(M)$ and $e\in E(J)$. Then

$$\begin{aligned} \dim J=2\dim G+\dim H_e-\dim C_G^r(e)-\dim C_G^l(e). \end{aligned}$$

In particular, if J is completely regular,

$$\begin{aligned} \dim J=\dim G-\dim G_e. \end{aligned}$$

Proof

The algebraic group $G\times G$ acts on M via left and right multiplication: $(g,h)\cdot x=gxh^{-1}$. From [22, proposition 6.1], we know that the orbit of the element $e\in E(M)$ under this action is just the $\mathcal {J}$-class of e, $J_e$. Consider $(G\times G)_e$, the isotropy subgroup of e under the action of $G\times G$. Then

$$\begin{aligned}&(G\times G)_e=\{(x,y)\in G\times G\mid xey^{-1}=e\}\nonumber \\&\quad =\{(x,y)\in G\times G\mid xe=exe=eye=ey\}. \end{aligned}$$

Hence $(G\times G)_e$ is a closed subgroup of $G\times G$, which contained in $C_G^r(e)\times C_G^l(e)$. By [22, Theorem 6.16] and its proof, we have

$$\begin{aligned} C_G^r(e)e= & {} eC_G^l(e)=eC_G(e)=H_e,\\ \dim C_G^r(e)= & {} \dim G^r(e)+\dim H_e,\ \dim C_G^l(e)=\dim G^l(e)+\dim H_e, \end{aligned}$$

and $\dim C_G(e)=\dim G_e+\dim H_e.$

Let $\phi : (G\times G)_e \rightarrow H_e\times H_e$, defined by $(x,y)\mapsto (xe,ey)$. Obviously, $\phi$ is a homomorphism of algebraic groups. Then

$$\begin{aligned} \phi ((G\times G)_e)=\{(xe,ey)\in H_e\times H_e\mid xe=ey\} \end{aligned}$$

is isomorphic to $H_e$ as algebraic groups. Hence $\dim \phi ((G\times G)_e)=\dim H_e$. Note that $\ker (\phi )=G^r(e)\times G^l(e)$. So

$$\begin{aligned} \dim (G\times G)_e=\dim \phi ((G\times G)_e)+\dim \ker (\phi )=\dim H_e+\dim G^r(e)+\dim G^l(e). \end{aligned}$$

Since J is the $G\times G$-orbit of e in M and $(G\times G)_e$ is the isotropy subgroup of e, we have $\dim J= \dim (G\times G)-\dim (G\times G)_e=2\dim G-\dim H_e-\dim G^r(e)-\dim G^l(e)=2\dim G+\dim H_e-\dim C_G^r(e)-\dim C_G^l(e).$

If J is completely regular and $e\in E(J)$, then by Theorem 2.2, $G=C_G^l(e)C_G^r(e)$. Hence $\dim G=\dim C_G^l(e)+\dim C_G^r(e)-\dim C_G(e)$. So

$$\begin{aligned} \dim J=\dim G-\dim C_G(e)+\dim H_e=\dim G-\dim G_e. \end{aligned}$$

$\square$

If M is a linear algebraic monoid with unit group G, then the unit group of $M_J$ is exactly the unit group G of M. Hence we can study the structure of the algebraic group G in term of the data of completely regular $\mathcal {J}$-classes of M. By Theorem 2.2, $e\in E(\overline{R(G)})$ if only if $J_e$ is a completely regular $\mathcal {J}$-class of M. Then using the results about the kernel [8, Theorem 2.1], [10, Theorem 5.5], [11, Proposition 2.3, Theorem 2.4], we obtain the followings directly.

Corollary 5.5

Let M be an irreducible linear algebraic monoid with unit group G, and let $e\in E(\overline{R(G)})$. Then

• (i) $\dim R(G)=\dim E(J_e) +\dim R(G_e) +\dim R(eGe)$;

• (ii) $\dim R_u(G) =\dim E(J_e)+ \dim R_u(G_e)+\dim R_u(eGe)$.

• (iii) G is reductive if and only if $G_e$ and $J_e$ are both reductive groups.

Corollary 5.6

Let M be an irreducible linear algebraic monoid with unit group G, and let $e\in E(\overline{R(G)})$. Let P be a parabolic subgroup of G. Then

• (i) $P=C_P(e)R(G)=C_P(e)R_u(G)$;

• (ii) $P=C_P(e)\ltimes R_u(G)$ if and only if $G_e$ and $H_e$ are both reductive groups;

• (iii) if $G_e$ is a reductive group and $J_e=E(J_e)$, then $P=P_e\ltimes R_u(G)$.

