Abstract. We provide a clarification of the classification of two-dimensional algebras over an arbitrary base field. In the finite field case, we compute the number of two-dimensional non-isomorphic algebras at least with one non-zero trace.

1 Introduction↩︎

The number of non-isomorphic two-dimensional algebras over a finite field \(\mathbb{F}\) with \(|\mathbb{F}|=q\) was computed in [1] as

For another approach to computing this number, see [2].

Our original aim was to derive this number using the classification result of [3]. However, the discrepancy between the above formula and the number obtained from the classification result forced us to re-examine the classification. In that classification, all two-dimensional algebras were presented as a disjoint union of five invariant subsets, and the algebras in each subset were classified up to isomorphism. In the finite field case a difficulty arises in computing the number of non-isomorphic algebras in the fifth subset.

In this paper we present a corrected version of classification of fifth subset algebras and then compute the number of two-dimensional algebras, at least with one non-zero trace, up to isomorphism over a finite field.

The classification result presented in [3] is as follows.

2 On the Classification of Two-Dimensional Algebras over an Arbitrary Base Field↩︎

In this section, we correct certain inaccuracies in the classification result presented in [3].

Recall that a two-dimensional algebra given by the matrix \[A=\left( \begin{array}{cccc} \alpha _1 & \alpha _2 & \alpha _3 & \alpha _4 \\ \beta _1 & \beta _2 & \beta _3 & \beta _4 \end{array} \right)\] means the algebra with multiplication \[e_1e_1=\alpha _1e_1+\beta _1e_2, \;e_1e_2=\alpha _2e_1+\beta _2e_2,\;e_2e_1=\alpha _3e_1+\beta _3e_2, \;e_2e_2=\alpha _4e_1+\beta _4e_2.\] By the traces of \(A\), we mean vectors \(tr_1(A)=(\alpha _1+\beta_3,\alpha _2+\beta_4), tr_2(A)=(\alpha _1+\beta_2,\alpha _3+\beta_4)\). Note that under the change of basis \(e'=eg^{-1}\), the corresponding matrices and traces transform according to \(A'=gA(g^{-1}\otimes g^{-1})\) and \(tr_i(A') =tr_i(A)g^{-1}\), respectively. Our corrections concern only the following fifth-subset algebras(non-trivial algebras with \(tr_1(A)=tr_2(A)=0\)) considered in that paper, and we follow the notation used there.

The fifth subset in the case \(char(\mathbb{F})\neq 2,3\) . In this case \[A=\left( \begin{array}{cccc} \alpha _1 & \alpha _2 & \alpha _2 & \alpha _4 \\ \beta _1 & -\alpha _1 & -\alpha _1 & -\alpha _2 \end{array} \right),\;where\;g^{-1}=\left(\begin{array}{cc} \xi_1& \eta_1\\ \xi_2& \eta_2\end{array}\right),\;\Delta=\xi_1\eta_2-\xi_2\eta_1,\;and\] \[A'=\left( \begin{array}{cccc} \alpha' _1 & \alpha' _2 & \alpha' _2 & \alpha' _4 \\ \beta' _1 & -\alpha' _1 & -\alpha' _1 & -\alpha' _2 \end{array} \right)=gA(g^{-1})^{\otimes 2},\;then we have\]

\(\alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\alpha _1 \eta _2 \xi _1^2+2 \alpha _1 \eta _1 \xi _1 \xi _2+2 \alpha _2 \eta _2 \xi _1 \xi _2+\alpha _2 \eta _1 \xi _2^2+\alpha _4 \eta _2 \xi _2^2\right),\)

\(\alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-2 \alpha _1 \eta _1 \eta _2 \xi _1-\alpha _2 \eta _2^2 \xi _1-\alpha _1 \eta _1^2 \xi _2-2 \alpha _2 \eta _1 \eta _2 \xi _2-\alpha _4 \eta _2^2 \xi _2\right),\)

\(\alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-3 \alpha _1 \eta _1^2 \eta _2-3 \alpha _2 \eta _1 \eta _2^2-\alpha _4 \eta _2^3\right),\)

\(\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-3 \alpha _1 \xi _1^2 \xi _2-3 \alpha _2 \xi _1 \xi _2^2-\alpha _4 \xi _2^3\right).\)

Case 1: \(\alpha _4\neq 0\). It follows that, if \(\xi _1=0\) and \(\eta _2=-\frac{\alpha_2}{\alpha_4}\), then \(\alpha' _1=0\). Therefore, it is sufficient to consider the case \(\alpha_1=0\).

