Solutions of the Yang–Baxter Equation and Strong Semilattices of Skew Braces

Abstract

We prove that any set-theoretic solution of the Yang–Baxter equation associated with a dual weak brace is a strong semilattice of non-degenerate bijective solutions. This fact makes use of the description of any dual weak brace S we provide in terms of strong semilattice Y of skew braces $B_\alpha$, with $\alpha \in Y$. Additionally, we describe the ideals of S and study its nilpotency by correlating it to that of each skew brace $B_\alpha$.

1 Introduction

The quantum Yang–Baxter equation takes its name from two independent works by Yang [35] and Baxter [3]. It is an important tool in several fields of research, among these are statistical mechanics, quantum field theory, and quantum group theory, whose study of solutions has been a major research area for the past 50 years. The challenging problem of determining all the set-theoretic solutions arose in 1992 in the paper by Drinfel’d [15] and is still open. Into the specific, given a set S, a map $r:S\times S\rightarrow S\times S$ is said to be a set-theoretic solution of the Yang–Baxter equation, or briefly a solution, if r satisfies the braid identity

$$\begin{aligned} \left( r\times {{\,\textrm{id}\,}}_S\right) \left( {{\,\textrm{id}\,}}_S\times r\right) \left( r\times {{\,\textrm{id}\,}}_S\right) = \left( {{\,\textrm{id}\,}}_S\times r\right) \left( r\times {{\,\textrm{id}\,}}_S\right) \left( {{\,\textrm{id}\,}}_S\times r\right) . \end{aligned}$$

Writing $r\left( x,y\right) = \left( \lambda _x\left( y\right) , \rho _y\left( x\right) \right)$, with $\lambda _x, \rho _x$ maps from S into itself, then r is left non-degenerate if $\lambda _x\in {{\,\textrm{Sym}\,}}_S$, right non-degenerate if $\rho _x\in {{\,\textrm{Sym}\,}}_S$, for every $x\in S$, and non-degenerate if r is both left and right non-degenerate.

Several techniques for constructing solutions starting from known solutions have been introduced over the years. For the purposes of this paper, we mention the strong semilattice of solutions [8], which is a method that allows for determining solutions starting from a semilattice Y, a family of disjoint sets $\left\{ X_{\alpha }\ |\ \alpha \in Y\right\}$, and solutions $r_\alpha$ defined on these sets.

In 2007, Rump [31] innovatively showed how an involutive non-degenerate solution can be obtained starting from the special algebraic structure of brace. This type of approach had a large following in the last few years and other similar structures have been studied. Among these, we mention the weak brace [9] that is a triple $\left( S,+,\circ \right)$ such that $\left( S,+\right)$ and $\left( S,\circ \right)$ are inverse semigroups satisfying

$$\begin{aligned} a\circ \left( b + c\right) = a\circ b - a + a\circ c \qquad \text {and} \qquad a\circ a^- = -a + a, \end{aligned}$$

for all $a,b,c\in S$, where $-a$ and $a^-$ denote the inverses of a with respect to $+$ and $\circ$, respectively. Clearly, the sets of the idempotents ${{\,\textrm{E}\,}}(S,+)$ and ${{\,\textrm{E}\,}}(S, \circ )$ coincide, so we will simply write ${{\,\textrm{E}\,}}(S)$. In particular, if $|{{\,\textrm{E}\,}}(S)|=1$, then $(S, +)$ and $(S, \circ )$ are groups having the same identity, and so $(S, +, \circ )$ is a skew brace [17] which is a brace if the group $(S,+ )$ is abelian. Necessarily, the additive structure is a Clifford semigroup, instead, in general, the multiplicative one is not. A class of weak braces having $(S, \circ )$ as a Clifford semigroup is obtained in [10, Theorem 16]. We call dual the weak braces for which $(S,\circ )$ is Clifford.

Any weak brace $(S, +, \circ )$ gives rise to a solution $r:S\times S\rightarrow S\times S$ defined by

$$\begin{aligned} r\left( a,b\right) = \left( -a + a\circ b, \ \left( -a + a\circ b\right) ^-\circ a\circ b\right) , \end{aligned}$$

for all $a,b\in S$ (see [9, Theorem 11]). Such a map r is close to being bijective, and, in the case of a dual one, r is also close to being non-degenerate (see [10, pp. 604–605]). Besides, a new family of solutions coming from dual weak braces has been investigated in [24]. For this kind of structure, a notion of ideal has been also introduced in [10, Definition 20]. Moreover, it turns out that the quotient structure is a new dual weak brace with semilattice of idempotents isomorphic to ${{\,\textrm{E}\,}}\left( S\right)$.

In this paper, we entirely describe the structure of a dual weak brace $(S, +, \circ )$, by showing that it is a strong semilattice Y of skew braces $\left( B_{\alpha }, +, \circ \right)$, for every $\alpha \in Y$, where $\left( B_\alpha , +\right)$ and $\left( B_\alpha , \circ \right)$ are the groups fulfilling the structure of $\left( S, +\right)$ and $\left( S, \circ \right)$, respectively, as Clifford semigroups (see [18, Theorem 4.2.1]). As a consequence, we prove that the solution r associated with S is the strong semilattice Y of the non-degenerate bijective solutions $r_\alpha$ on each $B_\alpha$. Any strong semilattice of skew braces is realized by combining skew brace homomorphisms $\phi _{\alpha ,\beta }$ from $B_\alpha$ to $B_\beta$ (whenever $\alpha \ge \beta$), thus this further motivates the study of such maps, a problem already emerged in literature (cf. [11, Problem 10.2], [34, Problem 2.18], [26, 29, 30]).

Despite the obtained description, the skew brace theory is not exhaustive for developing the theory of dual weak braces. In fact, for instance, although we show the ideals of any dual weak brace S are specific strong semilattices of ideals of every skew brace $B_\alpha$, if we consider known ideals, such as its socle ${{\,\textrm{Soc}\,}}(S)$, in general, it is not the strong semilattice Y of each ${{\,\textrm{Soc}\,}}\left( B_\alpha \right)$. This led us to deepen the theory of dual weak braces and not just reduce it to the study of every skew brace $B_\alpha$. As a first step, we introduce the binary operation $\cdot$ on S given by $a\cdot b:= -a + a\circ b -b$, for all $a,b\in S$, classically known in the context of radical Jacobson rings. We give some properties that are useful to characterize the ideals of S in terms of the operation $\cdot$. Furthermore, this has allowed us to investigate the right nilpotency and the nilpotency of S by relating them with those of each skew braces $B_\alpha$. We highlight that nilpotency in skew braces has been intensively studied over the years by many authors (see, for instance, [1, 2, 4,5,6, 12, 20, 32]) and it is still under investigation above all concerning multipermutation solutions [16].

2 Basics on Weak Braces

This section aims to give actual results on the structures of weak braces [9] paying particular attention to the behavior of the idempotents.

To make this paper self-contained and to set up the notation, throughout the paper where it will be needed, we will recall some notions contained in classical books on inverse semigroups, as [14, 18, 23, 27]. A semigroup S is inverse if for each $a\in S$, there exists a unique $a^{-1}\in S$ such that $a=aa^{-1}a$ and $a^{-1}=a^{-1}aa^{-1}$. We call such an element $a^{-1}$ the inverse of a. The class of inverse semigroups is very close to that of groups since $(a b)^{-1}=b^{-1} a^{-1}$ and $(a^{-1})^{-1}=a$, for all $a,b \in S$. Note that $aa^{-1}$ and $a^{-1}a$ are the idempotents of S, for every $a \in S$. An inverse semigroup S is called Clifford if its idempotents are central, or, equivalently, $aa^{-1}=a^{-1}a$, for every $a \in S$.

Definition 1

[9, Definition 5] Let S be a set endowed with two operations $+$ and $\circ$ such that $\left( S,+\right)$ and $\left( S,\circ \right)$ are inverse semigroups. Then, $(S, +, \circ )$ is a weak brace if

$$\begin{aligned} a\circ \left( b + c\right) = a\circ b -a + a\circ c\qquad \text {and}\qquad a\circ a^-=-a+a, \end{aligned}$$

for all $a,b,c\in S$, where $-a$ and $a^-$ denote the inverses of a with respect to $+$ and $\circ$, respectively.

Clearly, the sets of the idempotents $E(S,+)$ and $E(S,\circ )$ coincide, thus we will simply denote such a set by E(S). Obviously, if $|{{\,\textrm{E}\,}}(S)|=1$, then $(S,+, \circ )$ is a skew brace [17].

In [9, Theorem 8], it is proved that the additive semigroup of any weak brace is necessarily Clifford. In general, the multiplicative one is not (see [9, Example 3]). Any Clifford semigroup $\left( S, \circ \right)$ determines two trivial weak braces having both Clifford structures, by setting $a + b:= a\circ b$ or $a + b:= b\circ a$, for all $a,b\in S$. A bigger class of weak braces having $(S, \circ )$ Cifford is studied in [10, Theorem 16].

Definition 2

[10, Definition 2] A weak brace $(S, +, \circ )$ is called dual if $(S, \circ )$ is Clifford.

