# Algebras of length one

O.V. Markova<EMAIL_ADDRESS>C. Martínez<EMAIL_ADDRESS>R.L.
Rodrigues<EMAIL_ADDRESS>Department of Mechanics and Mathematics,
Lomonosov Moscow State University, Moscow, 119991, Russia.  Moscow Center for
Fundamental and Applied Mathematics, Moscow, 119991, Russia.  Moscow Institute
of Physics and Technology, Dolgoprudny, 141701, Russia.  Department of
Mathematics, University of Oviedo, Calvo Sotelo, s/n, 33007 Oviedo, Spain.
Department of Mathematics, Federal University of Ceará, Pici Campus, Block
914, 60455-760, Fortaleza, Brazil.  Department of Mathematics, University of
São Paulo, Butantã, 05508-090, São Paulo, Brazil.

###### Abstract

Let ${\cal A}$ be an algebra over a field ${\mathbb{F}}$ and let $\mathcal{S}$
be a generating set of ${\cal A}$. The length of $\mathcal{S}$ indicates the
maximal length needed to express an arbitrary element of ${\cal A}$ as a
linear combination of words in the elements of $\mathcal{S}$. The length of an
algebra ${\cal A}$ is defined as the maximum of lengths of its generating
sets. In this paper (not necessarily associative) algebras, over fields of
arbitrary characteristic, having length equal to 1 are determined.

###### keywords:

length, non-associative algebra, dimension, basis, generating set,
characteristic.

###### MSC:

[2020] 17A60, 17A01, 15A03, 17A05.

††journal: Journal of Pure and Applied Algebra

## Introduction

In the present paper, we will denote by ${\cal A}$ a unital (with identity
element $1_{{\cal A}}$) not necessarily associative algebra over a field
${\mathbb{F}}$. Information about non-associative algebras can be found, for
instance, in [16, 19, 21].

Any product of a finite number of elements of a finite subset
$\mathcal{S}\subset{\cal A}$ is a word in the letters from $\mathcal{S}$. The
length of a word is equal to the number of letters different of $1_{{\cal A}}$
in the corresponding product. By convention, $1_{{\cal A}}$ is a word in any
subset ${\mathcal{S}}$ of length 0.

If $\mathcal{S}$ is a generating system of the algebra ${\cal A}$, that is,
${\cal A}$ is the smallest subalgebra of ${\cal A}$ containing $\mathcal{S}$,
then any element of ${\cal A}$ can be expressed as a linear combination of
words in elements of $\mathcal{S}$. If we can express all elements of ${\cal
A}$ using words of length at most $k$, but we can not use only words of length
at most $k-1$, we say that the length of the generating system is $k$. The
length of a finitely generated algebra ${\cal A}$, $l({\cal A})$, is the
maximal length of its finite generating systems.

Notice that the unital algebra ${\cal A}$ has length equal to 0 if and only if
${\cal A}={\mathbb{F}}1_{{\cal A}}$. Otherwise length is a positive integer or
infinite.

If ${\cal A}$ is not unital, we can consider the unital hull of ${\cal A}$,
$\tilde{{\cal A}}={\mathbb{F}}1_{{\cal A}}+{\cal A}$, and clearly $l({\cal
A})=l(\tilde{{\cal A}})$.

A finite-dimensional associative algebra has always finite length and a
finitely generated associative algebra with finite length has finite
dimension. Algebras considered in this paper will be always finite-
dimensional.

Guterman and Kudryavtsev in [4, 5] started the study of length of non-
associative algebras. In particular, they showed that any finite-dimensional
non-associative algebra ${\cal A}$ has finite length, and if $\dim{\cal
A}=n>2$, then the length is bounded by $l({\cal A})\leq 2^{\dim{\cal A}-2}$.

The length of an algebra is an important invariant for the study of finite-
dimensional algebras. In some sense it measures the multiplicative complexity
of the algebra. The knowledge of a universal upper bound for the length gives
the maximal size of products in generators that we need to consider in order
to verify some property. First of all, the length of a subset in a given
algebra determines the complexity of the procedures verifying whether this set
generates the algebra and computing a basis of the algebra from the generating
set. For matrix algebras, properties under consideration include simultaneous
reduction (by similarity) of a set of matrices to a canonical, for example,
(block-)triangular form [3], or unitary similarity [2]. That is, the length
function has applications in computational methods of matrix theory.

Therefore, the study of algebras whose length is close to the minimal value is
of interest. Clearly the minimal non-trivial value of the length is $1$.
Matrix algebras of length $1$ were described in [13]. The complete
classification of associative algebras of length $1$ is also known (see [14]).

A similar question for non-associative algebras had not been addressed until
now, may be because the problem, in a totally general context, seems somehow
wild. It was unclear what kind of tools should be used. In this paper we
address this question and give a complete description of algebras of length
$1$ over an arbitrary field ${\mathbb{F}}$, mainly using methods of linear
algebra and linearization techniques.

Our main result gives a characterization of algebras of length 1 in terms of a
basis whose multiplication table satisfies some precise conditions.

##  1 Preliminaries

In order to place the problem in the proper context and to make it easier to
readers we will start with some general definitions and results.

######  Notation 1.1.

Given $\mathcal{S}\subseteq{\cal A}$, ${\mathbb{F}}\mathcal{S}$ (or
${\mathbb{F}}(a_{1},\ldots,a_{n})$, if
$\mathcal{S}=\left\\{a_{1},\ldots,a_{n}\right\\}$) denotes the vector space
linearly spanned by the elements of $\mathcal{S}$ and ${\rm alg}(\mathcal{S})$
(or ${\rm alg}(a_{1},\ldots,a_{n})$, if
$\mathcal{S}=\left\\{a_{1},\ldots,a_{n}\right\\}$) will represent the
subalgebra of ${\cal A}$ generated by $\mathcal{S}$, that is, ${\rm
alg}(\mathcal{S})=\cap\ {\cal B}$, where ${\cal B}$ runs over all subalgebras
of ${\cal A}$ with $\mathcal{S}\subseteq{\cal B}$.

The set of all words in $\mathcal{S}$ with length not greater than $i$ is
denoted by $\mathcal{S}^{i}$, for $i\geq 0$ and the set ${\cal
L}_{i}(\mathcal{S})$ stands for ${\mathbb{F}}\mathcal{S}^{i}$. Clearly
$\displaystyle\bigcup_{i\geq 0}\mathcal{L}_{i}(\mathcal{S})={\rm
alg}(\mathcal{S})$.

######  Remark 1.2.

A subset $\mathcal{S}\subset{\cal A}$ is a generating set of ${\cal A}$ if and
only if ${\cal A}={\rm alg}(\mathcal{S})$.

With these definitions and notations in mind we may now introduce formally the
concept of length, both for $\mathcal{S}$ and for ${\cal A}$, as follows.

