Complete Classification of Two-Dimensional Algebras

TL;DR

This paper provides a comprehensive classification of all two-dimensional algebras over algebraically closed fields, filling a fundamental gap in algebraic structure theory.

Contribution

It offers the first complete classification of two-dimensional algebras over algebraically closed fields, detailing their structure and types.

Findings

Complete list of two-dimensional algebra structures

Identification of isomorphism classes

Framework for future algebraic classifications

Abstract

A complete classification of two-dimensional algebras over algebraically closed fields is provided

Equations

$$\tau (P(\mathbf{u}),\mathbf{u})=\tau (P(\mathbf{v}),\mathbf{v}).$$

$$A=\left(\begin{array}[]{cccc}A^{1}_{1,1}&A^{1}_{1,2}&A^{1}_{2,1}&A^{1}_{2,2}\\ A^{2}_{1,1}&A^{2}_{1,2}&A^{2}_{2,1}&A^{2}_{2,2}\\ \end{array}\right)\in Mat(2\times 4;\mathbf{F})$$

$$\mathbf{u}\cdot \mathbf{v}=eA(u\otimes v)$$

$$\mathbf{u}\cdot \mathbf{v}=eA(u\otimes v)=e^{\prime }B(u^{\prime }\otimes v^{\prime })=e{g}^{-1}B(gu\otimes gv)=e{g}^{-1}B(g\otimes g)(u\otimes v)$$

$$B=gA(g^{-1}{)}^{\otimes 2}$$

$$(g^{-1})^{\otimes 2}=g^{-1}\otimes g^{-1}=\left(\begin{array}[]{cccc}\xi_{1}^{2}&\xi_{1}\eta_{1}&\xi_{1}\eta_{1}&\eta_{1}^{2}\\ \xi_{1}\xi_{2}&\xi_{1}\eta_{2}&\xi_{2}\eta_{1}&\eta_{1}\eta_{2}\\ \xi_{1}\xi_{2}&\xi_{2}\eta_{1}&\xi_{1}\eta_{2}&\eta_{1}\eta_{2}\\ \xi_{2}^{2}&\xi_{2}\eta_{2}&\xi_{2}\eta_{2}&\eta_{2}^{2}\end{array}\right).$$

$$T{r}_1(gA(g^{-1}{)}^{\otimes 2})=T{r}_1(A)g^{-1},T{r}_2(gA(g^{-1}{)}^{\otimes 2})=T{r}_2(A)g^{-1}$$

$$T{r}_1(A)=(A_{1,1}^1+A_{2,1}^2,A_{1,2}^1+A_{2,2}^2),T{r}_2(A)=(A_{1,1}^1+A_{1,2}^2,A_{2,1}^1+A_{2,2}^2)$$

$$A=\left(\begin{array}[]{cccc}\alpha_{1}&\alpha_{2}&\alpha_{3}&\alpha_{4}\\ \beta_{1}&\beta_{2}&\beta_{3}&\beta_{4}\end{array}\right),$$

$$A_{1}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&\alpha_{2}&\alpha_{2}+1&\alpha_{4}\\ \beta_{1}&-\alpha_{1}&-\alpha_{1}+1&-\alpha_{2}\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\alpha_{1},\alpha_{2},\alpha_{4},\beta_{1})\in\mathbf{F}^{4},$$

$$A_{2}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ \beta_{1}&\beta_{2}&1-\alpha_{1}&0\end{array}\right)\simeq\left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ -\beta_{1}&\beta_{2}&1-\alpha_{1}&0\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\alpha_{1},\beta_{1},\beta_{2})\in\mathbf{F}^{3},$$

$$A_{3}(\mathbf{c})=\left(\begin{array}[]{cccc}0&1&1&0\\ \beta_{1}&\beta_{2}&1&-1\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\beta_{1},\beta_{2})\in\mathbf{F}^{2},$$

$$A_{4}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\ 0&\beta_{2}&1-\alpha_{1}&0\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\alpha_{1},\beta_{2})\in\mathbf{F}^{2},$$

$$A_{5}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\ 1&2\alpha_{1}-1&1-\alpha_{1}&0\end{array}\right),\ \mbox{where}\ \mathbf{c}=\alpha_{1}\in\mathbf{F},$$

$$A_{6}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ \beta_{1}&1-\alpha_{1}&-\alpha_{1}&0\end{array}\right)\simeq\left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ -\beta_{1}&1-\alpha_{1}&-\alpha_{1}&0\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\alpha_{1},\beta_{1})\in\mathbf{F}^{2},$$

$$A_{7}(\mathbf{c})=\left(\begin{array}[]{cccc}0&1&1&0\\ \beta_{1}&1&0&-1\end{array}\right),\ \mbox{where}\ \mathbf{c}=\beta_{1}\in\mathbf{F},$$

$$A_{8}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\ 0&1-\alpha_{1}&-\alpha_{1}&0\end{array}\right),\ \mbox{where}\ \mathbf{c}=\alpha_{1}\in\mathbf{F},$$

$$A_{9}=\left(\begin{array}[]{cccc}\frac{1}{3}&0&0&0\\ 1&\frac{2}{3}&-\frac{1}{3}&0\end{array}\right),\ \ A_{10}=\left(\begin{array}[]{cccc}0&1&1&0\\ 0&0&0&-1\end{array}\right),$$

$$A_{11}=\left(\begin{array}[]{cccc}0&1&1&0\\ 1&0&0&-1\end{array}\right),\ \ A_{12}=\left(\begin{array}[]{cccc}0&0&0&0\\ 1&0&0&0\end{array}\right).$$

$$T{r}_1(A)=({\alpha }_1+{\beta }_3,{\alpha }_2+{\beta }_4),T{r}_2(A)=({\alpha }_1+{\beta }_2,{\alpha }_3+{\beta }_4).$$

$$P(gA(g^{-1}{)}^{\otimes 2})=P(A)g^{-1},$$

$$V_0=\{A\in Mat(2\times 4,\mathbf{F}):\det (P(A))={\alpha }_1({\alpha }_3-{\alpha }_2)+{\beta }_4({\beta }_3-{\beta }_2)+{\alpha }_3{\beta }_3-{\alpha }_2{\beta }_2\ne 0.\}$$