Now we want to generalize the results (Theorems 4.1, 4.2, 4.3) of the Schwarz radical to the case in terms of a completely regular $\mathcal {J}$-class J in an irreducible linear algebraic monoid M. First, we give some properties of $M_J$.

Lemma 5.7

Let M be an irreducible linear algebraic monoid. Let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class and $e\in E(J)$. Then $a\mid e$ if and only if $eae \mid e$ for any $a\in M$.

Proof

Let G be the unit group of M, and $a\in M$. If $eae\mid e$, then $a\mid eae \mid e$. So we only need to prove that if $a\mid e$, then $eae\mid e$. Suppose $a\mid e$. It follows from [22, Corollary 6.13] that $a\in GM_eG$, and thus there exist $x,y\in G$ and $b\in M_e$ such that $a=xby$. Thus $eae=exbye$. And $exbe=exe \mathcal {H}e$ following Theorem 2.2. So $exb \mid e$ which implies $exb\in J$. Clearly, $ye\in J$. Since J is completely regular, $J^2=J$. Hence $eae=exbye\in J$ and thus $eae\mid e$. $\square$

Proposition 5.8

Let M be an irreducible linear algebraic monoid with unit group G, and let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class, $e\in E(J)$. Let $S=M\setminus G=\cup _{i\in I} S_i$, where $S_i$ are the irreducible components of S. Then

$$\begin{aligned} M_J=\{a\mid a\in M, a\mid e\}=GM_eG=M\setminus (\cup _{i\in \Delta }S_i), \end{aligned}$$

where $\Delta =\{i\in I\mid \text {for all } a\in S_i,\ a\not \mid e\}$.

Proof

Given an element $a\in M$, by Lemma 5.7, we have $a\mid e$ if and only if $eae\mid e$. By the proof of Proposition 5.1, $M_J=\{a\mid a\in M,\ eae\mid e\}$. Hence

$$\begin{aligned} M_J=\{a\mid a\in M,\ a\mid e\}=GM_eG \end{aligned}$$

following [22, Corollary 6.13]. In particular,

$$\begin{aligned} a \not \mid e\iff eae\not \mid e \iff eae \notin H_e \iff det_e(a)=0. \end{aligned}$$

Let $I(e)=\{a\mid a\in M, a\not \mid e\}$. Then $I(e) =\{a\mid a\in M, det_e(a)=0\}$ and thus $M\setminus M_J=I(e)$.

If $M=G$, then $S=\varnothing$ and $M_J=G$. And if $J=\ker (M)$, then $M_J=M$. Suppose $M\ne G$ and $J\ne \ker (M)$. Now we claim that $I(e)=\cup _{i\in \Delta }S_i$, where $\Delta =\{i\in I\mid \text {for all } a\in S_i,\ a\not \mid e\}$. By [22, Theorem 3.15], we can assume that M is a closed submonoid of some $M_n(K)$. Consider $\phi : M\rightarrow K$ given by $\phi (a)=det_e(a)=det(eae+1-e)$. Thus $I(e)=\phi ^{-1}(0)$. By [22, Theorem 2.21], the dimension of every irreducible component of I(e) is $p-1$. Since J is completely regular, $\omega (e)=1$. From Theorem 2.2, we have eGe is the $\mathcal {H}$-class of e. Thus if $a\in G$, then $eae\in H_e$, which implies $a\notin I(e)$. So $I(e)\subseteq M\setminus G=S$. Moreover, by [22, Proposition 6.2], the dimension of every irreducible component of S is $p-1$. Hence every irreducible component of I(e) is also the irreducible component of S. Let $S_0$ be an irreducible component of S which satisfy that for any element $a\in S_0$, $a\not \mid e$. Then $S_0\subseteq I(e)$. Hence $I(e)=\cup _{i\in \Delta }S_i$. Therefore, $M_J=M\setminus I(e)=M\setminus (\cup _{i\in \Delta }S_i)$. $\square$

Corollary 5.9

Let M be an irreducible submonoid of $M_n(K)$ with unit group G, and let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class. Then

$$\begin{aligned} \left\{ a\in M\mid a^n\in J\}=\{a\in M_J\mid a^n\in J\right\} . \end{aligned}$$