Case 1-1: \(\alpha _2\neq 0\). If \(\xi_2=\eta_1=0\), then \(\alpha' _1=0,\ \alpha' _2=\alpha_2 \eta _2,\;\alpha'_4=\alpha_4\frac{\eta_2^2}{\xi_1},\;\beta' _1=\beta _1\frac{\xi_1^2}{\eta_2}\). Thus one can make \(\alpha' _2=1,\;\alpha'_4=1\). Therefore, without loss of generality, we may assume that \(\alpha_1=0,\alpha _2=1, \alpha_4=1\) and the above system becomes

\(\alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+2\eta _2 \xi _1 \xi _2+\eta _1 \xi _2^2+\eta _2 \xi _2^2\right),\)

\(\alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-\eta _2^2 \xi _1-2 \eta _1 \eta _2 \xi _2-\eta _2^2 \xi _2\right),\)

\(\alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-3\eta _1 \eta _2^2-\eta _2^3\right),\)

\(\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-3\xi _1 \xi _2^2- \xi _2^3\right).\)

In the case \(\xi_2\eta_1=0\), this yields the algebra \(\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ 4-\beta_1 &0 &0 &-1 \end{array} \right)\). Indeed, if \(\eta_1=0\), then \(\Delta= \xi_1\eta_2\) and

\(\alpha' _1=\xi _1\left(2 \xi _2/\xi _1+(\xi _2/\xi _1)^2\right)\) implies \(\xi _2=-2\xi _1,\)

\(\alpha' _2=\eta_2\left(1+\xi _2/\xi _1\right)=-\eta_2\xi _1=1,\)

\(\alpha' _4=\frac{1}{\xi_1}\left(\eta _2^2\right)=\frac{1}{\xi_1^3}=1.\) So, if \(\xi_1=a, \xi_2=-2a\) and \(\eta_2=-1/a\), where \(a^3=1\), then

\(\beta' _1=\frac{1}{-1}(\beta _1a^3-12a^3+8a^3)=4-\beta _1.\)

In the case \(\xi_2=0\), no new algebra appears.

In the \(\xi_2\eta_1\neq 0\) case \(\alpha' _1=0\) is equivalent to \(\frac{ \eta _2}{\eta_1}(2\frac{\xi _1}{\xi_2}+1)-\beta _1 (\frac{\xi _1}{\xi_2})^2+ 1=0\). If \(2\frac{\xi _1}{\xi_2}+1=0\), then this yields \(\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ 4 &0 &0 &-1 \end{array} \right)\simeq\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ 0 &0 &0 &-1 \end{array} \right)\).

If \(2\frac{\xi _1}{\xi_2}+1\neq 0\), then \(\frac{ \eta _2}{\eta_1}=\frac{\beta _1 (\frac{\xi _1}{\xi_2})^2-1}{2\frac{\xi _1}{\xi_2}+1}=\frac{\beta _1 \xi^2-1}{2\xi+1}\), where \(\xi=\frac{\xi _1}{\xi_2}\), and \[\alpha' _2=\frac{\eta _1^2\xi _2}{\Delta }\left(-\beta _1 \frac{\xi _1}{\xi_2}+2\frac{ \eta _2}{\eta_1} +(1+\frac{\xi _1}{\xi_2})(\frac{ \eta _2}{\eta_1})^2 \right)=\]\[\frac{\eta _1^2\xi _2}{\Delta }\left(-\beta _1 \xi+2\frac{\beta _1 \xi^2-1}{2\xi+1} +(1+\xi)(\frac{\beta _1 \xi^2-1}{2\xi+1})^2 \right)=\] \[\frac{\eta _1^2\xi _2}{\Delta }\frac{\beta _1^2\xi^5+\beta _1^2\xi^4-2\beta _1\xi^3-4\beta _1\xi^2-(3 +\beta_1)\xi-1}{(2\xi+1)^2}=\] \[\frac{\eta _1^2\xi _2}{\Delta }\frac{(\beta _1\xi^3-3\xi-1)(\beta _1\xi^2+\beta _1\xi+1)}{(2\xi+1)^2}.\] \[\beta'_1=\frac{\xi_2^3}{\Delta}(\beta _1\xi^3-3\xi-1)\] \[\alpha'_4=-\frac{\eta_1^3}{\Delta}(\beta _1 -3(\frac{\eta _2}{\eta _1})^2- (\frac{\eta _2}{\eta _1})^3)=\frac{\eta_1^3}{(2\xi+1)^3\Delta}(\beta _1^3\xi^6 +6\beta_1^2\xi^5-20\beta_1\xi^3-15\beta_1\xi^2+6(1-\beta_1)\xi+2-\beta_1)\] Note that the following equalities hold \[\Delta=\xi_2\eta_1(\xi\frac{\beta _1 \xi^2-1}{2\xi+1}-1)=\xi_2\eta_1\frac{\beta _1 \xi^3-3\xi-1}{2\xi+1},\]\[P(t)=\beta _1^3t^6 +6\beta_1^2t^5-20\beta_1t^3-15\beta_1t^2+6(1-\beta_1)t+2-\beta_1=(\beta_1t^3-3t-1)(\beta_1^2t^3+6\beta_1t^2+3\beta_1t+\beta_1-2).\]

To make \(\alpha'_2=\alpha'_4=1\), there must exist \(\xi_0\neq -1/2\) such that \(P_1(\xi_0)P_2(\xi_0)P_3(\xi_0)\neq 0\), where \(P_1(t)=\beta _1t^2+\beta _1t+1, P_2(t)=\beta _1t^3-3t-1, P_3(t)=\beta_1^2t^3+6\beta_1t^2+3\beta_1t+\beta_1-2\).