$$P(B)B(P(B{)}^{-1}\otimes P(B{)}^{-1})=P(A)A(P(A{)}^{-1}\otimes P(A{)}^{-1})$$

$$P(A)^{-1}=\left(\begin{array}[]{cc}\xi_{1}&\eta_{1}\\ \xi_{2}&\eta_{2}\end{array}\right)$$

$$\left(\begin{array}[]{c}-\frac{\beta_{1}\eta_{1}\xi_{1}^{2}}{\Delta}+\frac{\alpha_{1}\eta_{2}\xi_{1}^{2}}{\Delta}+\frac{2\alpha_{1}\eta_{1}\xi_{1}\xi_{2}}{\Delta}-\frac{2\beta_{4}\eta_{2}\xi_{1}\xi_{2}}{\Delta}-\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}}+\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}-\frac{\beta_{4}\eta_{1}\xi_{2}^{2}}{\Delta}+\frac{\alpha_{4}\eta_{2}\xi_{2}^{2}}{\Delta}+\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}}\\ \frac{\beta_{1}\xi_{1}^{3}}{\Delta}-\frac{3\alpha_{1}\xi_{1}^{2}\xi_{2}}{\Delta}+\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}+\frac{3\beta_{4}\xi_{1}\xi_{2}^{2}}{\Delta}+\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}}-\frac{2\xi_{1}^{2}\xi_{2}^{2}}{\Delta^{2}}-\frac{\alpha_{4}\xi_{2}^{3}}{\Delta}\end{array}\right),$$

$$\left(\begin{array}[]{c}-\frac{\beta_{1}\eta_{1}^{2}\xi_{1}}{\Delta}+\frac{2\alpha_{1}\eta_{1}\eta_{2}\xi_{1}}{\Delta}-\frac{\beta_{4}\eta_{2}^{2}\xi_{1}}{\Delta}-\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}+\frac{\alpha_{1}\eta_{1}^{2}\xi_{2}}{\Delta}-\frac{2\beta_{4}\eta_{1}\eta_{2}\xi_{2}}{\Delta}-\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}+\frac{\alpha_{4}\eta_{2}^{2}\xi_{2}}{\Delta}+\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}}\\ \frac{\beta_{1}\eta_{1}\xi_{1}^{2}}{\Delta}-\frac{\alpha_{1}\eta_{2}\xi_{1}^{2}}{\Delta}-\frac{2\alpha_{1}\eta_{1}\xi_{1}\xi_{2}}{\Delta}+\frac{2\beta_{4}\eta_{2}\xi_{1}\xi_{2}}{\Delta}+\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}}-\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}+\frac{\beta_{4}\eta_{1}\xi_{2}^{2}}{\Delta}-\frac{\alpha_{4}\eta_{2}\xi_{2}^{2}}{\Delta}-\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}}\end{array}\right),$$

$$\left(\begin{array}[]{c}-\frac{\beta_{1}\eta_{1}^{2}\xi_{1}}{\Delta}+\frac{2\alpha_{1}\eta_{1}\eta_{2}\xi_{1}}{\Delta}-\frac{\beta_{4}\eta_{2}^{2}\xi_{1}}{\Delta}-\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}+\frac{\eta_{2}^{2}\xi_{1}^{2}}{\Delta^{2}}+\frac{\alpha_{1}\eta_{1}^{2}\xi_{2}}{\Delta}-\frac{2\beta_{4}\eta_{1}\eta_{2}\xi_{2}}{\Delta}-\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}+\frac{\alpha_{4}\eta_{2}^{2}\xi_{2}}{\Delta}+\frac{\eta_{1}^{2}\xi_{2}^{2}}{\Delta^{2}}\\ \frac{\beta_{1}\eta_{1}\xi_{1}^{2}}{\Delta}-\frac{\alpha_{1}\eta_{2}\xi_{1}^{2}}{\Delta}+\frac{\eta_{2}^{2}\xi_{1}^{2}}{\Delta^{2}}-\frac{2\alpha_{1}\eta_{1}\xi_{1}\xi_{2}}{\Delta}+\frac{2\beta_{4}\eta_{2}\xi_{1}\xi_{2}}{\Delta}-\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}+\frac{\beta_{4}\eta_{1}\xi_{2}^{2}}{\Delta}+\frac{\eta_{1}^{2}\xi_{2}^{2}}{\Delta^{2}}-\frac{\alpha_{4}\eta_{2}\xi_{2}^{2}}{\Delta}-\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}}\end{array}\right),$$

$$\left(\begin{array}[]{c}-\frac{\beta_{1}\eta_{1}^{3}}{\Delta}+\frac{3\alpha_{1}\eta_{1}^{2}\eta_{2}}{\Delta}-\frac{3\beta_{4}\eta_{1}\eta_{2}^{2}}{\Delta}-\frac{2\eta_{1}^{2}\eta_{2}^{2}}{\Delta^{2}}+\frac{\alpha_{4}\eta_{2}^{3}}{\Delta}+\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}+\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}\\ \frac{\beta_{1}\eta_{1}^{2}\xi_{1}}{\Delta}-\frac{2\alpha_{1}\eta_{1}\eta_{2}\xi_{1}}{\Delta}+\frac{\beta_{4}\eta_{2}^{2}\xi_{1}}{\Delta}+\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}-\frac{\alpha_{1}\eta_{1}^{2}\xi_{2}}{\Delta}+\frac{2\beta_{4}\eta_{1}\eta_{2}\xi_{2}}{\Delta}+\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}-\frac{\alpha_{4}\eta_{2}^{2}\xi_{2}}{\Delta}-\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}}\end{array}\right).$$

$$A^{\prime}=\left(\begin{array}[]{cccc}\alpha^{\prime}_{1}&\alpha^{\prime}_{2}&\alpha^{\prime}_{2}+1&\alpha^{\prime}_{4}\\ \beta^{\prime}_{1}&-\alpha^{\prime}_{1}&-\alpha^{\prime}_{1}+1&-\alpha^{\prime}_{2}\end{array}\right).$$

$$A_{1}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&\alpha_{2}&\alpha_{2}+1&\alpha_{4}\\ \beta_{1}&-\alpha_{1}&-\alpha_{1}+1&-\alpha_{2}\end{array}\right),\ \mbox{where}\ \mathbf{c}=(\alpha_{1},\alpha_{2},\alpha_{4},\beta_{1})\in\mathbf{F}^{4}.$$

Full text

emails: 1houida_m7@yahoo.com; 2bekbaev@iium.edu.my; 3rakhimov@upm.edu.my.

