Combinatorics and structure of Hecke-Kiselman algebras

TL;DR

This paper investigates the structure of Hecke-Kiselman algebras associated with finite graphs, revealing unexpected matrix-like structures and characterizing when these algebras are Noetherian.

Contribution

It uncovers matrix-type structures within Hecke-Kiselman monoids for cycle graphs and characterizes Noetherian properties of their algebras based on graph structure.

Findings

$K[C_n]$ is a Noetherian algebra

$K[C_n]$ has Gelfand-Kirillov dimension one

Characterization of Noetherian algebras $K[HK_{\Theta}]$ based on graph properties

Abstract

Hecke-Kiselman monoids $\textrm{HK}_{\Theta}$ and their algebras $K[\textrm{HK}_{\Theta}]$, over a field $K$, associated to finite oriented graphs $\Theta$ are studied. In the case $\Theta $ is a cycle of length $n\geqslant 3$, a hierarchy of certain unexpected structures of matrix type is discovered within the monoid $C_n=\textrm{HK}_{\Theta}$ and it is used to describe the structure and the properties of the algebra $K[C_n]$. In particular, it is shown that $K[C_n]$ is a right and left Noetherian algebra, while it has been known that it is a PI-algebra of Gelfand-Kirillov dimension one. This is used to characterize all Noetherian algebras $K[\textrm{HK}_{\Theta}]$ in terms of the graphs $\Theta$. The strategy of our approach is based on the crucial role played by submonoids of the form $C_n$ in combinatorics and structure of arbitrary Hecke-Kiselman monoids $\textrm{HK}_{\Theta}$.

Equations

$$A_i=\mathrm{suff}(\{(x_{k_s}\cdots x_s)(x_{k_{s+1}}\cdots x_{s+1})\cdots (x_{k_{i+1}}\cdots x_{i+1})\}),$$

$$B_i=\mathrm{pref}(\{x_n(x_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1})\cdots x_n(x_1\cdots x_{i_r}{x}_{n-1}\cdots x_{j_r})x_n{x}_{n-1}\cdots x_{j_{r+1}}\}),$$

$$w=a(x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}{)}^{k}b$$

$$w=x_n{x}_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1}{x}_n{x}_1\cdots x_{i_2}{x}_{n-1}\cdots x_{j_2}.$$

$$u{x}_n(x_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1})x_n(x_1\cdots x_{i_2}{x}_{n-1}\cdots x_{j_2})\cdots x_n(x_1\cdots x_{i_r}{x}_{n-1}\cdots x_{j_r})x_{n}v$$

$$i_r⩽i_{r-1}⩽\cdots ⩽i_1<j_1⩽j_2⩽\cdots ⩽j_r.$$

$$w=v(x_{k_s}\cdots x_s)u{x}_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1},$$

$$w=u(x_{k_s}\cdots x_s)(x_{k_{s+1}}\cdots x_{s+1})\cdots (x_{k_{i+1}}\cdots x_{i+1})x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}$$

$$x_j(x_{k_s}\cdots x_s)(x_{k_{s+1}}\cdots x_{s+1})\cdots (x_{k_{i+1}}\cdots x_{i+1})x_n{x}_1\cdots x_{j-1}{x}_j=x_{j}z{x}_j$$

$$w=u{x}_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}$$

$$u=(x_{k_s}\cdots x_s)(x_{k_{s+1}}\cdots x_{s+1})\cdots (x_{k_{i+1}}\cdots x_{i+1}),$$

$$w=zv{x}_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1},$$

$$v=(x_{k_s}\cdots x_s)\cdots (x_{k_{i+1}}\cdots x_{i+1})$$

$$|(x_{k_s}\cdots x_s)(x_{k_{s+1}}\cdots x_{s+1})\cdots (x_{k_{i+1}}\cdots x_{i+1}){|}_p=0$$

$$u(x_n{x}_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1})\cdots (x_n{x}_1\cdots x_{i_m}{x}_{n-1}\cdots x_{j_m})x_{n}v,$$

$$u(x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}{)}^k{x}_n(x_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1})\cdots x_n(x_1\cdots x_{i_r}{x}_{n-1}\cdots x_{j_r})x_{n}v,$$

$$q=x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}{x}_n{x}_1\cdots x_{i_1}{x}_{n-1}\cdots x_{j_1}$$

$$\widetilde{M_i}=\{a(x_n{q}_i{)}^{k}b\in C_n:a\in A_i,b\in B_i,k⩾1\}.$$

$$I_i=\{w\in C_n:C_{n}w{C}_n\cap ⟨x_n{q}_i⟩=\varnothing \}$$

$$(x_{k+1}\cdots x_n{x}_1\cdots x_k)x_j{w}^{\prime }\stackrel{(4j)}{\to }(x_{k+1}\cdots x_n{x}_1\cdots x_k)w^{\prime }$$

$$\tau (x_i)=\{\begin{array}{ll}{x}_{n-i}& \text{ for }i\ne n\\ x_n& \text{ for }i=n\end{array}.$$

$$(x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1})x_m\cdots x_{i+2}\stackrel{(5m)}{\to }{x}_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{m+1}{x}_{m-1}\cdots x_{i+1}{x}_m\cdots x_{i+2}.$$

$$(x_n{q}_i)x_m\cdots x_{i+2}$$

$$\stackrel{(3)}{\to }(x_{m-1}\cdots x_{i+2})x_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}.$$

$$(x_n{q}_i{)}^k{x}_{i+2}$$

$$\stackrel{(2)}{\to }(x_n{q}_i{)}^{k-1}{x}_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}$$

$$\stackrel{(5(i+1))}{\to }(x_n{q}_i{)}^{k-2}{x}_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+2}{x}_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}$$

$$\stackrel{(5i)}{\to }(x_n{q}_i{)}^{k-2}{x}_n{x}_1\cdots x_{i-1}{x}_{n-1}\cdots x_{i+2}{x}_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}$$

$$\stackrel{(5(i-1))}{\to }\stackrel{(5(i-2))}{\to }\cdots \stackrel{(51)}{\to }\stackrel{(5n)}{\to }(x_n{q}_i{)}^{k-2}{x}_{n-1}\cdots x_{i+2}{x}_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}$$

$$\stackrel{(\ast )}{\to }{x}_{n-k+1}\cdots x_{i+2}(x_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}{)}^{k-1},$$

$$(x_n{q}_i{)}^k{x}_{i+2}\to (x_n{q}_{i+1}{)}^k.$$

$$(x_n{x}_1\cdots x_i{x}_{n-1}\cdots x_{i+1}{)}^m{x}_{i+2}\to \{\begin{array}{l}{x}_{n-k}\cdots x_{i+2}(x_n{x}_1\cdots x_{i+1}{x}_{n-1}\cdots x_{i+2}{)}^{m-l-1}\text{ for }k\ne 0\\ (x_n{q}_{i+1}{)}^{m-l}\text{ for }k=0.\end{array}$$

Full text

Combinatorics and structure of Hecke–Kiselman algebras

J. Okniński, M. Wiertel

Hecke–Kiselman monoids $\operatorname{HK}_{\Theta}$ and their algebras $K[\operatorname{HK}_{\Theta}]$, over a field $K$, associated to finite oriented graphs $\Theta$ are studied. In the case $\Theta$ is a cycle of length $n\geqslant 3$, a hierarchy of certain unexpected structures of matrix type is discovered within the monoid $C_{n}=\operatorname{HK}_{\Theta}$ and it is used to describe the structure and the properties of the algebra $K[C_{n}]$. In particular, it is shown that $K[C_{n}]$ is a right and left Noetherian algebra, while it has been known that it is a PI-algebra of Gelfand–Kirillov dimension one. This is used to characterize all Noetherian algebras $K[\operatorname{HK}_{\Theta}]$ in terms of the graphs $\Theta$. The strategy of our approach is based on the crucial role played by submonoids of the form $C_{n}$ in combinatorics and structure of arbitrary Hecke–Kiselman monoids $\operatorname{HK}_{\Theta}$.

2010 Mathematics Subject Classification: 16S15, 16S36, 16P40, 20M05, 20M25, 20C08, 05C25.

Key words: Hecke–Kiselman algebra, monoid, simple graph, reduced words, algebra of matrix type, Noetherian algebra

[TABLE]

Acknowledgment. This work was supported by grant 2016/23/B/ST1/01045 of the National Science Centre (Poland).

1 Introduction

For an arbitrary finite simple digraph $\Theta$ with $n$ vertices $\{1,\ldots,n\}$, a finitely generated monoid $\operatorname{HK}_{\Theta}$ was defined by Ganyushkin and Mazorchuk in [7] by specifying generators and relations. Namely,

• (i)

$\operatorname{HK}_{\Theta}$ is generated by elements $x_{i}=x_{i}^{2}$, where $1\leqslant i\leqslant n$,

• (ii)

if the vertices $i$, $j$ are not connected in $\Theta$, then $x_{i}x_{j}=x_{j}x_{i}$,

• (iii)

if $i$, $j$ are connected by an arrow $i\to j$ in $\Theta$, then $x_{i}x_{j}x_{i}=x_{j}x_{i}x_{j}=x_{i}x_{j}$,

• (iv)

if $i$, $j$ are connected by an (unoriented) edge in $\Theta$, then $x_{i}x_{j}x_{i}=x_{j}x_{i}x_{j}$.

If the graph $\Theta$ is unoriented (has no arrows), the monoid $\operatorname{HK}_{\Theta}$ is isomorphic to the so-called [math]-Hecke monoid $H_{0}(W)$, where $W$ is the Coxeter group of the graph $\Theta$. Because of its strong connection to the Coxeter group, [17], and to the corresponding Hecke algebra, [11], the latter monoid plays an important role in representation theory. One of the reasons for the interest in the Hecke–Kiselman monoids is that they are natural quotients of the Hecke monoids. It is worth mentioning that relations of the above types arise also in a natural way in certain other contexts of representation theory, [8]. If $K$ is a field, then by the Hecke–Kiselman algebra defined by $\Theta$ we mean the semigroup algebra $K[\operatorname{HK}_{\Theta}]$. In other words, this is the $K$–algebra defined by the above presentation. Several combinatorial properties of the Hecke–Kiselman monoids, and their representations, have been studied in [6],[7],[10]. The aim of this paper is to continue the study of the algebra $K[\operatorname{HK}_{\Theta}]$ in the case when $\Theta$ is an oriented graph, started in [12]. In particular, a version of growth alternative has been obtained there and algebras of finite Gelfand–Kirillov dimension have been characterized. If $\Theta$ is oriented, it is also well known that $\operatorname{HK}_{\Theta}$ is finite if and only if the graph $\Theta$ is acyclic. Because of these results, it seems that the structure of the algebras $K[C_{n}]$, where $C_{n}$ is the Hecke–Kiselman monoid corresponding to the oriented cycle of length $n$, is crucial for understanding the structure and properties of arbitrary algebras $K[\operatorname{HK}_{\Theta}]$. This is the starting point for the approach in the present paper. We propose an entirely new structural approach to Hecke–Kiselman monoids and their algebras.

We denote by $F$ the free monoid generated by the elements of the set $X=\{x_{1},\ldots,x_{n}\}$. However, the same notation will be used for the generators of the monoid $\operatorname{HK}_{\Theta}$, if unambiguous. For any words $w,v\in F$ we say that $w$ is a factor of $v$ if $v=v_{1}wv_{2}$ for some $v_{1},v_{2}\in F$.

Let $C_{n}$ denote the Hecke–Kiselman monoid associated to the oriented cycle of length $n$, namely: $x_{1}\rightarrow x_{2}\rightarrow\cdots\rightarrow x_{n}\rightarrow x_{1}$. We consider the deg-lex order on $F$ induced by $x_{1}<x_{2}<\cdots<x_{n}$. Let $|w|_{q}$ denote the degree of $w\in F$ in the generator $x_{q}$. The following result, proved in [13], will be crucial. We refer to [4] for basic facts concerning Gröbner bases and the diamond lemma.

Theorem 1.1**.**

Let $\Theta=C_{n}$. Let $S$ be the system of reductions in $F$ consisting of all pairs of the form

1. (1)

$(x_{i}x_{i},x_{i})$* for all $i\in\{1,\dotsc,n\}$,* 2. (2)

$(x_{j}x_{i},x_{i}x_{j})$* for all $i,j\in\{1,\dotsc,n\}$ such that $1<j-i<n-1$,* 3. (3)

$(x_{n}(x_{1}\dotsm x_{i})x_{j},x_{j}x_{n}(x_{1}\dotsm x_{i}))$* for all $i,j\in\{1,\dotsc,n\}$ such that $i+1<j<n-1$,* 4. (4)

$(x_{i}ux_{i},x_{i}u)$* for all $i\in\{1,\dotsc,n\}$ and $1\neq u\in F$ such that $|u|_{i}=|u|_{i-1}=0$. Here, we write $i-1=n$ if $i=1$, (we say, for the sake of simplicity, that the word $x_{i}ux_{i}$ is of type $(4i)$),* 5. (5)

$(x_{i}vx_{i},vx_{i})$* for all $i\in\{1,\dotsc,n\}$ and $1\neq v\in F$ such that $|v|_{i}=|v|_{i+1}=0$. Here we write $i+1=1$ if $i=n$, (and similarly, we say that the word $x_{i}vx_{i}$ is of type $(5i)$).*

Then the set $\{w-v\mid\text{ for }(w,v)\in S\}$ is a Gröbner basis of the algebra $K[C_{n}]$.

Corollary 1.2**.**

$C_{n}$* can be identified with the monoid $\mathcal{R}(S)$ of words in $F$ that are reduced with respect to the system $S$, with the operation defined for $u,w\in C_{n}$ by $u\cdot w=\mathcal{R}_{S}(uw)$, where $\mathcal{R}_{S}(uw)$ is the $S$–reduced form of the word $uw$. More precisely, $\mathcal{R}(S)$ is the set of words in $F$ that do not have factors of the form $w_{\sigma}$, where $\sigma=(w_{\sigma},v_{\sigma})\in S$.*

For $w,v\in F$, we write $w\xrightarrow{(\eta)}v$ in case $w=uw_{\sigma}z,v=uv_{\sigma}z$ for some $u,z\in F$ and an element $(w_{\sigma},v_{\sigma})$ of the set $S$ of reductions of type ($\eta$). Here $(\eta)$ may be one of: ($1$) – ($5$), or even more explicitly ($4i$) or ($5i$), for some $i$. More generally, $w\xrightarrow{(\eta)}v$ may also denote a sequence of consecutive reductions of type ($\eta$). If clear from the context, $w\rightarrow v$ will denote an unspecified sequence of reductions.

Let $K\langle X\rangle=K\langle x_{1},\ldots,x_{n}\rangle$ be the free algebra over a field $K$. The length of a word $w\in F$ is denoted by $|w|$. By $\operatorname{suff}_{m}(w)$ ($\operatorname{pref}_{m}(w)$, respectively) we mean the suffix (prefix, respectively) of length $m$ in $w$. For every subset $Z\subseteq F$ by $\operatorname{suff}(Z)$ ($\operatorname{pref}(Z)$, respectively) we denote the set of all suffixes (prefixes, respectively) of elements of $Z$. If $w\in F$ then $w^{\infty}$ denotes the infinite word $www\ldots$. For a subset $Z$ of a monoid $M$ by $\langle Z\rangle$ we denote the submonoid generated by $Z$. If $Z=\{w\}$, then we also write $\langle w\rangle$. If $S$ is a semigroup, then $S^{1}$ stands for the monoid obtained by adjoining an identity element to $S$. Also, $S^{0}$ denotes the semigroup $S$ with zero adjoined. The support $\operatorname{supp}(\alpha)$ of an element $\alpha=\sum_{i}m_{i}s_{i}$, where $m_{i}\in K,s_{i}\in M$, of the semigroup algebra $K[M]$ is defined as the set of all $s_{i}$ such that $m_{i}\neq 0$. If $M$ has a zero element $\theta$ then the contracted semigroup algebra $K_{0}[M]$ is defined as the factor algebra $K[M]/K\theta$.

If $S$ is a semigroup, $A,B$ are nonempty sets and $P=(p_{ba})$ is a $B\times A$ - matrix with entries in $S^{0}$, then the semigroup of matrix type ${\mathcal{M}}^{0}(S,A,B;P)$ over $S$ is the set of all triples $(s,a,b)$, where $s\in S,a\in A,b\in B$, with the zero element $\theta$, with operation $(s,a,b)(s^{\prime},a^{\prime},b^{\prime})=(sp_{ba^{\prime}}s^{\prime},a,b^{\prime})$ if $p_{ba^{\prime}}\in S$ and $\theta$ otherwise. Moreover ${\mathcal{M}}^{0}(K[S],A,B;P)$ denotes the corresponding algebra of matrix type. It is defined as $K_{0}[{\mathcal{M}}^{0}(S,A,B;P)]$ and (if $A,B$ are finite) it can be interpreted as the set of all $A\times B$ - matrices over $K[S]$ with operation $\alpha\beta=\alpha\circ P\circ\beta$, where $\circ$ stands for the standard matrix product. This construction plays a fundamental role in the structure and representations of semigroup algebras. We refer to [14], Chapter 5, for basic results. Structures of these types are crucial for the approach and the results of this paper.

The material is organized as follows. In Section 2 we derive the reduced form of almost all words in $F$ representing the elements of the monoid $C_{n}$. The main results are stated in Theorem 2.1 and Proposition 2.14. In Section 3 a natural ideal chain $I_{n-3}\subseteq I_{n-4}\subseteq\cdots\subseteq I_{0}\subseteq I_{-1}$ of $C_{n}$ is introduced. And it is shown that all factors $I_{j-1}/I_{j}$, and $(I_{n-3})^{0}$, have (modulo finitely many elements) a structure of a semigroup $M_{j}$, $j=0,1,\ldots,n-2$, of matrix type over an infinite cyclic semigroup. Using certain natural involutions on $C_{n}$ (Definition 3.11) it is shown that these matrix structures have a very regular form, see Corollary 3.23, and Remark 3.24. The results of this section are summarized in Corollary 3.25, which provides a very transparent structural tool for approaching the monoids $C_{n}$ and their algebras. In Section 4 these results are illustrated with the (misleadingly simple) cases of the monoids $C_{n}$ for $n=3$ and $n=4$. In Section 5 it is first shown in Theorem 5.8 that all algebras of matrix type resulting from the semigroups $M_{j}$ are prime. Then, in Theorem 5.9, this is used to prove that $K[C_{n}]$ is a right and left Noetherian algebra. Existence of an embedding of $K[C_{n}]$ into a matrix ring over a field follows, which is of interest in the context of several earlier results on faithful matrix representations of various classes of Hecke–Kiselman monoids, [6],[7],[10]. Section 5 culminates with an important application of the approach developed in this paper. Namely, a characterization of all oriented graphs $\Theta$ such that the corresponding Hecke–Kiselman algebra $K[\operatorname{HK}_{\Theta}]$ is Noetherian, Theorem 5.10. We prove that the latter is equivalent to saying that each of the connected components of $\Theta$ is either an oriented cycle or an acyclic graph. We conclude with some open problems.