Proof

If $a\in M$ with $a^n\in J$. Let $e\in E(J)$. Then $a\mid a^n \mid e$. So $a \notin I(e)$. By Proposition 5.8, $a\in M_J$. Hence $\left\{ a\in M\mid a^n\in J\right\} =\left\{ a\in M_J\mid a^n\in J\right\}$. $\square$

For an irreducible linear algebraic monoid M with unit group G. Let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class, we denote

$$\begin{aligned} \sqrt{J} =\{a\in M\mid a^i \in J \text { for some } i\in \mathbb {Z}^+\} \end{aligned}$$

and $\mathcal {R}(J)=\{a\in M\mid GaG\subseteq \sqrt{J}\}$. By Proposition 3.4 and Corollary 5.9, it is easy to see that the Schwarz radical $\mathcal {R}\ker (M_J)$ of $M_J$ is equal to $\mathcal {R}(J)$. Moreover, J, $\sqrt{J}$ and $\mathcal {R}(J)$ are both affine variety. This is because $M_J$ is an affine variety and J, $\sqrt{J}$, $\mathcal {R}(J)$ are both closed in $M_J$.

For an irreducible linear algebraic monoid M with unit group G, there is a natural algebraic group $G\times G$ acts on M via left and right multiplication:

$$\begin{aligned} (g,h)\cdot x=gxh^{-1} \text { for } g,h \in G\text { and } a\in M. \end{aligned}$$

From [22, Proposition 6.1], we know that the orbit of an element $a\in M$ under this action is just the $\mathcal {J}$-class of a, $J_a$. Now we can give a generalization of the results in Sect. 4 in the language of linear algebraic group actions.

Corollary 5.10

Let M be an irreducible linear algebraic monoid with unit group G, and let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class, $e\in E(J)$. Then

• (i) $G_e$ is reductive if and only if J is the unique $G\times G$-stable affine subvariety of M contained in $\sqrt{J}$ $(i.e.,\ \mathcal {R}(J)=J)$;

• (ii) $G_e$ is solvable if and only if $\sqrt{J}$ is a $G\times G$-stable affine subvariety of M $(i.e.,\ \mathcal {R}(J)=\sqrt{J})$

Proof

Let $M_J=\{a\in M\mid aJ\subseteq J\}$. By Proposition 5.1, $M_J$ is an irreducible linear algebraic monoid with kernel J. Obviously, the unit group of $M_J$ is G. Moreover, as we see above, the radical of the kernel of $M_J$ is equal to $\sqrt{J}$, and $\mathcal {R}\ker (M_J)=\mathcal {R}_{M_J}(J)=\mathcal {R}(J)$. By [22, Theorem 7.4], $M_J$ is regular if and only if $G_e$ is reductive since $e\in E(J)$. Apply Theorems 4.2 and 4.3 for $M_J$, we have $G_e$ is reductive if and only if $\mathcal {R}(J)=J$, and $G_e$ is solvable if and only if $\mathcal {R}(J)=\sqrt{J}$. Since $\mathcal {R}(J)=\{a\in M\mid GaG\subseteq \sqrt{J}\}$ and J is the $G\times G$-orbit of e in M, we have that that $\mathcal {R}(J)=J$ is equivalent to that J is the unique $G\times G$-stable affine subvariety of M contained in $\sqrt{J}$, and $\mathcal {R}(J)=\sqrt{J}$ is equivalent to that $\sqrt{J}$ is a $G\times G$-stable affine subvariety of M. $\square$

As an immediate consequence of Corollary 5.5(iii), 5.10 and [22, Proposition 6.25] we have,

Corollary 5.11

Let M be an irreducible linear algebraic monoid with unit group G, let $J\in \mathcal {U}(M)$ be a completely regular $\mathcal {J}$-class. Then

• (i) G is reductive if and only if J is both a reductive group and the unique $G\times G$-stable affine subvariety of M contained in $\sqrt{J}$;

• (ii) G is solvable if and only if $\sqrt{J}$ is a $G\times G$-stable affine variety and a maximal subgroup of J is solvable.

Remark 5.12

Let M be an irreducible solvable linear algebraic monoid. Since M is an $s\pi r$-semigroup, it is easy to verify that

$$\begin{aligned} M=\bigcup _{J\in \mathcal {U}(M)} \sqrt{J}, \end{aligned}$$

which is a semilattice of $\sqrt{J}$ $(J\in \mathcal {U}(M))$, and every $\sqrt{J}$ is an archimedean semigroup and a nil extension of the completely simple semigroup J.