1. Introduction

The classification problem of finite dimensional algebras is one of the important problems of modern algebra. So far two approaches are known to the solution of the problem. One of them is structural (basis free, invariant) approach. For instance, the classification of finite dimensional simple and semi-simple associative algebras by Wedderburn and simple and semi-simple Lie algebras by Cartan are examples of such approach. But it is observed that this approach becomes more difficult when one considers more general types of algebras. Another approach to the solution of the problem is coordinate based approach (see [2, 3, 4, 5, 7] for the latest results). These two approaches somehow are complementary to each other.

In two-dimensional case a complete classification, by basis free approach, is stated in [6] over any basic field. In this paper we follow the coordinate (basis, structure constants) based approach, we consider such problem for two-dimensional algebras over any algebraically closed field $\mathbf{F}$. We provide a list of algebras, given by their matrices of structure constants (MSC), such that any 2-dimensional algebra is isomorphic to only one algebra from the list. A similar result has been stated in [5]. The main difference between these two results is as follows. In [5] the authors can state the existence only whereas the uniqueness can not be guaranteed. Moreover, the approach to the problem followed in the present paper is totally different that in [5]. Some details of our approach for arbitrary dimensional case have been given in [2].

The paper is organized as follows. First we study the problem over algebraically closed fields of characteristics not 2 and 3, then the solution is given for algebras over algebraically closed fields of characteristics 2 and the last subsection of paper contains the result over algebraically closed fields of characteristics 3. In each of these cases we present the list of algebras via their matrices of structure constants.

2. Classification of two-dimensional algebras

To classify the main part of two-dimensional algebras we use a particular case of the following result from [2]. Let $n$, $m$ be any natural numbers, $\tau:(G,V)\rightarrow V$ be a fixed linear algebraic representation of an algebraic subgroup $G$ of $GL(m,\mathbf{F})$ on the $n$-dimensional vector space $V$ over $\mathbf{F}$. Assume that there exists a nonempty $G$-invariant subset $V_{0}$ of $V$ and an algebraic map $P:V_{0}\rightarrow G$ such that

$P(\tau(g,\mathbf{v}))=P(\mathbf{v})g^{-1}$,

whenever $\mathbf{v}\in V_{0}$ and $g\in G$. The following result holds true [2].

Theorem. * Elements $\mathbf{u},\mathbf{v}\in V_{0}$ are $G$-equivalent, that is $\mathbf{u}=\tau(g,\mathbf{v})$ for some $g\in G$, if and only if*

[TABLE]

Let $\mathbf{A}$ be any 2 dimensional algebra over $\mathbf{F}$ with multiplication $\cdot$ given by a bilinear map $(\mathbf{u},\mathbf{v})\mapsto\mathbf{u}\cdot\mathbf{v}$ whenever $\mathbf{u},\mathbf{v}\in\mathbf{A}$. If $e=(e^{1},e^{2})$ is basis for $\mathbf{A}$ as a vector space over $\mathbf{F}$ then one can represent this bilinear map by a matrix

[TABLE]

such that

[TABLE]

for any $\mathbf{u}=eu,\mathbf{v}=ev,$ where $u=(u_{1},u_{2}),$ and $v=(v_{1},v_{2})$ are column coordinate vectors of $\mathbf{u}$ and $\mathbf{v},$ respectively, $(u\otimes v)=(u_{1}v_{1},u_{1}v_{2},u_{2}v_{1},u_{2}v_{2})$, $e^{i}\cdot e^{j}=A^{1}_{i,j}e^{1}+A^{2}_{i,j}e^{2}$ whenever $i,j=1,2$. So the algebra $\mathbf{A}$ is presented by the matrix $A\in Mat(2\times 4;\mathbf{F})$ (called the matrix of MSC of $\mathbf{A}$ with respect to the basis $e$).

If $e^{\prime}=(e^{\prime 1},e^{\prime 2})$ is also another basis for $\mathbf{A}$, $g\in G=GL(2,\mathbf{F})$, $e^{\prime}g=e$ and $\mathbf{u}\cdot\mathbf{v}=e^{\prime}B(u^{\prime}\otimes v^{\prime})$, where $\mathbf{u}=e^{\prime}u^{\prime},\mathbf{v}=e^{\prime}v^{\prime}$, then

[TABLE]

as far as $\mathbf{u}=eu=e^{\prime}u^{\prime}=eg^{-1}u^{\prime},\mathbf{v}=ev=e^{\prime}v^{\prime}=eg^{-1}v^{\prime}$. Therefore the equality

[TABLE]

is valid, where for g^{-1}=\left(\begin{array}[]{cccc}\xi_{1}&\eta_{1}\\ \xi_{2}&\eta_{2}\end{array}\right) one has

[TABLE]

Definition. Two-dimensional algebras $\mathbf{A}$, $\mathbf{B}$, given by their matrices of structural constants $A$, $B$, are said to be isomorphic if $B=gA(g^{-1})^{\otimes 2}$ holds true for some $g\in GL(2,\mathbf{F})$.

Note that the following identities

[TABLE]

hold true whenever $A\in Mat(2\times 4,\mathbf{F}),g\in GL(2,\mathbf{F})$, where

[TABLE]

are row vectors.

We divide $Mat(2\times 4,\mathbf{F})$ into the following five disjoint subsets:

All $A$ for which the system $\{Tr_{1}(A),Tr_{2}(A)\}$ is linear independent.

• 2.

All $A$ for which the system $\{Tr_{1}(A),Tr_{2}(A)\}$ is linear dependent and $Tr_{1}(A),Tr_{2}(A)$ are nonzero vectors.

• 3.

All $A$ for which $Tr_{1}(A)$ is nonzero vector and $Tr_{2}(A)=(0,0)$.

• 4.

All $A$ for which $Tr_{1}(A)=(0,0)$ and $Tr_{2}(A)$ is nonzero vector.

• 5.

All $A$ for which$Tr_{1}(A)=Tr_{2}(A)=(0,0)$.

Due to (1) it is clear that algebras with matrices from these different classes can not be isomorphic. We deal with each of these subsets separately. Further, for the simplicity, we use the notation

[TABLE]

where $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\beta_{1},\beta_{2},\beta_{3},\beta_{4}$ stand for any elements of $\mathbf{F}$.