Given a weak brace $(S,+, \circ )$, let $\lambda : S\rightarrow {{\,\textrm{End}\,}}(S,+), \,a\mapsto \lambda _a$ and $\rho : S\rightarrow {{\,\textrm{Map}\,}}(S), \,b\mapsto \rho _b$ be the maps defined by

$$\begin{aligned} \lambda _a\left( b\right) = -a+a\circ b\qquad \text {and} \qquad \rho _b\left( a\right) = \left( -a+a\circ b\right) ^{-}\circ a \circ b, \end{aligned}$$

for all $a,b\in S$, respectively. One has that $\lambda _a(b)=a\circ \left( a^- +b \right)$, for all $a,b \in S$, and $\lambda _a(E(S)) \subseteq E(S)$. Besides, the map $\lambda$ is a homomorphism of $\left( S,\circ \right)$ into the endomorphism semigroup of $\left( S,+\right)$ and the map $\rho$ is an anti-homomorphism of $\left( S,\circ \right)$ into the monoid ${{\,\textrm{Map}\,}}(S)$. Following [9, Theorem 11], the map $r:S\times S\rightarrow S\times S$ defined by $r\left( a,b\right) = \left( \lambda _a(b), \ \rho _b(a)\right)$, for all $a,b\in S$, is a solution that has a behavior close to bijectivity: indeed, considered the solution $r^{op}$ associated with the opposite weak brace $S^{op}:=(S, +^{op}, \circ )$ of S they hold

$$\begin{aligned} r\, r^{op}\, r = r, \qquad r^{op}\, r\, r^{op} = r^{op}, \qquad \text {and}\qquad rr^{op} = r^{op}r. \end{aligned}$$

In addition, if S is dual, r has also a behavior close to the non-degeneracy since

$$\begin{aligned} \lambda _a\lambda _{a^-}\lambda _a = \lambda _{a}, \qquad \lambda _{a^-}\lambda _{a}\lambda _{a^-} = \lambda _{a^-}, \quad&\text {and} \qquad \lambda _a\lambda _{a^-} =\lambda _{a^-}\lambda _{a}, \\ \rho _{a}\rho _{a^-}\rho _{a} = \rho _a, \qquad \rho _{a^-}\rho _{a}\rho _{a^-} = \rho _{a^-}, \qquad&\text {and} \qquad \rho _{a}\rho _{a^-} = \rho _{a^-}\rho _{a}, \end{aligned}$$

for every $a \in S$. Clearly, if S is a skew brace, r is non-degenerate and bijective with $r^{-1} = r^{op}$ [21]. Moreover, S is a brace if and only if r is involutive.

In the lemma below, we collect some useful properties of weak braces provided in [9, 10].

Lemma 1

Let $(S,+, \circ )$ be a weak brace. Then, the following hold:

• 1. $\lambda _a\left( a^-\right) = - a$,

• 2. $a \circ \left( -b\right) =a-a\circ b+a$,

• 3. $a\circ b= a + \lambda _a\left( b\right)$,

• 4. $a+b=a \circ \lambda _{a^-}\left( b\right) ,$

for all $a,b \in S$.

The following key lemma highlights the behavior of idempotents in any weak brace.

Lemma 2

If $(S,+,\circ )$ is a weak brace and $e\in {{\,\textrm{E}\,}}(S)$, then $\rho _e\left( a\right) =a \circ e$ and

$$\begin{aligned} \lambda _e(a)=e \circ a=e+a, \end{aligned}$$

(1)

for every $a\in S$. In particular, if S is a dual weak brace, then $e\circ a = e + a = a\circ e = a + e = \lambda _e\left( a\right) = \rho _e\left( a\right)$, for every $a\in S$..

Proof

Let $a \in S$. Then, by Lemma 1-3., $e \circ a=e +\lambda _e\left( a\right) =e-e+e\circ a=\lambda _e\left( a\right)$. As a consequence, we obtain $\lambda _e(a) = e \circ e \circ \left( e+ a\right) = e \circ \lambda _e(a) = e + a,$ where the last equality follows from Lemma 1-4.. In addition, by [9, Proposition 9-3.], $\rho _e\left( a\right) ^- = e \circ a^--e=\lambda _e\left( a^-\right) =e \circ a^-$, hence $\rho _e\left( a\right) = a\circ e$, thus the claim is satisfied. $\square$

In light of Lemma 2, in any dual weak brace S the set of the idempotents ${{\,\textrm{E}\,}}(S)$ gives rise to a structure of trivial weak sub-brace of S, the general definition of which is given below.

Definition 3

Let $(S,+, \circ )$ be a weak brace and $H\subseteq S$. Then, H is said to be a weak sub-brace of $(S, +, \circ )$ if H both is an inverse semigroup of $(S,+)$ and $(S,\circ )$ such that ${{\,\textrm{E}\,}}(S)\subseteq H$.

Convention. From now on, we will only deal with dual weak braces and briefly write $a^0$ to denote the idempotent $a-a=a\circ a^-$, for every $a \in S$.

It follows by Lemma 2 that $a\circ b=a^0+a \circ b=a \circ b+b^0$, for all $a,b \in S$.

In the last part of this section, we introduce a new operation on a dual weak brace $(S, +, \circ )$ that will allow characterizing its ideals. Such an operation is the usual multiplication in the context of rings and is already known for skew braces [12, 22] and cancellative semi-braces [6]. Specifically, we define the operation $\cdot$ on S given by

$$\begin{aligned} a\cdot b:= -a + a\circ b - b, \end{aligned}$$

for all $a,b\in S$, that can be also written as $a\cdot b = \lambda _a\left( b\right) - b$ and, by Lemma 2, $a\cdot b=a^0+a \cdot b=a \cdot b+b^0$. As it is usual in brace theory (cf. [31, Definition 2]), by (1), one has that

$$\begin{aligned} \lambda _a\left( b\right) = - a + a\circ b+ b^0 = a\cdot b + b, \end{aligned}$$

(2)

for all $a,b\in S$.

Lemma 3

Let $(S, +, \circ )$ be a dual weak brace and $a,b \in S$. Then, the following are equivalent:

• (i) $a \cdot b \in {{\,\textrm{E}\,}}(S)$,

• (ii) $a\cdot b=a^0+b^0$,

• (iii) $a+b=a \circ b$.

Proof

If $a\cdot b\in {{\,\textrm{E}\,}}(S)$, then we have that

$$\begin{aligned} a\cdot b = a\cdot b - a\cdot b = - a + a\circ b + b^0 - a\circ b +a = a^0 + b^0 + \left( a\circ b\right) ^0 \underset{(1)}{=} a^0 + b^0. \end{aligned}$$

Now, if $a\cdot b = a^0 + b^0$, then $a + b = a + a^0 + b^0 + b = a + a\cdot b + b \underset{(1)}{=}a\circ b$. Finally, if $a+b=a \circ b$, we obtain that $a\cdot b = -a + a\circ b - b = -a + a + b - b = a^0 + b^0\in {{\,\textrm{E}\,}}(S)$, which completes the proof. $\square$

Using Lemma 3 and (1), we immediately obtain that, for all $a\in S$ and $e\in E\left( S\right)$,

$$\begin{aligned} a \cdot e = e \cdot a \in {{\,\textrm{E}\,}}(S). \end{aligned}$$

(3)

Proposition 1

Let $(S, +, \circ )$ be a dual weak brace. Then, the following are satisfied:

• 1. $a \circ b=a+a\cdot b+b$

• 2. $a\cdot \left( b + c\right) = a\cdot b + b + a\cdot c - b$,

• 3. $\left( a\circ b\right) \cdot c = a\cdot \left( b\cdot c\right) + b\cdot c + a\cdot c$,

for all $a,b,c\in S$.

Proof

Let $a, b, c \in S$. Firstly, by (1), we have $a \circ b= a^0+a \circ b + b^0=a + a\cdot b + b.$ Moreover, $a\cdot \left( b + c\right) =-a+a\circ b+b^0-a+a\circ c-c-b=a\cdot b + b + a\cdot c - b.$ Finally, using (1), we get

$$\begin{aligned} a&\cdot \left( b\cdot c\right) + b\cdot c + a\cdot c\\&=-a+a\circ \left( b \cdot c\right) +a \cdot c \\&=-a\circ b+a\circ (b \circ c-c)+a \cdot c&\text{ by } \text{ Lemma }\,1-2.\\&=-a\circ b+a\circ b \circ c-a\circ c+a\circ c-c&\text{ by } \text{ Lemma }\,2-2.\\&=-a \circ b + a \circ b \circ c \circ a^0 \circ c^0-c \\&=-a\circ b+a \circ b \circ c-c \\&=\left( a \circ b\right) \cdot c, \end{aligned}$$

which completes our claim. $\square$

Corollary 1

Let $(S, +, \circ )$ be a dual weak brace. Then, they hold:

• 1. $a\cdot \left( b + e\right) =a\cdot b + a\cdot e$,

• 2. $\left( e + a\right) \cdot b=e\cdot b + a\cdot b$,

for all $a,b \in S$ and $e \in E(S)$.