######  Definition 1.3.

The length of a finite subset $\mathcal{S}$ of ${\cal A}$ is

$l(\mathcal{S})={\rm min}\left\\{k\in\mathbb{Z}_{+}:{\rm
alg}(\mathcal{S})={\cal L}_{k}(\mathcal{S})\right\\}.$

######  Definition 1.4.

The length of a finitely generated algebra ${\cal A}$ is

$l({\cal A})={\rm max}\left\\{l(\mathcal{S}):{\cal A}={\rm
alg}(\mathcal{S})\hskip 14.22636pt\mathcal{S}\ {\rm finite}\right\\}.$

Given a set $\mathcal{S}$ its length evaluation is a standard linear algebraic
problem: the construction of a basis of ${\rm alg}(\mathcal{S})$. But its
computational complexity comes from the fact that the number of words in each
$\mathcal{S}^{i}$ grows exponentially. We would like to emphasize that the
different behaviour in associative and non-associative cases is linked to the
different properties of the sequence $\\{\dim{\cal L}_{i}(\mathcal{S})\\},\
i=0,1,\ldots$. In the associative case this sequence is strictly increasing,
that is, $\dim{\cal L}_{i}(\mathcal{S})<\dim{\cal L}_{i+1}(\mathcal{S})$,
until it stabilizes at step $l(\mathcal{S})$, that is, ${\cal
L}_{l(\mathcal{S})}(\mathcal{S})={\rm alg}(\mathcal{S})$ and $\dim{\cal
L}_{n}(\mathcal{S})=\dim{\rm alg}(\mathcal{S})$ for every $n\geq
l(\mathcal{S})$. In the non-associative case the sequence $\\{\dim{\cal
L}_{i}(\mathcal{S})\\},\ i=0,1,\ldots$ is non-decreasing, that is, $\dim{\cal
L}_{i}(\mathcal{S})\leq\dim{\cal L}_{i+1}(\mathcal{S})$ until it stabilizes.
But in the non-associative case the equality ${\cal L}_{i}(\mathcal{S})={\cal
L}_{i+1}(\mathcal{S})$ does not imply that ${\cal L}_{i}(\mathcal{S})={\rm
alg}(\mathcal{S})$. However, for a given $n\geq 1$, it is sufficient to have

$\dim{\cal L}_{n}(\mathcal{S})=\dim{\cal
L}_{n+1}(\mathcal{S})=\ldots=\dim{\cal L}_{2n}(\mathcal{S})$

to guarantee that the identity $\dim{\cal L}_{n}(\mathcal{S})=\dim{\cal
L}_{n+t}(\mathcal{S})=\dim{\rm alg}(\mathcal{S})$ holds for any $t\geq 1$, as
it was established in [5, Proposition 2.3]. Consequently, the equality ${\cal
L}_{1}(\mathcal{S})={\cal L}_{2}(\mathcal{S})$ implies that ${\cal
L}_{1}(\mathcal{S})={\cal L}_{t}(\mathcal{S})={\rm alg}(\mathcal{S})$ for all
$t\geq 2$.

Applications of the length function appear, for example, in computational
methods of matrix theory, since the length determines the complexity of some
rational procedures. Al’pin and Ikramov extended the classical Specht
criterion for unitary similarity between two complex $n\times n$ matrices to
unitary similarity between two normal matrix sets of cardinality $m$. In the
same way, the well-known result of Pearcy extending Specht’s theorem can be
used as a finiteness criterion. The complexity of this criterion depends both,
on $n$ and on the length of algebras under their analysis [2, 3].

The problem of computing lengths of the matrix algebra $M_{n}({\mathbb{F}})$,
as a function of the matrix size $n$, was first addressed by Paz [18] in 1984,
and still remains open. The known bounds are non-linear functions of $n$, see
[18, 17, 20]. Linear bounds for specific generating sets has been found in [7,
8, 9]. See also references therein.

In the case of arbitrary associative algebras Pappacena [17] provided an upper
bound for the length of any finite-dimensional associative algebra ${\cal A}$
as a function of two invariants of ${\cal A}$: its dimension and $m({\cal
A})$, the maximal degree of minimal polynomials of elements in the algebra. In
[12] a stronger bound, depending of the same two parameters, was obtained for
commutative algebras. A study of algebraic properties of the length function,
namely the length of the direct sum, can be found in [6].

Let us remember the classification of associative algebras of length 1.

######  Theorem 1.5 ([14, Theorem 3]).

Let ${\mathbb{F}}$ be an arbitrary field and let ${\cal A}$ be a finite-
dimensional associative unital ${\mathbb{F}}$-algebra. Then $l({\cal A})=1$ if
and only if it is one of the algebras in the following list:

  * $(1)$

${\cal A}\cong{\mathbb{F}}\oplus{\mathbb{F}}$;

  * $(2)$

${\cal A}$ is a field, $\dim{\cal A}=2$;

  * $(3)$

${\mathbb{F}}={\mathbb{F}}_{2}$, ${\cal
A}\cong{\mathbb{F}}\oplus{\mathbb{F}}\oplus{\mathbb{F}}$;

  * $(4)$

dim ${\cal A}/J({\cal A})=1$, $J({\cal A})\neq 0$, $J({\cal A})^{2}=0$, where
$J({\cal A})$ denotes the Jacobson radical of ${\cal A}$;

  * $(5)$

There exist elements $e,f\in{\cal A}$ such that
$e^{2}=e,f^{2}=f,ef=fe=0,e+f=1;{\rm dim}\ e{\cal A}e={\rm dim}\ f{\cal
A}f=1;f{\cal A}e=0,e{\cal A}f\neq 0.$

Algebras of different types are not isomorphic. Two algebras ${\cal
B},\mathcal{C}$ of type $(4)$ (respectively of type $(5)$) are isomorphic if
and only if ${\rm dim}\ {\cal B}={\rm dim}\ \mathcal{C}$.

From now on we will not assume associativity of the considered algebras. Our
aim is to determine algebras of length 1, without any additional assumption.

##  2 General results

In this section we give some general results about algebras of length 1. The
main result of the section gives a characterization of them that is essential
for the rest of results in this paper. Some consequences of this result are
given explicitly. In particular, algebras of length 1 are proved to be power-
associative.

To start, let us remark that $l({\cal A})=1$ means that if $\mathcal{S}$ is an
arbitrary generating set of ${{\cal A}}$ and
$\bar{\mathcal{S}}=\mathcal{S}\cup\\{1_{{\cal A}}\\}$, then
$\bar{\mathcal{S}}$ linearly spans ${\cal A}$, that is, ${\cal
A}=\rm{alg}(\mathcal{S})$ implies ${\cal A}={\mathbb{F}}\bar{\mathcal{S}}$. If
${\cal A}$ is a finitely generated algebra over a field ${\mathbb{F}}$ and
$l({\cal A})=1$, then ${\cal A}$ is finite-dimensional.

Next proposition will play the main role to get our description of algebras of
length 1 in terms of the existence of a basis with a known multiplication
table.

######  Proposition 2.1.

Let ${\cal A}$ be an ${\mathbb{F}}$-algebra of dimension $n$ with identity
element $1_{{\cal A}}$. ${\cal A}$ has length $1$ if and only if
$ab\in{\mathbb{F}}(1_{{\cal A}},a,b)$, for all $a,b\in{\cal A}$.