In this case \[\alpha' _2=\frac{\eta _1^2\xi _2}{\Delta }\frac{(\beta _1\xi_0^3-3\xi_0-1)(\beta _1\xi_0^2+\beta _1\xi_0+1)}{(2\xi_0+1)^2}=\eta _1\frac{\beta _1\xi_0^2+\beta _1\xi_0+1}{2\xi_0+1}.\]

Therefore if \(\eta _1=\frac{2\xi_0+1}{\beta _1\xi_0^2+\beta _1\xi_0+1}\), then \(\alpha' _2=1\).

Equality \(\alpha'_4=\frac{\eta_1^3}{(2\xi_0+1)^3\Delta}P(\xi_0)=1\) is equivalent to \(\xi_2=\frac{\beta_1^2\xi_0^3+6\beta_1\xi_0^2+3\beta_1\xi_0+\beta_1-2}{(\beta_1\xi_0^2+\beta_1\xi_0+1)^2}\).

In this case \[\beta' _1=\frac{\xi_2^3}{\Delta }(\beta _1 \xi_0^3-3\xi_0 -1)=\frac{(2\xi_0+1)\xi_2^2}{\eta_1}=\frac{(\beta_1^2\xi_0^3+6\beta_1\xi_0^2+3\beta_1\xi_0+\beta_1-2)^2}{(\beta_1\xi_0^2+\beta_1\xi_0+1)^3}\] and hence we obtain the matrix of structure constants (MSC) \[\left(\begin{array}{cccc} 0 &1 & 1 &1 \\ 4-\beta_1 &0 &0 &-1 \end{array} \right)\simeq A_{10}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta'_1(a) &0 &0 &-1 \end{array}\right),\] where \(\frac{-1}{2}\neq a\in \mathbb{F}\), \((\beta _1a^3-3a-1)(\beta _1a^2+\beta_1a+1)(\beta_1^2a^3+6\beta_1a^2+3\beta_1a+\beta_1-2)\neq 0\), and \[\beta' _1(t)=\frac{(\beta_1^2t^3+6\beta_1t^2+3\beta_1t+\beta_1-2)^2}{(\beta_1t^2+\beta_1t+1)^3}.\] It follows that \(\beta' _1(\frac{-1}{2})=4-\beta_1\).

Case 1-2: \(\alpha _2= 0\). The system becomes
\(\begin{array}{ll} \alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\alpha _4 \eta _2 \xi _2^2\right),& \alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-\alpha _4 \eta _2^2 \xi _2\right),\\ \alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-\alpha _4 \eta _2^3\right),&\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-\alpha _4 \xi _2^3\right)\end{array}.\)
To make \(\alpha' _1=\alpha' _2=0\), we must have \(\xi_1=\eta_2=0\) or \(\xi_2=\eta_1=0\). If \(\xi_2=\eta_1=0\) and \(\xi_1=\alpha _4 \eta _2^2\), then \(\alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-\alpha _4 \eta _2^3\right)=\frac{-1}{\xi_1}(-\alpha _4 \eta _2^2)=1\) and \(\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-\alpha _4 \xi _2^3\right)=\frac{\xi _1^2}{\eta _2}\beta _1= \beta _1\alpha _4^2\eta _2^3\) and hence it yields the algebra \[A_{11}(\mathbf{c})=\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1 &0 &0 &0 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &0 & 0 &1 \\ a^3\beta_1 &0 &0 &0 \end{array}\right),\] where \(\mathbf{c}=\beta_1\in \mathbb{F}\) and \(0\neq a\in \mathbb{F}\).

Case 2: \(\alpha _4= 0\). The system becomes

\(\alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\alpha _1 \eta _2 \xi _1^2+2 \alpha _1 \eta _1 \xi _1 \xi _2+2 \alpha _2 \eta _2 \xi _1 \xi _2+\alpha _2 \eta _1 \xi _2^2\right),\)

\(\alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-2 \alpha _1 \eta _1 \eta _2 \xi _1-\alpha _2 \eta _2^2 \xi _1-\alpha _1 \eta _1^2 \xi _2-2 \alpha _2 \eta _1 \eta _2 \xi _2\right),\)

\(\alpha' _4=\frac{-\eta _1}{\Delta }\left(\beta _1 \eta _1^2-3 \alpha _1 \eta _1 \eta _2-3 \alpha _2 \eta _2^2\right),\)

\(\beta' _1=\frac{\xi _1}{\Delta }\left(\beta _1 \xi _1^2-3 \alpha _1 \xi _1 \xi _2-3 \alpha _2\xi _2^2\right).\)

If \(\eta _1 = 0\), then \(\Delta=\xi_1\eta_2\), \(\alpha'_4=0\), and

\(\alpha' _1=\xi _1\left(\alpha _1+2\alpha _2\frac{\xi_2}{\xi_1}\right),\)

\(\alpha' _2=\alpha _2\eta_2,\)

\(\beta' _1=\frac{ \xi _1^2}{\eta_2 }\left(\beta _1-3 \alpha _1 \frac{\xi_2}{\xi_1}-3 \alpha _2(\frac{\xi_2}{\xi_1})^2\right).\)

