Counting Finite-Dimensional Algebras Over Finite Field

In this paper, we describe an elementary method for counting the number of non-isomorphic algebras of a fixed, finite dimension over a given finite field. We show how this method works in the case of 2-dimensional algebras over the field F2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {F}}_{2}$$\end{document}.


Introduction
Classifying finite-dimensional algebras over a given field is usually a very hard problem. The first general result was a classification by Hendersson and Searle of 2-dimensional algebras over the base field R, which appeared in 1992 ( [1]). This was generalised in 2000 by Petersson ([3]), who managed to give a full classification of 2-dimensional algebras over an arbitrary base field. The methods employed in these papers are quite involved and rely on a large amount of previous work by many illustrious authors.
Our aim in this paper is to give perhaps not a classification but at least a way to compute the exact number of non-isomorphic n-dimensional algebras over a fixed finite field by elementary means. Indeed, nothing more complicated than linear algebra and some very basic results about group actions will be needed: we describe isomorphism classes of n-dimensional K-algebras as orbits of a certain GL n (K)-action on Mat n (K) n and use a basic result about group actions to count these orbits. In the first three sections, we give a proof based on concrete calculations, while Sect. 4 is dedicated to a more abstract alternative which avoids all computations. In Sect. 5, we work out the concrete example n = 2, K = F 2 .

Notation and Basics
Fix a field K. In this article, an algebra is understood to be a K-vector space A equipped with a multiplication, i.e. a bilinear map A × A → A. If a, b are in A, we will write ab for the image of (a, b) under this map. We do not assume algebras to have a unit or to be associative. By the dimension of an algebra we mean its dimension as a K-vector space. Two algebras A and A will be called isomorphic if there exists a K-linear bijection f : The isomorphism class of an algebra A will be denoted by [A]. For n ∈ N, we define Alg n (K) to be the set of isomorphism classes of n-dimensional algebras.
Given a vector M = (M i ) i=1,...,n of n (n × n)-matrices over K, we can define an algebra alg(M) which is K n as a K-vector space and for which multiplication is defined to be the unique bilinear map K n × K n → K n with where the e i are the canonical basis vectors of K n . Intuitively, this means that multiplying an element a ∈ alg(M) on the left with e i is multiplying the coordinate vector of a (with respect to the canonical basis) with M i and interpreting the result again as a coordinate vector (with respect to the canonical basis). This allows us to define the map  Proof. Let A be an n-dimensional algebra with basis a 1 , . . . , a n . There are α ij,k in K such that a i a j = k α ik,j a k for all 1 ≤ i, j ≤ n. Define the matrix

A Group Action on Mat n (K) n
Recall that for a given set X and a group G with neutral element e, a (right) G-action on X is a map φ : The set of φ-orbits is denoted by X/G. The fixpoints of a g ∈ G are the elements of X g = {x ∈ X | φ(x, g) = x}.
which is the term on the left. , considered as K-vector spaces, are just K n , there must be a G ∈ GL n (K) such that f (x) is just Gx for all x ∈ alg(M ). As f is an isomorphism, we find for arbitrary x, y ∈ alg(M). In particular, if x = e l , we find GM l y =

Counting Orbits
From now on, we assume K to be a finite field with q elements. As a consequence of Lemma 2.2, we find that alg induces a well-defined, injective map which is also surjective by Lemma 1.1. The number of isomorphism classes of ndimensional K-algebras therefore equals the number of φ-orbits of Mat n (K) n .
The following well-known result from the theory of group actions will help us count the latter: Proof. Cf. e.g. [5], p.58.
To use this lemma, we need to know the number of fixpoints of a given invertible matrix M . For that, we need the following definition:  where Eig 1 (A) denotes the eigenspace of the matrix A with eigenvalue 1.
It is known (see e.g. [2]) that, for arbitrary A, B, C in Mat n (K), we have (B T ⊗ A)vec(C) = vec(ACB). From this, we conclude:

Theorem 3.4. The number of non-isomorphic n-dimensional K-algebras is
Proof. By Lemma 2.2, the number of non-isomorphic n-dimensional k-algebras is the number of φ-orbits. By 3.1 and 3.3, this is equal to the given formula.