As it has been stated above we consider three cases corresponding to $Char(\mathbf{F})\neq 2,3,$ $Char(\mathbf{F})=2$ and $Char(\mathbf{F})=3,$ respectively.

2.1. Characteristic not $2$, $3$ case

Theorem 1. Over an algebraically closed field $\mathbf{F}$ with the characteristic not 2 and 3, any non-trivial 2-dimensional algebra is isomorphic to only one of the following listed, by their matrices of structure constants, algebras:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof. The first subset case. In this case let $P(A)$ stand for a nonsingular matrix \left(\begin{array}[]{cc}\alpha_{1}+\beta_{3}&\alpha_{2}+\beta_{4}\\ \alpha_{1}+\beta_{2}&\alpha_{3}+\beta_{4}\end{array}\right) with rows

[TABLE]

Due to (2.1) we have the identity

[TABLE]

which means that one can apply the Theorem as $V=\mathbf{F}^{8}$ and $\tau(g,A)=gA(g^{-1})^{\otimes 2}.$ In this case for $V_{0}$ one can take

[TABLE]

Therefore, two-dimensional algebras $\mathbf{A}$, $\mathbf{B}$, given by their matrices of structure constants $A$, $B\in V_{0}$, are isomorphic if and only if the equality

[TABLE]

holds true.

For $\xi_{1}=\frac{\alpha_{3}+\beta_{4}}{\Delta}$, $\xi_{2}=-\frac{\alpha_{1}+\beta_{2}}{\Delta}$, $\eta_{1}=-\frac{\alpha_{2}+\beta_{4}}{\Delta}$, $\eta_{2}=\frac{\alpha_{1}+\beta_{3}}{\Delta}$, where $\Delta=\Delta(A)=\det(P(A))$, one has

[TABLE]

and A^{\prime}=\left(\begin{array}[]{cccc}\alpha^{\prime}_{1}&\alpha^{\prime}_{2}&\alpha^{\prime}_{3}&\alpha^{\prime}_{4}\\ \beta^{\prime}_{1}&\beta^{\prime}_{2}&\beta^{\prime}_{3}&\beta^{\prime}_{4}\end{array}\right)=P(A)A(P(A)^{-1}\otimes P(A)^{-1}) is matrix consisting of columns

[TABLE]

[TABLE]

[TABLE]

[TABLE]

From the above it can be easily seen that $\alpha^{\prime}_{1}=-\beta^{\prime}_{2}$, $\alpha^{\prime}_{3}=\alpha^{\prime}_{2}+1$, $\beta^{\prime}_{3}=\beta^{\prime}_{2}+1$ and $\beta^{\prime}_{4}=-\alpha^{\prime}_{2},$ i.e.

[TABLE]

Therefore, the main role in the finding of $A^{\prime}$ is played by the functions

$\alpha^{\prime}_{1}=-\frac{\beta_{1}\eta_{1}\xi_{1}^{2}}{\Delta}+\frac{\alpha_{1}\eta_{2}\xi_{1}^{2}}{\Delta}+\frac{2\alpha_{1}\eta_{1}\xi_{1}\xi_{2}}{\Delta}-\frac{2\beta_{4}\eta_{2}\xi_{1}\xi_{2}}{\Delta}-\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}}+\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}-\frac{\beta_{4}\eta_{1}\xi_{2}^{2}}{\Delta}+\frac{\alpha_{4}\eta_{2}\xi_{2}^{2}}{\Delta}+\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}},$

$\beta^{\prime}_{1}=\frac{\beta_{1}\xi_{1}^{3}}{\Delta}-\frac{3\alpha_{1}\xi_{1}^{2}\xi_{2}}{\Delta}+\frac{\eta_{2}\xi_{1}^{2}\xi_{2}}{\Delta^{2}}+\frac{3\beta_{4}\xi_{1}\xi_{2}^{2}}{\Delta}+\frac{\eta_{1}\xi_{1}\xi_{2}^{2}}{\Delta^{2}}-\frac{2\xi_{1}^{2}\xi_{2}^{2}}{\Delta^{2}}-\frac{\alpha_{4}\xi_{2}^{3}}{\Delta},$

$\alpha^{\prime}_{2}=-\frac{\beta_{1}\eta_{1}^{2}\xi_{1}}{\Delta}+\frac{2\alpha_{1}\eta_{1}\eta_{2}\xi_{1}}{\Delta}-\frac{\beta_{4}\eta_{2}^{2}\xi_{1}}{\Delta}-\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}+\frac{\alpha_{1}\eta_{1}^{2}\xi_{2}}{\Delta}-\frac{2\beta_{4}\eta_{1}\eta_{2}\xi_{2}}{\Delta}-\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}+\frac{\alpha_{4}\eta_{2}^{2}\xi_{2}}{\Delta}+\frac{2\eta_{1}\eta_{2}\xi_{1}\xi_{2}}{\Delta^{2}},$

$\alpha^{\prime}_{4}=-\frac{\beta_{1}\eta_{1}^{3}}{\Delta}+\frac{3\alpha_{1}\eta_{1}^{2}\eta_{2}}{\Delta}-\frac{3\beta_{4}\eta_{1}\eta_{2}^{2}}{\Delta}-\frac{2\eta_{1}^{2}\eta_{2}^{2}}{\Delta^{2}}+\frac{\alpha_{4}\eta_{2}^{3}}{\Delta}+\frac{\eta_{1}\eta_{2}^{2}\xi_{1}}{\Delta^{2}}+\frac{\eta_{1}^{2}\eta_{2}\xi_{2}}{\Delta^{2}}.$

It can be easily verified that these functions can take any values in $\mathbf{F}.$ Therefore, the values of $\alpha^{\prime}_{1},\beta^{\prime}_{1},\alpha^{\prime}_{2},\alpha^{\prime}_{4}$ also can be any elements in $\mathbf{F}$. Using obvious redenotions one can list the following non-isomorphic “canonical” algebras from the first subset, given by their MSCs as follows:

[TABLE]

For any algebra $\mathbf{A}$ from the first subset there exists a unique algebra from the class $A_{1}(\mathbf{c})$ isomorphic to $\mathbf{A}$. Note that the appearance of $A_{1}(\mathbf{c})$ for the first subset does not depend on either algebraically closeness of $\mathbf{F}$ or its characteristic.