Proof

If $a,b \in S$ and $e \in E(S)$, we obtain

$$\begin{aligned} a\cdot \left( b + e\right)&= a\cdot b + b + a\cdot e -b&\text{ by } \text{ Proposition }\,1-2.\\&= a\cdot b + b^0 + a\cdot e&\text{ by } (3)\\&= a\cdot b + a\cdot e&\text{ by } (1) \end{aligned}$$

and

$$\begin{aligned} \left( e + a\right) \cdot b&=\left( e \circ a\right) \cdot b&\text{ by } (1) \\&=e \cdot (a \cdot b) + a \cdot b+e \cdot b&\text{ by } \text{ Proposition }\,1-3.\\&=e \circ (a \cdot b)^0 + a \cdot b+e \cdot b&\text{ by } (1)\\&=e \cdot b+a \cdot b&\text{ by } (3)- (1) \end{aligned}$$

Therefore, the claim follows. $\square$

3 A Description of Dual Weak Braces and Their Solutions

In this section, we provide a description theorem for dual weak braces by showing that they are strong semilattices of specific skew braces. This description is consistent with the fact that Clifford semigroups are strong semilattices of groups (see [27, Theorem II.2.6]).

Let Y be a (lower) semilattice and $\{G_{\alpha }\ | \ \alpha \in Y\}$ a family of disjoint groups. For all $\alpha ,\beta \in Y$ such that $\alpha \ge \beta$, let $\phi _{\alpha ,\beta }:G_{\alpha }\rightarrow G_{\beta }$ be a homomorphism of groups such that

• 1. $\phi _{\alpha ,\alpha }$ is the identical automorphism of $G_{\alpha }$, for every $\alpha \in Y$,

• 2. $\phi _{\beta ,\gamma }{}\phi _{\alpha ,\beta }{} = \phi _{\alpha ,\gamma }{}$, for all $\alpha , \beta , \gamma \in Y$ such that $\alpha \ge \beta \ge \gamma$.

Then, endowed with the operation defined by $a\ b:= \phi _{\alpha ,\alpha \beta }(a)\ \phi _{\beta ,\alpha \beta }(b),$ for all $a\in G_{\alpha }$ and $b\in G_{\beta }$, is a Clifford semigroup, also called strong semilattice Y of groups $G_{\alpha }$, usually written as $S=[Y; G_{\alpha }; \phi _{\alpha ,\beta }]$. Conversely, any Clifford semigroup is of this form.

Theorem 1

Let Y be a (lower) semilattice, $\left\{ B_{\alpha }\ \left| \ \alpha \in Y\right. \right\}$ a family of disjoint skew braces. For each pair $\alpha ,\beta$ of elements of Y such that $\alpha \ge \beta$, let $\phi _{\alpha ,\beta }:B_{\alpha }\rightarrow B_{\beta }$ be a skew brace homomorphism such that

• 1. $\phi _{\alpha ,\alpha }$ is the identical automorphism of $B_{\alpha }$, for every $\alpha \in Y$,

• 2. $\phi _{\beta ,\gamma }{}\phi _{\alpha ,\beta }{} = \phi _{\alpha ,\gamma }{}$, for all $\alpha , \beta , \gamma \in Y$ such that $\alpha \ge \beta \ge \gamma$.

Then, $S = \bigcup \left\{ B_{\alpha }\ \left| \ \alpha \in Y\right. \right\}$ endowed with the addition and the multiplication, respectively, defined by

$$\begin{aligned} a+b:= \phi _{\alpha ,\alpha \beta }(a)+\phi _{\beta ,\alpha \beta }(b) \quad \text {and}\quad a\circ b:= \phi _{\alpha ,\alpha \beta }(a)\circ \phi _{\beta ,\alpha \beta }(b), \end{aligned}$$

for all $a\in B_{\alpha }$ and $b\in B_{\beta }$, is a dual weak brace with ${{\,\textrm{E}\,}}(S)$ isomorphic to Y, called strong semilattice S of skew braces $B_{\alpha }$ and denoted by $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$. Conversely, any dual weak brace is a strong semilattice of skew braces.

Proof

The proof of the sufficient condition is contained [10, Corollary 6]. Conversely, let $\left( S,+,\circ \right)$ be a dual weak brace and let $[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ and $[Z; H_i;\psi _{{i},{j}}]$ the two Clifford semigroups $\left( S,+\right)$ and $\left( S,\circ \right)$, respectively. Since ${{\,\textrm{E}\,}}\left( S,+\right)$ and ${{\,\textrm{E}\,}}\left( S,\circ \right)$ coincide and they are isomorphic to Y and Z, respectively, there exists a semilattice isomorphism $f:Y\rightarrow Z$, that, formally, can be written as $f\left( \alpha \right) = i_{\alpha }$, for every $\alpha \in Y$. Keeping in mind this fact, it is not restrictive to consider $Y = Z$. Now, let $\alpha \in Y$ and show that $B_\alpha = H_\alpha$. Thus, if $a\in B_\alpha$, then there exists $\beta \in Y$ such that $a\in H_\beta$. Denoted by $e_\alpha$ and $e_\beta$ the identities of the groups $\left( B_\alpha , +\right)$ and $\left( H_\beta , \circ \right)$, respectively, note that

$$\begin{aligned} e_\alpha = a - a = a\circ a^- = e_\beta , \end{aligned}$$

hence $a\in B_\alpha \cap H_\alpha \subseteq H_\alpha$. Reversing the role of $+$ and $\circ$, one similarly obtains that $H_\alpha \subseteq B_\alpha$. Finally, if $\alpha ,\beta \in Y$ and $\alpha \ge \beta$, observe that

$$\begin{aligned} \phi _{\alpha , \beta }\left( a\right) = a + e_\beta \underset{(1)}{=} a\circ e_\beta = \psi _{\alpha ,\beta }\left( a\right) , \end{aligned}$$

for every $a\in B_\alpha$. Therefore, $\left\{ B_{\alpha }\ \left| \ \alpha \in Y\right. \right\}$ is a family of disjoint skew braces and the claim follows. $\square$

The following result is a consequence of [27, Proposition II.2.8] and Theorem 1.

Proposition 2

Two dual weak braces $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ and $T=[Z; C_{i}; \psi _{i, j}]$ are isomorphic if and only if there exists a semilattice homomorphism $\eta : Y \rightarrow Z$ and a family of skew brace isomorphisms $\{\varTheta _\alpha : B_\alpha \rightarrow C_{\eta (\alpha )}\mid \alpha \in Y\}$ such that $\varTheta _\beta \phi _{\alpha , \beta }=\psi _{\eta (\alpha ), \eta (\beta )}\varTheta _\alpha ,$ for every $\alpha \ge \beta$.

The following example shows that Theorem 1 allows for obtaining new dual weak braces even if we start from trivial skew braces.

Example 1

Let us consider $Y=\{\alpha , \beta \}$, with $\alpha > \beta$, $B_\alpha$ and $B_\beta$ the trivial skew braces on the cyclic group $C_3$ and on the symmetric group ${{\,\textrm{Sym}\,}}_3$ of order 3, respectively, and $\phi _{\alpha , \beta }: C_3 \rightarrow {{\,\textrm{Sym}\,}}_3$ the homomorphism given by $\phi _{\alpha , \beta }(0)={{\,\textrm{id}\,}}_3$, $\phi _{\alpha , \beta }(1)=(123)$, $\phi _{\alpha , \beta }(2)=(132)$. Then, $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ is a dual weak brace.

The following natural question arises intending to concretely construct dual weak braces.

Question Determining all skew brace homomorphisms.

Note that the problem of studying homomorphisms between skew braces has emerged yet in the literature, such as [11, Problem 10.2], [34, Problem 2.18], [26, 29], and [30], and also in a recent conference talk held by Civino [13].

Now, by Theorem 1, we illustrate that the solution r associated with any dual weak brace $S=[Y; B_\alpha ;\phi _{\alpha , \beta }]$ is the strong semilattice of the solutions $r_{\alpha }$ on each skew brace $B_{\alpha }$. In this regard, we recall the more general result contained in [8, Theorem 4.1].

Theorem 2

Let Y be a (lower) semilattice, let $\left\{ r_{\alpha }\ \left| \ \alpha \in Y \right. \right\}$ be a family of disjoint solutions on $X_{\alpha }$ indexed by Y such that for each pair $\alpha , \beta \in Y$ with $\alpha \ge \beta$, there is a map $\phi _{\alpha , \beta }:X_{\alpha }\rightarrow X_{\beta }$. Let X be the union

$$\begin{aligned} X = \bigcup \left\{ X_{\alpha }\ \left| \ \alpha \in Y\right. \right\} \end{aligned}$$

and let $r:X\times X \longrightarrow X\times X$ be the map defined as

$$\begin{aligned} r\left( x, y\right) := r_{\alpha \beta }\left( \phi _{\alpha ,\alpha \beta }\left( x\right) , \phi _{\beta , \alpha \beta }\left( y\right) \right) , \end{aligned}$$

for all $x\in X_{\alpha }$ and $y\in X_{\beta }$. Then, if the following conditions are satisfied

• 1. $\phi _{\alpha , \alpha }$ is the identity map of $X_{\alpha }$, for every $\alpha \in Y$,

• 2. $\phi _{\beta , \gamma }\phi _{\alpha , \beta } = \phi _{\alpha , \gamma }$, for all $\alpha , \beta , \gamma \in Y$ such that $\alpha \ge \beta \ge \gamma$,

• 3. $\left( \phi _{\alpha ,\beta }\times \phi _{\alpha ,\beta }\right) r_{\alpha } = r_{\beta }\left( \phi _{\alpha , \beta }\times \phi _{\alpha , \beta }\right)$, for all $\alpha , \beta \in Y$ such that $\alpha \ge \beta$.

then r is a solution on X called a strong semilattice of solutions $r_{\alpha }$ indexed by Y.