2 The form of (almost all) reduced words in $C_{n}$

The main aim of this section is to prove that all elements of the monoid $C_{n}$, except for finitely many words, have a very special reduced form (with respect to the deg-lex order and the reduction system $S$ introduced in Theorem 1.1). This will be the key to describe the structure and properties of $C_{n}$ in the next sections. Because, in view of Corollary 1.2, we may identify the elements of $C_{n}$ with the reduced words in $F$.

We adopt the following notation. If $i,j\in\{1,\ldots,n\}$ then $x_{i}\cdots x_{j}$ denotes the product of all consecutive generators from $x_{i}$ up to $x_{j}$ if $i<j$, or down to $x_{j}$, if $i>j$.

Let $q_{i}=x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}\in F$, for $i=0,\ldots,n-2$. Here we agree that $q_{0}=x_{n-1}\cdots x_{1}$. From Corollary 1.2 it follows that the word $(x_{n}q_{i})^{k}$ is reduced for every $k\geqslant 0$.

For every $i=0,\ldots,n-2$ we define two subsets $A_{i}$ and $B_{i}$ of $F$, as follows. First,

[TABLE]

where $s\in\{0,\ldots,i+1\}$, $k_{s+1}<k_{s+2}<\cdots<k_{i+1}\leqslant n-1$, $k_{s}\leqslant s$ and $k_{q}>q$ for $q=s+1,\ldots,i+1$.

The convention is that the subset of $A_{i}$ corresponding to $s=i+1$ has the form $\operatorname{suff}(\{x_{k_{i+1}}\cdots x_{i+1}\})$, where $k_{i+1}\leqslant i+1$. Also, if $s=0$ then the corresponding subset of $A_{i}$ has the form

$\operatorname{suff}(\{(x_{k_{s+1}}\cdots x_{s+1})\cdots(x_{k_{i+1}}\cdots x_{i+1})\})$, where $k_{s+1}<k_{s+2}<\cdots<k_{i+1}\leqslant n-1$ and $k_{q}>q$ for $q=s+1,\ldots,i+1$.

The set $B_{i}$ is defined by

[TABLE]

where $r\geqslant 0$, $i_{r}<i_{r-1}<\cdots<i_{1}<i+1$ and $i+1<j_{1}<j_{2}<\cdots<j_{r+1}\leqslant n$.

Here, the subset of $B_{i}$ corresponding to $r=0$ has the form $\operatorname{pref}(\{x_{n}x_{n-1}\cdots x_{j_{1}}\})$.

The following result characterizes all reduced words that have a factor of the form $x_{n}q_{i}$.

Theorem 2.1**.**

Assume that $w$ is a reduced word that contains a factor of the form $x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$ for some $i=0,\ldots,n-2$. Then

[TABLE]

for some $a\in A_{i}$, $b\in B_{i}$ and some $k\geq 1$. Moreover, all words of this type are reduced.

We will use the following convention. By a block we mean a factor of the form $x_{k_{j}}\cdots x_{j}$, for some $j\in\{s,\ldots,i+1\}$, appearing in the elements of the set $A_{i}$ or a factor of the form $x_{n}(x_{1}\cdots x_{i_{k}}x_{n-1}\cdots x_{j_{k}})$, for $k\in\{1,\ldots,r\}$, or $x_{n}x_{n-1}\cdots x_{j_{r+1}}$, appearing in the elements of $B_{i}$.

The main idea of the proof of Theorem 2.1 is to analyze the possible forms of reduced words that satisfy certain additional restrictions. The proof will be preceded by a series of technical lemmas.

Lemma 2.2**.**

If $w=x_{n-1}u$ is a reduced word, where $u\in F$ is such that $|u|_{n}=0$, then $w=x_{n-1}\cdots x_{k}$ for some $k\geqslant 1$.

Proof.

Let $w=x_{n-1}\cdots x_{t}u^{\prime}$ for some $n-1\geqslant t>1$ and $\operatorname{pref}_{1}(u^{\prime})=x_{s}$. If $s<t-1$, then the word $x_{n-1}\cdots x_{t}x_{s}$ has a factor $x_{t}x_{s}$, with $s<t-1$, so it is not reduced. If $n>s\geqslant t$, then the word $x_{n-1}\cdots x_{t}x_{s}$ has a factor $x_{s}vx_{s}$, where $|v|_{s+1}=0$, whence it is not reduced. Therefore, we get that $s=t-1$ and $w$ is a prefix of the word of the form $x_{n-1}\cdots x_{1}v$ for some $v$. Notice that for every $k\neq n$ the word $x_{n-1}\cdots x_{1}x_{k}$ has a factor $x_{k}zx_{k}$, where $|z|_{k+1}=0$, so it is not reduced. It follows that $v$ must be the empty word. The assertion follows. ∎

Lemma 2.3**.**

If $w=x_{n}x_{1}u$ is a reduced word, where $|u|_{n}=0$, then $w$ is of one of the forms

$w=x_{n}x_{1}x_{2}\cdots x_{i}x_{n-1}\cdots x_{j}$* for some $1\leqslant i<j\leqslant n-1$;* 2. 2.

$w=x_{n}x_{1}x_{2}\cdots x_{i}$* for some $1\leqslant i<n-1$.*

Proof.

Let $w=x_{n}x_{1}\cdots x_{k}u^{\prime}$ for some $n-1>k\geqslant 1$ and let $\operatorname{pref}_{1}(u^{\prime})=x_{s}$ for some $s<n$. If $s<k$, then $w$ has a factor $x_{s}vx_{s}$, with $|v|_{s-1}=0$. On the other hand, if $n-1>s>k+1$ then we get a factor of the form $x_{n}x_{1}\cdots x_{k}x_{s}\xrightarrow{(3)}x_{s}x_{n}x_{1}\cdots x_{k}$. It follows that $\operatorname{pref}_{1}(u)\in\{x_{k+1},x_{n-1}\}$. This means that $w=x_{n}x_{1}\cdots x_{i}x_{n-1}v$ for some $i$ and some $v$ or $w=x_{n}x_{1}\cdots x_{i}$ for some $i<n-1$. In the former case Lemma 2.2 implies that $w=x_{n}x_{1}\cdots x_{i}x_{n-1}x_{n-2}\cdots x_{k}$ for some $k<n$. If $i\geqslant k$, then $w$ has a factor $x_{i}x_{n-1}\cdots x_{i}$ and $|x_{n-1}\cdots x_{i+1}|_{i-1}=0$, so $w$ is not a reduced word. ∎

Lemma 2.4**.**

If $w=x_{n}ux_{n}$ is a reduced word, where $|u|_{n}=0$, then $u$ is of one of the forms

$u=x_{n-1}\cdots x_{1}$; 2. 2.

$u=x_{1}\cdots x_{i}x_{n-1}\cdots x_{j}$* for $1\leqslant i<j\leqslant n-1$.*

Proof.

Notice that $\operatorname{pref}_{1}(u)\in\{x_{1},x_{n-1}\}$, since otherwise if $\operatorname{pref}_{1}(u)=j$, then $w$ has a factor $x_{n}x_{j}$ for some $1<j<n-1$, and thus it is not reduced.

Assume first that $\operatorname{pref}_{1}(u)=x_{n-1}$. Notice that $|u|_{1}\geqslant 1$. If $|u|_{1}=0$, then $x_{n}ux_{n}=x_{n}u$, so $w$ is not reduced. By Lemma 2.2 it now follows that $w$ must be of the form $x_{n}\cdots x_{1}vx_{n}$ for some $v$. Then $v$ must be the empty word, because for every $k$, if $\operatorname{pref}_{1}(v)=x_{k}$, then $w$ has a factor of the form $x_{k}x_{k-1}\cdots x_{1}x_{k}$, where $|x_{k-1}\cdots x_{1}|_{k+1}=0$, hence $w$ is not reduced.

Thus, assume that $\operatorname{pref}_{1}(u)=x_{1}$. Let $w=x_{n}x_{1}u^{\prime}x_{n}$ for some $u^{\prime}$ such that $|u^{\prime}|_{n}=0$. From Lemma 2.3 we know that $w=x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{j}x_{n}$ for some $1\leqslant i<j\leqslant n-1$ or $w=x_{n}x_{1}\cdots x_{i}x_{n}$ for some $1\leqslant i<n-1$. However, in the latter case $|x_{1}\cdots x_{i}|_{n-1}=0$, so that $w$ is not a reduced word. ∎

The following lemma shows that in the case where $i=0$ or $i=n-2$, the reduced words with a factor $x_{n}q_{i}$ have an extremely simple form.

Lemma 2.5**.**

If a reduced word has a factor of the form $x_{n}x_{1}\cdots x_{n-1}$ or $x_{n}x_{n-1}\cdots x_{1}$, then it must be a factor of the infinite word $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$ or $(x_{n}x_{n-1}\cdots x_{1})^{\infty}$.

Proof.

Define $x_{n+1}=x_{1}$ and $x_{0}=x_{n}$. Assume that $w=ux_{k+1}x_{k+2}\cdots x_{n}x_{1}\cdots x_{k}v$ for some $k=1,\ldots,n$ and some $u,v\in F$. We claim that $\operatorname{pref}_{1}(v)=x_{k+1}$ and $\operatorname{suff}_{1}(u)=x_{k}$. Then the first part of the assertion will follow.

If $\operatorname{pref}_{1}(v)=x_{s}$ for $s<k+1$, then the word $w$ has a factor $x_{s}x_{s+1}\cdots x_{k}x_{s}$, whence it cannot be reduced. Similarly, for $s>k+1$ the word $w$ has a factor of the form $x_{s}\cdots x_{n}x_{1}\cdots x_{k}x_{s}$ and $|x_{s+1}\cdots x_{n}x_{1}\cdots x_{k}|_{s-1}=0$, which is not possible. Suppose that $\operatorname{suff}_{1}(u)=x_{s}$ for some $s\neq k$. If $s<k$, then $w$ has a factor $x_{s}x_{k+1}\cdots x_{n}x_{1}\cdots x_{s-1}x_{s}$ and $|x_{k+1}\cdots x_{n}x_{1}\cdots x_{s-1}|_{s+1}=0$. Similarly, if $s>k$, then $w$ has a factor $x_{s}x_{k+1}\cdots x_{s}$, where $|x_{k+1}\cdots x_{s-1}|_{s+1}=0$, which leads to a contradiction again. This proves the claim.

The second part of the lemma follows by a symmetric argument. ∎

The next few lemmas will be used to determine the desired shape of the elements of $B_{i}$, which are the endings of the considered class of reduced words.

Lemma 2.6**.**

Let $w\in F$ be of the form $w=x_{n}x_{1}\cdots x_{i_{1}}x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j_{2}}$, where $1\leqslant i_{p}<j_{p}\leqslant n-1$ for $p=1,2$. If $w$ is a reduced word, then

$i_{2}<j_{1}$, 2. 2.

$i_{1}\geqslant i_{2}$, 3. 3.

$j_{1}\leqslant j_{2}$.

Proof.

Suppose that $i_{2}\geqslant j_{1}$. Then $j_{1}<n-1$. Thus, $w$ has a factor of the form $x_{j_{1}}x_{n}x_{1}\cdots x_{j_{1}-1}x_{j_{1}}$ and $|x_{n}x_{1}\cdots x_{j_{1}-1}|_{j_{1}+1}=0$, whence $w$ is not in a reduced form.

Suppose that $i_{1}<i_{2}$. Then $i_{1}<n-2$ and $w$ has a factor of the form $x_{i_{1}}x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{i_{1}}$ and $|x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{i_{1}-1}|_{i_{1}+1}=0$, because we know that $i_{1}+1<i_{2}+1\leqslant j_{1}$, so that $i_{1}+1<j_{1}$. Therefore, again $w$ cannot be in the reduced form.

Suppose that $j_{1}>j_{2}$. Then the hypothesis and the first part of the lemma imply that $i_{2}\leqslant j_{2}-1<j_{1}-1$. Notice that in this case $w$ has a factor of the form $x_{j_{1}}x_{n}x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j_{1}+1}x_{j_{1}}$ and $|x_{n}x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j_{1}+1}|_{j_{1}-1}=0$, because $i_{2}<j_{1}-1$; so $w$ is not in the reduced form. ∎

Lemma 2.7**.**

Let $w$ be a reduced word such that

[TABLE]

If $i_{1}=i_{2}=i$, then $j_{1}=i+1$ and if $j=j_{1}=j_{2}$, then $i_{1}=i_{2}=j-1$. Moreover, if $i_{1}+1=j_{1}$ and $i_{2}+1<j_{2}$, then $j_{2}>j_{1}$.

Proof.

Assume that $i_{1}=i_{2}=i$. Then $w$ contains a reduced factor $x_{i}x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{i}$. Hence, $|x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{i-1}|_{i+1}\geqslant 1$. Since $i<j_{1}$, this implies that $j_{1}=i+1$.

Consider the case where $j=j_{1}=j_{2}$. Then $w$ contains a reduced factor $x_{j}x_{n}x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j}$. This implies that $|x_{n}x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j+1}|_{j-1}\geqslant 1$. Since $i_{2}<j$, we must have $i_{2}=j-1$. By Lemma 2.6 $j-1=i_{2}\leqslant i_{1}<j$, so that $i_{1}=i_{2}=j-1$.

Finally, assume that $i_{1}+1=j_{1}$ and $i_{2}+1<j_{2}$. From Lemma 2.6 we know that $j_{2}\geqslant j_{1}$. Suppose that $j_{2}=j_{1}$. Then the part of the statement that has already been proved implies that $i_{2}=i_{1}=j_{1}-1=j_{2}-1$, which contradicts the hypothesis. This completes the proof. ∎

Corollary 2.8**.**

If a reduced word $w$ is of the form

[TABLE]

for some $u,v$ such that $|u|_{n}=|v|_{n}=0$, then it follows that

[TABLE]

*Furthermore, if $i_{k}+1=j_{k}$ for some $k$, then $i_{k}=i_{s}$ and $j_{k}=j_{s}$ for $s=1,\ldots,k-1$.

Moreover, if for some $l$ we have $i_{l}+1<j_{l}$, then $i_{r}<i_{r-1}<\cdots<i_{l}<i_{l}+1<j_{l}<\cdots<j_{r}$.

If $l>1$, then also $i_{l}\leqslant i_{l-1}<j_{l-1}<j_{l}$.*

Proof.

The inequalities of the first part of the assertion follow directly from Lemma 2.6.

Notice that if $i_{k}+1=j_{k}$, then Lemma 2.6 implies that $i_{k}\leqslant i_{k-1}<j_{k-1}\leqslant j_{k}$, so that $i_{k}=i_{k-1}$ and $j_{k}=j_{k-1}$. Repeating this argument, we get that if $i_{k}+1=j_{k}$ for some $k$, then $i_{k}=i_{s}$ and $j_{k}=j_{s}$ for $s=1,\ldots,k-1$.

Suppose that for some $l$ we have $i_{l}+1<j_{l}$. If $i_{l+1}=i_{l}$ or $j_{l+1}=j_{l}$, then Lemma 2.7 implies that $j_{l+1}=i_{l}+1$, contradicting the hypothesis. Now, repeating this argument and using the part of the assertion that has already been proved we get that $i_{r}<i_{r-1}<\cdots<i_{l}$ and $j_{l}<\cdots<j_{r}$.

Finally, assume that $l>1$ and also $i_{l}+1<j_{l}$. Then, by Lemma 2.6 we know that $i_{l}\leqslant i_{l-1}<j_{l-1}\leqslant j_{l}$. Suppose that $j_{l-1}=j_{l}$. Then Lemma 2.7 implies that $i_{l}=i_{l-1}=j_{l}-1$, a contradiction. Thus $j_{l-1}<j_{l}$. This completes the proof. ∎

The next few lemmas will be used to deal with the shape of the elements of the set $A_{i}$, which are the beginnings of the considered class of reduced words.

Lemma 2.9**.**

Let $w$ be a reduced word such that

[TABLE]

where $i=1,\ldots,n-3$, $k_{s}<s\leqslant i+1$, $|u|_{j}=0$ for $j=1,\ldots,s$ and $|u|_{n}=|v|_{n}=0$. Then $v=x_{r}x_{r+1}\cdots x_{k_{s}-1}$ for some $r\geqslant 1$.

Proof.

Assume that $k_{s}>1$ and $x_{j}=\operatorname{suff}_{1}(v)$. If $j>k_{s}+1$, then $w$ has a factor $x_{j}x_{k_{s}}$, which is not reduced. Similarly, if $j<k_{s}-1<i$, then $w$ has a factor $x_{j}(x_{k_{s}}\cdots x_{s})ux_{n}x_{1}\cdots x_{j}$. Since $|(x_{k_{s}}\cdots x_{s})ux_{n}x_{1}\cdots x_{j-1}|_{j+1}=0$, this also leads to a contradiction. It follows that $j=k_{s}-1$.

Suppose that $w=v^{\prime}x_{1}\cdots x_{s}ux_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$ for some nonempty word $v^{\prime}$. Let $x_{p}=\operatorname{suff}_{1}(v^{\prime})$ for some $p\neq n$. If $p\neq 2$, then $w$ has a factor $x_{p}x_{1}$, that is not reduced. Moreover, for $p=2$, since $1\leqslant k_{s}<s$, and hence $2\leqslant s$, we get a factor of the form $x_{2}x_{1}x_{2}$, that is not reduced. This contradiction completes the proof. ∎

Lemma 2.10**.**

Assume that a reduced word $w$ is of the form

[TABLE]

where $s\leqslant i+1$, $k_{s}\geqslant s$, $k_{q}>q$ for $q=s+1,\ldots,i+1$, $|u|_{n}=0$. Then $\operatorname{suff}_{1}(u)\in\{x_{k_{s}+1},x_{s-1}\}$.

Proof.

Let $\operatorname{suff}_{1}(u)=x_{j}$ for some $j\neq n$. We consider the following cases:

• •

if $j>k_{s}+1$ or $j=k_{s}$, then $w$ has a factor $x_{j}x_{k_{s}}$, which is not reduced,

• •

if $j=k_{s}-1$ and $k_{s}>s$, then $w$ has a factor $x_{j}x_{j+1}x_{j}$, which is not reduced,

• •

if $s\leqslant j<k_{s}-1\leqslant i$, then we have a factor $x_{j}x_{k_{s}}\cdots x_{j+1}x_{j}$ such that $|x_{k_{s}}\cdots x_{j+1}|_{j-1}=0$; whence $w$ is not a reduced word,

• •

if $j<s-1$, then $w$ has a factor of the form

[TABLE]

where $|z|_{j+1}=0$; so $w$ is not a reduced word.

We have thus proved that $\operatorname{suff}_{1}(u^{\prime})\in\{x_{k_{s}+1},x_{s-1}\}$, as desired. ∎

Lemma 2.11**.**

Assume that a reduced word $w$ is of the form

[TABLE]

for some $i\in\{1,\ldots,n-3\}$, where $|u|_{n}=0$ and $u$ has no factors of the form $x_{l}\cdots x_{j}$, where $l<j$. Then $u$ is of the form

[TABLE]

where $s\leqslant i+1$, $k_{s}<k_{s+1}<\cdots<k_{i+1}\leqslant n-1$, $k_{s}\geqslant s$, $k_{q}>q$ for $q=s+1,\ldots,i+1$.