Case 2-1: \(\alpha_2\neq 0\). If \(\frac{\xi_2}{\xi_1}=\frac{-\alpha _1}{2\alpha _2}\), then \(\alpha' _1=0\) \(\alpha' _2\neq 0\). Then \(\alpha' _2\) can be made equal to \(1\), yielding
\(A_{12}(\mathbf{c})= \begin{pmatrix} 0 & 1 & 1 & 0 \\ \beta_1 &0& 0 &-1 \end{pmatrix} \simeq \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ a^2\beta_1 &0& 0 &-1 \end{array} \right),\) where \(\mathbf{c}=\beta_1\in \mathbb{F}\) and \(0\neq a\in \mathbb{F}\).

Case 2-2: \(\alpha_2= 0\). The corresponding system is

\(\alpha' _1=\frac{\xi _1}{\Delta }\left(-\beta _1 \eta _1 \xi _1+\alpha _1 \eta _2 \xi _1+2 \alpha _1 \eta _1 \xi _2\right),\)

\(\alpha' _2=\frac{-\eta _1}{\Delta }\left(\beta _1 \eta _1 \xi _1-2 \alpha _1 \eta _2 \xi _1-\alpha _1 \eta _1 \xi _2\right),\)

\(\alpha' _4=\frac{-\eta _1^2}{\Delta }\left(\beta _1 \eta _1-3 \alpha _1 \eta _2\right),\)

\(\beta' _1=\frac{\xi _1^2}{\Delta }\left(\beta _1 \xi _1-3 \alpha _1 \xi _2\right).\)

During the re-examination, we found the following isomorphisms among the above classes of algebras:
\(A_{11}(\beta_1)\simeq A_{11}(\beta_1^2),\) since \(gA_{11}(\beta_1)(g^{-1}\otimes g^{-1})=A_{11}(\beta_1^2)\), where \(g=\begin{pmatrix}0 & 1/\beta_1\\ 1&0\end{pmatrix}, \beta_1\neq 0\).
Algebras \(A_{10}(\beta_1)\) and \(A_{11}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+2+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\). In this case \(gA_{10}(\beta_1)(g^{-1}\otimes g^{-1})=A_{11}(\beta'_1)\), where \(g=\begin{pmatrix}t^2+1/(\beta'_1t) & 1/(\beta'_1t)\\ t+1/(\beta'_1t^2)&t\end{pmatrix}\).
Algebras \(A_{10}(\beta_1)\) and \(A_{12}(\beta'_1)\) are isomorphic if and only if there exists \(t\neq \pm1/2, s\neq 0\) such that \(\beta_1=2(2t+1)^2(1-t), s^2\beta'_1=1-t^2\). In this case \(gA_{10}(\beta_1)(g^{-1}\otimes g^{-1})=A_{12}(\beta'_1)\), where \(g=\begin{pmatrix} (2t+1)s & s\\ (2t+1)(1-t)&t\end{pmatrix}\).
Algebras \(A_{11}(\beta_1)\) and \(A_{12}(\beta'_1)\) are isomorphic if and only if there exists \(s,t\in F^*\) such that \(\beta_1=-8t^3, s^2\beta'_1=-t^2\). In this case \(gA_{11}(\beta_1)(g^{-1}\otimes g^{-1})=A_{12}(\beta'_1)\), where \(g=\begin{pmatrix} 2ts & s\\ -2t^2&t\end{pmatrix}\).
The fifth subset in the case \(char(\mathbb{F})=2\).

The same approach applies.

Case 1: \(\alpha _4\neq 0\). If \(\xi _1=0\) and \(\eta _2=-\frac{\alpha_2}{\alpha_4}\), then \(\alpha' _1=0\). Therefore, we only consider the case \(\alpha_1=0\).

Case 1.1: \(\alpha _2\neq 0\). If \(\xi_2=\eta_1=0\), then \(\alpha' _1=0,\ \alpha' _2=\alpha_2 \eta _2,\;\alpha'_4=\alpha_4\frac{\eta_2^2}{\xi_1},\;\beta' _1=\beta _1\frac{\xi_1^2}{\eta_2}\). Therefore, without loss of generality, we may assume that \(\alpha_1=0,\ \alpha _2=1,\;\alpha_4=1\). The corresponding system becomes:

\(\alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\eta _1 \xi _2^2+\eta _2 \xi _2^2\right),\)

\(\alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-\eta _2^2 \xi _1-\eta _2^2 \xi _2\right),\)

\(\alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-\eta _1 \eta _2^2-\eta _2^3\right),\)

\(\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-\xi _1 \xi _2^2- \xi _2^3\right).\)

If \(\xi_2=\eta_1=0\), then \(\Delta= \xi_1\eta_2\) and \(\alpha' _1=0,\) \(\alpha' _2=\eta_2\xi _1,\) \(\alpha' _4=\frac{\eta _2^2}{\xi_1}\), \(\beta' _1=\frac{\xi _1^2}{\eta _2}\beta_1.\) Setting \(\eta_2=1\) and \(\xi _1=1\), we obtain \(\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\).