The second and third subset cases. In these cases one can make $Tr_{1}(A)g=(1,0)$. It implies that $Tr_{2}(A)g=(\lambda,0)$. Therefore, we can assume that

[TABLE]

Here if $\lambda$ is zero we can also cover the third subset. So let us consider

[TABLE]

with respect to $g\in GL(2,\mathbf{F})$ of the form

[TABLE]

as far as $(1,0)g^{-1}=(1,0)$ if and only if $g^{-1}$ is of the above form. In this case for the entries of $A^{\prime}$ we have A^{\prime}=\left(\begin{array}[]{cccc}\alpha^{\prime}_{1}&\alpha^{\prime}_{2}&\alpha^{\prime}_{2}&\alpha^{\prime}_{4}\\ \beta^{\prime}_{1}&\lambda-\alpha^{\prime}_{1}&1-\alpha^{\prime}_{1}&-\alpha^{\prime}_{2}\end{array}\right)=gA(g^{-1})^{\otimes 2}, where $\lambda$ is same for the $A$ and $A^{\prime}$, one has

$\alpha^{\prime}_{1}=\frac{1}{\Delta}(\alpha_{1}\eta_{2}+2\alpha_{2}\eta_{2}\xi_{2}+\alpha_{4}\eta_{2}\xi_{2}^{2})=\alpha_{1}+2\alpha_{2}\xi_{2}+\alpha_{4}\xi_{2}^{2}.$

$\alpha^{\prime}_{2}=\frac{-1}{\Delta}(-\alpha_{2}\eta^{2}_{2}-\alpha_{4}\eta_{2}^{2}\xi_{2})=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2}.$

$\alpha^{\prime}_{4}=\frac{-1}{\Delta}(\beta_{1}\eta_{1}^{3}+[(\lambda-2)\alpha_{1}+(1+\lambda)\beta_{3}]\eta_{1}^{2}\eta_{2}-[(1+\lambda)\alpha_{2}+(\lambda-2)\beta_{4}]\eta_{1}\eta_{2}^{2}-\alpha_{4}\eta_{2}^{3})=\alpha_{4}\eta_{2}^{2}.$

$\beta^{\prime}_{1}=\frac{1}{\Delta}(\beta_{1}\xi_{1}^{3}+[(\lambda-2)\alpha_{1}+(\lambda+1)\beta_{3}]\xi_{1}^{2}\xi_{2}-[(1+\lambda)\alpha_{2}+(\lambda-2)\beta_{4}]\xi_{1}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3})$

$=\frac{\beta_{1}+(1+\lambda-3\alpha_{1})\xi_{2}-3\alpha_{2}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

So it is enough to consider the system

$\alpha^{\prime}_{1}=\alpha_{1}+2\alpha_{2}\xi_{2}+\alpha_{4}\xi_{2}^{2},$

$\alpha^{\prime}_{2}=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2},$

$\alpha^{\prime}_{4}=\alpha_{4}\eta_{2}^{2},$

$\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda-3\alpha_{1})\xi_{2}-3\alpha_{2}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

Now we have to consider a few cases again.

Case 1: $\alpha_{4}\neq 0$. In this case one can make

[TABLE]

and once again using redenotion one can represent the corresponding “canonical” MSCs as follows

[TABLE]

as far as the expressions $\alpha_{1}-\frac{\alpha^{2}_{2}}{\alpha_{4}}$, $\sqrt{\alpha_{4}}(\beta_{1}-(1+\lambda-3\alpha_{1})\frac{\alpha_{2}}{\alpha_{4}}-2\frac{\alpha^{3}_{2}}{\alpha^{2}_{4}})$, $\lambda-(\alpha_{1}-\frac{\alpha^{2}_{2}}{\alpha_{4}})$ may have any values in $\mathbf{F}$. It can be checked that algebras \left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ \beta_{1}&\beta_{2}&1-\alpha_{1}&0\end{array}\right) and \left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ -\beta_{1}&\beta_{2}&1-\alpha_{1}&0\end{array}\right) are isomorphic. Note that the result $A_{2}(\mathbf{c})$, does not depend on the characteristic of $\mathbf{F}$.

Case 2: $\alpha_{4}=0.$

Subcase 2 - a): $\alpha_{2}\neq 0.$ If $\alpha_{2}\neq 0$ then one can make $\alpha^{\prime}_{1}=0$, $\alpha^{\prime}_{2}=1$, $\beta^{\prime}_{1}=\alpha_{2}\beta_{1}-\frac{2+2\lambda-3\alpha_{1}}{4}\alpha_{1}$ to get the following set of canonical matrices of structural constants

[TABLE]

Subcase 2 - b): $\alpha_{2}=0.$ If $\alpha_{2}=0$ then $\alpha^{\prime}_{1}=\alpha_{1},$ $\alpha^{\prime}_{2}=0,$ $\alpha^{\prime}_{4}=0,$ $\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda-3\alpha_{1})\xi_{2}}{\eta_{2}}.$

Subsubcase: 2 - b) - 1: $1+\lambda-3\alpha_{1}\neq 0.$ If $1+\lambda-3\alpha_{1}\neq 0$, that is $\lambda-\alpha_{1}\neq 2\alpha_{1}-1$, one can make $\beta^{\prime}_{1}=0$ to get

[TABLE]

Subsubcase: 2 - b) - 2: $1+\lambda-3\alpha_{1}=0.$ If $1+\lambda-3\alpha_{1}=0$ and $\beta_{1}\neq 0$ one can make $\beta^{\prime}_{1}=1$ to get

[TABLE]

If $\beta_{1}=0$ one has $\lambda-\alpha_{1}=2\alpha_{1}-1$ and therefore A^{\prime}=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\ 0&2\alpha_{1}-1&1-\alpha_{1}&0\end{array}\right)

which is $A_{4}(\mathbf{c})$ with $\beta_{2}=2\alpha_{1}-1$.