Given a dual weak brace $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$, clearly the conditions of Theorem 2 are satisfied by each solution $r_{\alpha }$ on each skew brace $B_\alpha$. In particular, the condition 3. follows by the fact that the map $\phi _{\alpha ,\beta }: B_\alpha \rightarrow B_\beta$ is a skew brace homomorphism, for every $\alpha \ge \beta$. Moreover, it is easy to see that also the converse is true. Thus, we can sum it all up in the following result.

Proposition 3

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be a dual weak brace. Then, the solution r associated with S is the strong semilattice of the solutions $r_\alpha$ associated with each skew brace $B_{\alpha }$.

To be thorough, we know from [33, Theorem 4.13] that in the finite case the order of each solution $r_{\alpha }$, or equivalently the period $p\left( r_{\alpha }\right)$ of $r_\alpha$, for every $\alpha \in Y$, is even and, by [8, Theorem 5.2], we can establish the order of the overall solution r. In the next result, for a finite dual weak brace $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ we mean that Y and $B_\alpha$ are finite, for every $\alpha \in Y$.

Corollary 2

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be a finite dual weak brace and r the solution associated with S. Then, $r^{2k+1}=r$ with $2k={{\,\textrm{lcm}\,}}\{p\left( r_{\alpha }\right) \, \mid \, \alpha \in Y\}$, where $r_\alpha$ is the solution on each skew brace $B_\alpha$, for every $\alpha \in Y$.

Corollary 3

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be a strong semilattice of braces such that Y is finite. Then, the solution r associated with S is cubic, namely, $r^3=r$.

4 Ideals of Dual Weak Braces

In this section, we entirely describe the structure of any ideal of a dual weak brace $S=[Y; B_\alpha ; \phi _{\alpha , \beta }]$ through ideals of each skew brace $B_\alpha$.

Let us start by recalling the notion of ideal contained in [10, Definition 20]. To this end, by a normal subsemigroup I of a Clifford semigroup S we will mean a subset I of S such that

• 1. ${{\,\textrm{E}\,}}(S) \subseteq I$ (i.e., I is full in S),

• 2. $\forall a, b \in I$    $ab \in I$    and    $a^{-1}\in I$,

• 3. $\forall a \in S$    $a^{-1}Ia\subseteq I$.

Note that such a definition is equivalent to that of normal subset contained in [28, Definition VI.1.2].

Definition 4

(cf. [10, Definition 20]) Let $(S,+,\circ )$ be a dual weak brace and I a subset of S. Then, I is an ideal of $(S,+,\circ )$ if they hold

• 1. I is a normal subsemigroup of $(S,+)$,

• 2. $\lambda _a(I)\subseteq I$, for every $a\in S$,

• 3. I is a normal subsemigroup of $(S,\circ )$.

It follows from the definition of every ideal I that ${{\,\textrm{E}\,}}(S) \subseteq I$. In particular, I is a dual weak sub-brace of S, thus S and ${{\,\textrm{E}\,}}(S)$ are trivial ideals of S. According to [10, Proposition 23], another known ideal of S is its socle (or left annihilator), i.e., the set

$$\begin{aligned} {{\,\textrm{Soc}\,}}\left( S\right) =\{ a \, | \,a \in S, \, \,\forall \ b \in S \quad a+b=a \circ b \quad \text {and} \quad a+b=b+a \}. \end{aligned}$$

In addition, note that if I and J are ideals of S, then, as proved in [10, Proposition 24], $I+J$ and $I \circ J$ also are, and one can easily see from Lemma 1 that they coincide.

In light of Theorem 1, we can give the following structure theorem for every ideal of a dual weak brace.

Theorem 3

Let $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ be a dual weak brace, $I_\alpha$ an ideal of each skew brace $B_\alpha$, and set $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{I_\alpha }}$, for all $\alpha \ge \beta$. If $\phi _{\alpha ,\beta }(I_\alpha )\subseteq I_\beta$, for any $\alpha > \beta$, then $I=[Y; I_{\alpha }; \psi _{\alpha , \beta }]$ is an ideal of S and, conversely, every ideal of S is of this form.

Proof

To show that $I=[Y; I_{\alpha }; \psi _{\alpha , \beta }]$ is an ideal of S, by [27, Exercises III.1.9(ii)], it is enough to prove the $\lambda$-invariance of I. If $a\in S$ and $x\in I$, there exist $\alpha ,\gamma \in Y$ such that $a\in B_{\alpha }$ and $x\in I_{\gamma }$, and so, since by the assumption $\phi _{\gamma , \alpha \gamma }\left( x\right) \in I_{\alpha \gamma }$, we have that

$$\begin{aligned} \lambda _a\left( x\right) = \phi _{\alpha , \alpha \gamma }\left( -a\right) + \phi _{\alpha , \alpha \gamma }\left( a\right) \circ \phi _{\gamma , \alpha \gamma }\left( x\right) = \lambda _{\phi _{\alpha ,\alpha \gamma }\left( a\right) }\phi _{\gamma ,\alpha \gamma }\left( x\right) \in I_{\alpha \gamma }, \end{aligned}$$

hence $\lambda _a\left( x\right) \in I$.

Vice versa, if I is an ideal of $(S,+, \circ )$, since I is a dual weak brace too, it follows by Theorem 1 and [27, Exercises III.1.9(ii)] that $I=[Y; I_{\alpha }; \psi _{\alpha , \beta }]$, with $I_\alpha =B_\alpha \cap I$, for every $\alpha \in Y$, $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{I_\alpha }}$, for all $\alpha \ge \beta$, such that $\phi _{\alpha ,\beta }(I_\alpha )\subseteq I_\beta$, for any $\alpha > \beta$. To get the claim, we prove the $\lambda$-invariance of each $I_\alpha$, i.e., $I_\alpha$ is an ideal of the skew brace $B_\alpha$, for every $\alpha \in Y$. Indeed, we have that $\lambda _a(x) \in B_\alpha \cap I=I_\alpha$, for all $a \in B_\alpha$ and $x \in I_\alpha$. Therefore, the claim follows. $\square$

By Theorem 3, the ideals of any dual weak brace $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ are strong semilattices Y of certain ideals $I_\alpha$ of each skew brace $B_\alpha$. However, the skew brace theory is not exhaustive for developing the theory of dual weak braces. In fact, for instance, if we consider the ideal ${{\,\textrm{Soc}\,}}(S)$, in general, it is not the strong semilattice Y of the socles ${{\,\textrm{Soc}\,}}\left( B_\alpha \right)$ of $B_{\alpha }$ (see Example 1).

Lemma 4

Let $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ be a dual weak brace. Then, denoting by $\phi ^{-}_{\alpha ,\beta }$ the inverse image of $\phi ^{}_{\alpha ,\beta }$, we have

$$\begin{aligned} {{\,\textrm{Soc}\,}}(S) = \big [Y; \bigcap _{\alpha \ge \beta }\phi ^{-}_{\alpha ,\beta }\left( {{\,\textrm{Soc}\,}}(B_\beta )\right) ; \psi _{\alpha ,\beta }\big ], \end{aligned}$$

where $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{{{\,\textrm{Soc}\,}}(B_\alpha )}}$, for all $\alpha \ge \beta$.

Proof

By the proof of Theorem 3, to prove the claim it is enough to show that ${{\,\textrm{Soc}\,}}(S)\cap B_{\alpha } = \bigcap \limits _{\alpha \ge \beta }\phi ^{-}_{\alpha ,\beta }\left( {{\,\textrm{Soc}\,}}(B_\beta )\right)$, for every $\alpha \in Y$. Let $x\in {{\,\textrm{Soc}\,}}(S)\cap B_\alpha$ and $\beta \in Y$ such that $\alpha \ge \beta$. Then, if $b\in B_\beta$,

$$\begin{aligned} \phi _{\alpha ,\beta }\left( x\right) + b&= x + b = x\circ b = \phi _{\alpha ,\beta }\left( x\right) \circ b\\ \phi _{\alpha ,\beta }\left( x\right) + b&= x+b=b+x= b+\phi _{\alpha ,\beta }\left( x\right) , \end{aligned}$$

hence $x\in \bigcap \limits _{\alpha \ge \beta }\phi ^{-}_{\alpha ,\beta }\left( {{\,\textrm{Soc}\,}}(B_\beta )\right)$. Conversely, if $x\in \bigcap \limits _{\alpha \ge \beta }\phi ^{-}_{\alpha ,\beta }\left( {{\,\textrm{Soc}\,}}(B_\beta )\right)$ and $y\in B_\gamma$, with $\gamma \in Y$, then $x\in B_{\alpha }$ and, since $\phi _{\alpha ,\alpha \gamma }\left( x\right) \in {{\,\textrm{Soc}\,}}\left( B_{\alpha \gamma }\right)$, we get

$$\begin{aligned} x + y&=\phi _{\alpha ,\alpha \gamma }\left( x\right) + \phi _{\gamma ,\alpha \gamma }\left( y\right) = \phi _{\alpha ,\alpha \gamma }\left( x\right) \circ \phi _{\gamma ,\alpha \gamma }\left( y\right) = x\circ y \\ x + y&= \phi _{\alpha ,\alpha \gamma }\left( x\right) + \phi _{\gamma ,\alpha \gamma }\left( y\right) = \phi _{\gamma ,\alpha \gamma }\left( y\right) + \phi _{\alpha ,\alpha \gamma }\left( x\right) = y + x, \end{aligned}$$

therefore $x\in {{\,\textrm{Soc}\,}}(S)\cap B_{\alpha }$. $\square$

As a direct consequence of Theorem 3 and Lemma 4, we have the following.