In the case \(\xi_2\eta_1\neq 0\), we have \(\alpha' _1=0\) if and only if \(\frac{ \eta _2}{\eta_1}=\beta _1 \xi^2-1\), where \(\xi=\frac{\xi _1}{\xi_2}\). In this case \[\alpha' _2=\frac{\eta _1^2\xi _2}{\Delta }\left(-\beta _1 \frac{\xi _1}{\xi_2}+(1+\frac{\xi _1}{\xi_2})(\frac{ \eta _2}{\eta_1})^2 \right)=\]\[\frac{\eta _1^2\xi _2}{\Delta }\left(-\beta _1 \xi +(1+\xi)(\beta _1 \xi^2-1)^2 \right)=\] \[\frac{\eta _1^2\xi _2}{\Delta }(\beta _1^2\xi^5+\beta _1^2\xi^4-(3 +\beta_1)\xi-1)=\] \[\frac{\eta _1^2\xi _2}{\Delta }(\beta _1\xi^3-\xi-1)(\beta _1\xi^2+\beta _1\xi+1).\] \[\beta'_1=\frac{\xi_2^3}{\Delta}(\beta _1\xi^3-\xi-1)\] \[\alpha'_4=-\frac{\eta_1^3}{\Delta}(\beta _1 -(\frac{\eta _2}{\eta _1})^2- (\frac{\eta _2}{\eta _1})^3)=\frac{\eta_1^3}{\Delta}(\beta _1^3\xi^6 -\beta_1\xi^2-\beta_1)\] Note that the following equalities hold: \[\Delta=\xi_2\eta_1(\xi(\beta _1 \xi^2-1)-1)=\xi_2\eta_1(\beta _1 \xi^3-\xi-1),\]\[P(t)=\beta _1^3t^6 -\beta_1t^2-\beta_1=(\beta_1t^3+t+1)^2\beta_1.\]

To make \(\alpha'_2=\alpha'_4=1\), we must have \(\beta_1\neq 0\), and there must exist \(\xi_0\) such that \(P_1(\xi_0)P_2(\xi_0)\neq 0\), where \(P_1(t)=\beta _1t^2+\beta _1t+1, P_2(t)=\beta _1t^3+t+1\).

In this case \[\alpha' _2=\frac{\eta _1^2\xi _2}{\Delta }(\beta _1\xi_0^3-\xi_0-1)(\beta _1\xi_0^2+\beta _1\xi_0+1)=\eta _1(\beta _1\xi_0^2+\beta _1\xi_0+1).\]

Therefore, if \(\eta _1=\frac{1}{\beta _1\xi_0^2+\beta _1\xi_0+1}\), then \(\alpha' _2=1\).

Equality \(\alpha'_4=\frac{\eta_1^3}{\Delta}P(\xi_0)=1\) is equivalent to \(\xi_2=\beta_1\frac{\beta_1\xi_0^3+\xi_0+1}{(\beta_1\xi_0^2+\beta_1\xi_0+1)^2}\).

In this case \[\beta' _1=\frac{\xi_2^3}{\Delta }(\beta _1 \xi_0^3-\xi_0 -1)=\beta_1^2\frac{(\beta_1\xi_0^3+\xi_0+1)^2}{(\beta_1\xi_0^2+\beta_1\xi_0+1)^3}\] and this yields the MSC \[A_{8.2}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta'_1(a) &0 &0 &-1 \end{array}\right),\] where \((\beta _1a^3+a+1)(\beta _1a^2+\beta_1a+1)\beta_1\neq 0\) and \[\beta' _1(t)=\beta_1^2\frac{(\beta_1t^3+t+1)^2}{(\beta_1t^2+\beta_1t+1)^3}.\]

Case 1.2: \(\alpha _2= 0\). The system becomes
\(\begin{array}{ll} \alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\alpha _4 \eta _2 \xi _2^2\right),& \alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-\alpha _4 \eta _2^2 \xi _2\right),\\ \alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-\alpha _4 \eta _2^3\right),&\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-\alpha _4 \xi _2^3\right)\end{array}\)
If \(\xi_2=\eta_1=0\), then \(\alpha' _4=\frac{-1}{\Delta }\left(\beta _1 \eta _1^3-\alpha _4 \eta _2^3\right)=\frac{\alpha _4 \eta _2^2}{\xi_1}\) and \(\beta' _1=\frac{1}{\Delta }\left(\beta _1 \xi _1^3-\alpha _4 \xi _2^3\right)=\frac{\xi _1^2}{\eta _2}\beta _1= \beta _1\alpha _4^2\eta _2^3\). Therefore one obtains \[A_{9.2}(\mathbf{c})=\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1 &0 &0 &0 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &0 & 0 &1 \\ a^3\beta_1 &0 &0 &0 \end{array}\right),\] where \(\mathbf{c}=\beta_1\in \mathbb{F}\) and \(0\neq a\in \mathbb{F}\).