The fourth subset case. By the similar justification as in the second and the third subsets case it is enough to consider

[TABLE]

$\alpha^{\prime}_{1}=\alpha_{1}+2\alpha_{2}\xi_{2}+\alpha_{4}\xi_{2}^{2},$

$\alpha^{\prime}_{2}=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2},$

$\alpha^{\prime}_{4}=\alpha_{4}\eta_{2}^{2},$

$\beta^{\prime}_{1}=\frac{\beta_{1}+\xi_{2}-3\alpha_{1}\xi_{2}-3\alpha_{2}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

Therefore we get the canonical MSCs as follows

[TABLE]

(it is $A_{2}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$) or

[TABLE]

(it is $A_{3}(\mathbf{\mathbf{c}})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\beta_{2}=0$) or

[TABLE]

(it is $A_{4}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$) or

[TABLE]

(it is $A_{5}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $3\alpha_{1}-1=0$).

The fifth subset case. In this case

[TABLE]

such that:

$\alpha^{\prime}_{1}=\frac{1}{\Delta}\left(-\beta_{1}\eta_{1}\xi_{1}^{2}+\alpha_{1}\eta_{2}\xi_{1}^{2}+2\alpha_{1}\eta_{1}\xi_{1}\xi_{2}+2\alpha_{2}\eta_{2}\xi_{1}\xi_{2}+\alpha_{2}\eta_{1}\xi_{2}^{2}+\alpha_{4}\eta_{2}\xi_{2}^{2}\right),$

$\alpha^{\prime}_{2}=\frac{-1}{\Delta}\left(\beta_{1}\eta_{1}^{2}\xi_{1}-2\alpha_{1}\eta_{1}\eta_{2}\xi_{1}-\alpha_{2}\eta_{2}^{2}\xi_{1}-\alpha_{1}\eta_{1}^{2}\xi_{2}-2\alpha_{2}\eta_{1}\eta_{2}\xi_{2}-\alpha_{4}\eta_{2}^{2}\xi_{2}\right),$

$\alpha^{\prime}_{4}=\frac{-1}{\Delta}\left(\beta_{1}\eta_{1}^{3}-3\alpha_{1}\eta_{1}^{2}\eta_{2}-3\alpha_{2}\eta_{1}\eta_{2}^{2}-\alpha_{4}\eta_{2}^{3}\right),$

$\beta^{\prime}_{1}=\frac{1}{\Delta}\left(\beta_{1}\xi_{1}^{3}-3\alpha_{1}\xi_{1}^{2}\xi_{2}-3\alpha_{2}\xi_{1}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3}\right).$

If $\alpha_{4}\neq 0$ by making $\frac{\eta_{2}}{\eta_{1}}$ equal to any root of the polynomial $p(t)=\beta_{1}-3\alpha_{1}t-3\alpha_{2}t^{2}-\alpha_{4}t^{3}$ one can make $\alpha^{\prime}_{4}=0$. Therefore, further it is assumed that $\alpha_{4}=0$.

Let us consider $g$ with $\eta_{1}=0$ to have $\alpha^{\prime}_{4}=0.$ In this case $\Delta=\xi_{1}\eta_{2}$ and

$\alpha^{\prime}_{1}=\xi_{1}\left(\alpha_{1}+2\alpha_{2}\frac{\xi_{2}}{\xi_{1}}\right),$

$\alpha^{\prime}_{2}=\alpha_{2}\eta_{2},$

$\beta^{\prime}_{1}=\frac{\xi_{1}^{2}}{\eta_{2}}\left(\beta_{1}-3\alpha_{1}\frac{\xi_{2}}{\xi_{1}}-3\alpha_{2}(\frac{\xi_{2}}{\xi_{1}})^{2}\right).$

Case a: $\alpha_{2}\neq 0$. One can consider $\frac{\xi_{2}}{\xi_{1}}=\frac{-\alpha_{1}}{2\alpha_{2}}$ to get $\alpha^{\prime}_{1}=0,\alpha^{\prime}_{2}=1$ and $\beta^{\prime}_{1}=\xi^{2}_{1}\frac{3\alpha^{2}_{1}+4\alpha_{2}\beta_{1}}{4}.$ Therefore one can make $\beta^{\prime}_{1}$ equal to [math] or $1,$ depending on $3\alpha^{2}_{1}+4\alpha_{2}\beta_{1}$ to have

[TABLE]

Case b: $\alpha_{2}=0$. Then $\alpha^{\prime}_{2}=\alpha^{\prime}_{4}=0$ and $\alpha^{\prime}_{1}=\xi_{1}\alpha_{1},\beta^{\prime}_{1}=\frac{\xi^{2}_{1}}{\eta_{2}}\left(\beta_{1}-3\alpha_{1}\frac{\xi_{2}}{\xi_{1}}\right).$ Therefore if $\alpha_{1}\neq 0$ one can make $\alpha^{\prime}_{1}=1,\beta^{\prime}_{1}=0$ to get A^{\prime}=\left(\begin{array}[]{cccc}1&0&0&0\\ 0&-1&-1&0\end{array}\right) , which isomorphic to $A_{10},$ if $\alpha_{1}=0$ then $\alpha^{\prime}_{1}=0$ and one can make $\beta^{\prime}_{1}=1$ to come to

[TABLE]

A routine check in each subset case shows that the corresponding algebras presented above are not isomorphic.

2.2. Characteristic $2$ case

In this case the result can be summarized in the following form.

Theorem 2.* Over an algebraically closed field $\mathbf{F}$ characteristic 2 any non-trivial 2-dimensional algebra is isomorphic to only one of the following listed, by their matrices of structure constants, algebras:*

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof. In this case $\alpha=-\alpha$ for any $\alpha\in\mathbf{F}$, but sometimes we use $-\alpha$ also to keep similarity with the previous case.

The first subset case. It has been noted earlier that in this case the result doesn’t depend either on algebraically closeness of $\mathbf{F}$ or its characteristics. Therefore, one has

[TABLE]

The second and third subset cases. Obviously in this case one can make $Tr_{1}(A)g=(1,0)$. This implies $Tr_{2}(A)g=(\lambda,0)$. Therefore, it can be assumed that $Tr_{1}(A)=(\alpha_{1}+\beta_{3},\alpha_{2}+\beta_{4})=(1,0)$ and $Tr_{2}(A)=(\lambda(\alpha_{1}+\beta_{3}),\lambda(\alpha_{2}+\beta_{4})=(\lambda,0)$. Here also we allow $\lambda$ to be zero to include the third subset’s case. Therefore, we consider

[TABLE]

$\alpha^{\prime}_{1}=\alpha_{1}+\alpha_{4}\xi_{2}^{2},$

$\alpha^{\prime}_{2}=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2},$

$\alpha^{\prime}_{4}=\alpha_{4}\eta_{2}^{2},$

$\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda-3\alpha_{1})\xi_{2}-3\alpha_{2}\xi_{2}^{2}-\alpha_{4}\xi_{2}^{3}}{\eta_{2}}=\frac{\beta_{1}+(1+\lambda+\alpha_{1})\xi_{2}+\alpha_{2}\xi_{2}^{2}+\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

As we have observed earlier in $\alpha_{4}\neq 0$ case one gets

[TABLE]

• 2.