Proposition 4

Let $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ be a dual weak brace, $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{{{\,\textrm{Soc}\,}}(B_\alpha )}}$, for all $\alpha \ge \beta$, and assume that $I:=[Y; {{\,\textrm{Soc}\,}}\left( B_{\alpha }\right) ; \psi _{\alpha , \beta }]$ is an ideal of S. Then, $I={{\,\textrm{Soc}\,}}(S)$.

According to [10, Theorem 21], ideals allow for obtaining quotient structures. Into the specific, if I is an ideal of a dual weak brace $(S,+,\circ )$, then the relation $\sim _{I}$ on S given by

$$\begin{aligned} \forall \ a,b\in S\qquad a\sim _I b \ \iff \ a^0 = b^0 \ \ \text {and} \ \ -a + b\in I \end{aligned}$$

is an idempotent separating congruence of $(S,+, \circ )$. Furthermore, $S/I:=S/\sim _I$ is a dual weak brace with semilattice of idempotents isomorphic to ${{\,\textrm{E}\,}}\left( S\right)$.

As it is natural to expect, we can consider the canonical epimorphism $\chi : S\rightarrow S/I, \ a\mapsto a+I$ and give the usual homomorphism theorems.

Theorem 4

Let $\phi : S\rightarrow T$ be a homomorphism between two dual weak braces $(S,+,\circ )$ and $(T,+,\circ )$. Then, the following statements hold:

• 1. ${{\,\textrm{Im}\,}}\,\phi$ is a dual weak sub-brace of $(T,+,\circ )$;

• 2. $\ker \phi =\{a \mid a \in S \quad \exists \, e \in {{\,\textrm{E}\,}}(S) \quad \phi (a)=\phi (e) \}$ is an ideal of $(S, +, \circ )$;

• 3. there exists a monomorphism $\widetilde{\phi }:S/{\ker \phi }\rightarrow T$ of dual weak braces such that ${{\,\textrm{Im}\,}}\,\widetilde{\phi } = {{\,\textrm{Im}\,}}\,\phi$ and the diagram

  

  commutes.

Remark 1

Let us observe that any ideal I of a dual weak brace S is the kernel of the canonical epimorphism $\chi$. Indeed, if $a\in I$, then $a + I = a^0 + I$, hence $a\in \ker \chi$. Conversely, if $a\in \ker \chi$, then there exists $e\in E(S)$ such that $\chi \left( a\right) = \chi \left( e\right)$, hence $a = a + a^0\in a + I = e+I\subseteq I$.

Corollary 4

Let $(S,+,\circ )$ be a dual weak brace, I an ideal of S, and H a dual weak sub-brace of S. Then,

• 1. $I+H$ is a dual weak sub-brace of S;

• 2. I is an ideal of $I+H$;

• 3. $I \cap H$ is an ideal of H;

• 4. $H/ {\left( I\cap H\right) }$ is isomorphic to $\left( I+H\right) / I$.

Corollary 5

Let $(S, +, \circ )$ be a dual weak brace and I, J ideals of S such that $J \subset I$. Then, S/I is isomorphic to $\left( S/J\right) /\left( I/J\right)$.

Corollary 6

Let I be an ideal of a dual weak brace $(S,+, \circ )$. There is a one-to-one correspondence between the set of ideals of S containing I and the set of ideals of S/I. Moreover, ideals of S containing I correspond to ideals of S/I.

5 Left Ideals of Dual Weak Braces

In this section, we provide a characterization of ideals of any dual weak brace which makes use of the concept of left ideals and the operation $\cdot$.

Let us start by introducing the notions of the left ideal and strong left ideal of a dual weak brace, consistently with [12] and [19, Definition 2.3].

Definition 5

Let $(S,+,\circ )$ be a dual weak brace. Then, a subset I of S is a left ideal of $(S,+,\circ )$ if

• 1. I is a full inverse subsemigroup of $(S,+)$,

• 2. $\lambda _a(I)\subseteq I$, for every $a\in S$.

A left ideal I is a strong left ideal if I is a normal subsemigroup of $(S,+)$.

Note that any left ideal I of a dual weak brace S is a full inverse subsemigroup of $\left( S,\circ \right)$. Indeed, by Lemma 1, if $a,b\in I$, then $a\circ b = a + \lambda _a\left( b\right) \in I$ and $a^- = \lambda _{a^-}\left( -a\right) \in I$.

Similar to Theorem 3, we can describe the structure of any strong left ideal, as follows.

Proposition 5

Let $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ be a dual weak brace, $I_\alpha$ a strong left ideal of each skew brace $B_\alpha$, and set $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{I_\alpha }}$, for all $\alpha \ge \beta$. If $\phi _{\alpha ,\beta }(I_\alpha )\subseteq I_\beta$, for any $\alpha > \beta$, then $I=[Y; I_{\alpha }; \psi _{\alpha , \beta }]$ is a strong left ideal of S and, conversely, every strong left ideal of S is of this form.

Example 2

Let $(S, +, \circ )$ be a dual weak brace. Every full endomorphism-invariant subsemigroup I of $(S,+)$, i.e., $\varphi (I) \subseteq I$, for every $\varphi \in {{\,\textrm{End}\,}}(S, +)$, is a strong left ideal (cf. [19, cf. Example 2.4]).

In skew braces theory [12], a known left ideal of a skew brace $(B, +, \circ )$ is ${{\,\textrm{Fix}\,}}(B)$, that is the set of the elements of B that are fixed by the map $\lambda _a$, for every $a \in B$. For a dual weak brace S, it can be defined similarly, as we show below.

Proposition 6

Let $(S,+, \circ )$ be a dual weak brace. Then, the following set

$$\begin{aligned} {{\,\textrm{Fix}\,}}(S):&=\{b \ | \ b \in S, \, \,\forall \ a \in S \quad a+b=a \circ b\}\\&=\{b \ | \ b \in S, \, \,\forall \ a \in S \quad \lambda _a(b)=a^0+b\} \end{aligned}$$

is a left ideal of S.

Proof

Initially, by (1), $E(S) \subseteq {{\,\textrm{Fix}\,}}(S)$. Moreover, $({{\,\textrm{Fix}\,}}(S), +)$ is trivially closed with respect to $+$ and, by Lemma 1-2., $a \circ (-b)=a-a\circ b+a=a-b+a^0=a-b$, for all $a \in S$ and $b \in {{\,\textrm{Fix}\,}}(S)$. Finally, if $a,c \in S$ and $b \in {{\,\textrm{Fix}\,}}(S)$, it follows that

$$\begin{aligned} c +\lambda _a(b)&=c+c^0+a^0+b&b\in {{\,\textrm{Fix}\,}}(S)\\&=c+c^0\circ a^0+b&\text{ by } (1)\\&=c-c\circ a+c\circ a+b\\&=c-c\circ a+c\circ a\circ b&b\in {{\,\textrm{Fix}\,}}(S)\\&=c+\lambda _{c \circ a}(b)\\&=c \circ \lambda _a(b)&\text{ by } \text{ Lemma }\,1-3. \end{aligned}$$

Therefore, ${{\,\textrm{Fix}\,}}(S)$ is a left ideal of S. $\square$

Definition 6

If $(S, +, \circ )$ is a dual weak brace, then we call the set $Zl(S):={{\,\textrm{Fix}\,}}(S) \cap \zeta (S,+)$ the left center (or right annihilator) of S, where $\zeta (S,+)$ denotes the center of $(S,+)$.

It is easy to check that Zl(S) is a strong left ideal of any dual weak brace S. In general, Zl(S) is not an ideal as shown in the context of skew braces in [2, Example 5.7].

In the following, if X and Y are subsets of a dual weak brace $(S, +, \circ )$, we denote by $X \cdot Y$ the additive inverse subsemigroup of S generated by the elements of the form $x \cdot y$, with $x \in X$ and $y \in Y$ (cf. [27, Definition II.1.11]).

