New Structure of Skew Braces and Their Ideals ^†

Abstract

The primary motivation of this article is to introduce the definition of a strongly belonging element of a brace, presenting some results concerning the case where I acts as an ideal of a brace, $(A,+,\circ )$. We find that an element $a\in I$ acts as strongly belonging element of I when $a+b$ yields a for all b in I.

1. Introduction

The motivation for writing this paper came from the publication of Konovalov et al. [1]. Historically, the Yang–Baxter formula appeared in 1944 as one of the earliest in the field of statistical mechanics. The Yang–Baxter formula has been calculated extensively, and its applications can be established in classical statistical mechanics, quantum groups and other fields. Several authors have investigated the solution of the Yang–Baxter formula via establishing some algebraic concepts. In 2007, Rump [2] introduced one of such efforts, employing the concept of braces as an expansion of the concept of $rad\left(R\right)$, the Jacobson radical of a ring R, so as to provide support for the research involution found in some results of the Yang–Baxter relationship. The notion of braces has been calculated merely for its algebraic characteristics, and they have been attached to several sectors. In 2018, the author T. Gateva-Ivanova [3] suggested that the braces are in agreement with braided groups and an involution braiding factor. Moreover, the braces have also been seen to be much the same as some objects in the concept of group theory. For example, bijective 1 cocycles together with regular subgroups of the holomorph. Indeed, they have been found in links to quantum integrable systems, flat manifolds, etc. In addition to that, the contribution in [4] considers skew braces as a generalization of braces, while the author D. Bachiller [5] recognized that there is a relationship between the two concepts of Hopf–Galois theory and skew braces, which was supplementary material in [6]. These affirmations confirm the importance of distinguishing braces, which is an orientation many authors have taken in recent years.

Both kinds of the concept of cyclic additive groups with braces were categorized in references [7,8]. The main ideas concern braces which have the size $pq$ and $p^{2}q$, where the two primes p and q were categorized in [9,10]. In actuality, all kinds of braces of the cardinality $p^3$ have been found to be substantive via Bachiller [5], while other kinds, which include skew braces, have the same size as that found by Nejabati [11]. The concept of skew braces was provided in [4]. The authors deal with the left braces as a non-commutative generalization, which appeared in [2]. This is used to generalize radical algebras. As a matter of fact, the connotation of skew braces was calculated in [6,12]. The primary motivation in the investigation of both the notion of skew braces and that of braces was pushed by the search for the results of dismissals of the Yang–Baxter identity. However, there there also exists the nearest link between the ideas of the skew braces and Hopf–Galois structures, which is discussed in the Galois extensions of fields. This correlation derived from the invention [13] of the interaction between the concepts of radical algebra and regular subgroups acts on the affine group, with its dependent generalization and employment status in relation to the abelian Hopf–Galois. This construction of an elementary abelian Galois expansion of fields is explored in reference [14].

The main motivation of this paper is to explore the action of a new concept, which is the strongly belonging element of the brace $(A,+,\circ )$ with the ideal I.

2. Background Information

Throughout this paper, A, as a set providing + and ∗ as two binary operations, refers to a left brace when $(A,+)$ represents a group-specific abelian group. Furthermore, the distributive combined with associativity is satisfied with the following relationships:

Taking the elements $x,y,z\in A$,

(i)

$(x+y+x\ast y)\ast z=x\ast z+y\ast z+x\ast (y\ast z)$;

(ii)

$x\ast (y+z)=x\ast y+x\ast z$.

In addition to that, we define the operation ∘ by $x\circ y=x+y+x\ast y$, where $(A,\circ )$ forms a group. For more information about the original definition, view [2]. Furthermore, view reference [15] to obtain a summarized equivalent definition via the employment of group theory.

Now, we will depend on the following definition in expressions of the procedure ‘∘’ (see [15]): A is a set providing the two binary operations of addition + and multiplication ∘, forming a brace under the conditions that

(i)

$(A,+)$ forms an abelian group;

(ii)

$(A,\circ )$ forms a group;

(ii)

For every $x,y,z\in A$, $x\circ (y+z)+x=x\circ y+x\circ z.$

We employ the following common determination of an ideal in a brace.

Definition 1 

([16]). Suppose the triple $(A,+,\circ )$ is a brace. Then, I acts as an ideal in a brace $(A,+,\circ )$ when $I\subseteq A$. Additionally, each $x,y\in I$ and each $r\in A$ yield $x+y$ and $x-y$ in I, and $r\ast x$ is used to gather $x\ast r$ in I.