###### Proof.

Let ${\cal A}$ be an ${\mathbb{F}}$-algebra of length 1. We will prove first
that $a^{2}\in{\mathbb{F}}(1_{A},a)$. If $a\in{\mathbb{F}}(1_{{\cal A}})$, the
result is obvious. Therefore we can assume w.l.g. that $\left\\{1_{{\cal
A}},a\right\\}$ is a linearly independent set, and extend it to a basis
$\left\\{s_{1}=a,s_{2}=1_{{\cal A}},s_{3},\ldots,s_{n}\right\\}$ of ${\cal
A}$. So $a^{2}=\alpha_{1}a+\alpha_{2}1_{{\cal A}}+\cdots+\alpha_{n}s_{n}$ for
some $\alpha_{1},\ldots,\alpha_{n}\in{\mathbb{F}}$. If
$a^{2}\notin{\mathbb{F}}(1_{{\cal A}},a)$, then there is some $\alpha_{i}\neq
0$, $i>2$. We can assume (reordering the basis if needed) that $\alpha_{3}\neq
0$. This implies that $s_{3}\in{\rm alg}(s_{1}=a,s_{2}=1_{{\cal
A}},s_{4},\ldots,s_{n})$. Consequently ${\cal A}={\rm
alg}(\mathcal{S}^{\prime})\neq{\mathbb{F}}\mathcal{S}^{\prime}$, where
$\overline{\mathcal{S}^{\prime}}=\mathcal{S}^{\prime}=\mathcal{S}\setminus\left\\{s_{3}\right\\}$,
a contradiction.

If $a,b\in{\cal A}$, let us consider $\left\\{1_{{\cal A}},a,b\right\\}$. For
$\left\\{1_{{\cal A}},a,b\right\\}$ linearly dependent the result directly
follows from what has been proved above. So we can assume that
$\left\\{1_{{\cal A}},a,b\right\\}$ are linearly independent and extend this
set (as we did before) to a basis $\left\\{1_{{\cal
A}},a,b,s_{4},\ldots,s_{n}\right\\}$. Then $ab=\lambda_{1}1_{{\cal
A}}+\lambda_{2}a+\lambda_{3}b+\lambda_{4}s_{4}+\cdots+\lambda_{n}s_{n}$ for
some
$\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\ldots,\lambda_{n}\in{\mathbb{F}}$.
If there is some $\lambda_{i}\neq 0$, $i\geq 4$ (we can assume
$\lambda_{4}\neq 0)$, then considering
$\mathcal{S}^{\prime\prime}=\mathcal{S}\setminus\left\\{s_{4}\right\\}$ we
have, as before, ${\cal A}={\rm
alg}(\mathcal{S}^{\prime\prime})\neq{\mathbb{F}}\mathcal{S}^{\prime\prime}$, a
contradiction.

The converse is clear. ∎

######  Corollary 2.2.

Let ${\cal A}$ be a finite dimensional algebra with identity element $1_{{\cal
A}}$ over a field ${\mathbb{F}}$. If $l({\cal A})=1$, then
$a^{2}\in{\mathbb{F}}(1_{{\cal A}},a)$, for all $a\in{\cal A}$.

Next results immediately follow from Proposition 2.1.

######  Corollary 2.3.

Every unital algebra of dimension $2$ over an arbitrary field has length $1$.

Let us notice that any 2-dimensional unital algebra is associative.

A power-associative algebra over a field ${\mathbb{F}}$ is an algebra in which
every subalgebra generated by a single element is associative. Examples
include associative, Lie, Jordan and alternative algebras.

Since the unital subalgebra generated by an element of an algebra ${\cal A}$
with $l({\cal A})=1$ has dimension at most 2, it is associative. Now next
corollary immediately follows.

######  Corollary 2.4.

An algebra ${\cal A}$ of length $1$ over a field ${\mathbb{F}}$ is power-
associative.

######  Corollary 2.5.

Let $\mathcal{B}$ be a subalgebra of a unital finite-dimensional
${\mathbb{F}}$-algebra ${\cal A}$. If $l({\cal A})=1$, then
$l(\mathcal{B})=1$.

Notice that the above Corollaries 2.2 and 2.5 extend results of Corollary 2.2
and Lemma 2.1 of [13] from associative case.

In general, it is not true that $l({\cal B})\leq l({\cal A})$ for any ${\cal
B}$ subalgebra of ${\cal A}$, even in the associative case (see [8]).

##  3 Characteristic different from 2

In this section we will address the aim of the paper in the case of
characteristic different from 2. So in this section ${\mathbb{F}}$ will denote
a field with ${\rm char}({\mathbb{F}})\neq 2$.

######  Lemma 3.1.

Let ${\cal A}$ be an ${\mathbb{F}}$-algebra of dimension $n$ and length $1$
with identity element $1_{{\cal A}}$. Then there is a basis ${\cal B}$ of
${\cal A}$ such that $b^{2}\in{\mathbb{F}}1_{{\cal A}}$, for every $b\in{\cal
B}$.

###### Proof.

Let $\left\\{a_{1}=1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ be a basis of
${\cal A}$. According to Corollary 2.2, we can write
$a_{i}^{2}=\alpha_{i}1_{{\cal A}}+\gamma_{i}a_{i}$, for some
$\alpha_{i},\gamma_{i}\in{\mathbb{F}}$, for all $2\leq i\leq n$. It is easy to
check that the basis ${\cal B}=\left\\{a_{1}=1_{{\cal
A}},a_{2}-\frac{1}{2}\gamma_{2}1_{{\cal
A}},\ldots,a_{n}-\frac{1}{2}\gamma_{n}1_{{\cal A}}\right\\}$ satisfies the
required conditions. ∎

######  Definition 3.2.

A basis ${\cal B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ of a
unital algebra ${\cal A}$ satisfying that $a_{i}^{2}=\mu_{i}1_{{\cal A}}$ for
some $\mu_{i}\in{\mathbb{F}}$, $2\leq i\leq n$ will be called canonical basis.

Notice that Lemma 3.1 says that an algebra of length 1 has some canonical
basis.

######  Lemma 3.3.

Let ${\cal A}$ be an ${\mathbb{F}}$-algebra of length 1 with identity element
$1_{{\cal A}}$ and let ${\cal B}=\left\\{a_{1}=1_{{\cal
A}},a_{2},a_{3},\ldots,a_{n}\right\\}$ be a canonical basis of ${\cal A}$.
Then $a_{i}a_{j}+a_{j}a_{i}\in{\mathbb{F}}1_{{\cal A}}$, for every $2\leq
i\neq j\leq n$.

###### Proof.

Since ${\cal B}$ is a canonical basis, $a_{i}^{2}=\mu_{i}1_{{\cal A}}$, for
some $\mu_{i}\in{\mathbb{F}}$, $2\leq i\leq n$.

By Proposition 2.1, $a_{i}a_{j}\in{\mathbb{F}}(1_{{\cal A}},a_{i},a_{j})$ for
every pair of indices $i\neq j\in\\{2,\ldots,n\\}$. So, there are elements
$\alpha_{ij},\beta_{ij},\beta_{ji}^{*},\alpha_{ji},\beta_{ji},\beta_{ij}^{*}\in{\mathbb{F}}$,
$2\leq i<j\leq n$, such that $a_{i}a_{j}=\alpha_{ij}1_{{\cal
A}}+\beta_{ij}a_{i}+\beta_{ji}^{*}a_{j}$ and $a_{j}a_{i}=\alpha_{ji}1_{{\cal
A}}+\beta_{ji}a_{j}+\beta_{ij}^{*}a_{i}$.