Let now assume that $\alpha_{4}=0.$

• 2-a).

If $\alpha_{2}\neq 0$ then one can make $\alpha^{\prime}_{1}=\alpha_{1}$, $\alpha^{\prime}_{2}=1$, $\alpha^{\prime}_{4}=0$ and $\beta^{\prime}_{1}=0$ to get

[TABLE]

• 2-b).

If $\alpha_{2}=0$ then $\alpha^{\prime}_{1}=\alpha_{1},$ $\alpha^{\prime}_{2}=0,$ $\alpha^{\prime}_{4}=0,$ and $\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda+\alpha_{1})\xi_{2}}{\eta_{2}}.$

• 2-b)-1.

If $1+\lambda+\alpha_{1}\neq 0$, that is $\lambda-\alpha_{1}\neq 1$, one can make $\beta^{\prime}_{1}=0$ to get

[TABLE]

• 2-b)-2.

Let $1+\lambda+\alpha_{1}=0$. In this case

• ·

if $\beta_{1}\neq 0$ one can make $\beta^{\prime}_{1}=1$ to have

[TABLE]

• ·

if $\beta_{1}=0$ then

[TABLE]

The fourth subset case. For this case we have

[TABLE]

therefore

$\alpha^{\prime}_{1}=\alpha_{1}+\alpha_{4}\xi_{2}^{2},$

$\alpha^{\prime}_{2}=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2},$

$\alpha^{\prime}_{4}=\alpha_{4}\eta_{2}^{2},$

$\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\alpha_{1})\xi_{2}+\alpha_{2}\xi_{2}^{2}+\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

Hence, one gets A_{6,2}(\mathbf{c})=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&1\\ \beta_{1}&1-\alpha_{1}&-\alpha_{1}&0\end{array}\right)\ \ \mbox{where}\ \mathbf{c}=(\alpha_{1},\beta_{1})\in\mathbf{F}^{2}

(it is $A_{2,2}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$),

[TABLE]

(it is $A_{3,2}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$),

[TABLE]

(it is $A_{4,2}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$) and

[TABLE]

(it is $A_{5,2}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+1=0$).

The fifth subset case. In this case

[TABLE]

and for the entries of A^{\prime}=\left(\begin{array}[]{cccc}\alpha^{\prime}_{1}&\alpha^{\prime}_{2}&\alpha^{\prime}_{2}&\alpha^{\prime}_{4}\\ \beta^{\prime}_{1}&-\alpha^{\prime}_{1}&-\alpha^{\prime}_{1}&-\alpha^{\prime}_{2}\end{array}\right) one has

$\alpha^{\prime}_{1}=\frac{1}{\Delta}\left(\beta_{1}\eta_{1}\xi_{1}^{2}+\alpha_{1}\eta_{2}\xi_{1}^{2}+\alpha_{2}\eta_{1}\xi^{2}_{2}+\alpha_{4}\eta_{2}\xi_{2}^{2}\right),$

$\alpha^{\prime}_{2}=\frac{1}{\Delta}\left(\beta_{1}\eta_{1}^{2}\xi_{1}+\alpha_{1}\eta_{1}^{2}\xi_{2}+\alpha_{2}\eta_{2}^{2}\xi_{1}+\alpha_{4}\eta_{2}^{2}\xi_{2}\right),$

$\alpha^{\prime}_{4}=\frac{1}{\Delta}\left(\beta_{1}\eta_{1}^{3}+\alpha_{1}\eta_{1}^{2}\eta_{2}+\alpha_{2}\eta_{1}\eta_{2}^{2}+\alpha_{4}\eta_{2}^{3}\right),$

$\beta^{\prime}_{1}=\frac{1}{\Delta}\left(\beta_{1}\xi_{1}^{3}+\alpha_{1}\xi_{1}^{2}\xi_{2}+\alpha_{2}\xi_{1}\xi_{2}^{2}+\alpha_{4}\xi_{2}^{3}\right).$

As it was observed one can assume that $\alpha_{4}=0$.

Let us consider $g$ with $\eta_{1}=0$ to have $\alpha^{\prime}_{4}=0.$ In this case $\Delta=\xi_{1}\eta_{2}$ and

$\alpha^{\prime}_{1}=\xi_{1}\alpha_{1},$

$\alpha^{\prime}_{2}=\alpha_{2}\eta_{2},$

$\beta^{\prime}_{1}=\frac{\xi_{1}^{2}}{\eta_{2}}\left(\beta_{1}+\alpha_{1}\frac{\xi_{2}}{\xi_{1}}+\alpha_{2}(\frac{\xi_{2}}{\xi_{1}})^{2}\right).$

If one of $\alpha_{1},\alpha_{2}$ is not zero one can make $\beta^{\prime}_{1}=0$ and depending on $\alpha_{1},\alpha_{2}$ one can have $\alpha^{\prime}_{1}=0,\alpha^{\prime}_{2}=1$ or $\alpha^{\prime}_{1}=1,\alpha^{\prime}_{2}=1$ or $\alpha^{\prime}_{1}=1,\alpha^{\prime}_{2}=0$ to get A_{10,2}=\left(\begin{array}[]{cccc}0&1&1&0\\ 0&0&0&-1\end{array}\right) or A_{11,2}=\left(\begin{array}[]{cccc}1&1&1&0\\ 0&-1&-1&-1\end{array}\right) or A^{\prime}=\left(\begin{array}[]{cccc}1&0&0&0\\ 0&-1&-1&0\end{array}\right) which is isomorphic to $A_{10,2}$

If $\alpha_{1}=\alpha_{2}=0$ then one can make $\beta^{\prime}_{1}=1$ to get

[TABLE]

2.3. Characteristic $3$ case

In this case we summarize the result as follows.