Proposition 7

Let $(S,+, \circ )$ be a dual weak brace. Then, the following hold:

• 1. a full inverse subsemigroup I of $\left( S,+\right)$ is a left ideal of S if and only if $S \cdot I \subseteq I$;

• 2. if I is an ideal of S, then $S\cdot I\subseteq I$ and $I\cdot S\subseteq I$;

• 3. a normal subsemigroup I of $\left( S,+\right)$ is an ideal of S if and only if $I\cdot S\subseteq I$ and $\lambda _a\left( I\right) \subseteq I$, for every $a\in S$.

Proof

1. If I is a left ideal of S, $a \in S$ and $x \in I$, then $a\cdot x=\lambda _a\left( x\right) -x \in I$. Conversely, if $S\cdot I\subseteq I$, $a\in S$ and $x\in I$, by (2), we obtain that $\lambda _a\left( x\right) = a\cdot x + x\in I$.

2. By 1., it is enough to see that if $a \in S$ and $x \in I$, then

$$\begin{aligned} x \cdot a=-x+a-a+a \circ a^-\circ x\circ a-a=-x+\underbrace{a+\lambda _a\left( a^-\circ x \circ a\right) -a}_{\in I} \in I. \end{aligned}$$

3. It is enough to show that if $a\in S$ and $x\in I$, then by Lemma 1

$$\begin{aligned} a\circ x\circ a^-&= a + \lambda _a\left( x + \lambda _x\left( a^-\right) \right) =a + \lambda _a\left( x \right) + \lambda _{a\circ x}\left( a^-\right) + a - a\\&= a + \lambda _a\left( x + \lambda _x\left( a^-\right) - a^-\right) - a = a + \lambda _a\underbrace{\left( x +x\cdot a^-\right) }_{\in I} - a\in I, \end{aligned}$$

which completes the proof. $\square$

As a consequence of Proposition 7, one can characterize the ideals of a dual weak brace as is usual in ring theory.

Corollary 7

Let $(S, +, \circ )$ be a dual weak brace. Then, a normal subsemigroup I of $\left( S,+\right)$ is an ideal of S if and only if $S\cdot I\subseteq I$ and $I\cdot S\subseteq I$.

To conclude this section, similarly to [4, Proposition 1.3], we show how to obtain instances of left ideals starting from ideals. If X, Y are ideals of a dual weak brace S, we denote by $\left[ X,\, Y\right] _+ = \bigl \langle \, \left[ x,\, y\right] _+ \ | \ x\in X,\ y\in Y\, \bigr \rangle$.

Proposition 8

Let $\left( S,+,\circ \right)$ be a dual weak brace and I, J ideals of S. Then, $I\cdot J$, $I\cdot J + J\cdot I$, and $\left[ I,\, J\right] _+$ are left ideals of S.

Proof

Initially, note that $I\cdot J$ and $I\cdot J + J\cdot I$ trivially contain ${{\,\textrm{E}\,}}\left( S\right)$. Moreover, $I\cdot J$ clearly is an inverse subsemigroups of $\left( S, +\right)$ and if $s\in S$, $x\in I$, and $h\in J$, then, by Lemma 1-2. and (3),

$$\begin{aligned} s\cdot \left( x\cdot h\right)&= \lambda _{s\circ x}\left( h\right) - \lambda _{s}\left( h\right) - x\cdot h = \lambda _{s\circ x\circ s^-}\lambda _{s}\left( h\right) - \lambda _{s}\left( h\right) - x\cdot h\\&= \left( s\circ x\circ s^-\right) \cdot \lambda _{s}\left( h\right) - x\cdot h\in I\cdot J, \end{aligned}$$

hence, by Proposition 7-1., $I\cdot J$ is a left ideal of S.

To prove that $I\cdot J + J\cdot I$ is an inverse subsemigroup of $\left( S,+\right)$, let us check that $J\cdot I + I\cdot J\subseteq I\cdot J + J\cdot I$. If $x,y\in I$ and $h,k\in J$, since $\left( y\cdot \left( h\cdot x\right) \right) ^0 = y^0 + y^0\circ \left( h\cdot x\right) ^0 + \left( h\cdot x\right) ^0 = \left( h\cdot x\right) ^0 + y^0$ and $y\cdot k = y^0 + y\cdot k$, it follows that

$$\begin{aligned} h\cdot x + y\cdot k&= \left( y\cdot \left( h\cdot x\right) \right) ^0 + h\cdot x + y\cdot k + \left( h\cdot x\right) ^0\\&= - y\cdot \left( h\cdot x\right) + y\cdot \left( h\cdot x + k\right) + h\cdot x \in I\cdot J + J\cdot I, \end{aligned}$$

where in the last equality we use Proposition 1-2.. Furthermore, by the first part of the proof, we trivially get $\lambda _a\left( I\cdot J + J\cdot I\right) \subseteq I\cdot J + J\cdot I$, for every $a\in S$.

Finally, $\left[ I,\, J\right] _+$ trivially is a left ideal of S. Therefore, the claim is proved. $\square$

6 Right Nilpotent Dual Weak Braces

Following [12, 31], if $(S,+,\circ )$ is a dual weak brace, set $S^{(1)}:=S$, we inductively define

$$\begin{aligned} S^{(n+1)}=S^{(n)}\cdot S, \end{aligned}$$

for every $n\ge 1$, and obtain that it is an ideal, whose proof can be proved similarly to [12, Proposition 2.1]. However, unlike what happens for the socle, here we can use the characterization of ideals in Theorem 3.

Proposition 9

If $(S,+,\circ )$ is a dual weak brace, then $S^{(n)}$ is an ideal of S, for every $n \ge 1$.

Proof

Initially, by Theorem 1, we have that $S=[Y; B_\alpha ; \phi _{\alpha ,\beta }]$ and one clearly obtains . The claim follows by Theorem 3, by showing that $\phi _{\alpha ,\beta }\left( B_\alpha ^{(n)}\right) \subseteq B_\beta ^{(n)}$, for any $\alpha > \beta$, by proceeding by an easy induction on n. $\square$

Definition 7

A dual weak brace $(S,+,\circ )$ is said to be right nilpotent if $S^{(n)}=E(S)$ for some $n\ge 1$. The smallest positive integer m such that $S^{(m+1)}={{\,\textrm{E}\,}}(S)$ is called right nilpotency index of S.

It is easy to check that if S is a right nilpotent dual weak brace, then any ideal I and quotient S/I are right nilpotent as well. At the present state of our knowledge, we do not know whether the vice versa is true in general. Surely, it is true when I is the socle of any dual weak brace, as we show in the following result. At first, note that ${{\,\textrm{Soc}\,}}\left( S\right)$ is right nilpotent since, using Lemma 3, it can be rewritten in terms of the $\cdot$ operation, as

$$\begin{aligned} {{\,\textrm{Soc}\,}}\left( S\right) = \{ a \, | \,a \in S, \, \,\forall \ b \in S \quad a\cdot b \in {{\,\textrm{E}\,}}(S) \quad \text {and} \quad a+b=b+a \}. \end{aligned}$$

Proposition 10

Let $(S,+, \circ )$ be a dual weak brace such that $S/{{\,\textrm{Soc}\,}}\left( S\right)$ is right nilpotent. Then, S is right nilpotent.

Proof

Since there exists $m\in \mathbb {N}$ such that $\left( S/{{\,\textrm{Soc}\,}}\left( S\right) \right) ^{\left( m\right) } = E\left( S/{{\,\textrm{Soc}\,}}\left( S\right) \right)$, it follows that $S^{\left( m\right) }\subseteq {{\,\textrm{Soc}\,}}\left( S\right)$. Hence, $S^{\left( m+1\right) } = S^{\left( m\right) }\cdot S\subseteq {{\,\textrm{Soc}\,}}\left( S\right) \cdot S\subseteq E\left( S\right)$. Thus, the claim is proved. $\square$

Definition 8

Let $(S,+, \circ )$ be a dual weak brace. Set ${{\,\textrm{Soc}\,}}_0\left( S\right) := E\left( S\right)$, we define ${{\,\textrm{Soc}\,}}_n\left( S\right)$ to be the ideal of S containing ${{\,\textrm{Soc}\,}}_{n-1}\left( S\right)$ such that

$$\begin{aligned} {{\,\textrm{Soc}\,}}_n\left( S\right) /{{\,\textrm{Soc}\,}}_{n-1}\left( S\right) ={{\,\textrm{Soc}\,}}\left( S/{{\,\textrm{Soc}\,}}_{n-1}\left( S\right) \right) , \end{aligned}$$

for every positive integer n.

In particular, ${{\,\textrm{Soc}\,}}_n\left( S\right) :=\{a \, | \, a\in S, \, \, \forall \, b \in S \quad a \cdot b\in {{\,\textrm{Soc}\,}}_{n-1}\left( S\right) , \,\, [a,b]_{+} \in {{\,\textrm{Soc}\,}}_{n-1}\left( S\right) \},$ for every positive integer n (cf. [6, Remark 28]).

Definition 9

Let $(S,+,\circ )$ be a dual weak brace. An s-series (or left annihilator series) of S is a sequence of ideals of S

$$\begin{aligned} {{\,\textrm{E}\,}}(S) = I_0\subseteq I_1\subseteq I_2\subseteq \cdots \subseteq I_m = S \end{aligned}$$

such that $I_{j}/I_{j-1}\subseteq {{\,\textrm{Soc}\,}}\left( S/{I_{j-1}}\right)$, for each $j\in \{1,\ldots ,m\}$.