By Corollary 2.2 we know that for an arbitrary $\lambda\in{\mathbb{F}}$,
$(a_{i}+\lambda a_{j})^{2}\in{\mathbb{F}}(1_{{\cal A}},a_{i}+\lambda a_{j}).$
Since $(a_{i}+\lambda a_{j})^{2}=a_{i}^{2}+\lambda a_{i}a_{j}+\lambda
a_{j}a_{i}+\lambda^{2}a_{j}^{2}=$

$\displaystyle=(\mu_{i}+\lambda^{2}\mu_{j})1_{{\cal
A}}+\lambda(\alpha_{ij}1_{{\cal
A}}+\beta_{ij}a_{i}+\beta_{ji}^{*}a_{j})+\lambda(\alpha_{ji}1_{{\cal
A}}+\beta_{ji}a_{j}+\beta_{ij}^{*}a_{i}),$

we conclude that the coefficient of $a_{j}$ is $\lambda$ times the coefficient
of $a_{i}$, that is,
$\lambda(\beta_{ji}^{*}+\beta_{ji})=\lambda^{2}(\beta_{ij}+\beta_{ij}^{*}).$

Since $|{\mathbb{F}}|\geq 3$ and every $\lambda\in{\mathbb{F}}$ is a root of
$x(\beta_{ji}^{*}+\beta_{ji})=x^{2}(\beta_{ij}+\beta_{ij}^{*}),$ we conclude
that $\beta_{ji}^{*}+\beta_{ji}=0$ for every distinct
$i,j\in\\{2,\ldots,n\\}$. That is, $a_{i}a_{j}=\alpha_{ij}1_{{\cal
A}}+\beta_{ij}a_{i}-\beta_{ji}a_{j}$, $a_{j}a_{i}=\alpha_{ji}1_{{\cal
A}}+\beta_{ji}a_{j}-\beta_{ij}a_{i}$.

This implies that $a_{i}a_{j}+a_{j}a_{i}=(\alpha_{ij}+\alpha_{ji})1_{{\cal
A}}$, and completes the proof. ∎

We will consider first the case ${\rm dim}\ {\cal A}=3$, since it behaves in a
different way.

######  Proposition 3.4.

Let ${\cal A}$ be a unital ${\mathbb{F}}$-algebra of dimension $3$. If there
is a canonical basis $\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ such
that $a_{2}a_{3}+a_{3}a_{2}\in{\mathbb{F}}1_{{\cal A}}$ then $l({\cal A})=1$.

###### Proof.

Let $a$ and $b$ be arbitrary elements of ${\cal A}$. If
$\mathcal{S}=\left\\{1_{{\cal A}},a,b\right\\}$ is a linearly independent set,
then it is a basis of ${\cal A}$ and so
$ab\in{\mathbb{F}}\mathcal{S}={\mathbb{F}}(1_{{\cal A}},a,b)$.

Otherwise, there are scalars $\alpha,\beta,\gamma\in{\mathbb{F}}$, not all of
them equal to zero, such that $\alpha 1_{{\cal A}}+\beta a+\gamma b=0$.

If $\gamma=0$, then $\alpha 1_{{\cal A}}+\beta a=0$, that is,
$a\in{\mathbb{F}}1_{{\cal A}}$. This implies that
$ab\in{\mathbb{F}}b\subseteq{\mathbb{F}}(1_{{\cal A}},b)$.

If $\gamma\neq 0$, $b\in{\mathbb{F}}(1_{{\cal A}},a)$, so
$ab\in{\mathbb{F}}(a,a^{2})$. So, in order to prove that $l({\cal A})=1$, we
only need to show that $a^{2}\in{\mathbb{F}}(1_{{\cal A}},a)$, for every
$a\in{\cal A}$.

Given $a=\lambda_{1}1_{{\cal A}}+\lambda_{2}a_{2}+\lambda_{3}a_{3}\in{\cal
A}$, using that ${\cal B}$ is a canonical basis it follows that $a^{2}=$

$=\lambda_{1}(\lambda_{1}1_{{\cal
A}}+\lambda_{2}a_{2}+\lambda_{3}a_{3})+\lambda_{2}a_{2}(\lambda_{1}1_{{\cal
A}}+\lambda_{2}a_{2}+\lambda_{3}a_{3})+\lambda_{3}a_{3}(\lambda_{1}1_{{\cal
A}}+\lambda_{2}a_{2}+\lambda_{3}a_{3})$

$=2\lambda_{1}(\lambda_{1}1_{{\cal
A}}+\lambda_{2}a_{2}+\lambda_{3}a_{3})-\lambda_{1}^{2}1_{{\cal
A}}+\lambda_{2}^{2}a_{2}^{2}+\lambda_{3}^{2}a_{3}^{2}+\lambda_{2}\lambda_{3}(a_{2}a_{3}+a_{3}a_{2})$

$=2\lambda_{1}a-\lambda_{1}^{2}1_{{\cal
A}}+\lambda_{2}^{2}a_{2}^{2}+\lambda_{3}^{2}a_{3}^{2}+\lambda_{2}\lambda_{3}(a_{2}a_{3}+a_{3}a_{2})\in{\mathbb{F}}(1_{{\cal
A}},a)$,

since $a_{2}a_{3}+a_{3}a_{2},a_{2}^{2},a_{3}^{2}\in{\mathbb{F}}1_{{\cal A}}$.
∎

######  Corollary 3.5.

Let ${\cal A}$ be a $3$-dimensional ${\mathbb{F}}$-algebra with identity
element $1_{{\cal A}}$. Then $l({\cal A})=1$ if and only if there is a
canonical basis $\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ of ${\cal
A}$ such that $a_{2}a_{3}+a_{3}a_{2}\in{\mathbb{F}}1_{{\cal A}}$.

######  Definition 3.6.

Let ${\cal A}$ be a unital ${\mathbb{F}}$-algebra. A canonical basis of ${\cal
A}$, ${\cal B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$
$(a_{i}^{2}=\mu_{i}1_{{\cal A}},\mu_{i}\in{\mathbb{F}})$ will be called
special if there are elements $\alpha_{ij},\beta_{i}\in{\mathbb{F}}$, for
every $2\leq i\neq j\leq n$ satisfying $a_{i}a_{j}=\alpha_{ij}1_{{\cal
A}}+\beta_{j}a_{i}-\beta_{i}a_{j}$.

We have proved in Lemma 3.1 that a unital algebra ${\cal A}$ of length 1 has
always a canonical basis. Furthermore, Lemma 3.3 gives a way to get a
canonical basis from any basis of ${\cal A}$ that contains $1_{{\cal A}}$.
Next proposition will show that an arbitrary canonical basis of ${\cal A}$ is
special.

######  Proposition 3.7.

Let ${\cal A}$ be a unital algebra of length $1$. Then an arbitrary canonical
basis ${\cal B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ is special.