We use the operation $x\circ y=x\ast y+x+y$ for all $x,y\in A.$ In the next step, we will depend on the following notation: Suppose I and J are subsets of a left brace A. Then, $I\ast J$ forms the additive subgroup of A created by elements $i\ast j$, where $i\in I$ and $j\in J.$

Definition 2 

([6]). Let A be a set providing the two binary operations + and . such that for all x, y, and z in $A.$,

(i)

$(A,+)$ forms an abelian group;

(ii)

$(A,.)$ forms a group;

(iii)

$x(y+z)+x=xy+xz$.

The above structure form a left brace where $(A,+)$ is the additive group and $(A,.)$ the multiplicative group of the left brace.

Under the same axiomatic, we can define a right brace in similar fashion, writing the previous Condition (iii) using the formula $(x+y)z+z=xz+yz.$

It is simple to review and determine A in a left brace. Here, it is necessary to confirm the important relationship between the multiplicative group of A, which has the multiplicative identity 1, and the neutral element 0 for the additive group of A, being equal to each other. In this paper, brace A acts as a left brace.

3. The Main Definition

Definition 3.

Suppose the triple $(A,+,\circ )$ is a brace and I is an ideal of A. An element $a\in I$ is named a strongly belonging element if $a+b=a$ for all $b\in I$.

Our motivation and a closer look at the previous definition are supplied by the following example.

Example 1.

Assuming A forms a brace, the elements are complex variables. Suppose $\delta =x+iy$ (where x and y are real numbers) is a complex number. Then, $\overline{\delta }=x-iy$ is a conjugate of δ

Hence, $\delta +\overline{\delta }=x+iy+x-iy=2x$, and then the real part is a strongly belonging element.

4. Results

We begin with this result.

Let us recall from ([1], Lemma 2.1) that A is a skew brace. Then, $\pi :(A,\circ )⟶Aut(A,+)$ can be written as $a⟼{\pi }_a$, where ${\pi }_a\left(b\right)$ is equal to $-a+a\circ b$. Then, this forms a well-defined group homomorphism.

Proposition 1.

Assume A is a skew brace. Let I and J be ideals of A. If J has a strongly belonging membership, then $I+J=J$.

Proof. 

Suppose a in A, u in I, and v in $J.$ Then, based on this assumption, we have ${\pi }_a(u+v)\in I+J$. After that, ${\pi }_a(I+J)\subseteq I+J.$ Consequently,

$(u+v)\ast a=(u\circ {\pi }_a^{-1}u\left(v\right))\ast a$, which yields $u\ast ({\pi }_a^{-1}u\left(v\right)\ast a)+{\pi }_a^{-1}u\left(v\right)\ast a+u\ast a$, belonging to $I+J.$

Consequently, the previous formula yields

$a\circ (u+v)\circ a^{{}^{\prime }}=a+{\pi }_a((u+v)+(u+v)\ast a^{{}^{\prime }})-a\in I+J.$

Thus, it follows that $a\circ (I+J)\circ a^{{}^{\prime }}\subseteq I+J.$ Finally, $I+J$ forms a normal subgroup of $(A,+)$, since the identity

$a+({\sum }_k{u}_k+v_k)-a$ is equal to the identity $({\sum }_k{u}_k+v_k)$, after removing the term a.

Due to $u_k$ in I and $v_k$ in J for all k, using the fact that J has strongly belonging membership, we deduce that

$a+({\sum }_k{u}_k+v_k)-a$ is equal to the identity $({\sum }_k{u}_k+v_k)=v_k=J$. □

Theorem 1.

The sum of any number of N-ideals in A is equal to an ideal that has strongly belonging membership in A, where A acts as a brace.

Proof. 

Suppose I and J are two N-ideals belonging to A. First of all, we move to provide the relationship $I+J$ equal to the set $\{a+b:a\in I,b\in J\}$, forming an n-ideal in I. Suppose $a_1,a_2,...$ is an N-sequence with the relationship $a_1=a+b$, where a in I and b in $J.$

Take into consideration the two results. For the first, the factor brace is $A/J$, while the second is the N-sequence,

$a_1^{{}^{\prime }}=a_1+J=a+J,a_2^{{}^{\prime }}=a_2+J,...$ in $A/J$. We observe that every N-sequence in A starting with $a\in I$ will approach zero. Consequently, every n-sequence in $A/J$ will approach zero based on $(r+J)\ast (s+J)$ equal to $r\ast s+J$ for all $r,s\in A$.