Proof.

Let $x_{s}=\operatorname{suff}_{1}(u)$ for some $s\neq n$. If $s>i+1$, then $w$ has a factor $x_{s}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{s+1}x_{s}$ and $|x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{s+1}|_{s-1}=0$, whence $w$ is not reduced. If $s<i$, then $w$ has a factor $x_{s}x_{n}x_{1}\cdots x_{s-1}x_{s}$, so it is not reduced. It follows that $\operatorname{suff}_{1}(u)=x_{i+1}$.

Applying Lemma 2.10, we deduce that if $|u|>1$, then $w=u^{\prime}x_{i+2}x_{i+1}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$ or $w=u^{\prime}x_{i}x_{i+1}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$ for some $u^{\prime}\in F$. In the latter case, $w$ has a factor $x_{i}x_{i+1}$, which contradicts the hypothesis. Repeating this argument we get the desired form of $u$, because an inequality $k_{j}\geqslant k_{j+1}$ would allow a reduction of type (5). ∎

Lemma 2.12**.**

If $w$ is a reduced word and $w=ux_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$ for some $i=1,\ldots,n-3$ and some $u\in F$ such that $|u|_{n}=0$, then $u$ is of the form

[TABLE]

where $s\leqslant i+1$, $k_{s+1}<k_{s+2}<\cdots<k_{i+1}\leqslant n-1$ and $k_{q}>q$ for $q=s+1,\ldots,i+1$ (but perhaps $k_{s}\leqslant s$).

Proof.

If $u$ has no factors of the form $x_{l}x_{l+1}\cdots x_{j}$, where $l<j$, then the desired form of the word follows from Lemma 2.11.

Thus, assume otherwise. Let

[TABLE]

where the word $v$ does not have increasing factors, $\operatorname{suff}_{2}(z)=x_{j-1}x_{j}$ for some $j$ and $\operatorname{pref}_{1}(v)=x_{k_{s}}$ for $k_{s}\neq j+1$. Then Lemma 2.11 implies that

[TABLE]

and $s\leqslant i+1$, $k_{s}<k_{s+1}\cdots<k_{i+1}\leqslant n-1$, $k_{q}>q$ for $q=s+1,\ldots,i+1$, $k_{s}\geqslant s$.

Moreover, by Lemma 2.10, we have $x_{j}\in\{x_{k_{s}+1},x_{s-1}\}$. If $j=k_{s}+1$, then $w$ has a factor $x_{j-1}x_{j}x_{j-1}$, which is impossible. Hence $j=s-1$, and then

[TABLE]

for $p=1,\ldots,s-1$, and the assertion follows from Lemma 2.9. ∎

Proof of Theorem 2.1. By Corollary 1.2 it is clear that all words described in the statement are reduced.

Let $w\in F$ be a reduced word that contains a factor $x_{n}q_{i}$. By Lemma 2.5 the assertion holds for $i=0,n-2$. Notice that if the word $w$ has the form $x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}v$ for some $v\in C_{n}$, then we must have $\operatorname{pref}(v)=x_{n}$. Indeed, if $\operatorname{pref}(v)=x_{j}$ for $j\leqslant i$, then $x_{n}q_{i}x_{j}\xrightarrow{(4j)}x_{n}q_{i}$. Similarly, if $i+1\leqslant j\leqslant n-1$, then $x_{n}q_{i}x_{j}\xrightarrow{(5j)}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{j+1}x_{j-1}\cdots x_{i+1}$.

From Lemma 2.4 and Lemma 2.6 we know that if $i=1,\ldots,n-3$ then $w$ is of the form

[TABLE]

for some $m$, where $1\leqslant i_{k}<j_{k}\leqslant n-1$ for every $k$ and $|u|_{n}=|v|_{n}=0$.

In view of Corollary 2.8 this implies that $w$ is of the form

[TABLE]

where $i_{r}<i_{r-1}<\cdots<i_{1}<i+1$ and $i+1<j_{1}<j_{2}<\cdots<j_{r}$ and $|u|_{n}=|v|_{n}=0$, where the factor of the form $(x_{1}\cdots x_{i_{1}}x_{n-1}\cdots x_{j_{1}})\cdots x_{n}(x_{1}\cdots x_{i_{r}}x_{n-1}\cdots x_{j_{r}})$ does not have to occur in $w$ (that is, $w=u(x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1})^{k}x_{n}v$) and then we put $r=0$.

Notice that $\operatorname{pref}_{1}(v)\in\{x_{1},x_{n-1}\}$, since otherwise $w$ contains a factor $x_{n}x_{s}$ for $s<n$, which is not reduced.

If $\operatorname{pref}_{1}(v)=x_{n-1}$, Lemma 2.2 implies that $v=x_{n-1}\cdots x_{j_{r+1}}$. Moreover, we must have $j_{r}<j_{r+1}$, as otherwise $w$ has a factor $x_{j_{r}}x_{n}\cdots x_{j_{r}+1}x_{j_{r}}$ such that $|x_{n}\cdots x_{j_{r}+1}|_{j_{r}-1}=0$, which is not reduced.

If $\operatorname{pref}_{1}(v)=x_{1}$, then by Lemma 2.3 and Corollary 2.8 we get $v=x_{1}\cdots x_{i_{r+1}}x_{n-1}\cdots x_{j_{r+1}}$ for $i_{r+1}<i_{r}$ and $j_{r+1}>j_{r}$, if $r>0$. If $r=0$, then in view of (1) we have $w=pq$, where $p\in F$ and

[TABLE]

Corollary 2.8 implies that $i_{1}\leqslant i$ and $j_{1}\geqslant i+1$. The desired form of the elements of the set $B_{i}$ follows.

Since $k\geqslant 1$, the desired form of the elements of the set $A_{i}$ follows by Lemma 2.12. This completes the proof of Theorem 2.1. $\square$

Our next aim is to show that Theorem 2.1 characterizes reduced forms of almost all elements of $C_{n}$.

Definition 2.13**.**

For every $i=0,\ldots,n-2$ we denote by $\tilde{M_{i}}$ the following set

[TABLE]

(the set of reduced forms of elements of $C_{n}$ that have a factor $x_{n}q_{i}$). Define also $\tilde{M}=\bigcup_{i=0}^{n-2}\tilde{M_{i}}$.

Corollary 1.2 ensures that two elements $w,w^{\prime}\in\tilde{M}$ are equal in $C_{n}$ if and only if the equality $w=w^{\prime}$ holds in the free monoid $F$ generated by $x_{1},\ldots,x_{n}$. In particular, we can write $\tilde{M}\subseteq C_{n}$. This identification will be often used without further comment.

Proposition 2.14**.**

$C_{n}\setminus\tilde{M}$* is a finite set.*

Proof.

Let $w$ be a reduced word that does not contain any factor of the form $x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$. Assume that $|w|_{n}=k$, so that $w=u_{0}x_{n}u_{1}x_{n}\cdots x_{n}u_{k}$ for some words $u_{j}$, $j=0,\ldots,k$, such that $|u_{j}|_{n}=0$.

If $k=0$, then $|w|_{1}\leqslant 1$, since otherwise $w$ contains a factor $x_{1}yx_{1}$, where $|y|_{n}=0$. We will prove by induction on $m=1,\ldots,n-1$, that $|w|_{m}\leqslant m$. We know that the assertion holds for $m=1$. Suppose that $|w|_{m-1}\leqslant m-1$ and also that $|w|_{m}>m$. Then $w$ has a factor of the form $x_{m}yx_{m}$, where $|y|_{m-1}=0$ (because $|w|_{m-1}\leqslant m-1$), whence $w$ is not reduced, a contradiction. This proves the inductive assertion. It follows that if $|w|_{n}=0$ then $|w|\leq 1+\cdots+(n-1)=\frac{n(n-1)}{2}=C_{1}(n)$. So the number of such possible words does not exceed $n^{C_{1}(n)}$.

Assume now that $|w|_{n}=k\geqslant 1$. By Lemma 2.4 for $s=1,\ldots,k-1$ we get $u_{s}=x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}$ for some $1\leqslant i_{s}<j_{s}\leqslant n-1$. Hence, in particular it follows that the length of every $u_{s}$, for $s=1,\ldots,k-1$, is at most $n-1$. Moreover, the assumption implies that $u_{1}=x_{1}\cdots x_{i_{1}}x_{n-1}\cdots x_{j_{1}}$ where $i_{1}+1<j_{1}$. From Corollary 2.8 it follows that we have the following inequalities: $1\leqslant i_{k-1}<i_{k-2}<\cdots<i_{1}+1<j_{1}<\cdots<j_{k-1}\leqslant n-1$. In particular, this implies that $k\leqslant\frac{n}{2}+1$, and hence the length of the reduced word $v=x_{n}u_{1}x_{n}\cdots x_{n}$ (such that $w=u_{0}vu_{k}$) is at most $k+(k-1)(n-1)<kn\leqslant(\frac{n}{2}+1)n=C_{2}(n)$. By the first part of the proof we get that that the length of $u_{0}$ and of $u_{k}$ is at most $C_{1}(n)$. We have proved that every reduced word of the form $w=u_{0}x_{n}u_{1}x_{n}\cdots x_{n}u_{k}$ has length at most $C_{2}(n)+2C_{1}(n)$; so there are at most $(n+1)^{C_{2}(n)+2C_{1}(n)}$ such words.

Consequently, the cardinality of $C_{n}\setminus\tilde{M}$ is bounded by $n^{C_{1}(n)}+(n+1)^{C_{2}(n)+2C_{1}(n)}$, so it is finite. ∎

3 An ideal chain and related structures of matrix type

Our next goal is to introduce a special ideal chain in the monoid $C_{n}$ that is strongly related to certain structures of matrix type. This will be essential when dealing with the structure and properties of the algebra $K[C_{n}]$, and consequently of every Hecke–Kiselman algebra, in Section 5.

In view of Corollary 1.2 we identify elements of $C_{n}$ with the (unique) reduced forms of words in $F$.

3.1 An ideal chain

We will introduce two special families of ideals of $C_{n}$. First, let

[TABLE]

for $i=0,\ldots,n-2$. We also define $I_{-1}=I_{0}\cup C_{n}x_{n}q_{0}C_{n}$. It is clear that every $I_{i}$ is an ideal in $C_{n}$, if it is nonempty. We show that $I_{n-2}=\emptyset$. This is a consequence of the following observation.

Lemma 3.1**.**

Let $w\in C_{n}$. If $k=1,\ldots,n$ then the reduced form of $(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})w$ is a factor of the infinite word $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$. Moreover $(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})w$ has a prefix of the form $x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}$.

Proof.

Let $w\in C_{n}$. We proceed by induction on the length of $w$. The assertion is clear for the empty word. Assume that $|w|>1$. Let $w=x_{j}w^{\prime}$ for some $j$. If $j=k+1$ (for $k=n$ we put $j=1$), then the assertion follows by the induction hypothesis for $k+1$ and $w^{\prime}$. If $j=k$, the assertion is also clear by induction. Otherwise,

[TABLE]

and the assertion again follows by the induction hypothesis. ∎

Corollary 3.2**.**

$I_{n-2}=\emptyset$**

Proof.

Let $w\in C_{n}$. By Lemma 3.1, $(x_{n}x_{1}\cdots x_{n-1})w$ is a factor of $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$. Hence, in particular $(x_{n}x_{1}\cdots x_{n-1})w\notin I_{n-2}$, so that $w\notin I_{n-2}$. The assertion follows. ∎

A dual version of Lemma 3.1 also holds. In order to prove this, we introduce a natural involution of the monoid $C_{n}$ that will be useful also later.

Definition 3.3**.**

Let $\tau:\langle x_{1},\ldots,x_{n}\rangle\longrightarrow\langle x_{1},\ldots,x_{n}\rangle$ be the involution such that

[TABLE]

It is easy to see that $\tau$ preserves the set of defining relations of $C_{n}$. Hence, it determines an involution of $C_{n}$, also denoted by $\tau$.

Lemma 3.4**.**

Let $w\in C_{n}$. If $k=0,\ldots,n-1$ then the reduced form of $w(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})$ is a factor of the infinite word $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$. Moreover, $w(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})$ has a suffix of the form $x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}$.

Proof.

Notice that for every $k\in\{0,\ldots,n-1\}$ we have $\tau(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})=x_{n-k}\cdots x_{n}x_{1}\cdots x_{n-k-1}$, where for $k=0$ we put $x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}=x_{1}\cdots x_{n}$. In particular, the image under $\tau$ of every factor of the infinite word $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$ is a factor of this word. Let $w\in C_{n}$. The above shows that $\tau(w(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}))=x_{n-k}\cdots x_{n}x_{1}\cdots x_{n-k-1}\tau(w)$. Lemma 3.1 yields that $\tau(w(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}))$ is a factor of $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$, with a prefix of the form $x_{n-k}\cdots x_{n}x_{1}\cdots x_{n-k-1}$. Applying the involution $\tau$ again we get that $w(x_{k+1}\cdots x_{n}x_{1}\cdots x_{k})$ is a factor of $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$, with a suffix of the form $x_{k+1}\cdots x_{n}x_{1}\cdots x_{k}=\tau(x_{n-k}\cdots x_{n}x_{1}\cdots x_{n-k-1})$. The assertion follows. ∎

We can now return to the ideal chain $I_{j}$.

Lemma 3.5**.**

$I_{i+1}\subseteq I_{i}$* for $i=0,\ldots,n-3$.*

Proof.

We claim that for every $i=0,\ldots,n-3$ and $l\geqslant 1$ we have $(x_{n}q_{i})^{l}\notin I_{i+1}$; in other words, there exist $u,v\in C_{n}$ such that $u(x_{n}q_{i})^{l}v\in\langle x_{n}q_{i+1}\rangle$.

By Corollary 3.2, the assertion holds for $i=n-3$. So, assume that $i\in\{0,\ldots,n-4\}$. First, notice that for every $i+2\leqslant m\leqslant n-1$ we have

[TABLE]

Using the inequality $i+1<i+2<n-1$ we conclude that for $i\neq n-3$

[TABLE]

It follows that for $1<k\leqslant n-i-1$

[TABLE]

where $\xrightarrow{(*)}$ stands for applying $k-2$ times the observation made before.

Moreover, the first two steps of the above computation show that for $k=1$ we get

[TABLE]

Similarly, for every positive integer $m$, if $m-1=l(n-i-1)+k$, where $l\geqslant 0$ and $n-i-2\geqslant k\geqslant 0$, then

[TABLE]

Hence, for every $i$ and $m-1=l(n-i-1)+k$, defining $u=x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{n-k+1}$ if $k\neq 0$, and $u=1$ if $k=0$, we get $u(x_{n}q_{i})^{m}x_{i+2}\in\langle x_{n}q_{i+1}\rangle$. Thus, $(x_{n}q_{i})^{m}\notin I_{i+1}$, which proves our claim.

Let $i=0,\ldots,n-3$. Suppose that there exists $w\in I_{i+1}\setminus I_{i}$. Let $x,y\in C_{n}$ and $k\geqslant 1$ be such that $xwy=(x_{n}q_{i})^{k}$. By the first part of the proof, there exist $u,v\in C_{n}$ such that $uxwyv\in\langle x_{n}q_{i+1}\rangle$. Thus, $w\notin I_{i+1}$, which leads to a contradiction. ∎

As a consequence, we get the following ideal chain in $C_{n}$

[TABLE]

In order to prove that certain elements of $C_{n}$ are contained in $I_{i}$ we introduce another useful family of ideals $Q_{i}$. Let $\textrm{Map}(\mathbb{Z}^{n},\mathbb{Z}^{n})$ denote the monoid of all functions $\mathbb{Z}^{n}\longrightarrow\mathbb{Z}^{n}$, under composition. Consider the homomorphism $f:C_{n}\longrightarrow\textrm{Map}(\mathbb{Z}^{n},\mathbb{Z}^{n})$, introduced in [2], which is defined on generators $x_{i}$ of $C_{n}$ as follows.

[TABLE]

If $w\in C_{n}$ then the components of $f(w)(m_{1},\ldots,m_{n})$ are polynomials in the variables $m_{1},\ldots,m_{n}$. Let $\operatorname{supp}(f(w))$ be the minimal subset $N$ of the set $M=\{1,\ldots,n\}$ such that for every $(m_{1},\ldots,m_{n})\in\mathbb{Z}^{n}$ the components of $f(w)(m_{1},\ldots,m_{n})$ are polynomials depending on the variables with indices from the set $N$. So $|\operatorname{supp}(f(w))|$ denotes the number of variables on which the value of $f(w)$ depends. For example, if $f(w)(m_{1},\ldots,m_{n})=(m_{1},\ldots,m_{i-1},m_{i+1},m_{i+1},\ldots,m_{n})$, then $\operatorname{supp}(f(w))=\{1,\ldots,i-1,i+1,\ldots,n\}$ and $|\operatorname{supp}(f(w))|=n-1$.

We now show that the value of $|\operatorname{supp}(f((x_{n}q_{i})^{k}))|$ does not depend on $k$.

Lemma 3.6**.**

For every $k\geqslant 1$ and $i=0,\ldots,n-2$ we have $|\operatorname{supp}(f((x_{n}q_{i})^{k}))|=n-i-1$.

Proof.

We show by induction that for every $k\geqslant 1$, if $k=m(n-i)+r$, where $0\leqslant r\leqslant n-i-1$, then $f((x_{n}q_{i})^{k})(m_{1},\ldots,m_{n})$ has the form

[TABLE]

where for $r=n-i-1$ the above expression is interpreted as

[TABLE]

This immediately yields the assertion.

First, notice that

[TABLE]

In particular, for $k=1$ we have $|\operatorname{supp}(f((x_{n}q_{i})^{k}))|=n-i-1$. Assume that the claim holds for $k-1$. Then, applying (3) and the induction hypothesis we see that $f((x_{n}q_{i})^{k})(m_{1},\ldots,m_{n})=f(x_{n}q_{i})(f((x_{n}q_{i})^{k-1}(m_{1},\ldots,m_{n})))$ has the form

[TABLE]

if $r\neq n-i-1$ and

[TABLE]

otherwise. This proves the inductive step. ∎

For every $i=-1,\ldots,n-2$ consider the following set

[TABLE]

Then $Q_{i}$ is an ideal in $C_{n}$ for $i<n-2$ because for every $x,y,w\in C_{n}$ we have $\operatorname{supp}(f(xw))\subseteq\operatorname{supp}(f(w))$ and $\operatorname{supp}(f(wy))\subseteq\operatorname{supp}(f(w))$. Thus, we get the following chain of ideals

[TABLE]

We will show that this chain is strongly related to the ideals $I_{j}$ introduced in this section.

Lemma 3.7**.**

For every $i=0,\ldots,n-2$ we have $Q_{i}\subseteq I_{i}$.