###### Proof.

Assume $l({\cal A})=1$. The case ${\rm dim}(A)\leq 3$ is obvious, by Lemma
3.3. So we will assume ${\rm dim}({\cal A})\geq 4$. Using Lemma 3.3 we know
the existence of scalars $\mu_{i}$, $i\in\left\\{2,\ldots,n\right\\}$ and
$\alpha_{ij},\beta_{ij}$ for arbitrary distinct indices
$i,j\in\left\\{2,\ldots,n\right\\}$ such that $a_{i}^{2}=\mu_{i}1_{{\cal A}}$,
$a_{i}a_{j}=\alpha_{ij}1_{{\cal A}}+\beta_{ij}a_{i}-\beta_{ji}a_{j}$.

Consider arbitrary distinct indices $i,j,k\in$ $\left\\{2,\ldots,n\right\\}$.
Then for every $\mu\in{\mathbb{F}}$ we have

$a_{i}(a_{j}+\mu a_{k})\in{\mathbb{F}}(1_{{\cal A}},a_{i},a_{j}+\mu a_{k}).$

As we have argued above, from $a_{i}(a_{j}+\mu a_{k})=$

$=\alpha_{ij}1_{{\cal
A}}+\beta_{ij}a_{i}-\beta_{ji}a_{j}+\mu\alpha_{ik}1_{{\cal
A}}+\mu\beta_{ik}a_{i}-\mu\beta_{ki}a_{k}=\alpha 1_{{\cal A}}+\beta
a_{i}+\gamma(a_{j}+\mu a_{k}),$

it follows that the coefficient of $a_{k}$ is equal to $\mu$ times the
coefficient of $a_{j}$, that is, $\beta_{ki}=\beta_{ji}$, for $i,j,k$ distinct
indices. Taking $\beta_{i}=\beta_{ki}$ for any $k\neq i$ in
$\left\\{2,\ldots,n\right\\}$ we get the result. ∎

We can already prove the main theorem of the paper, that says that algebras of
length 1 are characterized by the existence of an special algebra.

######  Theorem 3.8.

Let ${\cal A}$ be a unital finite-dimensional ${\mathbb{F}}$ algebra. Then
$l({\cal A})=1$ if and only if there is a special basis ${\cal
B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ of ${\cal A}$.

###### Proof.

We already know the existence of a special basis in any algebra of length 1,
${\cal B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ of ${\cal A}$. We
will keep the notation of Definition 3.6

Take $a^{\prime},b^{\prime}$ arbitrary elements of ${\cal A}$, then there
exist $\lambda,\mu\in{\mathbb{F}}$, $a,b\in{\mathbb{F}}(a_{2},\ldots,a_{n})$
such that $a^{\prime}=\lambda 1_{{\cal A}}+a$ and $b^{\prime}=\mu 1_{{\cal
A}}+b$. Therefore,

$a^{\prime}b^{\prime}=(\lambda 1_{{\cal A}}+a)(\mu 1_{{\cal A}}+b)=\lambda\mu
1_{{\cal A}}+\lambda b+\mu a+ab.$

In order to prove that $a^{\prime}b^{\prime}\in{\mathbb{F}}(1_{{\cal
A}},a^{\prime},b^{\prime})$, it is sufficient to prove that
$ab\in{\mathbb{F}}(1_{{\cal A}},a,b)$.

So, let $a=\lambda_{2}a_{2}+\cdots+\lambda_{n}a_{n}$ and
$b=\xi_{2}a_{2}+\cdots+\xi_{n}a_{n}$ be arbitrary elements of
${\mathbb{F}}(a_{2},\ldots,a_{n})$. Then, there are elements $\mu_{1}$ and
$\mu_{2}$ $\in{\mathbb{F}}$ such that

$\displaystyle ab$
$\displaystyle=\sum_{i,j=2}^{n}\lambda_{i}\xi_{j}a_{i}a_{j}=\sum_{\begin{subarray}{c}i=2\\\
j\neq
i\end{subarray}}^{n}\lambda_{i}\xi_{j}(\beta_{j}a_{i}-\beta_{i}a_{j})+\mu_{1}1_{{\cal
A}}$ $\displaystyle=\sum_{j=2}^{n}\sum_{i\neq
j}(\beta_{j}a_{i})\lambda_{i}\xi_{j}-\sum_{l=2}^{n}\sum_{h\neq
l}(\beta_{l}a_{h})\lambda_{l}\xi_{h}+\mu_{1}1_{{\cal A}}$
$\displaystyle=\sum_{j=2}^{n}\beta_{j}\xi_{j}\left(\sum_{i\neq
j}\lambda_{i}a_{i}\right)-\sum_{l=2}^{n}\beta_{l}\lambda_{l}\left(\sum_{h\neq
l}\xi_{h}a_{h}\right)+\mu_{1}1_{{\cal A}}$
$\displaystyle=\sum_{j=2}^{n}\beta_{j}\xi_{j}(a-\lambda_{j}a_{j})-\sum_{l=2}^{n}\beta_{l}\lambda_{l}(b-\xi_{l}a_{l})+\mu_{2}1_{{\cal
A}}$
$\displaystyle=\sum_{j=2}^{n}\beta_{j}\xi_{j}a-\sum_{j=2}^{n}\beta_{j}\xi_{j}\lambda_{j}a_{j}-\sum_{l=2}^{n}\beta_{l}\lambda_{l}b+\sum_{l=2}^{n}\beta_{l}\lambda_{l}\xi_{l}a_{l}+\mu_{2}1_{{\cal
A}}$
$\displaystyle=\left(\sum_{j=2}^{n}\beta_{j}\xi_{j}\right)a-\left(\sum_{j=2}^{n}\beta_{l}\lambda_{l}\right)b+\mu_{2}1_{{\cal
A}}\in{\mathbb{F}}(1_{{\cal A}},a,b),$

which completes the proof. ∎

Notice that the previous characterization of unital algebras of length one
easily allows the construction of an algorithm to decide if an arbitrary
unital finite dimensional ${\mathbb{F}}$-algebra has length one. For that we
only need to know a basis and the corresponding multiplication table and make
use of Lemma 3.1, Proposition 3.7 and Theorem 3.8.

So let us assume, without loss of generality, that we know a basis ${\cal
B}=\left\\{a_{1}=1_{{\cal A}},a_{2},...,a_{n}\right\\}$ of a unital algebra
${\cal A}$ over ${\mathbb{F}}$ that contains the identity $1_{{\cal A}}$.

The algorithm proceeds as follows:

ALGORITHM

Step 1. For any element $a_{i}\in{\cal B}$ check if
$a_{i}^{2}\in{\mathbb{F}}(1_{{\cal A}},a_{i})$.

  * (i)

If it fails for some $2\leq i\leq n$, then ${\cal A}$ has length $>1$;

  * (ii)

If it is satisfied for any $2\leq i\leq n$, then go to Step 2.

Step 2. Construct, from ${\cal B}$, a canonical basis ${\cal
B}_{c}=\left\\{b_{1}=1_{{\cal A}},b_{2},\ldots,b_{n}\right\\}$
($b_{i}^{2}=\mu_{i}1_{{\cal A}}$) (we know how to do it by Lemma 3.1).