In [17] (Question 2.1(2)), for any brace such that the operation defined as $a\ast b=-a+a\circ b-b$ is associative, it is a two-sided brace. This fact is true.

We employ this relationship in $(r+J)\ast (s+J)=r\ast s+J$, where

$r\ast s+J=-r+r\circ s-s+J$ is equal to $-r+(r\circ s)-s+J$. This relationship can be written as

$-r+(r\circ s)-s+J=-r+(r\circ (s+0\left)\right)-s+J$.

Now, applying the relationship $a\circ (b+c)=a\circ b-a+a\circ c$ to the term $(r\circ (s+0\left)\right)$, we arrive at

where $-r+(r\circ (s+0\left)\right)-s+J$ is equal to $-r+(r\circ s)-r+(r\circ 0)-s+J.$

Moreover,

$-r+(r\circ s)-s+J$ yields $-r+(r\circ s)-r+(r\circ 0)-s+J.$

Obviously, we can remove some terms from this relationship

$r=(r\circ 0).$ Based on the certainty of the relationship $a\circ b=a\ast b+a+b$, we find that

$r=r\ast 0+r+0.$ Thus, we conclude that

$A\ast 0=0.$ Hence, if $s=0$, then the relationship $(r+J)\ast (0+J)$ equals the relationship $r\ast 0+J=J.$ □

The following result represents the immediate consequences of the previous theorem.

Corollary 1.

Assume A is a brace, such that A has the sum of any number of N-ideals; then, $A\ast 0=0.$

Proposition 2.

Assume A is a skew brace, and J is an ideal that has strongly belonging membership in A. Suppose I is an ideal of skew brace $A/J$ such that $I^{★}$ is equal to $\{a\in A:a+J\in I\}$, which is an ideal in A. Then, ${\pi }_J\left(J\right)$ in I.

Proof. 

Observe that $a\in I^{★}$ if and only if $a+J\in I$. Let $a,b\in I^{★}.$ Due to the fact that $a+J$ in I and $b+J$ in I, then $a+b+J$ is equal to $(a+J)+(b+J)$, which belongs to $I.$ Consequently, $a+b\in I^{★}.$ In this manner, if $a\in I^{★}$ and $c\in A,$ then $a+J$ in $I.$ Therefore, ${\pi }_c\left(a\right)+J={\pi }_{c+J}(a+J)$ in I. Based on J having strongly belonging membership in A, then ${\pi }_J\left(J\right)\in I$. This step completes the proof. □

Corollary 2.

Assume $J\subseteq I$ are ideals in a skew brace A such that J has strongly belonging membership in A. Then, $I/J$ forms an ideal in $A/J$ and $J\subseteq I/J$.

Proof. 

Suppose $a+J$ and $b+J$ in $I/J$ and c in $A.$ Then, $a\in I$ and $b\in I.$ Consequently, because J is an ideal, $(a+J)+(b+J)$ is equal to $a+b+J$, which belongs to $I/J$. Then, ${\pi }_{c+J}(a+J)$ is equal to ${\pi }_{c\left(a\right)}+J$, which lies in $I/J$. Due to the hypothesis that J has strongly belonging membership in A, then the last expression can be modified as $J\subseteq I/J$. This completes the proof. □

Depending on the fact that $I+J$ forms an ideal, we find the following corollary.

Corollary 3.

Assume A is a skew brace, and I and J are ideals in A such that J has strongly belonging membership in A. Then, $(I+J)/J$ is an ideal in $A/J$.

Note: In [18], the authors suppose that A acts as a left brace. For any arbitrary element $a\in A$, we define a function ${\pi }_a:A⟶A$ via ${\pi }_a\left(b\right)=ab-a$, where $b\in A$. Based on this function, we find the following result.

Proposition 3. 

A is a left brace. Then,

(i)

${\pi }_a(x+y)=a(x+y-2)$, which holds for all $a,x,y\in A$;

(ii)

${\pi }_a{\pi }_b=a(b(x-1)-1)$, which holds for all $a,b,x\in A$.

Proof. 

(i) Let a, x, and y in A. Then, ${\pi }_a(x+y)=a(x+y)-a=ax+ay-a-a$, which yields $a(x+y)-2a=a(x+y-2).$

(ii) Let a, b and x in $A.$ Then, ${\pi }_a{\pi }_b\left(x\right)={\pi }_a(bx-b)$, which yields $a(bx-b)-a=a\left(b\right(x-1)-1)$.

With this step, the proof is complete. □