A Computation-Free Road to Rome
In this section, we will outline a version of the proof which avoids all concrete computations. Grateful use has been made of an anonymous referee's report.
Suppose A is a K-algebra. We can express a choice of basis for A as a K-vector space isomorphism b : K n → A. An algebra with basis can then be seen as a pair (A, b). We call two such pairs (A, b), We denote the isomorphism class of (A, b) as (A, b) and the set of isomorphism classes of n-dimensional K-algebras with basis as AlgBas n (K).
For a K-algebra A, we write μ A : A ⊗ A → A, x ⊗ y → xy. Similarly, if M is an element of Hom(K n ⊗ K n , K n ), we write alg(M ) for the algebra which is K n as a K-vector space and with multiplication given by xy = M (x ⊗ y) for all x, y in K n . We can identify the set AlgBas n (K) with Hom(K n ⊗ K n , K n ) by the following maps Hom(K n ⊗ K n , K n ) → AlgBas n (K), M → (alg(M ), Id) which can be checked to be well-defined and inverse to each other.
We can define a GL n (K)-action φ on AlgBas n (K) by φ ((A, b), g) = (A, b• g −1 ). The GL n (K)-orbits correspond to the fibers of the forgetful functor so we can count the number of non-isomorphic n-dimensional K-algebras by counting the GL n (K)-orbits of φ. By the above correspondence, we get a GL n (K)-action on Hom(K n ⊗ K n , K n ) as well and we can count the orbits of that action instead. In order to apply Burnside's lemma, we need to find the M ∈ Hom(K n ⊗ K n , K n ) fixed by a given element g ∈ GL n (K). These are precisely those M which satisfy g −1 M (g ⊗ g) = M .
Let us recall a few results from basic linear algebra. For any two finitedimensional K-vector spaces V, W we have, if we write V * for the dual space Similarly, a map where we have written b * for the dual of b.
In particular, we can apply this with V = K n ⊗K n and W = K n . Writing vec for the isomorphism Hom K (K n ⊗ K n , K n ) → (K n ⊗ K n ) * ⊗ K n , we find that g −1 M (g ⊗ g) = M is equivalent to We conclude that vec induces an isomorphism between the subvector space of Hom K (K n ⊗ K n , K n ) consisting of elements fixed under g on the one hand and Eig 1 ((g ⊗ g) T ⊗ g −1 ) on the other hand.

Example: the Case n = 2, q = 2
Any element of GL 2 (K) has, counting (algebraic) multiplicities, two eigenvalues in the algebraic closure K of K. Clearly, either both or none are elements of K which makes counting invertible matrices with eigenvalues in K considerably easier. Indeed, the only possible Jordan normal forms for a 2 × 2 matrix are 1 Here and later we only write the non-zero entries in our matrices, as is usual.
for some α, β ∈ K. Note that dim Eig 1 (N 1 ) = 3 unless the characteristic of K is 2, in which case dim Eig 1 (N 1 ) = 4. If α = 1, we obviously have dim Eig 1 (N α ) = 0. For M α,β , the dimension of the eigenspace associated to 1 depends heavily on α and β, ranging from 8 if α = β = 1 to 0 if 1 / ∈ α, β, α 2 β −1 , α −1 β 2 . We will do the computations explicitly for the concrete example of K = F 2 . There are 6 invertible matrices, namely The identity obviously yields a contribution of 2 8 . The next three are conjugate to J 2 with α = 1, therefore yielding a contribution of 2 4 each. The last two have no eigenvalues over K. Their eigenvalues are the roots t 1 , t 2 of the polynomial x 2 +x+1. As these roots satisfy t 2 1 = t 2 , t 2 2 = t 1 , both matrices give a contribution of 2 2 . This gives a total of 2 8 + 3 · 2 4 + 2 · 2 2 = 312 which divided by the total number of invertible matrices gives 312/6 = 52. This number fits the formulae which were obtained, using completely different methods, by Petersson and Scherer in [4]. for a given k ∈ N? If q and n are fixed, this is a finite problem and can therefore be calculated, but this is rather tedious and time-consuming. Having a closed formula in q and n would be nice.

Outlook
On the algebraic side, it would be interesting to see whether the method described in this paper can also be used to count certain subclasses of algebras, like alternating algebras, associative algebras, or division algebras.