Step 3. Check if this basis ${\cal B}_{c}$ is special, that is, if
$b_{i}b_{j}\in{\mathbb{F}}(1_{{\cal A}},b_{i},b_{j})$ and
$b_{i}b_{j}+b_{j}b_{i}\in{\mathbb{F}}1_{{\cal A}}$ for any $2\leq i\neq j\leq
n$.

  * (I)

If ${\cal B}_{c}$ is special, then $l({\cal A})=1$;

  * (II)

If ${\cal B}_{c}$ is not special, then $l({\cal A})>1$.

Jordan algebras are an important class of non-associative algebras. They were
introduced by P. Jordan, J. von Neumann and E. Wigner to treat with the
formalism of Quantum Mechanics [11]. For results about Jordan algebras we
refer readers to [10], [15] and [21].

######  Definition 3.9.

A Jordan algebra is an algebra ${\cal A}$ over a field ${\mathbb{F}}$, ${\rm
char}\ {\mathbb{F}}\neq 2$, satisfying the following 2 identities:

  * (J.1)

$xy=yx$ (commutativity);

  * (J.2)

$x^{2}(yx)=(x^{2}y)x$ (Jordan identity).

Examples of Jordan algebras include the algebra ${\cal A}^{+}$, obtained from
an associative algebra ${\cal A}$ changing the original associative product to
the new product given by $x.y=\frac{1}{2}(xy+yx)$, and ${\cal
A}={\mathbb{F}}1_{{\cal A}}+\mathcal{V}$, the Jordan algebra of a bilinear
form, where $\mathcal{V}$ is a vector space over ${\mathbb{F}}$ with a
symmetric bilinear form
$\varphi\colon\mathcal{V}\times\mathcal{V}\rightarrow{\mathbb{F}}$ and the
multiplication in ${\cal A}$ is given by

$(\alpha 1_{{\cal A}}+x)(\beta 1_{{\cal
A}}+y)=(\alpha\beta+\varphi(x,y))1_{{\cal A}}+\alpha y+\beta x,$

for all $\alpha,\beta\in{\mathbb{F}}$ and $x,y\in\mathcal{V}$.

######  Theorem 3.10.

Let ${\cal A}$ be a unital algebra with $l({\cal A})=1$ and let ${\cal
B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ be a special basis of
${\cal A}$. Then the following conditions are equivalent:

  * $(1)$

${\cal A}$ is a Jordan algebra;

  * $(2)$

${\cal A}$ is commutative;

  * $(3)$

$\alpha_{ij}=\alpha_{ji}$, $\beta_{j}=0$, for all $i\neq
j\in\left\\{2,\ldots,n\right\\}$.

###### Proof.

It is clear that $(1)\Rightarrow(2)$.

$(2)\Rightarrow(3)$. Let ${\cal A}$ be commutative. Then, for all $i\neq
j\in\left\\{2,\ldots,n\right\\}$ $a_{i}a_{j}=a_{j}a_{i}$, that is,
$\beta_{j}a_{i}-\beta_{i}a_{j}+\alpha_{ij}1_{{\cal
A}}=\beta_{i}a_{j}-\beta_{j}a_{i}+\alpha_{ji}1_{{\cal A}}$. So
$\alpha_{ij}=\alpha_{ji}$ and $\beta_{j}=0$.

$(3)\Rightarrow(1)$. If $\alpha_{ij}=\alpha_{ji}$ and $\beta_{j}=0$, for every
$i\neq j$, then the product of ${\cal A}$ with respect to the special basis
${\cal B}=\left\\{1_{{\cal A}},a_{2},\ldots,a_{n}\right\\}$ is given by
$a_{i}^{2}=\mu_{i}1_{{\cal A}},a_{i}a_{j}=a_{j}a_{i}=\alpha_{ij}1_{{\cal A}}.$

Let us denote $\mathcal{V}={\mathbb{F}}(a_{2},\ldots,a_{n})$ and
$\left\langle\cdot,\cdot\right\rangle\colon\mathcal{V}\times\mathcal{V}\rightarrow{\mathbb{F}}$
the symmetric bilinear form on $\mathcal{V}$ given by:

$\left\langle a_{i},a_{i}\right\rangle=\mu_{i},\left\langle
a_{i},a_{j}\right\rangle=\left\langle a_{j},a_{i}\right\rangle=\alpha_{ij},\
{\rm if}\ i\neq j.$

So ${\cal A}={\mathbb{F}}1_{{\cal A}}+\mathcal{V}$ is the Jordan algebra of
the bilinear form $\left\langle\cdot,\cdot\right\rangle$. ∎

######  Corollary 3.11.

Jordan algebras of a symmetric bilinear form are the only unital Jordan
algebras of length one.

Let us remember that an algebra is called flexible if $x(yx)=(xy)x$ for every
$x,y\in{\cal A}$. Clearly associative algebras (resp. commutative algebras)
are flexible.

######  Theorem 3.12.

Let ${\cal A}$ be an algebra of length $1$ and let ${\cal B}=\left\\{1_{{\cal
A}},a_{2},\ldots,a_{n}\right\\}$ be an special basis of ${\cal A}$. Theorem
3.7. Then ${\cal A}$ is flexible if and only if the following conditions are
satisfied:

$\displaystyle\alpha_{ij}$ $\displaystyle=\alpha_{ji},$
$\displaystyle\beta_{j}\mu_{i}$ $\displaystyle=\beta_{i}\alpha_{ij},$
$\displaystyle\beta_{h}\alpha_{ij}+\beta_{i}\alpha_{hj}$
$\displaystyle=2\beta_{j}\alpha_{ih},$ (F)

for arbitrary distinct i, j, h $\in\left\\{2,\ldots,n\right\\}$.

###### Proof.

If the algebra ${\cal A}$ is flexible then using the flexible identity and its
linearization we get

$\displaystyle a_{i}(a_{j}a_{i})$ $\displaystyle=(a_{i}a_{j})a_{i}$
$\displaystyle a_{i}(a_{j}a_{h})+a_{h}(a_{j}a_{i})$
$\displaystyle=(a_{i}a_{j})a_{h}+(a_{h}a_{j})a_{i}$

for arbitrary $i,j,h\in\left\\{2,\ldots,n\right\\}$.