One can check that S admits an s-series if and only if there exists a positive integer n such that $S={{\,\textrm{Soc}\,}}_n\left( S\right)$, as in the case of the skew braces (cf. [12, Lemma 2.15]). In light of this fact, we give the following definitions.

Definition 10

A dual weak brace $(S,+,\circ )$ is called left annihilator nilpotent if S admits an s-series. Consequently, if S is left annihilator nilpotent, we call socle series (or upper left annihilator series) the series introduced in Definition 8. The smallest non-negative integer n such that $S={{\,\textrm{Soc}\,}}_n\left( S\right)$ is called left annihilator nilpotency index of S.

Proposition 11

Let $(S,+,\circ )$ be a left annihilator nilpotent dual weak brace. Then, S is right nilpotent.

Proof

Let ${{\,\textrm{E}\,}}(S) = I_0\subseteq I_1\subseteq I_2\subseteq \cdots \subseteq I_m = S$ an s-series of S. Let $k\in \{1,\ldots ,m\}$, we prove that $S^{(k)}\subseteq I_{m-k+1}$ by proceeding by induction on k. If $k=1$, then $S=I_m$. Now, let $k \in \mathbb {N}$ and assume that $S^{(k)}\subseteq I_{m-k+1}$. Then, considered the canonical epimorphism $\chi : S \rightarrow S/I_{m-k}$, it holds that $\chi \left( I_{m-k+1}\right) \cdot \chi \left( S\right) \subseteq E\left( S/I_{m-k}\right)$, and so $I_{m-k+1} \cdot S \subseteq I_{m-k}$. Hence,

$$\begin{aligned} S^{(k+1)}=S^{(k)}\cdot S \subseteq I_{m-k+1} \cdot S \subseteq I_{m-k} \end{aligned}$$

and the claim follows by induction. Therefore, $S^{(m+1)}=E\left( S\right)$ and so S is right nilpotent of right nilpotency index less or equal to m. $\square$

The converse of Proposition 11 is true in the particular case in which $(S,+)$ is a nilpotent Clifford semigroup. The notion of nilpotent Clifford semigroup which we adopt in this work is consistent with the one given in [25].

Proposition 12

Let $(S,+, \circ )$ be a dual weak brace such that $(S,+)$ is nilpotent. Then, S is right nilpotent if and only if S is left annihilator nilpotent.

Proof

The necessary condition follows by Proposition 11. The sufficient one is similar to the proof of [12, Lemma 2.16] by using lower central series of $\left( S, +\right)$ in [25, Definition 3.4]. $\square$

Below, we relate the left annihilator nilpotency of a dual weak brace $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ with those of each skew brace $B_\alpha$. It will be a direct consequence of the following lemma.

Lemma 5

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be a dual weak brace. Then, $B_\alpha \cap \, {{\,\textrm{Soc}\,}}_k\left( S\right) \subseteq {{\,\textrm{Soc}\,}}_k\left( B_\alpha \right)$, for all $\alpha \in Y$ and $k \in \mathbb {N}$.

Proof

Let us check that $B_\alpha \cap \, {{\,\textrm{Soc}\,}}_k\left( S\right) \subseteq {{\,\textrm{Soc}\,}}_k\left( B_\alpha \right)$, by proceeding by induction on k. If $k = 1$ the claim follows by Theorem 3. Now, suppose that the statement holds for $k\in \mathbb {N}$ and let $a\in B_\alpha \cap {{\,\textrm{Soc}\,}}_{k+1}\left( S\right)$. Then, for any $b\in B_\alpha$, $a\cdot b\in {{\,\textrm{Soc}\,}}_{k}\left( S\right)$, $[a,b]_+\in {{\,\textrm{Soc}\,}}_{k}\left( S\right)$, and $a\cdot b\in B_\alpha$, and so, by the inductive hypothesis, $a\in {{\,\textrm{Soc}\,}}_{k+1}\left( B_\alpha \right)$. $\square$

Proposition 13

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be a left annihilator nilpotent dual weak brace of index n. Then, for every $\alpha \in Y$ each skew brace $B_\alpha$ is left annihilator nilpotent of index less or equal to n.

Vice versa, if each skew brace $B_\alpha$ is left annihilator nilpotent of index $k_\alpha$, then S is left annihilator nilpotent of index equal to the maximum between $k_\alpha$, for every $\alpha \in Y$.

7 Annihilator Nilpotency of Dual Weak Braces

Let us start by introducing a further ideal of a dual weak brace $(S, +,\circ )$, that is the annihilator of S. Denoted by $\zeta \left( S,\circ \right)$ the center of $\left( S,\circ \right)$, it is the set

$$\begin{aligned} {{\,\textrm{Ann}\,}}\left( S\right) := {{\,\textrm{Soc}\,}}\left( S\right) \cap \zeta \left( S,\circ \right) , \end{aligned}$$

namely, ${{\,\textrm{Ann}\,}}\left( S\right) = \{a \ | \ a\in S\quad \forall \, b\in B\quad a + b = b + a = a\circ b = b\circ a\}$. Note that the definition is consistent with that given in [7, Definition 7] for skew braces.

Proposition 14

Let $(S, +, \circ )$ be a dual weak brace. Then, ${{\,\textrm{Ann}\,}}\left( S\right)$ is an ideal of S.

Proof

Clearly, $E\left( S\right)$ is contained in ${{\,\textrm{Ann}\,}}\left( S\right)$. Moreover, if $x\in {{\,\textrm{Ann}\,}}\left( S\right)$ and $a\in S$,

$$\begin{aligned} b\circ \left( -a + x + a\right)&= b - b\circ a + b^0 + b\circ x - b + b\circ a&\text{ by } \text{ Lemma }\,1-2.\\&= b - b\circ a + x\circ b - b + b\circ a&x\in \zeta \left( S,\circ \right) \\&= b - b\circ a + x + b^0 + b\circ a&x\in {{\,\textrm{Soc}\,}}\left( S\right) \\&= b + x + \left( b\circ a\right) ^0&x\in {{\,\textrm{Soc}\,}}\left( S\right) \\&= b + x + a^0+ b^0\\&= \left( - a + x + a \right) + b&x\in {{\,\textrm{Soc}\,}}\left( S\right) \\&= \left( - a + x + a \right) \circ b.&- a + x + a\in {{\,\textrm{Soc}\,}}\left( S\right) \end{aligned}$$

for every $b\in S$. Besides, one has that $\lambda _a\left( x\right) = - a + x\circ a =-a+x+a\in {{\,\textrm{Ann}\,}}\left( S\right)$. Finally, $a^-\circ x\circ a = a^0\circ x=\lambda _a(x)\in {{\,\textrm{Ann}\,}}\left( S\right)$, which completes the proof. $\square$

As it happens for the socle, in general, the annihilator of a dual weak brace S does not coincide with the union of the annihilators of each skew brace $B_\alpha$ (see, for instance, Example 1). Similar to Proposition 4, we obtain the following result.

Proposition 15

Let $S=[Y; B_{\alpha }; \phi _{\alpha ,\beta }]$ be a dual weak brace, $\psi _{\alpha ,\beta }:= {\phi _{\alpha ,\beta }}_{|_{{{\,\textrm{Ann}\,}}(B_\alpha )}}$, for all $\alpha \ge \beta$, and assume that $I:=[Y; {{\,\textrm{Ann}\,}}\left( B_{\alpha }\right) ; \psi _{\alpha , \beta }]$ is an ideal of S. Then, $I={{\,\textrm{Ann}\,}}(S)$.

Remark 2

As a consequence of Lemma 3, if $(S, +, \circ )$ is a dual weak brace, it is also easy to see that

$$\begin{aligned} {{\,\textrm{Ann}\,}}\left( S \right)&=\{a \, | \, a \in S,\,\,\, \forall \ b \in S \quad a\cdot b=b \cdot a =[a,b]_{+}=a^0+b^0\}. \end{aligned}$$

Similar to [20], as is usual in ring theory, we define the kth annihilator of a dual weak brace $(S,+, \circ )$.

Definition 11

Let $(S,+, \circ )$ be a dual weak brace. Set ${{\,\textrm{Ann}\,}}_0\left( S\right) := E\left( S\right)$, we define ${{\,\textrm{Ann}\,}}_k\left( S\right)$ to be the ideal of S containing ${{\,\textrm{Ann}\,}}_{k-1}\left( S\right)$ such that

$$\begin{aligned} {{\,\textrm{Ann}\,}}_k\left( S\right) /{{\,\textrm{Ann}\,}}_{k-1}\left( S\right) ={{\,\textrm{Ann}\,}}\left( S/{{\,\textrm{Ann}\,}}_{k-1}\left( S\right) \right) , \end{aligned}$$

for every positive integer k.

Note that ${{\,\textrm{Ann}\,}}_k\left( S\right) =\{a \, | \, a\in S, \, \, \forall \, b \in S \quad a \cdot b, \, b \cdot a, \, [a,b]_{+} \in {{\,\textrm{Ann}\,}}_{k-1}\left( S\right) \},$ for every positive integer k.