Proof.

Suppose that for some $i\in\{0,\ldots,n-2\}$ there exists $w$ such that $w\in Q_{i}\setminus I_{i}$. Then, by the definition of the ideals $I_{i}$ we get that for some $x,y\in C_{n}$ the element $xwy$ has the form $(x_{n}q_{i})^{k}$ for some $k\geqslant 1$. Then Lemma 3.6 implies that $|\operatorname{supp}(f(xwy))|=n-i-1$. On the other hand, $xwy\in Q_{i}$, and thus $|\operatorname{supp}(f(xwy))|\leqslant n-i-2$, which leads to a contradiction. ∎

The following technical lemma will be used in the proof of Theorem 3.10.

Lemma 3.8**.**

For all $n-1\geqslant j>i+1\geqslant 1$ we have

$x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}\in I_{i-1}$; 2. 2.

$w=x_{j}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{j+1}\in I_{i}$;

where for $j=n-1$ we put $w=x_{n-1}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}$.

Proof.

For $i\neq 0$, the first part is a direct consequence of Lemma 3.6 and Lemma 3.7. Moreover, $x_{n}x_{n-1}\cdots x_{1}\in I_{-1}$ by the definition of $I_{-1}$.

We will prove the second part. A direct computation shows that for every $(m_{1},\ldots,m_{n})\in\mathbb{Z}^{n}$ and for $j<n-1$ the element $f(x_{j}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{j+1})(m_{1},\ldots,m_{n})$ has the form

[TABLE]

Similarly, if $j=n-1$ then it is easy to see that $f(w)(m_{1},\ldots m_{n})$ has the form

[TABLE]

In both cases $|\operatorname{supp}(f(w))|=n-i-2$, so Lemma 3.7 implies that $w\in I_{i}$. ∎

The following is a direct consequence of the definition of the ideals $I_{i}$ and of Lemma 3.8.

Corollary 3.9**.**

For every $i\in\{0,\ldots,n-2\}$ we have $(x_{n}q_{i})^{k}\in I_{i-1}\setminus I_{i}$. Moreover, for all $i\in\{1,\ldots,n-2\}$ and $j\in\{i+1,\ldots,n-1\}$ we have $x_{j}\cdots x_{i+1}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{j+1}\in I_{i-1}\setminus I_{i}$.

Next, we prove a result implying that, after factoring the ideal $I_{i}$, the set $\tilde{M_{i}}\cup\{\theta\}$ becomes a semigroup. This will be crucial for the results of the next section.

Theorem 3.10**.**

Let $w\in C_{n}$. Then for every $i\in\{0,\ldots,n-2\}$ we have $(x_{n}q_{i})w(x_{n}q_{i})\in\{(x_{n}q_{i})^{k}:k\geqslant 2\}\cup I_{i}$.

Proof.

We proceed by induction on the length of the (reduced) word $w$. If $w$ is the empty word, the assertion is clear. Assume that $|w|>0$.

Let $\operatorname{pref}_{1}(w)=x_{j}$, so $w=x_{j}w^{\prime}$. If $j\leqslant i$ then $(x_{n}q_{i})w(x_{n}q_{i})\xrightarrow{(4j)}(x_{n}q_{i})w^{\prime}(x_{n}q_{i})$ and the assertion follows by induction. Hence, assume next that $n>j>i$. Then

[TABLE]

Lemma 3.8 implies that $x_{j-1}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{j}w^{\prime}(x_{n}q_{i})\in I_{i}$, so the assertion follows.

Hence, for the rest of the proof we may assume that $\operatorname{pref}_{1}(w)=x_{n}$. Then $w=x_{n}u_{1}x_{n}u_{2}\cdots x_{n}u_{k}$ for some $k\geqslant 1$ and some $u_{j}$ such that $|u_{j}|_{n}=0$.

If $u_{k}$ is the empty word, then $w(x_{n}q_{i})\xrightarrow{(1)}x_{n}u_{1}\cdots x_{n}u_{k-1}(x_{n}q_{i})$, and the assertion follows by the induction hypothesis. Hence, assume that $|u_{k}|\geqslant 1$ and the word $x_{n}u_{k}$ is in the reduced form. If $\operatorname{pref}_{1}(u_{k})=x_{j}$ ($u_{k}=x_{j}u_{k}^{\prime}$), then $j\in\{1,n-1\}$, since otherwise $x_{n}u_{k}\xrightarrow{(2)}x_{j}x_{n}u_{k}^{\prime}$, which contradicts the assumption that $w$ is in the reduced form. Therefore, we consider two cases:

$u_{k}=x_{n-1}u$, where $|u|_{n}=0$. Then, by Lemma 2.2 it follows that $u_{k}=x_{n-1}x_{n-2}\cdots x_{j}$ for some $n-1\geqslant j\geqslant 1$. Then $x_{n}u_{k}(x_{n}q_{i})=x_{n}x_{n-1}\cdots x_{j}(x_{n}q_{i})$. If $j>1$ then $x_{n}u_{k}(x_{n}q_{i})\xrightarrow{(5n)}u_{k}(x_{n}q_{i})$ and $(x_{n}q_{i})w(x_{n}q_{i})$ after this reduction still contains at least two factors of the form $x_{n}q_{i}$; so the assertion follows by the induction hypothesis. Similarly, if $j=1$ and $i\neq 0$ then we get $x_{n}u_{k}(x_{n}q_{i})\xrightarrow{(51)}x_{n}x_{n-1}\cdots x_{2}(x_{n}q_{i})$ and then the word $x_{n-1}\cdots x_{2}$ is shorter than $u_{k}$. On the other hand, if $j=1$ and $i=0$, then $x_{n}u_{k}(x_{n}q_{i})=(x_{n}q_{i})^{2}$ and the assertion also follows from the induction hypothesis. 2. 2.

$u_{k}=x_{1}u$, where $|u|_{n}=0$. Then by Lemma 2.3 we know that $x_{n}u_{k}=x_{n}x_{1}\cdots x_{j}$, where $1\leqslant j<n-1$ or $x_{n}u_{k}=x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{s}$ for some $1\leqslant r<s\leqslant n-1$.

Let $x_{n}u_{k}=x_{n}x_{1}\cdots x_{j}$ and consider the case where $j\leqslant i$. Then $x_{n}u_{k}(x_{n}q_{i})\xrightarrow{(5j)}x_{n}x_{1}\cdots x_{j-1}(x_{n}q_{i})$, the number of factors $x_{n}q_{i}$ after this reduction is at least $2$ and the assertion follows from the induction hypothesis. If $n-1>j>i+1$, then

[TABLE]

Similarly, if $j=i+1$, then

[TABLE]

and the assertion also follows.

Let $x_{n}u_{k}=x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{s}$, where $1\leqslant r<s\leqslant n-1$.

• •

If $s\leqslant i$, then $x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{s}(x_{n}q_{i})\xrightarrow{(5s)}x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{s+1}(x_{n}q_{i})$, and the obtained word also has at least two factors $x_{n}q_{i}$; so the assertion follows by induction.

• •

If $s=i+1$, then $r<i+1$. If additionally $r<i$, then $x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{s}(x_{n}q_{i})\xrightarrow{(5r)}x_{n}x_{1}\cdots x_{r-1}x_{n-1}\cdots x_{s}(x_{n}q_{i})$, and the obtained word still contains at least two factors $x_{n}q_{i}$; so the assertion follows by induction.

If $r=i$, then $x_{n}u_{k}(x_{n}q_{i})=(x_{n}q_{i})^{2}$ and we are done by induction.

• •

If $n-1>s>i+1$, then

[TABLE]

Similarly, if $n-1=s>i+1$, then

[TABLE]

Lemma 3.8 implies that $x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}\in I_{i}$. Therefore we also get $x_{n}x_{1}\cdots x_{r}x_{n-1}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{s+1}\in I_{i}$, and $w(x_{n}q_{i})\in I_{i}$.

This completes the proof. ∎

3.2 Structures of matrix type

Our next aim is to refine the information on the ideal chain

[TABLE]

of $C_{n}$ defined in the previous section. We will show that every factor $I_{j-1}/I_{j}$, for $j=0,\ldots,n-2$, is, up to finitely many elements, a semigroup of matrix type over a cyclic semigroup and also that $C_{n}/I_{-1}$ is finite. Namely, the elements of the family $\tilde{M}_{j}$, described in Definition 2.13, with a zero element adjoined, treated as elements of the Rees factor $I_{j-1}/I_{j}$, form a semigroup of matrix type. Using certain natural involutions on $C_{n}$, we will also show that the corresponding sandwich matrices are square matrices and they are symmetric. In particular, this means that, for every $j$, there is a bijection between the sets $A_{j}$ and $B_{j}$, which is not clear directly from the description obtained in Theorem 2.1.

Recall the definition of the sets $\tilde{M_{i}}$ and $\tilde{M}$ (Definition 2.13). For every $i=0,\ldots,n-2$ we write

[TABLE]

This is the set of elements considered in Theorem 2.1. In what follows, we identify elements of $C_{n}$ with the corresponding reduced words. Hence, $\tilde{M}=\bigcup_{i=0}^{n-2}\tilde{M_{i}}$ consists of elements of $C_{n}$ that have (in the reduced form) a factor of the form $x_{n}q_{i}$, for some $i$. Moreover, from Proposition 2.14 we know that almost all elements of $C_{n}$ are in this set.

Certain involutions that preserve the ideals $I_{i-1}$ and sets $\tilde{M}_{i}$, for $i\in\{0,\ldots,n-2\}$, will be useful in this context. In particular, they can be used to establish an internal symmetry of every set $\tilde{M}_{i}$.

Definition 3.11**.**

Let $\tau:C_{n}\longrightarrow C_{n}$ be the involution defined in Definition 3.3. So $\tau(x_{i})=x_{n-i}$ for $i\neq n$ and $\tau(x_{n})=x_{n}$. Let $\sigma:\langle x_{1},\ldots,x_{n}\rangle\longrightarrow\langle x_{1},\ldots,x_{n}\rangle$ be the automorphism such that $\sigma(x_{i})=x_{i+1}$ for every $i=1,\ldots,n$, where we put $x_{n+1}=x_{1}$. It is easy to check that $\sigma$ preserves the set of defining relations of $C_{n}$. Hence, $\sigma$ can be viewed as an automorphism of $C_{n}$. Therefore, the map $\sigma^{i}\tau$ also is an involution of $C_{n}$, for $i=0,\ldots,n-1$.

Corollary 3.12**.**

The involution $\chi_{i}=\sigma^{i+1}\tau$ satisfies: $\chi_{i}((x_{n}q_{i})^{k})=(x_{n}q_{i})^{k}$ for every $k\geqslant 1$, $\chi_{i}(A_{i})=B_{i},\chi_{i}(B_{i})=A_{i}$, and in particular $\chi_{i}(\tilde{M}_{i})=\tilde{M}_{i}$.

Proof.

Notice that $\chi_{i}(x_{n}x_{1}\cdots x_{i})=x_{1}\cdots x_{i+1}$ and $\chi_{i}(x_{n-1}\cdots x_{i+1})=x_{n}x_{n-1}\cdots x_{i+2}$. Hence $\chi_{i}(x_{n}q_{i})=(x_{n}x_{n-1}\cdots x_{i+2})(x_{1}\cdots x_{i+1})=x_{n}q_{i}$ in $C_{n}$, so for $k\geqslant 1$ we get $\chi_{i}((x_{n}q_{i})^{k})=(x_{n}q_{i})^{k}$ for every $i$. Let $w=(x_{n}q_{i})^{m}b\in\tilde{M}_{i}$, where $b\in B_{i}$ and $m\geqslant 1$. Then $\chi_{i}(w)=\chi_{i}(b)(x_{n}q_{i})^{m}$. Moreover, the reduced form of $\chi_{i}(b)(x_{n}q_{i})^{m}$ is obtained by moving certain generators to the left (other reductions are not possible because they reduce the length of the word, while $\chi_{i}^{2}=id$). But $x_{n}$ cannot be moved to the left, so the consecutive generators in $(x_{n}q_{i})^{m}$ cannot be moved to the left either. It follows that the reduced form of $\chi_{i}(w)$ is equal to $a(x_{n}q_{i})^{m}$, where $a$ is the reduced form of $\chi_{i}(b)$. If $a=u(x_{n}q_{i})^{r}$ for some $r\geqslant 1$ and some $u$, then $b=\chi_{i}(a)=(x_{n}q_{i})^{r}\chi_{i}(u)$ and $w=\chi_{i}(a(x_{n}q_{i})^{m})=(x_{n}q_{i})^{r+m}\chi_{i}(u)$. And again, since no generator can be moved into the word $x_{n}q_{i}$ from the right (without making the word shorter), it follows that we cannot have $r>0$. Thus, $a\in A_{i}$. Therefore, $\chi_{i}(B_{i})\subseteq A_{i}$. A symmetric argument shows that $\chi_{i}(A_{i})\subseteq B_{i}$. Since $\chi_{i}^{2}=id$, we get $\chi_{i}(B_{i})=A_{i}$ and $\chi_{i}(A_{i})=B_{i}$. The result follows. ∎

As noticed in Lemma 2.5, if $i=0$ or $i=n-2$, then reduced words in $C_{n}$ that have a factor of the form $x_{n}q_{i}$ must come from the infinite word $(x_{n}q_{i})^{\infty}$. It is then clear that for such a word $s$ we can find $w$,$z\in C_{n}$ such that $wsz\in\langle x_{n}q_{i}\rangle$. We will show that the latter property remains valid for all $i$.

Theorem 3.13**.**

Let $i\in\{0,\ldots,n-2\}$. Then:

for every $a\in A_{i}$ there exists $w\in C_{n}$ such that $wa\in\langle x_{n}q_{i}\rangle$; 2. 2)

for every $b\in B_{i}$ there exists $w\in C_{n}$ such that $bw\in\langle x_{n}q_{i}\rangle$.

Proof.

It is clear that the assertion holds for $i=0,n-2$. So, we will further assume that $i\in\{1,\ldots,n-3\}$. In view of Corollary 3.12 it is enough to prove assertion 2). We will use notation of Theorem 2.1. This includes the notion of blocks, introduced directly after the formulation of this theorem.

The description of the set $B_{i}$ implies directly that every $b\in B_{i}$ is a prefix of the following element

[TABLE]

Clearly, if the assertion holds for some word then it also holds for every prefix of this word. Thus, it is enough to prove the assertion in the case where $b=b^{\prime}$. We proceed by induction on the number of blocks in the word $b$ of this type; so on $r+1$. Let $r=0$. Then $b=x_{n}x_{n-1}\cdots x_{j_{1}}$ for some $j_{1}>i+1$. If we define $w=x_{1}\cdots x_{i}x_{j_{1}-1}\cdots x_{i+1}$ then $bw\xrightarrow{(2)}x_{n}x_{1}\cdots x_{i}x_{n-1}\cdots x_{i+1}$, as desired.

So, assume that the assertion holds for all words in $B_{i}$, that are of the form

[TABLE]

where $k<r$ (the number of blocks is $k+1$). Let $b$ be the following word with $r+1$ blocks:

[TABLE]

Let $w^{\prime}=x_{i_{1}+1}\cdots x_{i}x_{j_{1}-1}\cdots x_{i+1}$, where for $i_{1}=i$, we put $w^{\prime}=x_{j_{1}-1}\cdots x_{i+1}$. Since for $k=r,r+1$ we have inequalities $1<j_{k}-i<n-1$, using reduction (2) applied several times, it is easy to see that the word $bw^{\prime}$ can be reduced to

[TABLE]

Moreover, we know that $i_{r}+1<i_{1}+1<n-1$, so applying reduction (3) several times, we can rewrite the word $bw^{\prime}$ to the form

[TABLE]

Repeating this process, alternately using the inequalities $1<j_{k}-i<n-1$ for $k>1$ (and applying reduction (2)), and using the inequalities $i_{k}+1<i_{1}+1<n-1$ for $k>1$ (and applying reduction (3)), we come to the following form of the element $bw^{\prime}$:

[TABLE]

Similarly, applying alternately inequalities $1<m-(j_{1}-1)<n-1$ for $m=j_{2},\ldots,n$ (and reduction (2)) and inequalities $i_{k}+1<m+1<n-1$ for $m=i+1,\ldots,j_{1}-1$, $k=2,\ldots,r$ (and reduction (3) applied several times) we can finally rewrite the word $bw^{\prime}$ to the form

[TABLE]

Then $b^{\prime}=x_{n}(x_{1}\cdots x_{i_{2}}x_{n-1}\cdots x_{j_{2}})\cdots x_{n}(x_{1}\cdots x_{i_{r}}x_{n-1}\cdots x_{j_{r}})x_{n}x_{n-1}\cdots x_{j_{r+1}}\in B_{i}$. Thus, by the inductive hypothesis, there exists $v\in C_{n}$ such that $b^{\prime}v\in\langle x_{n}q_{i}\rangle$. Hence, for $w=w^{\prime}v$ we have $bw\in\langle x_{n}q_{i}\rangle$. This completes the proof. ∎

Next we show that for every $i$ the set $\tilde{M_{i}}$ is contained in $I_{i-1}\setminus I_{i}$.

Proposition 3.14**.**

For every $i\in\{0,\ldots,n-2\}$ we have $\tilde{M_{i}}\subseteq I_{i-1}\setminus I_{i}$.

Proof.

Since $I_{i}$ is an ideal of $C_{n}$, Lemma 3.8 implies that $\tilde{M_{i}}\subseteq I_{i-1}$. Let $a(x_{n}q_{i})^{k}b\in\tilde{M_{i}}$. Theorem 3.13 implies that there exist $w,v\in C_{n}$ such that $wa,bv\in\langle x_{n}q_{i}\rangle$. So, $wa(x_{n}q_{i})^{k}bv\in\langle x_{n}q_{i}\rangle$ and $a(x_{n}q_{i})^{k}b\notin I_{i}$. The assertion follows. ∎

Corollary 3.15**.**

For every $i\in\{0,\ldots,n-2\}$ the set $(I_{i-1}\setminus I_{i})\setminus\tilde{M_{i}}$ is finite.

Proof.

By Proposition 3.14, $\tilde{M_{i}}\subseteq I_{i-1}\setminus I_{i}$. Let $j>i$. Then $I_{j-1}\subseteq I_{i}$, so that $(I_{i-1}\setminus I_{i})\cap(I_{j-1}\setminus I_{j})=\emptyset$. In particular, for $j\neq i$ we get $\tilde{M_{j}}\cap(I_{i-1}\setminus I_{i})=\emptyset$. Since from Proposition 2.14 we know that $C_{n}\setminus\tilde{M}$ is finite, it follows that $(I_{i-1}\setminus I_{i})\setminus\tilde{M_{i}}$ is also finite. ∎

By Proposition 2.14 we know that $C_{n}\setminus\tilde{M}$ is a finite set. Moreover, Proposition 3.14 implies that for every $i=0,\ldots,n-2$ we have $\tilde{M_{i}}\subseteq I_{i-1}\setminus I_{i}\subseteq I_{-1}$, so that also $\tilde{M}\subseteq I_{-1}$. Our next observation follows.