Case 2: \(\alpha _4= 0\). The system becomes

\(\alpha' _1=\frac{1}{\Delta }\left(-\beta _1 \eta _1 \xi _1^2+\alpha _1 \eta _2 \xi _1^2+\alpha _2 \eta _1 \xi _2^2\right),\)

\(\alpha' _2=\frac{-1}{\Delta }\left(\beta _1 \eta _1^2 \xi _1-\alpha _2 \eta _2^2 \xi _1-\alpha _1 \eta _1^2 \xi _2\right),\)

\(\alpha' _4=\frac{-\eta _1}{\Delta }\left(\beta _1 \eta _1^2- \alpha _1 \eta _1 \eta _2- \alpha _2 \eta _2^2\right),\)

\(\beta' _1=\frac{\xi _1}{\Delta }\left(\beta _1 \xi _1^2- \alpha _1 \xi _1 \xi _2- \alpha _2\xi _2^2\right).\)

If \(\eta _1 = 0\), then \(\alpha'_4=0\), \(\Delta=\xi_1\eta_2\), and

\(\alpha' _1=\xi _1\alpha _1,\)

\(\alpha' _2=\alpha _2\eta_2,\)

\(\beta' _1=\frac{ \xi _1^2}{\eta_2 }\left(\beta _1- \alpha _1 \frac{\xi_2}{\xi_1}- \alpha _2(\frac{\xi_2}{\xi_1})^2\right).\)

Case 2-1: \(\alpha_1\neq 0, \alpha_2\neq 0\). We may assume \(\alpha' _1=1\) and \(\alpha' _2= 1\), obtaining
\(A_{10,2}(\mathbf{c})=\left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ \beta_1 &1& 1 &1 \end{array} \right)\simeq \left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ a^2+a+\beta_1 &1& 1 &1 \end{array} \right),\) where \(\mathbf{c}=\beta_1\in \mathbb{F}\) and \(a\in \mathbb{F}\).

Case 2-2: \(\alpha_1=0, \alpha_2\neq 0\). In this case, we obtain
\(A_{11,2}(\mathbf{c})=\left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ \beta_1 &0& 0 &1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ b^2(\beta_1+a^2) &0&0 &1 \end{array} \right),\) where \(\mathbf{c}=\beta_1\in \mathbb{F}\) and \(a,b\in \mathbb{F}, b\neq 0\).

Case 2-3: \(\alpha_1\neq 0, \alpha_2= 0\). This yields the algebra
\(A'=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 &-1& -1 &0 \end{array} \right)\), which is isomorphic to \(A_{11,2}(0)\).

Case 2-4: \(\alpha_1= 0, \alpha_2= 0\). In this case one has \(A=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 &0&0 &0\end{array}\right)\simeq A_{9,2}(0).\)

In the case \(char(\mathbb{F})=2\), among the above class of algebras there are the following isomorphisms:
Algebras \(A_{8,2}(\beta_1)\) and \(A_{9,2}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\). In this case \(gA_{8,2}(\beta_1)(g^{-1}\otimes g^{-1})=A_{9,2}(\beta'_1)\), where \(g=\begin{pmatrix}t^2+1/(\beta'_1t) & 1/(\beta'_1t)\\ t+1/(\beta'_1t^2)&t\end{pmatrix}\).
Algebras \(A_{8,2}(\beta_1)\) and \(A_{10,2}(\beta'_1)\) are isomorphic if and only if there exists \(s\in F^*\setminus \{1\}\) such that \(\beta_1=s^3+s\) and \(\beta'_1=(1+s)/s\). In this case \(gA_{8,2}(\beta_1)(g^{-1}\otimes g^{-1})=A_{10,2}(\beta'_1)\), where \(g=\begin{pmatrix}s^2+s& s\\ s^2+s&1\end{pmatrix}\).
Algebras \(A_{8,2}(\beta_1)\) and \(A_{11,2}(\beta'_1)\) are isomorphic if and only if \(\beta_1=0=\beta'_1\). In this case \(gA_{8.2}(\beta_1)(g^{-1}\otimes g^{-1})=A_{11,2}(\beta'_1)\), where \(g=\begin{pmatrix}1& 1\\ 0&1\end{pmatrix}\).
Algebras \(A_{9,2}(\beta_1)\) and \(A_{10,2}(\beta'_1)\) are isomorphic if and only if there exists \(s\neq 0,t\in F\) such that \(\beta_1=s^3\), \(t^2+st+(\beta'_1+1)s^2=0\). In this case \(gA_{9,2}(\beta_1)(g^{-1}\otimes g^{-1})=A_{10,2}(\beta'_1)\), where \(g=\begin{pmatrix}s^2&s\\ s^2+st&t\end{pmatrix}\).
Algebras \(A_{9,2}(\beta_1)\) and \(A_{11,2}(\beta'_1)\) are not isomorphic.
Algebras \(A_{10,2}(\beta_1)\) and \(A_{11,2}(\beta'_1)\) are not isomorphic.
The fifth subset in the case \(char(\mathbb{F})=3\).
The proof is similar to that of the case \(char(\mathbb{F})\neq 2,3\), except for the following situation.
Case2-2:\(\alpha_2= 0\). The corresponding system is