Definition 12

A dual weak brace $(S, +, \circ )$ is said to be annihilator nilpotent if there exists $n \in \mathbb {N}$ such that ${{\,\textrm{Ann}\,}}_n\left( S\right) = S$.

Clearly, any ideal I and quotient S/I of an annihilator nilpotent dual weak brace S are annihilator nilpotent. The next example shows that, unlike what happens in ring theory, the converse is not true.

Example 3

Let $B:= \left( \mathbb {Z}_6,+,\circ \right)$ be the brace having as an additive group the cyclic group of 6 elements and multiplication defined by $a\circ b:= a + \left( -1\right) ^{a}b$, for all $a,b\in B$. Then, $I:=\{0,2,4\}$ is an ideal of B such that ${{\,\textrm{Ann}\,}}\left( I\right) = I$, so I is annihilator nilpotent. Clearly, ${{\,\textrm{Ann}\,}}\left( B/I\right) = B/I$, so B/I also is annihilator nilpotent. However, B is not annihilator nilpotent since ${{\,\textrm{Ann}\,}}\left( B\right) = \{0\}$.

Following [4, Definition 2.3], given a dual weak brace $(S, +, \circ )$ and an ideal I of S, set $\varGamma _0\left( I\right) := I$, we can inductively define

$$\begin{aligned} \varGamma _k\left( I\right) := \bigl \langle \,\varGamma _{k-1}\left( I\right) \cdot S, \ S \cdot \varGamma _{k-1}\left( I \right) , \ [\varGamma _{k-1}\left( I\right) , S]_+ \,\bigr \rangle _+. \end{aligned}$$

Observe that, set $\varGamma (I):=\varGamma _1(I)$, then $\varGamma _k(I)=\varGamma \left( \varGamma _{k-1}(I)\right)$, for every $k\in \mathbb {N}$.

Proposition 16

Let $(S, +, \circ )$ be a dual weak brace and I an ideal of S. Then, $\varGamma _k(I)$ is an ideal of S, for every $k \in \mathbb {N}$.

Proof

Assuming $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$, by an easy induction, one can show that $\phi _{\alpha , \beta }\left( \varGamma _k(B_\alpha )\right) \subseteq \varGamma _k\left( B_\beta \right)$, for any $\alpha > \beta$. Hence, the proof follows by Theorem 3. $\square$

One can show that if $(B, +, \circ )$ is a skew brace and I an ideal of B, then $\varGamma (I) \subseteq [I, B]$, where [I, B] denotes the commutator introduced in [5], which is the smallest ideal of B containing $[I, B]_+, [I, B]_\circ$, and the set $\{i \circ b-(i+b) \ \mid \ i \in I, \ b \in B\}$. We can not establish whether $\varGamma (I) = [I, B]$.

We give the following lemmas, which is useful to prove the main result of this section and can be compared with [1, Proposition 15] regarding commutators in the context of skew braces.

Lemma 6

Let M, N ideals of a dual weak brace $(S,+, \circ )$ with $M\subseteq N$. The following statements are equivalent:

• (i) $N/M \subseteq {{\,\textrm{Ann}\,}}(S/M)$,

• (ii) $\varGamma (N)\subseteq M$.

Proof

Initially, assume that $N/M \subseteq {{\,\textrm{Ann}\,}}(S/M)$. Hence, $M+(n \circ s)=M+(n+s)$, for all $n \in N$ and $s \in S$. Then,

$$\begin{aligned} n \cdot s=-n+\underbrace{n^0+\left( n\circ s\right) }_{M+(n \circ s)}-s=\underbrace{(-n+m+n)}_{M}+s-s \in M, \end{aligned}$$

for some $m \in M$. Similarly, $s \cdot n \in M$. Furthermore, since $M+(n+s)=(s+n)+M$,

$$\begin{aligned} {[}n,s]_+=-n-s+\underbrace{n^0+n+s}_{M+(n+s)}=\underbrace{-n-s+s+n}_{M}+m \in M, \end{aligned}$$

for some $m \in M$. Therefore, $\varGamma (N)\subseteq M$. The rest of the claim directly follows by Remark 2. $\square$

Lemma 7

Let $(S, +, \circ )$ be a dual weak brace and M, N ideals of S. Then, it holds that $\varGamma (M+N)=\varGamma (M)+\varGamma (N)$.

Proof

It is a direct consequence of the definition of $\varGamma (I)$, for every ideal I of S, and Proposition 1- 2 and 3. $\square$

Definition 13

Let $(S,+, \circ )$ be a dual weak brace. An ascending series

$$\begin{aligned} {{\,\textrm{E}\,}}(S) = I_0\subseteq I_1\subseteq \cdots \subseteq I_k=S \end{aligned}$$

of ideals of S is called an annihilator series of S if $I_{j+1}/I_j\subseteq {{\,\textrm{Ann}\,}}\left( S/I_j\right)$, for $0 \le j \le k$.

Theorem 5

Let $(S,+,\circ )$ be a dual weak brace. If ${{\,\textrm{E}\,}}(S) = I_0\subseteq I_1\subseteq \cdots \subseteq I_k=S$ is an annihilator series of S, then

$$\begin{aligned} \varGamma _{k-j}(S) \subseteq I_j\subseteq {{\,\textrm{Ann}\,}}_j(S), \end{aligned}$$

for $0\le j\le k$.

Proof

First, we set $i:=k-j$ and show that $\varGamma _i(S)\subseteq I_{k-i}$, by proceeding by induction on i. If $i=0$, then $\varGamma _0(S)=S=I_k$. Now, let $i\in \mathbb {N}$ and assume that $\varGamma _i(S)\subseteq I_{k-i}$. Thus, by Lemma 6,

$$\begin{aligned} \varGamma _{i+1}(S)=\varGamma (\varGamma _i(S))\subseteq \varGamma (I_{k-i})\subseteq I_{k-(i+1)}, \end{aligned}$$

which completes the first part of the proof.

Now, we prove that $I_j \subseteq {{\,\textrm{Ann}\,}}_j(S)$ by induction on j. If $j=0$, then $I_0={{\,\textrm{E}\,}}(S)={{\,\textrm{Ann}\,}}_0(S)$. Now, let $j \in \mathbb {N}$ and assume that $I_j \subseteq {{\,\textrm{Ann}\,}}_j(S)$. By Lemma 7 and Lemma 6, we have that

$$\begin{aligned} \varGamma \left( I_j+{{\,\textrm{Ann}\,}}_{j-1}(S)\right) =\varGamma (I_j)+\varGamma \left( {{\,\textrm{Ann}\,}}_{j-1}(S)\right) \subseteq I_{j-1}+{{\,\textrm{Ann}\,}}_{j-1}(S) \subseteq {{\,\textrm{Ann}\,}}_{j-1}(S), \end{aligned}$$

where the last inclusion follows by the inductive hypothesis. Thus, since by Lemma 6

$$\begin{aligned} \left( I_j+{{\,\textrm{Ann}\,}}_{j-1}(S)\right) / {{\,\textrm{Ann}\,}}_{j-1}(S) \subseteq {{\,\textrm{Ann}\,}}\left( S/{{\,\textrm{Ann}\,}}_{j-1}(S)\right) ={{\,\textrm{Ann}\,}}_j(S)/{{\,\textrm{Ann}\,}}_{j-1}(S), \end{aligned}$$

we obtain that $I_j+ {{\,\textrm{Ann}\,}}_{j-1}(S) \subseteq {{\,\textrm{Ann}\,}}_{j}(S)$. Consequently, $I_j \subseteq {{\,\textrm{Ann}\,}}_{j}(S)$ and the claim follows. $\square$

In light of Theorem 5 we can give the following definitions.

Definition 14

Let $(S, +, \circ )$ be a dual weak brace. The smallest non-negative integer c such that $\varGamma _{c+1}(S)={{\,\textrm{E}\,}}(S)$ and ${{\,\textrm{Ann}\,}}_c(S)=S$ is called nilpotency index (or annihilator nilpotency index) of S. The series

$$\begin{aligned} {{\,\textrm{E}\,}}(S) = {{\,\textrm{Ann}\,}}_0(S)\subseteq {{\,\textrm{Ann}\,}}_1(S)\subseteq \cdots \subseteq {{\,\textrm{Ann}\,}}_c(S) = S \end{aligned}$$

is called upper annihilator series of S. The series

$$\begin{aligned} {{\,\textrm{E}\,}}(S) = \varGamma _{c+1}(S) \subseteq \varGamma _c(S)\subseteq \cdots \subseteq \varGamma _0(S) = S \end{aligned}$$

is named lower annihilator series of S.

Similarly to Lemma 5 and Proposition 13, one can prove the following result.

Proposition 17

Let $S=[Y; B_\alpha ;\phi _{\alpha ,\beta }]$ be an annihilator nilpotent dual weak brace of nilpotency index n. Then, for every $\alpha \in Y$ each skew brace $B_\alpha$ is annihilator nilpotent of index less or equal to n.

Vice versa, if each skew brace $B_\alpha$ is annihilator nilpotent of index $k_\alpha$, then S is annihilator nilpotent of index equal to the maximum between $k_\alpha$, for every $\alpha \in Y$.