Corollary 3.16**.**

$C_{n}/I_{-1}$* is a finite semigroup.*

The following two simple lemmas will be also useful.

Lemma 3.17**.**

For every $i\in\{0,\ldots,n-3\}$ we have $\sigma(I_{i})=I_{i}$. Moreover, $\sigma(w)\in\tilde{M}_{i}$ for almost all $w\in\tilde{M}_{i}$, if $i\in\{0,\ldots,n-2\}$.

Proof.

Clearly, $\sigma^{n}=id$ and $\sigma:C_{n}\longrightarrow C_{n}$ is an automorphism. Hence, for the first assertion, it is enough to prove that $\sigma(C_{n}\setminus I_{i})\subseteq C_{n}\setminus I_{i}$. Let $w\in C_{n}\setminus I_{i}$. Then there exist $u,v\in C_{n}$ such that $uwv=(x_{n}q_{i})^{m}$, where $m\geqslant 1$. A direct computation shows that $\sigma((x_{n}q_{i})^{m})=(x_{1}x_{2}\cdots x_{i+1}x_{n}x_{n-1}\cdots x_{i+2})^{m}\xrightarrow{(2)}(x_{1}\cdots x_{i+1})(x_{n}q_{i})^{m-1}(x_{n}\cdots x_{i+2})$. In particular, $x_{1}\cdots x_{i+1}\in A_{i}$, $x_{n}\cdots x_{i+2}\in B_{i}$, so from Theorem 3.13 we know that there exist $u^{\prime},v^{\prime}\in C_{n}$ such that $u^{\prime}(x_{1}\cdots x_{i+1})(x_{n}q_{i})^{m-1}(x_{n}\cdots x_{i+2})v^{\prime}\in\langle x_{n}q_{i}\rangle$. Then for $p=u^{\prime}\sigma(u)$ and $q=\sigma(v)v^{\prime}$ we get $p\sigma(w)q\in\langle x_{n}q_{i}\rangle$, so that $\sigma(w)\in C_{n}\setminus I_{i}$, as desired.

The second assertion follows now directly from Corollary 3.15 (if $i>0$) and from Corollary 3.16 (if $i=0$). ∎

Lemma 3.18**.**

For every $i\in\{1,\ldots,n-2\}$ and every nonnegative integer $m$ we have $\sigma^{m}\tau(I_{i-1}\setminus I_{i})=I_{i-1}\setminus I_{i}$ and $\sigma^{m}\tau(I_{i-1})=I_{i-1}$.

Proof.

We have seen in the proof of Corollary 3.12 that $\sigma^{i}\tau((x_{n}q_{i-1})^{k})=(x_{n}q_{i-1})^{k}$ for $k\geqslant 1$.

Let $w\in C_{n}$ be such that $\sigma^{i}\tau(w)\notin I_{i-1}$. Then, by the definition of $I_{i-1}$ there exist $x,y\in C_{n}$ such that $x\sigma^{i}\tau(w)y=(x_{n}q_{i-1})^{j}$, for some $j$. Then $\sigma^{i}\tau(y)w\sigma^{i}\tau(x)=(x_{n}q_{i-1})^{j}$, so in particular $w\notin I_{i-1}$. This shows that $\sigma^{i}\tau(I_{i-1})\subseteq I_{i-1}$.

Since $\sigma^{n}=id$, from Lemma 3.17 it now follows that $\tau(I_{i-1})\subseteq I_{i-1}$, and thus $\tau(I_{i-1})=I_{i-1}$ because $\tau$ is an involution. Therefore, $\sigma^{m}\tau(I_{i-1})=I_{i-1}$ for every nonnegative $m$. The assertion follows. ∎

In the two extreme cases, namely for $i=0$ and $i=n-2$, the description of $M_{i}$ is quite simple (see Lemma 2.5). In particular, $\tilde{M}_{n-2}$ coincides with the set of all factors of the word $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$, that contain a factor $x_{n}x_{1}\cdots x_{n-1}$. Moreover, our earlier results lead to the following consequence.

Corollary 3.19**.**

$\tilde{M}_{n-2}=C_{n}(x_{n}x_{1}\cdots x_{n-1})C_{n}$.

Proof.

The definition implies that $\tilde{M}_{n-2}\subseteq C_{n}(x_{n}x_{1}\cdots x_{n-1})C_{n}$. By Lemmas 3.1 and 3.4 we know that every element of $C_{n}(x_{n}x_{1}\cdots x_{n-1})C_{n}$ has reduced form that is a factor of $(x_{n}x_{1}\cdots x_{n-1})^{\infty}$. Moreover, the second parts of both lemmas imply that if $w(x_{n}x_{1}\cdots x_{n-1})v\in C_{n}(x_{n}x_{1}\cdots x_{n-1})C_{n}$, then the reduced form of this word has a factor $x_{n}x_{1}\cdots x_{n-1}$. ∎

In the second extreme case, namely when $i=0$, we have $\tilde{M}_{0}\subseteq C_{n}(x_{n}q_{0})C_{n}\subseteq I_{-1}$. Moreover, equality holds modulo the ideal $I_{0}$, as proved in the following lemma.

Lemma 3.20**.**

$I_{-1}=\tilde{M_{0}}\cup I_{0}$.

Proof.

By the definition, $I_{-1}=I_{0}\cup C_{n}(x_{n}q_{0})C_{n}$. It is clear that $\tilde{M}_{0}\subseteq C_{n}(x_{n}q_{0})C_{n}$, so also $\tilde{M}_{0}\cup I_{0}\subseteq I_{-1}$. Since $x_{n}q_{0}\in\tilde{M}_{0}$, it is enough to prove that $\tilde{M}_{0}\cup I_{0}$ is an ideal in $C_{n}$. Since $\sigma\tau(\tilde{M}_{0})=\tilde{M}_{0}$ and $\sigma\tau(I_{0})=I_{0}$ by Corollary 3.12 and Lemma 3.18, it is enough to check that $\tilde{M}_{0}\cup I_{0}$ is a right ideal of $C_{n}$. So, for every $w\in\tilde{M}_{0}\cup I_{0}$ and $x_{j}\in\{x_{1},\ldots,x_{n}\}$ we have $wx_{j}\in\tilde{M}_{0}\cup I_{0}$. If $w\in I_{0}$, then $wx_{j}\in I_{0}$, because $I_{0}$ is an ideal of $C_{n}$.

So, let $w=a(x_{n}x_{n-1}\cdots x_{1})^{m}b\in\tilde{M_{0}}$, where $a\in A_{0},b\in B_{0}$. From Theorem 2.1 we know that $b=x_{n}x_{n-1}\cdots x_{l}$ for some $l\in\{2,\ldots,n\}$ or $b$ is the empty word. First, notice that by Lemma 3.8, for $j\neq 1,n$, we have

[TABLE]

Clearly, for $j=1,n$ we have $(x_{n}x_{n-1}\cdots x_{1})x_{j}\in\tilde{M_{0}}$. This implies that the assertion holds if $b$ is the empty word.

Hence, assume that $b=x_{n}x_{n-1}\cdots x_{l}$ for some $l$. We will show that $(x_{n}x_{n-1}\cdots x_{1})x_{n}x_{n-1}\cdots x_{l}x_{j}\in I_{0}$ for every $j\neq l-1,l$. This will prove the result because $I_{0}$ is an ideal in $C_{n}$. Notice that for $j\neq l-1,l$ we have

[TABLE]

where $\xrightarrow{(*)}$ denotes $\xrightarrow{(5j)}$, $\xrightarrow{(2)}$ and $\xrightarrow{(5j)}$, respectively.

In each case, $(x_{n}q_{0})x_{n}x_{n-1}\cdots x_{l}x_{j}\in I_{0}$. In the first case, this is a consequence of the fact that $I_{0}$ is an ideal in $C_{n}$ and $j-1\neq 1,n$, so the previous computation implies that $(x_{n}q_{0})x_{j}\in I_{0}$. Similarly, in the second case $1<j<l-1\leqslant n-1$, so that also $(x_{n}q_{i})x_{j}\in I_{0}$. In the third case, from Lemma 3.8 and since $I_{0}$ is an ideal we know that $x_{n-1}\cdots x_{2}x_{n}x_{1}x_{n-1}\cdots x_{l}\in I_{0}$. The result follows. ∎

We are now in a position to improve the assertion of Theorem 3.13.

Corollary 3.21**.**

Let $i\in\{0,\ldots,n-2\}$. Then

for every $a\in A_{i}$ there exists $w\in\tilde{M_{i}}$ such that $wa\in\langle x_{n}q_{i}\rangle$, 2. 2.

for every $b\in B_{i}$ there exists $v\in\tilde{M_{i}}$ such that $bv\in\langle x_{n}q_{i}\rangle$.

Consequently, $x\mapsto wx$ is an injective map $a\langle x_{n}q_{i}\rangle B_{i}\longrightarrow\langle x_{n}q_{i}\rangle B_{i}$, and $x\mapsto xv$ is an injective map $A_{i}\langle x_{n}q_{i}\rangle b\longrightarrow A_{i}\langle x_{n}q_{i}\rangle$.

Proof.

We only prove the first part; the second part then follows from Corollary 3.12. Let $a\in A_{i}$. From Theorem 3.13 we know that there exists $u\in C_{n}$ such that $ua=(x_{n}q_{i})^{k}$ for some $k\geqslant 1$. The elements $(x_{n}q_{i})^{m}u\in C_{n}$, where $m\geqslant 1$ are pairwise different. Indeed, suppose that for some $m_{1},m_{2}\geqslant 1$ in $C_{n}$ we have $(x_{n}q_{i})^{m_{1}}u=(x_{n}q_{i})^{m_{2}}u$. Then also $(x_{n}q_{i})^{m_{1}}ua=(x_{n}q_{i})^{m_{2}}ua$, so that $(x_{n}q_{i})^{m_{1}+k}=(x_{n}q_{i})^{m_{2}+k}$, and consequently $m_{1}=m_{2}$. By Proposition 2.14 there exists $m\geqslant 1$ such that $(x_{n}q_{i})^{m}u\in\tilde{M}$. According to Lemma 3.8, for every $m\geqslant 1$ we have $(x_{n}q_{i})^{m}\in I_{i-1}\setminus I_{i}$. Since $I_{i-1}$ is an ideal of $C_{n}$, we get $(x_{n}q_{i})^{m}u\in I_{i-1}$. If $(x_{n}q_{i})^{m}u\in I_{i}$, then also $(x_{n}q_{i})^{m}ua=(x_{n}q_{i})^{m+k}\in I_{i}$, a contradiction. Therefore, by Proposition 3.14, $w=(x_{n}q_{i})^{m}u\in\tilde{M}\cap(I_{i-1}\setminus I_{i})=\tilde{M}_{i}$, and the assertion follows. ∎

Notice that $\tilde{M_{i}}\subseteq C_{n}$, for $i<n-2$, is not closed under multiplication. For example, consider $u=(x_{n}q_{i})x_{n}x_{n-1}$, $w=x_{n}q_{i}$ for some $1\leqslant i\leqslant n-3$. It is easy to check that the reduced form of $uw$ is equal to $x_{n-2}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}x_{n-1}\cdots x_{i+2}x_{n}x_{1}\cdots x_{i+1}$, so that indeed $uw\notin\tilde{M_{i}}$. However, from Theorem 3.10 we know that if $u=a(x_{n}q_{i})^{k}b,w=a^{\prime}(x_{n}q_{i})^{k^{\prime}}b^{\prime}\in\tilde{M_{i}}$, then either $uw=a(x_{n}q_{i})^{m}b^{\prime}\in\tilde{M}_{i}$ for some $m\geqslant 2$ or $uw\in I_{i}$.

We will define semigroups $M_{i}$, with properties described in the beginning of this section.

Definition 3.22**.**

Let $i\in\{0,\ldots,n-2\}$. Consider the set $M_{i}=\tilde{M_{i}}\cup\{\theta\}$ with operation, defined for any $u=a(x_{n}q_{i})^{k}b,w=a^{\prime}(x_{n}q_{i})^{k^{\prime}}b^{\prime}\in\tilde{M_{i}}$ by

[TABLE]

and $w\theta=\theta w=\theta$ for every $w\in M_{i}$. Then, by Theorem 3.10 the definition is correct and $M_{i}$ is a semigroup under this operation.

These semigroups can be interpreted as Rees factor semigroups. Namely, for $i\leqslant n-3$, $I_{i}$ is an ideal of $C_{n}$, and we may consider the factor semigroup $C_{n}/I_{i}$. In other words, $C_{n}/I_{i}$ is the semigroup $(C_{n}\setminus I_{i})\cup\{\theta\}$ with zero $\theta$ and with operation

[TABLE]

While $I_{n-2}=\emptyset$, for every subsemigroup $J$ of $C_{n}$ we define $J/I_{n-2}=J^{0}$; the semigroup $J$ with zero adjoined. Notice that $J_{i}=\tilde{M}_{i}\cup I_{i}$ is a subsemigroup of $I_{i-1}$ by Theorem 3.10 and Proposition 3.14. Thus, our definition yields $M_{i}=J_{i-1}/I_{i}\subseteq C_{n}/I_{i}$.

From Lemma 3.20 we know that $I_{-1}/I_{0}=M_{0}$. Moreover, Lemma 3.1 implies that $\tilde{M}_{n-2}=M_{n-2}\setminus\{\theta\}$ is an ideal in $C_{n}$.

We will show that all $M_{i}$ defined above are semigroups of matrix type, see [14], Section 5. Let $S_{i}$ denote the cyclic semigroup generated by $x_{n}q_{i}$.

Corollary 3.23**.**

$M_{i}$* is a semigroup of matrix type. Namely, $M_{i}\cong\mathcal{M}^{0}(S_{i},A_{i},B_{i};P_{i})$, where $P_{i}$ is a matrix of size $B_{i}\times A_{i}$ with coefficients in $\langle x_{n}q_{i}\rangle\cup\{\theta\}$.*

Proof.

Let $P_{i}=(p_{ba})$ be the matrix defined as follows

[TABLE]

We define $\phi:M_{i}\longrightarrow\mathcal{M}^{0}(S_{i},A_{i},B_{i};P_{i})$ by the formula $\phi(a(x_{n}q_{i})^{k}b)=((x_{n}q_{i})^{k};a,b)$ and $\phi(\theta)=\theta$. From the uniqueness of the reduced forms of elements of $M_{i}$ and by Proposition 3.14 this function is well defined. It is clear that $\phi$ is bijective. Moreover, we claim that $\phi$ is a homomorphism. Indeed, let $u=a(x_{n}q_{i})^{k}b,w=a^{\prime}(x_{n}q_{i})^{k^{\prime}}b^{\prime}\in M_{i}$.

If $uw\notin I_{i}$, then $(x_{n}q_{i})ba^{\prime}(x_{n}q_{i})=(x_{n}q_{i})^{\alpha}$ for some $\alpha\geqslant 2$ and then

[TABLE]

If $uw\in I_{i}$ then it is easy to see that $\phi(uw)=\theta=\phi(u)\phi(w)$, because $p_{ba^{\prime}}=\theta$. This completes the proof. ∎

Remark 3.24**.**

Assume that $x_{n}q_{i}bax_{n}q_{i}=(x_{n}q_{i})^{\alpha}$ for some $a\in A_{i},b\in B_{i}$. Then

[TABLE]

By Corollary 3.12, $\chi_{i}$ determines a bijection between the sets $A_{i}$ and $B_{i}$. Hence, from (4) in the proof of Corollary 3.23 it follows that the matrix $P_{i}$ is symmetric, if the ordering of the elements of the set $A_{i}$ corresponds to the ordering of their images under $\chi_{i}$ (see the examples in Section 4).

The main results of this section can be now summarized as follows.

Corollary 3.25**.**

$C_{n}$* has a chain of ideals*

[TABLE]

with the following properties

for $i=0,\ldots,n-2$ there exist semigroups of matrix type $M_{i}=\mathcal{M}^{0}(S_{i},A_{i},B_{i};P_{i})$, such that $M_{i}\subseteq I_{i-1}/I_{i}$, where $S_{i}$ is the cyclic semigroup generated by $x_{n}q_{i}$, $P_{i}$ is a square symmetric matrix of size $B_{i}\times A_{i}$ and with coefficients in $\langle x_{n}q_{i}\rangle\cup\{\theta\}$; 2. 2.

for $i=1,\ldots,n-2$ the sets $(I_{i-1}/I_{i})\setminus M_{i}$ are finite; 3. 3.

$I_{-1}/I_{0}=M_{0}$; 4. 4.

$\tilde{M}_{n-2}=M_{n-2}\setminus\{\theta\}\vartriangleleft C_{n}$; 5. 5.

$C_{n}/I_{-1}$* is a finite semigroup.*

4 Examples

In this section we illustrate our results with the simplest cases of Hecke–Kiselman monoids $C_{n}$ associated to cyclic graphs, for $n=3$ and $4$.

4.1 Cycle of length $3$

For simplicity, write $x_{1}=a$, $x_{2}=b$, $x_{3}=c$. Recall that

[TABLE]

The following is a direct consequence of Corollary 1.2.

Lemma 4.1**.**

The reduced form of every element of $C_{3}$ is a factor of one of the following infinite words: $(cab)^{\infty}$, $(cba)^{\infty}$.

Taking $n=3$ in Theorem 2.1, we get that $\tilde{M_{0}}$ consists of all factors of $(cba)^{\infty}$, that have $cba$ as a subfactor. Similarly, $\tilde{M_{1}}$ consists of all factors of $(cab)^{\infty}$ that have a subfactor $cab$.

According to Corollary 3.25, $C_{3}$ has an ideal chain

[TABLE]

where $I_{0}=\{w\in C_{3}:C_{3}wC_{3}\cap\langle cba\rangle=\emptyset\}$ and $I_{-1}=I_{0}\cup\tilde{M_{0}}$.

Lemma 4.2**.**

Let $T$ be the cyclic semigroup generated by $t=cab$. Then $M_{1}=(C_{3}cabC_{3})^{0}$ is a semigroup of matrix type $\mathcal{M}^{0}(T,A_{1},B_{1};P_{1})$, where $A_{1}=\{1,b,ab\}$, $B_{1}=\{1,c,ca\}$, with sandwich matrix (with coefficients in $T^{1}$)

[TABLE]

Similarly, for $I_{-1}$ we have

Lemma 4.3**.**

Let $S$ be the cyclic semigroup generated by $s=cba$. Then the semigroup $M_{0}=I_{-1}/I_{0}$ is a semigroup of matrix type $\mathcal{M}^{0}(S,A_{0},B_{0};P_{0})$, $A_{0}=\{1,a,ba\}$, $B_{0}=\{1,c,cb\}$, with sandwich matrix (with coefficients in $S^{1}\cup\{\theta\}$)

[TABLE]

Recall that the rows of $P_{i}$ are indexed by the set $B_{i}$, and columns by the set $A_{i}$. For simplicity, we identify the elements of these sets with $1,2,3$, in the order in which these elements were listed. For example, the $(3,2)$-entry of the sandwich matrix $P_{0}$ corresponds to the pair $(cb,a)$.