\(\alpha' _1=\frac{\xi _1}{\Delta }\left(-\beta _1 \eta _1 \xi _1+\alpha _1 \eta _2 \xi _1+2 \alpha _1 \eta _1 \xi _2\right),\)

\(\alpha' _2=\frac{-\eta _1}{\Delta }\left(\beta _1 \eta _1 \xi _1-2 \alpha _1 \eta _2 \xi _1-\alpha _1 \eta _1 \xi _2\right),\)

\(\alpha' _4=\frac{-\eta _1^2}{\Delta }\left(\beta _1 \eta _1-3 \alpha _1 \eta _2\right),\)

\(\beta' _1=\frac{\xi _1^2}{\Delta }\left(\beta _1 \xi _1-3 \alpha _1 \xi _2\right).\)

If \(\eta_1=0\), then \(\alpha'_2=\alpha'_4=0\) and \(\alpha'_1=\xi_1\alpha_1, \beta' _1=\frac{\xi^2_1}{\eta_2}\beta _1.\) Therefore, one obtains \(A'=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 &-1& -1 &0 \end{array} \right)\simeq A_{11,3}(0)\), or \(A_{12,3}=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 &-1& -1 &0 \end{array} \right)\simeq A_{9,3}(0)\) or \(A=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 &0& 0 &0 \end{array} \right)\simeq A_{10,3}(0)\).

Moreover, the following statements hold.
Algebras \(A_{9,3}(\beta_1)\) and \(A_{10,3}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+2+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\). In this case \(gA_{9,3}(\beta_1)(g^{-1}\otimes g^{-1})=A_{10,3}(\beta'_1)\), where \(g=\begin{pmatrix}t^2+1/(\beta'_1t) & 1/(\beta'_1t)\\ t+1/(\beta'_1t^2)&t\end{pmatrix}\).
Algebras \(A_{9,3}(\beta_1)\) and \(A_{11,3}(\beta'_1)\) are isomorphic if and only if there exists \(t\in \mathbb{F}\setminus \{\pm 1\}\) and \(s\neq 0\) such that \(\beta_1=t^3-1, s^2\beta'_1=1-t^2\). In this case \(gA_{9,3}(\beta_1)(g^{-1}\otimes g^{-1})=A_{11,3}(\beta'_1)\), where \(g=\begin{pmatrix} (1-t)s & s\\ t^2+t+1&t\end{pmatrix}\).
Algebras \(A_{10,3}(\beta_1)\) and \(A_{11,3}(\beta'_1)\) are isomorphic if and only if there exists \(s,t\in F^*\) such that \(\beta_1=t^3, s^2\beta'_1=-t^2\). In this case \(gA_{10,3}(\beta_1)(g^{-1}\otimes g^{-1})=A_{12,3}(\beta'_1)\), where \(g=\begin{pmatrix} -ts & s\\ t^2&t\end{pmatrix}\).

Conclusion. In Theorem 1.1 the items \(A_{13}, A_{12,2}, A_{12,3}, A_{13,3}\) can be omitted and the items \(A_{10}, A_{11}, A_{12}\), \(A_{8,2}, A_{9,2}, A_{10,2}, A_{11,2}\), \(A_{9,3},A_{10,3}, A_{11,3}\) should be replaced by as follows:

• \(A_{10}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta'_1(a) &0 &0 &-1 \end{array}\right),\) where \(\mathbf{c}=\beta_1\),
  \((\beta _1a^3-3a-1)(\beta _1a^2+\beta_1a+1)(\beta_1^2a^3+6\beta_1a^2+3\beta_1a+\beta_1-2)\neq 0,\\ \beta' _1(t)=\frac{(\beta_1^2t^3+6\beta_1t^2+3\beta_1t+\beta_1-2)^2}{(\beta_1t^2+\beta_1t+1)^3}\) if \(t\neq \frac{-1}{2}\), \(\beta'(\frac{-1}{2})=4-\beta_1\).
  

• \(\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1^2 &0 &0 &0 \end{array} \right)\simeq A_{11}(\mathbf{c})=\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1 &0 &0 &0 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &0 & 0 &1 \\ a^3\beta_1 &0 &0 &0 \end{array}\right),\;where\\ \;\mathbf{c}=\beta_1, a\in \mathbb{F}^*.\)
  Algebras \(A_{10}(\beta_1)\) and \(A_{11}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+2+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\).
  

• \(A_{12}(\mathbf{c})=\left( \begin{array}{cccc} 0 & 1 & 1 &0 \\ \beta_1 &0& 0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ a^2\beta_1 &0& 0 &-1 \end{array} \right),\;where\;\mathbf{c}=\beta_1\in \mathbb{F},\;a\in \mathbb{F}^*.\)
  Algebras \(A_{10}(\beta_1)\) and \(A_{12}(\beta'_1)\) are isomorphic if and only if there exists \(t\neq \pm1/2, s\neq 0\) such that \(\beta_1=2(2t+1)^2(1-t), s^2\beta'_1=1-t^2\).
  Algebras \(A_{11}(\beta_1)\) and \(A_{12}(\beta'_1)\) are isomorphic if and only if there exists \(s,t\in F^*\) such that \(\beta_1=t^3, \beta'_1=-s^2\).
  