In particular, taking $i,j,h$ distinct we have:

$\displaystyle a_{i}(a_{j}a_{i})$
$\displaystyle=(\beta_{i}\alpha_{ij}-\beta_{j}\mu_{i})1_{{\cal
A}}+(\alpha_{ji}+\beta_{i}\beta_{j})a_{i}-{\beta_{i}}^{2}a_{j}$
$\displaystyle(a_{i}a_{j})a_{i}$
$\displaystyle=(\beta_{j}\mu_{i}-\beta_{i}\alpha_{ij})1_{{\cal
A}}+(\alpha_{ij}+\beta_{i}\beta_{j})a_{i}-{\beta_{i}}^{2}a_{j}$ $\displaystyle
a_{i}(a_{j}a_{h})+a_{h}(a_{j}a_{i})$
$\displaystyle=(\beta_{j}(\alpha_{ih}+\alpha_{hi})-\beta_{i}\alpha_{hj}-\beta_{h}\alpha_{ji})1_{{\cal
A}}+$
$\displaystyle+(\alpha_{hj}+\beta_{h}\beta_{j})a_{i}+(\alpha_{ij}+\beta_{i}\beta_{j})a_{h}-2\beta_{i}\beta_{h}a_{j}$

and

$\displaystyle(a_{i}a_{j})a_{h}+(a_{h}a_{j})a_{i}$
$\displaystyle=(-\beta_{j}(\alpha_{ih}+\alpha_{hi})+\beta_{h}\alpha_{ij}+\beta_{i}\alpha_{hj})1_{{\cal
A}}+$
$\displaystyle+(\alpha_{jh}+\beta_{h}\beta_{j})a_{i}+(\alpha_{ji}+\beta_{i}\beta_{j})a_{h}-2\beta_{i}\beta_{h}a_{j}$

So ${\cal A}$ flexible implies:

$\displaystyle\beta_{i}\alpha_{ij}-\beta_{j}\mu_{i}$
$\displaystyle=\beta_{j}\mu_{i}-\beta_{i}\alpha_{ji}$
$\displaystyle\alpha_{ji}+\beta_{i}\beta_{j}$
$\displaystyle=\alpha_{ij}+\beta_{i}\beta_{j}$
$\displaystyle\beta_{j}(\alpha_{ih}+\alpha_{hi})-\beta_{i}\alpha_{hj}-\beta_{h}\alpha_{ji}$
$\displaystyle=-\beta_{j}(\alpha_{ih}+\alpha_{hi})+\beta_{h}\alpha_{ij}+\beta_{i}\alpha_{hj}$

that is $\alpha_{ij}=\alpha_{ji}$, $\beta_{j}\mu_{i}=\beta_{i}\alpha_{ji}$ and
$\beta_{h}\alpha_{ij}+\beta_{i}\alpha_{hj}=2\beta_{j}\alpha_{ih}$ for any
distinct $i,j,h\in\left\\{2,3,\ldots,n\right\\}$.

Conversely, let us assume that ${\cal A}$ has an special basis ${\cal
B}=\left\\{1_{{\cal A}}=a_{1},a_{2},\ldots,a_{n}\right\\}$ with a
multiplication table:

$a_{i}^{2}=\mu_{i}1_{{\cal A}},\hskip 14.22636pta_{i}a_{j}=\alpha_{ij}1_{{\cal
A}}+\beta_{j}a_{i}-\beta_{i}a_{j}$

satisfying the identities (F). In order to prove that ${\cal A}$ is flexible
we have to prove that
$a_{i}(a_{j}a_{h})+a_{h}(a_{j}a_{i})=(a_{i}a_{j})a_{h}+(a_{h}a_{j})a_{i}$ for
any $i,j,h\in\left\\{2,\ldots,n\right\\}$.

The proof of the previous implication shows that this is true if $i,j,h$ are
distinct or if $i=h\neq j$.

When $i=j=h$ the result is trivial since
$a_{i}.{a_{i}}^{2}={a_{i}}^{2}.a_{i}=\mu_{i}a_{i}$.

If $i=j\neq h$, then
$a_{i}(a_{i}a_{h})+a_{h}.{a_{i}}^{2}={a_{i}}^{2}.a_{h}+(a_{h}a_{i})a_{i}$
since

$\displaystyle a_{h}.{a_{i}}^{2}$
$\displaystyle={a_{i}}^{2}.a_{h}=\mu_{i}a_{h}$

and

$\displaystyle a_{i}(a_{i}a_{h})$
$\displaystyle=(\beta_{h}\mu_{i}-\beta_{i}\alpha_{ih})1_{{\cal
A}}+(\alpha_{ih}-\beta_{i}\beta_{h})a_{i}+\beta_{i}^{2}a_{h}$
$\displaystyle=(\alpha_{ih}-\beta_{i}\beta_{h})a_{i}+\beta_{i}^{2}a_{h}$
$\displaystyle(a_{h}a_{i})a_{i}$
$\displaystyle=(\beta_{i}\alpha_{hi}-\beta_{h}\mu_{i})1_{{\cal
A}}+\beta_{i}^{2}a_{h}+(\alpha_{hi}-\beta_{i}\beta_{h})a_{i}$
$\displaystyle=a_{i}(a_{i}a_{h})$

using the identities (F).

The remaining case, $i\neq j=h$ is similar to the previous one. ∎

Let us notice that the proof of Theorem 3.12 shows that we can take any
special basis of ${\cal A}$ ($l({\cal A})=1)$ to check if ${\cal A}$ is
flexible.

######  Theorem 3.13.

Let ${\cal A}$ be an ${\mathbb{F}}$-algebra with $1_{{\cal A}}$ and $l({\cal
A})=1$. Let ${\cal B}=\left\\{1_{{\cal A}}=a_{1},a_{2},\ldots,a_{n}\right\\}$
be an special basis of ${\cal A}$. Then ${\cal A}$ is associative if and only
if the following conditions are satisfied:

  * (A1)

$\mu_{i}=\beta_{i}^{2}$, for every $i\in\left\\{2,\ldots,n\right\\}$;

  * (A2)

$\alpha_{ij}=\beta_{j}\beta_{i}=\alpha_{ji}$, for all $i\neq
j\in\left\\{2,\ldots,n\right\\}$.

###### Proof.

Let us assume that ${\cal A}$ is associative. By simplicity, we will consider
first the case ${\rm dim}\ {\cal A}=3$. Then ${\cal A}$ is associative if and
only if $a_{i}(a_{j}a_{k})=(a_{i}a_{j})a_{k}$ for every
$\left\\{i,j,k\right\\}\subseteq\left\\{2,3\right\\}$. By symmetry we only
need to consider the identities: $a_{2}(a_{2}a_{3})=a_{2}^{2}a_{3}$ and
$a_{2}(a_{3}a_{2})=(a_{2}a_{3})a_{2}$. But $a_{2}^{2}a_{3}=\mu_{2}a_{3}$ and

$\displaystyle a_{2}(a_{2}a_{3})$ $\displaystyle=a_{2}(\alpha_{23}1_{{\cal
A}}+\beta_{3}a_{2}-\beta_{2}a_{3})$
$\displaystyle=\alpha_{23}a_{2}+\beta_{3}\mu_{2}1_{{\cal
A}}-\beta_{2}(\alpha_{23}1_{{\cal A}}+\beta_{3}a_{2}-\beta_{2}a_{3})$
$\displaystyle=(\beta_{3}\mu_{2}-\beta_{2}\alpha_{23})1_{{\cal
A}}+(\alpha_{23}-\beta_{2}\beta_{3})a_{2}+\beta_{2}^{2}a_{3}.$

So associativity of ${\cal A}$ implies
$0=\beta_{3}\mu_{2}-\beta_{2}\alpha_{23}$, $\alpha_{23}=\beta_{2}\beta_{3}$,
$\mu_{2}=\beta_{2}^{2}$, and the first identity follows from the other two:
$\beta_{3}\mu_{2}=\beta_{3}\beta_{2}^{2}=\alpha_{23}\beta_{2}$. Consequently
associativity of ${\cal A}$ implies (by symmetry)
$\alpha_{23}=\alpha_{32}=\beta_{2}\beta_{3}$ and $\mu_{2}=\beta_{2}^{2}$,
$\mu_{3}=\beta_{3}^{2}$.