The above two lemmas follow directly from Corollary 3.23. To indicate computations that are used to determine the coefficients of the sandwich matrices, let us focus on $P_{0}$. For simplicity, if $\alpha\in A_{0}$, $\beta\in B_{0}$, then we write

[TABLE]

that is if $\overline{p_{\beta\alpha}}=s^{k}$, then $p_{\beta\alpha}=s^{k-2}$ and if $\overline{p_{\beta\alpha}}=\theta$, then also $p_{\beta\alpha}=\theta$. Then, for example $\overline{p_{(cb)(ba)}}=(cba)cbba(cba)\xrightarrow{(1)}(cba)ca(cba)\xrightarrow{(2)}(cba)^{3}$. So, $p_{(cb)(ba)}=s$.

We derive the following consequence for the algebras $K_{0}[M_{0}]$ and $K_{0}[M_{1}]$.

Corollary 4.4**.**

Algebras $K_{0}[M_{0}]$ and $K_{0}[M_{1}]$ are of matrix type. Namely, $K_{0}[M_{1}]=\mathcal{M}(K[T],A_{1},B_{1};P_{1})$ and $K_{0}[M_{0}]=\mathcal{M}(K[S],A_{0},B_{0};P_{0})$, where $T$ and $S$ are the cyclic semigroups generated by $t=cab$ and $s=cba$, respectively.

It is easy to see that $\det P_{1}=-(t-1)^{2}\neq 0$ and $\det P_{0}=-s(s+1)\neq 0$, whence $P_{1}$ and $P_{0}$ are not zero divisors in $M_{3}(K[T])$ and $M_{3}(K[S])$. Using standard results (see [14], Chapter 5), we get

Corollary 4.5**.**

Semigroup algebras $K_{0}[M_{0}]$ and $K_{0}[M_{1}]$ are prime.

Due to the very special form of all reduced words in $C_{3}$ one can also prove the following result.

Theorem 4.6**.**

$K[C_{3}]$* is a semiprime algebra.*

Proof.

By $F$ we denote the free monoid generated by $a,b,c$. Write $R=K[C_{3}]$ and suppose that $x\in R$ is a nonzero element such that $xRx=0$. Then $x$ can be uniquely written in the form

[TABLE]

for some disjoint index sets $I$, $J$, where $\sigma_{i},\tau_{j}\neq 0$ are elements of the field $K$, $n_{i},l_{j}\geqslant 0$, $u_{i}\in\{1,a,ba\}$, $v_{i}\in\{1,c,cb\}$, $w_{j}\in\{1,b,ab\}$, $z_{j}\in\{1,c,ca\}$. In $F$ consider the deg-lex order induced by $a<b<c$. Let $u_{0}\alpha^{m}v_{0}$ ($\alpha\in\{cba,cab\}$) be the leading term in the support of $x$. We may assume that its coefficient is equal to $1$. Notice that $u_{0}$ and $v_{0}$ must be a suffix, and a prefix respectively, of $\alpha$. Hence, there exist words $p,q$ such that $pu_{0}=v_{0}q=\alpha$ holds in the free monoid $F$. Then for all elements $w\neq u_{0}\alpha^{m}v_{0}$ in the support of $x$ we have $\alpha^{m+2}=pu_{0}\alpha^{m}v_{0}q>pwq$ in $F$. If $xRx=0$, then also $pxqRpxq=0$. In particular, $(pxq)^{2}=0$. On the other hand, we know that in $K[C_{3}]$

[TABLE]

where $i\in(I\sqcup J)\setminus\{0\}$, $\rho_{i}\in K$, $y_{i}$ is the reduced form of the word $pu_{i}(cba)^{n_{i}}v_{i}q$ if $i\in I$, and if $i\in J$, then $y_{i}$ is the reduced form of $pw_{i}(cab)^{l_{i}}z_{i}q$. In particular, for every $y_{i}\neq\alpha^{k+2}$ we have $y_{i}<\alpha^{m+2}$.

Since $\alpha^{m+2}\alpha^{m+2}$ has reduced form $\alpha^{2(m+2)}$, for every $l,n\in I\sqcup J$ such that $(y_{l},y_{n})\neq(\alpha^{m+2},\alpha^{m+2})$ the reduced form $y_{ln}$ of the word $y_{l}y_{n}$ satisfies $y_{ln}<\alpha^{2(m+2)}$. In particular, the leading term of $(pxq)^{2}$ is equal to $\alpha^{2(m+2)}$ and it is nonzero. This contradiction shows that $K[C_{3}]$ is semiprime. ∎

Remark 4.7**.**

The argument used in the above proof also shows that for every $n\geqslant 3$ the algebras of matrix type $K_{0}[M_{0}]$ and $K_{0}[M_{n-2}]$ defined for $K[C_{n}]$ are semiprime (actually, prime). Indeed, $A_{0},B_{0}$ consist of all suffixes and prefixes, respectively, of the word $x_{n}x_{n-1}\cdots x_{1}$. While $A_{n-2},B_{n-2}$ consist of all suffixes and prefixes, respectively, of $x_{n}x_{1}\cdots x_{n-1}$, so the argument used in the proof can be applied.**

4.2 Cycle of length $4$

For simplicity, we write $x_{1}=a$, $x_{2}=b$, $x_{3}=c$, $x_{4}=d$. Recall that $C_{4}$ has the following presentation

[TABLE]

The form of the sets $A_{0},B_{0},A_{1},B_{1},A_{2},B_{2}$ follows directly from Theorem 2.1.

Lemma 4.8**.**

If an element of $C_{4}$ has a factor of the form $w_{0}=dcba$, $w_{1}=dacb$ or $w_{2}=dabc$ then it is of the form $\alpha_{i}w_{i}^{k}\beta_{i}$, with $k\geqslant 1$, $\alpha_{i}\in A_{i}$, $\beta_{i}\in B_{i}$, where

$A_{0}=\{1,a,ba,cba\}$, $B_{0}=\{1,d,dc,dcb\}$; 2. 2.

$A_{1}=\{1,b,cb,acb,ab,bacb\}$, $B_{1}=\{1,d,da,dac,dc,dacd\}$; 3. 3.

$A_{2}=\{1,c,bc,abc\}$, $B_{2}=\{1,d,da,dab\}$.

From Corollary 3.25 we know that $C_{4}$ has a chain of ideals

[TABLE]

with semigroups of matrix type $M_{0}$, $M_{1}$ and $M_{2}$, coming from Corollary 3.23, such that

$M_{2}=(C_{4}dabcC_{4})^{0}\subseteq I_{1}/I_{2}$ and the set $(I_{1}/I_{2})\setminus M_{2}$ is finite, 2. 2.

$M_{1}=\{\alpha(dacb)^{k}\beta:\alpha\in A_{1},\beta\in B_{1},k\geqslant 1\}\cup\{\theta\}\subseteq I_{0}/I_{1}$ and the set $(I_{0}/I_{1})\setminus M_{1}$ is finite, 3. 3.

$M_{0}=\{\alpha(dcba)^{k}\beta:\alpha\in A_{0},\beta\in B_{0},k\geqslant 1\}\cup\{\theta\}=I_{-1}/I_{0}$, 4. 4.

$C_{4}\setminus I_{-1}$ is finite, where $I_{-1}=I_{0}\cup C_{4}dcbaC_{4}$.

We present these structures of matrix type below. A simple verification is left to the reader.

Lemma 4.9**.**

Let $S_{2}$ be the cyclic semigroup generated by $s=dabc$. Then the ideal generated by $s$ in $C_{4}$, with a zero adjoined, that is $M_{2}=(C_{4}dacbC_{4})^{0}$, is a semigroup of matrix type $\mathcal{M}^{0}(S_{2},A_{2},B_{2};P_{2})$, where $A_{2}=\{1,c,bc,abc\}$, $B_{2}=\{1,d,da,dab\}$, with sandwich matrix (with coefficients in $S_{2}^{1}$)

[TABLE]

Lemma 4.10**.**

Let $S_{1}$ be the cyclic semigroup generated by $s=dacb$. Then $M_{1}$ is a semigroup of matrix type $\mathcal{M}^{0}(S_{1},A_{1},B_{1};P_{1})$ where $A_{1}=\{1,b,cb,acb,ab,bacb\}$, $B_{1}=\{1,d,dc,dac,da,dacd\}$, with sandwich matrix (with coefficients in $S_{1}^{1}\cup\{\theta\}$)

[TABLE]

Lemma 4.11**.**

Let $S$ be the cyclic semigroup generated by $s=dcba$. Then $M_{0}$ is a semigroup of matrix type $\mathcal{M}^{0}(S_{0},A_{0},B_{0};P_{0})$, where $s=dcba$, $A_{0}=\{1,a,ba,cba\}$, $B_{0}=\{1,d,dc,dcb\}$, with sandwich matrix

[TABLE]

We get the following consequence.

Corollary 4.12**.**

Algebras $K_{0}[M_{2}]$, $K_{0}[M_{1}]$ and $K_{0}[M_{0}]$ are algebras of matrix type. Namely, $K_{0}[M_{2}]=\mathcal{M}(K[S_{2}],A_{2},B_{2};P_{2})$, $K_{0}[M_{1}]=\mathcal{M}(K[S_{1}],A_{1},B_{1};P_{1})$ and $K_{0}[M_{0}]=\mathcal{M}(K[S_{0}],A_{0},B_{0};P_{0})$, where $S_{2},S_{1},S_{0}$ are the cyclic semigroups generated by $s_{2}=dabc$, $s_{1}=dacb$, and by $s_{0}=dcba$, respectively.

A direct computation shows that $\det P_{2}=-(s_{2}-1)^{3}\neq 0.$ Similarly, one can see that $\det P_{1}=-s_{1}^{3}(s_{1}+1)^{3}\neq 0$ and $\det P_{0}=-s_{0}^{2}(s_{0}-1)\neq 0$, so that the matrices $P_{i}$ are not zero divisors in the corresponding matrix rings $M_{n_{i}}(K[S_{i}])$, for $i\in\{0,1,2\}$. Therefore, by [14], Chapter 5, we get

Corollary 4.13**.**

Semigroup algebras $K_{0}[M_{2}]$, $K_{0}[M_{1}]$ and $K_{0}[M_{0}]$ are prime.

The assertion of Remark 4.7, Corollary 4.5 and Corollary 4.13 will be extended in Theorem 5.8 to all algebras $K_{0}[M_{t}]$ coming from monoids $C_{n}$, $n\geqslant 3$. This will be crucial for our main results in Section 5. However, the proof is much more complicated since the determinants of the corresponding matrices cannot be easily computed.

5 Noetherian Hecke–Kiselman algebras

In this section we characterize Noetherian Hecke–Kiselman algebras $K[\operatorname{HK}_{\Theta}]$ of arbitrary oriented graphs $\Theta$. The main difficulty is in proving that all algebras $K[C_{n}]$ are Noetherian. We start, however, with describing some simple obstacles to the Noetherian property. For $w,w^{\prime}\in F$ we write $w\sim w^{\prime}$ if $w,w^{\prime}$ represent the same element of $\operatorname{HK}_{\Theta}$.

Lemma 5.1**.**

Let $\Theta$ be the graph obtained by adjoining the arrow $y\rightarrow x_{1}$ to the cyclic graph $C_{n}$: $x_{1}\rightarrow x_{2}\rightarrow\cdots\rightarrow x_{n}\rightarrow x_{1}$. Then the monoid $\operatorname{HK}_{\Theta}$ does not satisfy the ascending chain condition on left ideals, and it does not satisfy the ascending chain condition on right ideals.

Proof.

Write $w_{k}=(x_{n}x_{n-1}\cdots x_{1})^{k}y$, for $k=1,2,\ldots$. It is clear that $w_{k}$ cannot be rewritten in the monoid $\operatorname{HK}_{\Theta}$ except for applying relations of the form $x_{i}^{2}=x_{i},y^{2}=y$. Therefore $w_{k}\notin\bigcup_{i=1}^{k-1}w_{i}\operatorname{HK}_{\Theta}$ for $k\geqslant 2$. Hence, $\operatorname{HK}_{\Theta}$ does not satisfy acc on right ideals.

Let $\phi:\langle x_{1},\ldots,x_{n},y\rangle\longrightarrow\langle x_{1},\ldots,x_{n}\rangle$ be the homomorphism such that $\phi(w)$ is obtained from $w$ by erasing all occurrences of $y$. Consider the following subsets of the free monoid $F=\langle x_{1},\ldots,x_{n},y\rangle$: $Z_{k}=\{\phi^{-1}((x_{1}^{i_{1}}x_{n}^{i_{n}}\cdots x_{2}^{i_{2}})^{k})\mid i_{j}\geqslant 1\mbox{ for every }j\}$, for $k=1,2,\ldots$, and

[TABLE]

We claim that $R_{k}$ is closed under relation $\sim$. It is easy to see that $R_{k}$ is closed under $x\sim x^{2}$ and under $xz\sim zx$ for generators $x,z$ not connected in the graph $\Theta$ (the only such factors of a word $u\in R_{k}$ can be of the form $yx_{j},x_{j}y$, where $j\geqslant 2$). Moreover, $u$ does not have factors of the form $x_{j}x_{i}$ with $i=3,\ldots,n$ and $j\neq i+1$ (modulo $n$). So we do not have to consider relations $x_{i}x_{i+1}x_{i}\sim x_{i+1}x_{i}x_{i+1}\sim x_{i}x_{i+1}$ for $i=2,\ldots,n$. It is also easy to see that every relation $yx_{1}y\sim yx_{1}$, $x_{1}yx_{1}\sim yx_{1}$ and $x_{1}yx_{1}\sim yx_{1}y$ leaves $R_{k}$ invariant. Finally, every relation $x_{2}x_{1}x_{2}\sim x_{1}x_{2}$, $x_{1}x_{2}x_{1}\sim x_{1}x_{2}$ and $x_{1}x_{2}x_{1}\sim x_{2}x_{1}x_{2}$ leaves $R_{k}$ invariant. This proves the claim.

Define $v_{k}=x_{1}x_{2}y(x_{1}x_{n}\cdots x_{2})^{k}$, for $k=1,2,\ldots$. Notice that $v_{k}\in R_{k}$ but $v_{k}\notin FR_{i}$ for $i<k$. It follows that $v_{k}\notin\bigcup_{i=1}^{k-1}\operatorname{HK}_{\Theta}v_{i}$, for every $k\geqslant 2$. Therefore $\operatorname{HK}_{\Theta}$ does not satisfy acc on left ideals. ∎

We continue with a useful observation of independent interest. In order to avoid confusion, we denote by $[w]$ the equivalence class of a word $w$ in $F$, with respect to the equivalence relation $\sim$.

Lemma 5.2**.**

Assume that $y_{1},y_{2},\ldots,y_{n-1}$ are the consecutive vertices of a cyclic graph $C_{n-1}$. Consider an epimorphism $\phi$ from the free monoid $Y=\langle y_{1},\ldots,y_{n-1}\rangle$ to the submonoid $\langle x_{2},\ldots,x_{n-1},x_{n}x_{1}\rangle$ of $F$ defined by

[TABLE]

Then $\phi$ induces a homomorphism $\overline{\phi}:C_{n-1}\longrightarrow C_{n}$ given by the formula $\overline{\phi}([w])=[\phi(w)]$, for every $w\in\langle y_{1},\ldots,y_{n-1}\rangle$. Moreover, $\overline{\phi}$ determines an isomorphism $C_{n-1}\cong\langle x_{2},\ldots,x_{n-1},x_{n}x_{1}\rangle\subseteq C_{n}$.

Proof.

It is verified in [12], Lemma 4, that $\overline{\phi}$ is a homomorphism. We claim that if a word $w=w(y_{1},\ldots,y_{n-1})$ is reduced in the sense of the reduction system $S^{\prime}$ as in Theorem 1.1, defined with respect to the deg-lex order extending $y_{1}<\cdots<y_{n-1}$ in the free monoid $Y=\langle y_{1},\ldots,y_{n-1}\rangle$, then the word $w(x_{2},\ldots,x_{n-1},x_{n}x_{1})$ is reduced with respect to the system $S$ in the free monoid $F=\langle x_{1},\ldots,x_{n}\rangle$.

If $w\in Y$ then it is clear that if $\phi(w)$ contains a factor (that is the leading term of a reduction) of type (1) in Theorem 1.1, then also $w$ contains such a factor. Assume that $\phi(w)$ contains a factor $x_{j}x_{i}$ of type (2). Then $w$ contains a factor $y_{j-1}y_{i-1}$. Assume that $\phi(w)$ has a factor $x_{i}ux_{i}$ that is of type (4) or (5). If $i=1$ or $i=n$ then $\phi(w)$ has a factor $x_{n}x_{1}vx_{n}x_{1}$. If $v$ does not contain $x_{2}$ ($x_{n-1}$, respectively) then $\phi^{-1}(v)$ does not contain $y_{1}$ ($y_{n-2}$, respectively), and we are done. If $i\neq 1,n$, and $u$ does not contain $x_{i+1}$ ($x_{i-1}$, respectively) then $\phi^{-1}(u)$ does not contain $y_{i}$ (respectively, $y_{i-2}$ if $i>2$; and if $i=2$ then $\phi^{-1}(u)$ does not contain $y_{n-1}$), as desired. Assume that $\phi(w)$ contains a factor of the form $x_{n}(x_{1}\dotsm x_{i})x_{j}$ for $i,j\in\{1,\dotsc,n\}$ such that $i+1<j<n-1$. Then $w$ contains a factor $y_{n-1}(y_{1}\cdots y_{i-1})y_{j-1}$ or $y_{n-1}y_{j-1}$, and the assertion follows as well. This proves the claim.

Therefore $\overline{\phi}$ is injective. The result follows. ∎

The crucial step in the proofs of the main results of this section is based on the following observation.

Proposition 5.3**.**

Assume that $t\in\{1,\ldots,n-3\}$ and $\alpha\in K_{0}[M_{t}]$ is such that $\alpha x_{i}=0$ in $K_{0}[M_{t}]$ for every $i\in\{1,\ldots,n\}$. Assume also that every $w\in\operatorname{supp}(\alpha)$ is of the form $(x_{n}q_{t})^{k}b$, where $k\geqslant 1$ and $b\in B_{t}$. Then $\alpha=0$.

In order to prove it, we need some preparatory technical lemmas. We assume that $t\in\{1,\ldots,n-3\}$. Moreover, we will suppose that a nonzero $\alpha\in K_{0}[M_{t}]$ is given that satisfies the hypotheses of the proposition. The aim is to come to a contradiction.

Roughly speaking, the first lemma describes the reduced form of any word of type $wx_{r}$ for $w$ in block form (see the convention introduced after Theorem 2.1), namely

[TABLE]

with certain conditions on indices, and $x_{r}$ such that $n-1\geqslant r\geqslant j_{k}-1$ or $r\leqslant i_{k}+1$. This means that $x_{r}$ cannot be pushed to the left by using only reductions (2) or (3) in such a way that $wx_{r}=(x_{n}q_{t})x_{n}(x_{1}\cdots x_{i_{1}}x_{n-1}\cdots x_{j_{1}})\cdots x_{n}x_{r}(x_{1}\cdots x_{i_{k}}x_{n-1}\cdots x_{j_{k}})$ in $C_{n}$.