Theorem 3.* Over an algebraically closed field $\mathbf{F}$ characteristics 3 any non-trivial 2-dimensional algebra is isomorphic to only one of the following listed, by their matrices of structure constant matrices, algebras:*

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proof.  The first subset case. One has

[TABLE]

The second and third subset cases. One can consider

[TABLE]

where $\lambda$ may be zero as well, with respect to g^{-1}=\left(\begin{array}[]{cc}1&0\\ \xi_{2}&\eta_{2}\end{array}\right) in that case for the entries of $A^{\prime}$ we get

$\alpha^{\prime}_{1}=\alpha_{1}+2\alpha_{2}\xi_{2}+\alpha_{4}\xi_{2}^{2},$

$\alpha^{\prime}_{2}=(\alpha_{2}+\alpha_{4}\xi_{2})\eta_{2},$

$\alpha^{\prime}_{4}=\alpha_{4}\eta_{2}^{2},$

$\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda)\xi_{2}-\alpha_{4}\xi_{2}^{3}}{\eta_{2}}.$

$\alpha_{4}\neq 0$. In this case we obtain

[TABLE]

• 2.

$\alpha_{4}=0.$

• 2-a)

If $\alpha_{2}\neq 0$ one can make $\alpha^{\prime}_{1}=0$, $\alpha^{\prime}_{2}=1$ to get

[TABLE]

• 2-b)

If $\alpha_{2}=0$ then $\alpha^{\prime}_{1}=\alpha_{1},$ $\alpha^{\prime}_{2}=0,$ $\alpha^{\prime}_{4}=0,$ and $\beta^{\prime}_{1}=\frac{\beta_{1}+(1+\lambda)\xi_{2}}{\eta_{2}}.$

• 2-b)-1.

Let $1+\lambda\neq 0$, that is $\lambda\neq-1$, one can make $\beta^{\prime}_{1}=0$ to get

[TABLE]

• 2-b)-2.

Let $1+\lambda=0.$

• ·

If $\beta_{1}\neq 0$ one can make $\beta^{\prime}_{1}=1$ to get

[TABLE]

• ·

If $\beta_{1}=0$ one has $\lambda=-1$ and therefore A^{\prime}=\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\ 0&-1-\alpha_{1}&1-\alpha_{1}&0\end{array}\right) which is $A_{4,3}(\mathbf{c})$ with $\beta_{2}=-1-\alpha_{1}$.

The fourth subset case. It is easy to see that in this case the result can be derived from the second and third subsets case. So we get

[TABLE]

(it is $A_{2,3}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$),

[TABLE]

(it is $A_{3,3}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\beta_{2}=0$),

[TABLE]

(it is $A_{4,3}(\mathbf{c})$, where the $2^{nd}$ and $3^{rd}$ columns are interchanged and $\alpha_{1}+\beta_{2}=0$).

The fifth subset case.

Here it is enough to consider $\alpha_{4}=0$ case. Therefore we have

$\alpha^{\prime}_{1}=\frac{1}{\Delta}\left(-\beta_{1}\eta_{1}\xi_{1}^{2}+\alpha_{1}\eta_{2}\xi_{1}^{2}-\alpha_{1}\eta_{1}\xi_{1}\xi_{2}-\alpha_{2}\eta_{2}\xi_{1}\xi_{2}+\alpha_{2}\eta_{1}\xi_{2}^{2}\right),$

$\alpha^{\prime}_{2}=\frac{-1}{\Delta}\left(\beta_{1}\eta_{1}^{2}\xi_{1}+\alpha_{1}\eta_{1}\eta_{2}\xi_{1}-\alpha_{2}\eta_{2}^{2}\xi_{1}-\alpha_{1}\eta_{1}^{2}\xi_{2}+\alpha_{2}\eta_{1}\eta_{2}\xi_{2}\right),$

$\alpha^{\prime}_{4}=\frac{-1}{\Delta}\beta_{1}\eta_{1}^{3},$

$\beta^{\prime}_{1}=\frac{1}{\Delta}\beta_{1}\xi_{1}^{3}.$

Making $\eta_{1}=0$ results in $\alpha^{\prime}_{1}=\alpha_{1}\xi_{1}-\alpha_{2}\xi_{2},$ $\alpha^{\prime}_{4}=0$, $\alpha^{\prime}_{2}=\alpha_{2}\eta_{2}$, $\beta^{\prime}_{1}=\frac{\xi^{2}_{1}}{\eta_{2}}\beta_{1}.$

• a)

If $\alpha_{2}\neq 0$ one can make $\alpha^{\prime}_{1}=0$ and $\alpha^{\prime}_{2}=1$.

• a)-1.

If $\beta_{1}\neq 0$ one can reduce $\beta^{\prime}_{1}=1$ to get A_{9,3}=\left(\begin{array}[]{cccc}0&1&1&0\\ 1&0&0&-1\end{array}\right).

• a)-2.

If $\beta_{1}=0$ then $\beta^{\prime}_{1}=0$ and one gets A_{10,3}=\left(\begin{array}[]{cccc}0&1&1&0\\ 0&0&0&-1\end{array}\right).

• b)

If $\alpha_{2}=0$ then $\alpha^{\prime}_{2}=0$.

• b)-1.

If $\beta_{1}\neq 0$ we reduce $\beta^{\prime}_{1}=1$ and $\alpha^{\prime}_{1}$ may take one or zero depending on $\alpha_{1}$. Therefore one has

[TABLE]

• b)-2.

If $\beta_{1}=0$ then $\beta^{\prime}_{1}=0$, $\alpha^{\prime}_{4}=0$ and one can make , $\alpha^{\prime}_{1}=1$ to get

[TABLE]

Remark 1. From the proof of Theorems it is clear that the same results remain be true if one assumes the existence of a root in $\mathbf{F}$ of every second and third order polynomial over $\mathbf{F}$ (instead of assumption that $\mathbf{F}$ to be algebraically closed). The results can be used for getting complete classification of different classes, for example, associative, Jordan, alternative and etc., 2-dimensional algebras.

Remark 2. In [1] the class $A_{3,2}(\mathbf{c})$ should be understood as it is in this paper.

Acknowledgment. The first author thanks M.A.A. Ahmed for the help in computer computations, the second author’s research is supported by FRGS14-153-0394, MOHE and the third author acknowledges MOHE for a support by grant 01-02-14-1591FR.