• \(A_{8,2}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta'_1(a) &0 &0 &-1 \end{array}\right),\)where \(\mathbf{c}=\beta_1\),
  \((\beta _1a^3+a+1)(\beta _1a^2+\beta_1a+1)\beta_1\neq 0\), and \(\beta' _1(t)=\beta_1^2\frac{(\beta_1t^3+t+1)^2}{(\beta_1t^2+\beta_1t+1)^3},\)
  

• \(A_{9,2}(\mathbf{c})=\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1 &0 &0 &0 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &0 & 0 &1 \\ a^3\beta_1 &0 &0 &0 \end{array}\right),\)where \(\mathbf{c}=\beta_1\in \mathbb{F}\), \(a\in \mathbb{F}^*\).
  Algebras \(A_{8,2}(\beta_1)\) and \(A_{9,2}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\).

• \(A_{10,2}(\mathbf{c})=\left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ \beta_1 &1& 1 &1 \end{array} \right)\simeq \left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ \beta_1+a+a^2 &1& 1 &1 \end{array} \right),\;where\;\mathbf{c}=\beta_1\in \mathbb{F},\;a\in \mathbb{F},\)
  Algebras \(A_{8,2}(\beta_1)\) and \(A_{10,2}(\beta'_1)\) are isomorphic if and only if there exists \(1\neq s\in F^*\) such that \(\beta_1=s^3+s\), \(\beta'_1=(1+s)/s\).
  Algebras \(A_{9,2}(\beta_1)\) and \(A_{10,2}(\beta'_1)\) are isomorphic if and only if there exists \(0\neq s,t\in F\) such that \(\beta_1=s^3\), \(t^2+st+(\beta'_1+1)s^2=0\).

• \(A_{11,2}(\mathbf{c})=\left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ \beta_1 &0& 0 &1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ b^2(\beta_1+a^2) &0& 0 &1 \end{array} \right),\) where \(\mathbf{c}=\beta_1\), \(0\neq b\), \(a,b\in \mathbb{F}\).
  Algebras \(A_{8,2}(\beta_1)\) and \(A_{11,2}(\beta'_1)\) are isomorphic if and only if \(\beta_1=0=\beta'_1\).
  

• \(A_{9,3}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &1 \\ \beta'_1(a) &0 &0 &-1 \end{array}\right),\) where \(\mathbf{c}=\beta_1\), \(a\in \mathbb{F}\),
  \((\beta _1a^3-1)(\beta _1a^2+\beta_1a+1)(\beta_1^2a^3+\beta_1+1)\neq 0\), \(\beta' _1(t)=\frac{(\beta_1^2t^3+\beta_1+1)^2}{(\beta_1t^2+\beta_1t+1)^3}\) if \(t\neq 1\), \(\beta'(1)=2\beta_1+1\).

• \(\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1^2 &0 &0 &0 \end{array} \right)\simeq A_{10,3}(\mathbf{c})=\left( \begin{array}{cccc} 0 &0 & 0 &1 \\ \beta_1 &0 &0 &0 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &0 & 0 &1 \\ a^3\beta_1 &0 &0 &0 \end{array}\right),\)where \(\mathbf{c}=\beta_1\), \(a\in \mathbb{F}^*\).
  Algebras \(A_{9,3}(\beta_1)\) and \(A_{10,3}(\beta'_1)\) are isomorphic if and only if there exists \(t\in F^*\) such that \(\beta_1=\beta'_1t^3+2+1/(\beta'_1t^3)\) and \((\beta'_1)^2t^6\neq 1\).
  

• \(A_{11,3}(\mathbf{c})=\left( \begin{array}{cccc} 0 &1 & 1 &0 \\ \beta_1 &0 &0 &-1 \end{array} \right)\simeq \left( \begin{array}{cccc} 0 &1 & 1 &0 \\ a^2\beta_1 &0 &0 &-1 \end{array}\right),\)where \(\mathbf{c}=\beta_1\), \(a\in \mathbb{F}^*\).
  Algebras \(A_{9,3}(\beta_1)\) and \(A_{11,3}(\beta'_1)\) are isomorphic if and only if there exists \(\pm 1\neq t\in F, s\neq 0\) such that \(\beta_1=t^3-1, s^2\beta'_1=1-t^2\).
  Algebras \(A_{10,3}(\beta_1)\) and \(A_{11,3}(\beta'_1)\) are isomorphic if and only if there exists \(s,t\in F^*\) such that \(\beta_1=t^3, \beta'_1=-s^2\).

3 On the Number of Non-Isomorphic Two-Dimensional Algebras over a Finite Field↩︎

In this section, we present a more moderate result concerning the number of non-isomorphic two-dimensional algebras. More precisely, we compute the number of non-isomorphic two-dimensional algebras for which at least one of the traces is non-zero.

We denote by \(\vert A_i\vert\) the number of non-isomorphic algebras in the class \(A_i\).

References↩︎