Lemma 5.4**.**

Let $t\in\{1,\ldots,n-3\}$. Consider the word $w$ of the form

[TABLE]

where $i_{k}<i_{k-1}<\cdots<i_{1}<t+1<j_{1}<\cdots<j_{k}$. The word $x_{n}q_{t}$ is also assumed to be of the above type for $k=0$. Moreover, in every $w$ we use the convention that $i_{0}=t,j_{0}=t+1$. Let $r\geqslant t$ be such that $n-1\geqslant r\geqslant j_{k}-1$ or $r\leqslant i_{k}+1$ (so in the latter case $t\leqslant r\leqslant i_{k}+1\leqslant t+1$). Then the following holds:

if $n-1\geqslant r>j_{k}$, then $wx_{r}\in I_{t}$; 2. 2.

if $r=j_{k}$, then $wx_{r}=w$ in $C_{n}$; 3. 3.

if $j_{k}=r+1$, then either $wx_{r}=w$ in $C_{n}$ or the word $wx_{r}$ is reduced; 4. 4.

if $j_{k}>r+1$, $r=t$, $i_{k}=t-1$, then either (for $k=1$) the word $wx_{r}$ has the reduced form $(x_{n}q_{t})x_{n}x_{1}\cdots x_{t}x_{n-1}\cdots x_{j_{k}}$ or (for $k\geqslant 2$) $wx_{r}\in I_{t}$; 5. 5.

if $j_{k}>r+1$, $r=t$, $i_{k}=t$, then $wx_{r}=w$ in $C_{n}$; 6. 6.

if $j_{k}>r+1$, $r=t+1$, $i_{k}=t$, then $wx_{r}\in I_{t}$.

Proof.

Parts 2. and 5. are clear.

To prove part 1., we proceed by induction on $k$ (the number of blocks in the word $w$). Let $n-1\geqslant r>t+1$. If $k=0$ then $w=x_{n}q_{t}$ and

[TABLE]

From Lemma 3.8 we obtain $(x_{r-1}\cdots x_{t+2})(x_{n}x_{1}\cdots x_{t+1})(x_{n-1}\cdots x_{r})\in I_{t}$, as desired.

So, assume that the assertion holds for every $m<k$, where $k\geqslant 1$. Consider

[TABLE]

for $r>j_{k}$. Then we have

[TABLE]

From the assumptions we know that $j_{k}>i_{k}+1$ and $r-1<n-1$, so the following holds:

[TABLE]

By the assumptions $j_{k}<r\leqslant n-1$ and $j_{k-1}<j_{k}$ so $j_{k-1}<r-1\leqslant n-1$. $I_{t}$ is an ideal in $C_{n}$, so from the above calculation and the induction hypothesis for the element

[TABLE]

(a word with $k-1$ blocks) the following holds in $C_{n}$

[TABLE]

Hence part 1. follows.

To prove part 3., assume that $j_{k}=r+1$. Recall that $i_{0}=t$ and $j_{0}=t+1$. It follows that for $k=0$ we have $r=t$. In this case $(x_{n}q_{t})x_{r}\xrightarrow{(4t)}w$. Hence, we can assume that $k\geqslant 1$. Then

[TABLE]

If $j_{k-1}<j_{k}-1$, then we see that the above word is reduced.

Hence, assume now that $j_{k-1}=j_{k}-1$. Then the word $wx_{r}$ has a factor $x_{j_{k}-1}x_{n}x_{1}\cdots x_{i_{k}}x_{n-1}\cdots x_{j_{k}}x_{j_{k}-1}$. If $i_{k}+1<j_{k}-1$, then

[TABLE]

It follows that $wx_{r}=w$.

Finally, if $i_{k}+1\geqslant j_{k}-1$, then $i_{k}<j_{k-1}=j_{k}-1\leqslant i_{k}+1$, so that $j_{k-1}=i_{k}+1$. Hence $i_{k}\leqslant i_{k-1}<t+1\leqslant j_{k-1}$ implies that $i_{k-1}=i_{k}=t$, $j_{k}=t+2$. It follows that $wx_{r}$ is reduced. This proves part 3.

In the proof of the remaining assertions (parts 4. and 6.) we can assume that $k\geqslant 1$, because for $k=0$ it is impossible to have $t+1=j_{k}>r+1$ and $r\in\{t,t+1\}$.

To prove part 4., assume that $j_{k}>r+1$, $r=t$, $i_{k}=t-1$. Then from the definition of $w$ we obtain that $k\in\{1,2\}$ and either $w=(x_{n}q_{t})x_{n}x_{1}\cdots x_{t}x_{n-1}\cdots x_{j_{1}}x_{n}x_{1}\cdots x_{t-1}x_{n-1}\cdots x_{j_{2}}$, where $j_{2}>j_{1}>t+1$ or $w=(x_{n}q_{t})x_{n}x_{1}\cdots x_{t-1}x_{n-1}\cdots x_{j_{1}}$. In the first case

[TABLE]

From part 1. applied to $x_{n}q_{t}$ and $r=n-1$ we get $wx_{r}\in I_{t}$.

In the second case $wx_{t}\xrightarrow{(2)}(x_{n}q_{t})x_{n}x_{1}\cdots x_{t}x_{n-1}\cdots x_{j_{1}}$ and the last word is reduced.

To prove part 6., assume that $j_{k}>r+1$, $r=t+1$, $i_{k}=t$. Then from the definition of $w$ it follows that $k=1$ and

[TABLE]

by Lemma 3.8. ∎

We continue under the assumptions of Proposition 5.3. By Theorem 2.1, every $w\in\operatorname{supp}(\alpha)$ must satisfy one of the following conditions:

1. (i)

$x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}\in\operatorname{suff}(w)$, where $i_{s}<i_{s-1}<t+1<j_{s-1}<j_{s}\leqslant n-1$, or $i_{s}=i_{s-1}=t$ and $j_{s-1}=t+1<j_{s}$, 2. (ii)

$x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}x_{n}x_{1}\cdots x_{i_{s+1}}\in\operatorname{suff}(w)$, where

$i_{s+1}<i_{s}<i_{s-1}<t+1<j_{s-1}<j_{s}\leqslant n-1$,

or $i_{s}=i_{s-1}=t>i_{s+1}$ and $j_{s-1}=t+1<j_{s}$;

or $(x_{n}q_{t})x_{n}x_{1}\cdots x_{i_{s+1}}\in\operatorname{suff}(w)$ with $i_{s+1}\leqslant t$, 3. (iii)

$x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{n-1}\cdots x_{j_{s}}\in\operatorname{suff}(w)$, where $2\leqslant i_{s-1}<t+1<j_{s-1}<j_{s}\leqslant n$ 4. (iv)

$x_{n}x_{1}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{n-1}\cdots x_{j_{s}}\in\operatorname{suff}(w)$, where $2\leqslant t+1<j_{s-1}<j_{s}\leqslant n$, 5. (v)

$b=x_{n}x_{n-1}\cdots x_{j_{s}}$, where $t+1<j_{s}\leqslant n$, 6. (vi)

$b=1$, i.e. $w=(x_{n}q_{t})^{k}$.

Hence, we can write $\alpha=\alpha_{(i)}+\alpha_{(ii)}+\alpha_{(iii)}+\alpha_{(iv)}+\alpha_{(v)}+\alpha_{(vi)}$, where $\operatorname{supp}(\alpha_{k})$ consists of all words of the form $(k)$ listed above, which are in the support of the element $\alpha$. We will prove that for every $k\in\{(i),\ldots,(vi)\}$ the element $\alpha_{(k)}$ is zero, which will contradict the supposition that $\alpha\neq 0$.

First, we prove the following result concerning $\alpha x_{1}$.

Lemma 5.5**.**

Let $\alpha$ be as described above. Then

$\alpha_{(vi)}=0$; 2. 2.

$\alpha_{(i)}=\alpha_{(i),i_{s}=1}$; 3. 3.

$\alpha_{(ii)}=\alpha_{(ii),i_{s+1}=1}$,

where $\alpha_{(i)}=\alpha_{(i),i_{s}=1}+\alpha_{(i),i_{s}>1}$ and $\operatorname{supp}(\alpha_{(i),i_{s}=1})$ consists of all words from the support of $\alpha_{(i)}$ with $i_{s}=1$, while $\operatorname{supp}(\alpha_{(i),i_{s}>1})$ does not contain such words; similarly $\alpha_{(ii),i_{s+1}=1}$ involves all words from the support of $\alpha_{(ii)}$ with $i_{s+1}=1$ (see the description of $\alpha_{(i)}$, $\alpha_{(ii)}$).

Proof.

We know that $\alpha x_{1}=0$ in $K_{0}[M_{t}]$. We calculate the reduced forms of $wx_{1}$ for all $w\in\operatorname{supp}(\alpha_{k})$, for $k\in\{(i),\ldots,(vi)\}$. It will be more convenient to consider certain suffixes of the given word $w$.

• •

$x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}x_{1}\xrightarrow{(41)}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}$, so $\alpha_{(i)}x_{1}=\alpha_{(i)}$;

• •

$x_{n}x_{1}\cdots x_{i_{s+1}}x_{1}\xrightarrow{(41)}x_{n}x_{1}\cdots x_{i_{s+1}}$, whence $\alpha_{(ii)}x_{1}=\alpha_{(ii)}$;

• •

$x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{n-1}\cdots x_{j_{s}}x_{1}\xrightarrow{(*)}\begin{cases}x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\textrm{ for }j_{s}=n\\ x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}x_{n-1}\cdots x_{j_{s}}\textrm{ for }j_{s}<n,\end{cases}$ where $(*)$ denotes equality in the first case and reduction $(2)$ in the second case. We see that in the first case $(j_{s}=n)$ the obtained word is reduced of type $(ii)$ with $i_{s+1}=1$. In the second case $(j_{s}<n)$ the word is reduced of type $(i)$ with $i_{s}=1$.

• •

$\displaystyle\begin{aligned} x_{n}x_{1}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{n-1}\cdots x_{j_{s}}x_{1}&\xrightarrow{(2)}x_{n}x_{1}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}x_{n-1}\cdots x_{j_{s}}\\ &\xrightarrow{(51),(5n)}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}x_{n-1}\cdots x_{j_{s}}.\end{aligned}$

In this case the obtained form of the element $wx_{1}$ has a factor of the form

[TABLE]

where $k=s-2$, $j_{k}<n-1$ (notice that $s\geqslant 2$). Assertion 1. of Lemma 5.4 implies that $wx_{1}=0$ in $K_{0}[M_{t}]$ for every $w\in\operatorname{supp}(\alpha_{(iv)})$.

• •

$x_{n}x_{n-1}\cdots x_{j_{s}}x_{1}\xrightarrow{(*)}\begin{cases}x_{n}x_{n-1}\cdots x_{j_{s}}x_{1}\textrm{ for }j_{s}=n\\ x_{n}x_{1}x_{n-1}\cdots x_{j_{s}}\textrm{ for }j_{s}<n,\end{cases}$

where $(*)$ denotes equality in the first case and reduction $(2)$ in the second case. We see that in the first case the word $wx_{1}$ is of the reduced form $(ii)$ with $i_{s+1}=1$, whereas in the second case we obtain a reduced word of type $(i)$ with $i_{s}=1$.

• •

$(x_{n}q_{t})x_{1}\xrightarrow{(41)}x_{n}q_{t}$, so $\alpha_{(vi)}x_{1}=\alpha_{(vi)}$.

From the above calculations we see that in $K_{0}[M_{t}]$

[TABLE]

It is clear that the terms from $\alpha_{(vi)}$ are the only terms of type $(vi)$ in the above sum, so $\alpha_{(vi)}=0$. Moreover, reduced forms of elements from $\alpha_{(i)}+\alpha_{(iii),j_{s}=n}x_{1}+\alpha_{(v),j_{s}=n}x_{1}$ are of type $(i)$, whereas reduced forms of words in the sum $\alpha_{(ii)}+\alpha_{(iii),j_{s}<n}x_{1}+\alpha_{(v),j_{s}<n}x_{1}$ are of type $(ii)$. It follows that these sums are [math] in $K_{0}[M_{t}]$. It is not difficult to see that every word from $\operatorname{supp}(\alpha_{(iii),j_{s}=n}x_{1})$ and $\operatorname{supp}(\alpha_{(v),j_{s}=n}x_{1})$ has reduced form ending with $x_{n}x_{1}$, so $\alpha_{(ii)}=\alpha_{(ii),i_{s+1}=1}$. Similarly, every (reduced) word from $\operatorname{supp}(\alpha_{(iii),j_{s}<n}x_{1})$ and $\operatorname{supp}(\alpha_{(v),j_{s}<n}x_{1})$ has a suffix of the form $x_{n}x_{1}x_{n-1}\cdots x_{j}$ for some $j$, so $\alpha_{(i)}=\alpha_{(i),i_{s}=1}$.∎

It follows that $\operatorname{supp}(\alpha)=\operatorname{supp}(\alpha_{(i),i_{s}=1})\cup\operatorname{supp}(\alpha_{(ii),i_{s+1}=1})\cup\operatorname{supp}(\alpha_{(iii)})\cup\operatorname{supp}(\alpha_{(iv)})\cup\operatorname{supp}(\alpha_{(v)})$. Let $m=\textrm{min}\{j_{s}:w\in\operatorname{supp}(\alpha)\}$, with $j_{s}$ defined for every word $w$ as in cases (i)-(vi) listed before Lemma 5.5. Then $n\geqslant m\geqslant t+1\geqslant 2$. By our assumption, also $\alpha x_{m-1}=0$. We calculate the reduced form of words $wx_{m-1}$, where $w\in\operatorname{supp}(\alpha)$. By $s_{k}$ we mean an appropriately chosen suffix of the word from the support of $\alpha_{k}$. We consider the following two cases.

Case I. Assume that $m=j_{s}$.

First, suppose that $j_{s-1}=j_{s}-1$. Then

1. (a)

$s_{(i)}x_{m-1}=x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s}-1}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}x_{j_{s}-1}$

If $i_{s}+1<j_{s}-1$, then $s_{(i)}x_{m-1}\xrightarrow{(4j_{s})}s_{(i)}$.

Otherwise we have $i_{s}+1=j_{s}-1$, which implies that $j_{s}-2\leqslant i_{s-1}<j_{s}-1$, so it follows easily that $i_{s-1}=t,j_{s-1}=t+1$. In this case $s_{(i)}x_{m-1}=x_{n}q_{t}x_{n}x_{1}\cdots x_{t}x_{n-1}\cdots x_{t+1}=(x_{n}q_{t})^{2}$. 2. (b)

$s_{(ii)}x_{m-1}=x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}x_{n}x_{1}x_{j_{s}-1}$.

Suppose that $j_{s-1}+1=j_{s}\leqslant 3$. Since $j_{s-1}\geqslant t+1$, it follows that $t=1$ and $s_{(ii)}$ must be the word $x_{n}x_{1}x_{n-1}\cdots x_{2}x_{n}x_{1}x_{n-1}\cdots x_{3}x_{n}x_{1}$, which is not reduced. Therefore we can assume that $n-1\geqslant j_{s}>3$ and

[TABLE]

It is clear that the reduced word $x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}$ is of the form $(ii)$. From the previous case we obtain

[TABLE] 3. (c)

$s_{(iii),(iv),(v)}x_{m-1}=x_{j_{s}-1}x_{n}x_{n-1}\cdots x_{j_{s}}x_{j_{s}-1}\xrightarrow{(4(j_{s}-1))}x_{j_{s}-1}x_{n}x_{n-1}\cdots x_{j_{s}}$. It follows that for every $w$ of the form $(iii)$, $(iv)$ or $(v)$ we have $wx_{m-1}=w$. 2. 2.

Secondly, assume that $j_{s-1}<j_{s}-1$. Then

1. (a)

for every $w$ of the form $(i)$ $wx_{m-1}$ is reduced; 2. (b)

since $j_{s}-1>j_{s-1}\geqslant 2$, then

[TABLE]

It follows that the reduced form of $wx_{m-1}$, where $w\in\operatorname{supp}(\alpha_{(ii),i_{s+1}=1})$, has a suffix as above; 3. (c)

similarly, it is clear that $wx_{m-1}$ is reduced for every $w\in\operatorname{supp}(\alpha_{k})$, where $k\in\{(iii),(iv),(v)\}$. 3. 3.

Assume that $s_{(ii)}x_{m-1}=(x_{n}q_{t})x_{n}x_{1}x_{m-1}$. In this case $m=j_{s}=t+1$. If $t=1$, then $s_{(ii)}x_{t}=s_{(ii)}$ in $C_{n}$. Moreover, if $t\geqslant 3$ then $(x_{n}q_{t})x_{n}x_{1}x_{t}\xrightarrow{(4t)}(x_{n}q_{t})x_{n}x_{1}$, so also $s_{(ii)}x_{t}=s_{(ii)}$. Finally, if $t=2$, it is easy to see that $wx_{t}$ is in the reduced form.

We summarize the foregoing observations as follows.

Corollary 5.6**.**

Let $m=j_{s}$ be as described above. Consider an element $w$ from the support of $\alpha_{(i),i_{s}=1}$, $\alpha_{(iii)}$, $\alpha_{(iv)}$ or $\alpha_{(v)}$.

If $j_{s-1}=j_{s}-1$, then $wx_{m-1}=w$ in $C_{n}$ or $wx_{m-1}$ is of the form $(vi)$. 2. 2.

If $j_{s-1}<j_{s}-1$, then $wx_{m-1}$ is reduced.

Assume now that $w$ is in the support of $\alpha_{(ii),i_{s+1}=1}$.

If $(x_{n}q_{t})x_{n}x_{1}\in\operatorname{suff}(w)$, then either (for $t\neq 2$) $wx_{m-1}=w$ in $C_{n}$ or $wx_{m-1}$ has the reduced form $(x_{n}q_{2})^{M}x_{n}x_{1}x_{2}$, for some $M\geqslant 1$. 2. 4.

If $j_{s-1}=j_{s}-1$, then either $wx_{m-1}=w$ in $C_{n}$ or $wx_{m-1}$ is of the form $(x_{n}q_{t})^{M}x_{n}x_{1}$, where $M\geqslant 1$. 3. 5.

If $j_{s-1}<j_{s}-1$, then $w=vx_{n}x_{1}$ for some reduced word $v$ and $wx_{m-1}$ has the reduced form $vx_{m-1}x_{n}x_{1}$.

In particular, words from the supports of $\alpha_{(i),i_{s}=1}$, $\alpha_{(ii),i_{s+1}=1}$, $\alpha_{(iii)}$, $\alpha_{(iv)}$ and $\alpha_{(v)}$ multiplied by $x_{m-1}$ have reduced forms ending with $x_{n-1}\ldots x_{j}$, $x_{j}x_{n}x_{1}$ or $x_{j}x_{n}x_{1}x_{2}$, where $j\leqslant m$.

Case II. Now assume that $m<j_{s}$. In particular $m-1<n-1$.

We claim that if $w$ is a word in the support of $\alpha_{(i),i_{s}=1}$, $\alpha_{(iii)}$, $\alpha_{(iv)}$ or $\alpha_{(v)}$, then $wx_{m-1}$ is [math] in $K_{0}[M_{t}]$ or its reduced form has a suffix of the form $x_{n-1}\cdots x_{j}$ for some $j>m$. Moreover, if $w$ is in $\operatorname{supp}(\alpha_{(ii),i_{s+1}=1})$, then $wx_{m-1}$ is [math] or $\operatorname{suff}(wx_{m-1})=x_{n-1}\cdots x_{j}x_{n}x_{1}$ for some $j>m$.

The idea is to reduce words by pushing $x_{m-1}$ to the left and then to use Lemma 5.4. As before, by $w_{k}$ we denote a suffix of a word of type $(k)$.

1. (a)

$w_{(i)}x_{m-1}=(x_{n}q_{t})x_{n}(x_{1}\cdots x_{i_{1}}x_{n-1}\cdots x_{j_{1}})\cdots x_{n}(x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}})x_{m-1}$. As long as $j_{k}-1>m-1>i_{k}+1$ ($k=1,\ldots,s+1$) we use reductions $(2)$ and $(3)$ to push $x_{m-1}$ to the left. After this procedure we obtain a word with a prefix $vx_{m-1}$, where $v$ is exactly a word from Lemma 5.4, for some $k_{0}$ and $r=m-1$. By the assumption $j_{s}\neq m$ (hence, it is impossible that $k_{0}=s$ and $j_{k_{0}}=r+1$), so applying Lemma 5.4 we obtain that $w_{(i)}x_{m-1}$ is either in $I_{t}$ or its reduced form ends with $x_{n-1}\cdots x_{j_{s}}$, $j_{s}>m$. 2. (b)

Since $t+1\leqslant m<j_{s}$, we must have $w\neq(x_{n}q_{t})^{M}x_{n}x_{1}$. Then

[TABLE]

If $m-1=1$, then $w_{(ii)}x_{m-1}=w_{(ii)}$ in $C_{n}$, and hence it has a suffix $x_{n-1}\cdots x_{j_{s}}x_{n}x_{1}$, $j_{s}>m$.

If $m-1=2$, then $j_{s}>3$ and $t\leqslant 2$. From the form of $w_{(ii)}$ we see that in this case $i_{s}>1$ and of course $i_{s}\leqslant 2$. It follows that $i_{s}=2$. Then

[TABLE]

Applying assertion 1. of Lemma 5.4, it follows that this word is in $I_{t}$.

If $m-1>2$ then $s_{(ii)}x_{m-1}\xrightarrow{(3)}x_{n}x_{1}\cdots x_{i_{s-1}}x_{n-1}\cdots x_{j_{s-1}}x_{n}x_{1}\cdots x_{i_{s}}x_{n-1}\cdots x_{j_{s}}x_{m-1}x_{n}x_{1}$. Using the observation made in the previous case and Lemma 5.4, we get that either $w_{(ii)}x_{m-1}\in I_{t}$ or its reduced form has a suffix $x_{n-1}\cdots x_{j_{s}}x_{n}x_{1}$, for $j_{s}>m$. 3. (c)

Every word $w\in\operatorname{supp}(\alpha_{(iii)})\cup\operatorname{supp}(\alpha_{(iv)})\cup\operatorname{supp}(\alpha_{(v)})$ can be written as $w=vx_{n}x_{n-1}\cdots x_{j_{s}}$, where $v$ has a block form as in Lemma 5.4. Then $wx_{m-1}\xrightarrow{(2)}vx_{m-1}x_{n}x_{n-1}\ldots x_{j_{s}}$. Pushing $x_{m-1}$ to the left by using reductions (2) and (3) we can apply Lemma 5.4. It follows that either $wx_{m-1}\in I_{t}$ or its reduced form has a suffix $x_{n}x_{n-1}\cdots x_{j_{s}}$, for $j_{s}>m$.

This completes the proof of our claim in Case II.

By our assumptions (of Proposition 5.3), we know that $\alpha x_{m-1}=0$ in $K_{0}[M_{t}]$. From the above discussion it follows that for every $w\in\operatorname{supp}(\alpha)$ either $wx_{m-1}$ is [math] (and it is possible only if $m<j_{s}$) or a suffix of the reduced form of $wx_{m-1}$ is equal to $x_{j}x_{n}x_{1}$, $x_{j}x_{n}x_{1}x_{2}$ (only if $w\in\operatorname{supp}(\alpha_{(ii)}))$, or to $x_{n-1}\cdots x_{j}$. Moreover, $j\leqslant m$ if and only if in the word $w$ we have $j_{s}=m$ (see the description of possible types of words). It follows that after multiplying by $x_{m-1}$ the sum of all elements with $j_{s}=m$ vanishes.

Assume that $v$, $z$ are reduced words such that $j_{s_{v}}=j_{s_{z}}=m$ (here $j_{s_{v}},j_{s_{z}}$ are defined for $v$ and $z$ as in the list of possible types (i)-(vi) listed before Lemma 5.5) and $vx_{m-1}=zx_{m-1}$ holds in $C_{n}$. We use the proof of Corollary 5.6 to conclude that $v=z$. Let $u$ be the reduced form of $vx_{m-1}=zx_{m-1}$.

• •

Assume $u$ is of type $(i)$, $(iii)$, $(iv)$ or $(v)$. If $u$ has a suffix $x_{m-1}$, then it follows that $j_{s-1}<j_{s}-1$ and $vx_{m-1}$, $zx_{m-1}$ are reduced, so that $v=z$. Otherwise $vx_{m-1}=v$ and $zx_{m-1}=z$, so also $v=z$.

• •

Assume $u$ is of type $(vi)$. Then $j_{s-1}=j_{s}-1$ and $z=v$ are of the form $(x_{n}q_{t})^{M}x_{n}x_{1}\cdots x_{t}x_{n-1}\ldots x_{t+2}$.

• •

Assume $u$ is of type $(ii)$.

If $m\neq t+1$, then it follows that $\operatorname{suff}(u)=x_{j}x_{n}x_{1}$ for $j\in\{m-1,m\}$. If $j=m$, then $v=vx_{m-1}$ and $zx_{m-1}=z$ in $C_{n}$, so the assertion holds. If $j=m-1$, it follows that for $v=v_{0}x_{n}x_{1}$, we have $vx_{m-1}=v_{0}x_{m-1}x_{n}x_{1}$ in $C_{n}$ and the latter word is the reduced form. It is clear that $v=z$ also in this case.

Otherwise $m=t+1$. Then we consider only part 3. in Corollary 5.6. It is clear that in this case if $vx_{m-1}=zx_{m-1}$, then $v=z$.

We have shown that for every pair of words $v,z$ with $j_{s_{v}}=j_{s_{z}}=m$ if $vx_{m-1}=zx_{m-1}$, then $v=z$. This implies that $\operatorname{supp}(\alpha)$ has no words with $j_{s}=m$, which contradicts the definition of $m$. Hence, the assertion of Proposition 5.3 has been proved.

We will also need another technical observation.

Lemma 5.7**.**

We have $x_{1}M_{t}\subseteq M_{t}\cup I_{t}$.

Proof.

Using the involution $\chi_{t}$ from Corollary 3.12, since $\chi_{t}(I_{t})\subseteq I_{t}$ by Lemma 3.18, it is enough to show that $M_{t}x_{t}\subseteq M_{t}\cup I_{t}$, or in other words $(x_{n}q_{t})bx_{t}\in\langle x_{n}q_{t}\rangle B_{t}\cup I_{t}$ for every $b\in B_{t}$. This is easily shown by applying the method used in the proof of Case II (pushing $x_{t}$ to the left in the considered word and then applying cases 4. and 5. of Lemma 5.4). ∎

For any $K$-algebra $A$, let ${\mathcal{P}}(A)$ denote the prime radical of $A$.

Theorem 5.8**.**

For every $t=0,1,\ldots,n-2$, the algebra $K_{0}[M_{t}]$ is prime.

Proof.

In view of Remark 4.7, $K_{0}[M_{n-2}]$ and $K_{0}[M_{0}]$ are prime. In particular, the result holds for $n=3$. We proceed by induction on $n$. Assume that $n>3$. Moreover, we may assume that $1\leqslant t\leqslant n-3$.

First, we show that $K_{0}[M_{t}]$ is semiprime. Suppose that $\alpha K_{0}[M_{t}]\alpha=0$ for some nonzero $\alpha\in K_{0}[M_{t}]$. Then, by Theorem 3.10, for every $u,w\in M_{t}$ we have $\operatorname{supp}(u\alpha w)\subseteq a\langle x_{n}q_{t}\rangle b$ for some $a\in A_{t},b\in B_{t}$. By Corollary 3.21, if $u\alpha w\neq 0$, then there exist $u^{\prime},w^{\prime}\in M_{t}$ such that $0\neq u^{\prime}u\alpha ww^{\prime}\in\langle x_{n}q_{t}\rangle$, and $u^{\prime}u\alpha ww^{\prime}\in{\mathcal{P}}(K[\langle x_{n}q_{t}\rangle])=0$. It follows that $u\alpha w=0$ for every $u,w\in M_{t}$. Thus, either $\alpha M_{t}=0$ or $M_{t}\alpha w=0$ for some $w\in M_{t}$ such that $\alpha w\neq 0$. This means that $\alpha\circ P_{t}=0$ or $P_{t}\circ\alpha w=0$ ($\circ$ stands for the ordinary matrix multiplication, where $K_{0}[M_{t}]$ is interpreted as a subset of the matrix algebra $M_{|A_{t}|}(K[\langle x_{n}q_{t}\rangle])$. Since $P_{t}$ is a symmetric matrix by Remark 3.24, we may assume that $\alpha\circ P_{t}=0$ for some nonzero $\alpha\in K_{0}[M_{t}]$. Then $\alpha$ can be chosen so that $\operatorname{supp}(\alpha)\subseteq a\langle x_{n}q_{t}\rangle B_{t}$ for some $a\in A_{t}$. Hence, Corollary 3.21 allows us to assume that $\operatorname{supp}(\alpha)\subseteq\langle x_{n}q_{t}\rangle B_{t}$. Finally, we may assume that $|\operatorname{supp}(\alpha)|$ is minimal possible.

We claim that $\alpha x_{1}=0$ in $K_{0}[M_{t}]$. By Lemma 5.7, $\alpha x_{1}M_{t}=0$ in $K_{0}[M_{t}]$. From the proof of Lemma 5.5 we know that $\alpha_{(iv)}x_{1}=0$ in $K_{0}[M_{t}]$ and $vx_{1}\in\langle x_{n}q_{t}\rangle B_{t}$ for every $v\in\operatorname{supp}(\alpha)\setminus\operatorname{supp}(\alpha_{(iv)})$. So, $\alpha x_{1}$ inherits the hypotheses on $\alpha$. Therefore, the minimal choice of $\alpha$ allows us to assume that $\alpha_{(iv)}=0$. Moreover, $\alpha x_{1}\in K[\langle x_{2},\ldots,x_{n-1},x_{n}x_{1}\rangle]$. But, from Lemma 5.2 we know that the latter is isomorphic to $K[C_{n-1}]$. Moreover, under this identification, $\operatorname{supp}(\alpha x_{1})$ is contained in a single row of the matrix structure $M_{t-1}^{(n-1)}$ defined for the monoid $C_{n-1}$ as in Definition 3.22. It is easy to see that $\alpha x_{1}M_{t-1}^{(n-1)}=0$ in $K_{0}[M_{t-1}^{(n-1)}]$. The inductive hypothesis implies that $\alpha x_{1}=0$. This proves the claim.

From Lemma 3.17 it follows that replacing $\alpha$ by $(x_{n}q_{t})^{k}\alpha$, for some $k\geqslant 1$, if necessary, we may assume that $\sigma(\alpha)\in M_{t}$ and hence we get that $\sigma(\alpha)$ lies in a single row of the matrix structure $K_{0}[M_{t}]$. In other words, there exists $a\in A_{t}$ such that $\operatorname{supp}(\alpha)\subseteq a\langle x_{n}q_{t}\rangle B_{t}$. Then, by Corollary 3.21, there exists $z\in M_{t}$ such that $\operatorname{supp}(z\sigma(\alpha))\subseteq\langle x_{n}q_{t}\rangle B_{t}$. The proof of Lemma 5.5 implies that for every $w\in\operatorname{supp}(\alpha)$ either $wx_{1}\in a\langle x_{n}q_{t}\rangle B_{t}$ or $wx_{1}=0$ in $K_{0}[M_{t}]$. Therefore, by the previous paragraph, $z\sigma(\alpha)x_{1}=0$. Hence, Lemma 3.21 implies that also $\sigma(\alpha x_{n})=\sigma(\alpha)x_{1}=0$. Consequently, $\alpha x_{n}=0$.

Repeating this argument, we get that $\alpha x_{i}=0$ in $K_{0}[M_{t}]$ for every $i=1,\ldots,n$. From Proposition 5.3 it now follows that $\alpha=0$, a contradiction. Thus, we have proved that $K_{0}[M_{t}]$ is semiprime. This implies that the sandwich matrix $P_{t}$ is not a zero divisor (in the corresponding matrix ring $M_{n_{t}}(K[\langle x_{n}q_{t}\rangle])$, where $n_{t}=|A_{t}|$). Since $K[\langle x_{n}q_{t}\rangle]$ is a domain, it is known that $K_{0}[M_{t}]$ must be prime, see [14], Chapter 5. ∎

It follows that every matrix $P_{t},t=0,\ldots,n-2$, has a nonzero determinant, which seems to be inaccessible by a direct proof. We are now ready for the main results of this section.

Theorem 5.9**.**

For every $n\geqslant 3$ the algebra $K[C_{n}]$ is right and left Noetherian.

Proof.

From [16] we know that the algebra $K[C_{n}]/{\mathcal{P}}(K[C_{n}])$ is right and left Noetherian because (by [12]) it is a semiprime algebra of Gelfand–Kirillov dimension $1$.

Suppose that $0\neq\beta\in{\mathcal{P}}(K[C_{n}])\cap K[\tilde{M}]$, where $\tilde{M}=\bigcup_{i=0}^{n-2}\tilde{M}_{i}$, as defined in Section 3. Let $i$ be the minimal integer such that $\operatorname{supp}(\beta)\cap\tilde{M}_{i}\neq\emptyset$. Passing to $K[C_{n}]/I_{i}$, we get a nonzero element $\overline{\beta}\in{\mathcal{P}}(K_{0}[C_{n}/I_{i}])\cap K_{0}[M_{i}]\subseteq{\mathcal{P}}(K_{0}[M_{i}])$ (see Corollary 3.25). This contradicts Theorem 5.8. Hence, ${\mathcal{P}}(K[C_{n}])\cap K[\tilde{M}]=0$. Since $C_{n}\setminus\tilde{M}$ is finite by Proposition 2.14, it follows that ${\mathcal{P}}(K[C_{n}])$ is finite dimensional. Therefore, $K[C_{n}]$ also is right and left Noetherian. ∎

Theorem 5.10**.**

Let $\Theta$ be a finite oriented graph. Then the following conditions are equivalent

$K[\operatorname{HK}_{\Theta}]$* is right Noetherian,* 2. 2)

$K[\operatorname{HK}_{\Theta}]$* is left Noetherian,* 3. 3)

each of the connected components of $\Theta$ is either an oriented cycle or an acyclic graph.

Proof.

Assume that condition 3) is satisfied. From [13] we know that $\operatorname{HK}_{\Theta}$ is a PI–algebra. In order to prove conditions 1) and 2) we proceed by induction on the number $k$ of connected components of $\Theta$. If $k=1$ then the assertion follows from Theorem 5.9 and from the fact that $\operatorname{HK}_{\Theta}$ is finite if $\Theta$ is an acyclic graph. Assume that $k>1$. Let $\Theta_{1}$ be a connected component of $\Theta$ and let $\Theta_{2}=\Theta\setminus\Theta_{1}$. Clearly, $\operatorname{HK}_{\Theta}$ is a direct product of $\operatorname{HK}_{\Theta_{1}}$ and $\operatorname{HK}_{\Theta_{2}}$, so that $K[\operatorname{HK}_{\Theta}]\cong K[\operatorname{HK}_{\Theta_{1}}]\otimes K[\operatorname{HK}_{\Theta_{2}}]$. By the induction hypothesis, $\operatorname{HK}_{\Theta_{i}}$ is (right and left) Noetherian and it is a PI–algebra, for $i=1,2$. Then $K[\operatorname{HK}_{\Theta}]$ is a Noetherian algebra by [3], Proposition 4.4 (which says that every finitely generated right Noetherian PI–algebra is a universally right Noetherian algebra).

Assume that 3) is not satisfied. Then $\Theta$ contains a subgraph $\Theta^{\prime}$ that is of the form described in Lemma 5.1 or the graph $\Theta^{\prime\prime}$ obtained from $\Theta^{\prime}$ by inverting all arrows. It is easy to see that in this case $K[\operatorname{HK}_{\Theta^{\prime}}]$, respectively $K[\operatorname{HK}_{\Theta^{\prime\prime}}]$, is a homomorphic image of $K[\operatorname{HK}_{\Theta}]$, as noticed in [6]. Moreover, $\Theta^{\prime}$ and $\Theta^{\prime\prime}$ are antiisomorphic. Therefore, Lemma 5.1 implies that $K[\operatorname{HK}_{\Theta}]$ is neither right nor left Noetherian. The result follows. ∎

From the proof it actually follows that the conditions in Theorem 5.10 are satisfied if and only if the monoid $\operatorname{HK}_{\Theta}$ has acc on right (left) ideals.

Since $K[C_{n}]$ is a PI–algebra [12], we derive the following direct consequence from the result of Anan’in [1]. It is of interest in view of the results on faithful representations of various special classes of Hecke–Kiselman monoids, obtained in [6],[7],[10].

Corollary 5.11**.**

$K[C_{n}]$* embeds into the matrix algebra $M_{r}(L)$ over a field $L$, for some $r\geqslant 1$.*

We conclude with some open questions.

Question 5.12**.**

Is $K[C_{n}]$ semiprime for every $n\geqslant 3$?

If this is the case, the main result of [16] allows to strengthen the assertion of Theorem 5.9: in this case $K[C_{n}]$ is a finitely generated module over its Noetherian center; and hence also of Theorem 5.10. Notice that this is the case if $n=3$, by Theorem 4.6. Moreover, the proof of Theorem 5.9 shows that ${\mathcal{P}}(K[C_{n}])$ is finite dimensional.

Because of the main results of Section 3, algebras $K[C_{n}]$ share the flavor of affine cellular algebras, introduced in [9], also see [5]. This motivates our second question.

Question 5.13**.**

Does $K[C_{n}]$ admit a structure of a cellular algebra?