Conversely, if the previous identities are satisfied then
$a_{2}(a_{2}a_{3})=a_{2}^{2}a_{3}$ and $a_{3}(a_{3}a_{2})=a_{3}^{2}a_{2}$. We
only need to check that $a_{2}(a_{3}a_{2})=(a_{2}a_{3})a_{2}$ and
$a_{3}(a_{2}a_{3})=(a_{3}a_{2})a_{3}$. This follows immediately from previous
result, since our conditions guarantee that the algebra ${\cal A}$ is
flexible.

So conditions (A1) + (A2) imply associativity of ${\cal A}$ when ${\rm dim}\
{\cal A}=3$.

Now let us consider the general case. If ${\cal A}$ is associative, ${\rm
dim}\ {\cal A}=n$, $l({\cal A})=1$ and $\left\\{1_{{\cal
A}},a_{2},\ldots,a_{n}\right\\}$ is an special basis of ${\cal A}$, then for
all $i\neq j\in\left\\{2,\ldots,n\right\\}$ the algebra ${\mathbb{F}}(1_{{\cal
A}},a_{i},a_{j})$ is 3-dimensional and associative. As we have just seen, this
implies that $\alpha_{ij}=\alpha_{ji}=\beta_{j}\beta_{i}$ and
$\mu_{i}=\beta_{i}^{2}$.

Conversely, let us assume that $\alpha_{ij}=\alpha_{ji}=\beta_{j}\beta_{i}$,
for all $i\neq j\in\left\\{2,\ldots,n\right\\}$, and $\mu_{i}=\beta_{i}^{2}$
for any $i\in\left\\{2,\ldots,n\right\\}$. So for any $i\neq j$ in
$\left\\{2,\ldots,n\right\\}$ we already know, using the 3-dimensional case
studied above, that $a_{i}(a_{k}a_{j})=(a_{i}a_{k})a_{j}$ and
$a_{j}(a_{k}a_{i})=(a_{j}a_{k})a_{i}$, when $k=i$ or $j$. Now suppose that
$i,j,k$ are 3 distinct elements in $\left\\{2,\ldots,n\right\\}$. We need to
prove that $(a_{i}a_{j})a_{k}=a_{i}(a_{j}a_{k})$ in order to get
associativity.

$\displaystyle(a_{i}a_{j})a_{k}$ $\displaystyle=(\alpha_{ij}1_{{\cal
A}}+\beta_{j}a_{i}-\beta_{i}a_{j})a_{k}$
$\displaystyle=\alpha_{ij}a_{k}+\beta_{j}(\alpha_{ik}1_{{\cal
A}}+\beta_{k}a_{i}-\beta_{i}a_{k})-\beta_{i}(\alpha_{jk}1_{{\cal
A}}+\beta_{k}a_{j}-\beta_{j}a_{k})$
$\displaystyle=(\beta_{j}\alpha_{ik}-\beta_{i}\alpha_{jk})1_{{\cal
A}}+\beta_{j}\beta_{k}a_{i}-\beta_{i}\beta_{k}a_{j}+(\alpha_{ij}-\beta_{i}\beta_{j}+\beta_{i}\beta_{j})a_{k}$
$\displaystyle=(\beta_{j}\beta_{i}\beta_{k}-\beta_{i}\beta_{j}\beta_{k})1_{{\cal
A}}+\alpha_{jk}a_{i}-\alpha_{ik}a_{j}+\alpha_{ij}a_{k}$
$\displaystyle=\alpha_{jk}a_{i}-\alpha_{ik}a_{j}+\alpha_{ij}a_{k}.$

In the same way

$a_{i}(a_{j}a_{k})=\alpha_{jk}a_{i}-\alpha_{ik}a_{j}+\alpha_{ij}a_{k}.$

So the algebra ${\cal A}$ is associative. ∎

######  Remark 3.14.

Theorem 3.13 allows to get examples of unital algebras of length 1 that are
not associative. For instance, the algebra ${\mathbb{F}}(1_{{\cal A}},a,b)$
with the product given by $a^{2}=0$, $b^{2}=b$, $ab=2(1_{{\cal A}})+a+b$,
$ba=-a-b$ is non-associative and has length equal to 1.

##  4 Characteristic 2

In this section we will address the case of characteristic 2. So
${\mathbb{F}}$ will denote an extension of the field of two elements
${\mathbb{F}}_{2}$. We will study first, in a separate way, the case of
dimension 3 that behaves in a different way.

In what follows, given $a,b$ arbitrary elements of ${\cal A}$, we denote by
$a\equiv b\ ({\rm mod}\ 1_{{\cal A}})$ (or simply by $a\equiv b$) if and only
if $a-b\in{\mathbb{F}}1_{{\cal A}}$.

######  Lemma 4.1.

Let ${\cal A}$ be a unital $3$-dimensional ${\mathbb{F}}$-algebra and
$\left\\{b_{1}=1_{{\cal A}},b_{2},b_{3}\right\\}$ an arbitrary basis of ${\cal
A}$. Then $l({\cal A})=1$ implies that the multiplication table of ${\cal A}$
satisfies the following conditions:

$\displaystyle b_{2}^{2}$ $\displaystyle\equiv\delta_{2}b_{2},$ $\displaystyle
b_{3}^{2}$ $\displaystyle\equiv\delta_{3}b_{3},$ $\displaystyle b_{2}b_{3}$
$\displaystyle\equiv\beta_{2}b_{2}+\beta_{3}^{\ast}b_{3},$ $\displaystyle
b_{3}b_{2}\equiv$ $\displaystyle\beta_{3}b_{3}+\beta_{2}^{\ast}b_{2},$ (C2)

where
$\delta_{2},\beta_{2},\beta_{2}^{\ast},\delta_{3},\beta_{3},\beta_{3}^{\ast}\in{\mathbb{F}}$,
and
$\beta_{2}+\beta_{2}^{\ast}+\delta_{2}=\beta_{3}+\beta_{3}^{\ast}+\delta_{3}.$

###### Proof.

Suppose that $l({\cal A})=1$. By Proposition 2.1, there exist
$\beta_{2},\beta_{3}^{*},\beta_{3},\beta_{2}^{*}\in{\mathbb{F}}$ such that
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{3}^{*}b_{3}$ and
$b_{3}b_{2}\equiv\beta_{3}b_{3}+\beta_{2}^{*}b_{2}$.

By Corollary 2.2, there exist $\delta_{2},\delta_{3}\in{\mathbb{F}}$ such that
$b_{2}^{2}\equiv\delta_{2}b_{2}$, $b_{3}^{2}\equiv\delta_{3}b_{3}$.

Corollary 2.2 also gives that

$(b_{2}+b_{3})^{2}=b_{2}^{2}+b_{2}b_{3}+b_{3}b_{2}+b_{3}^{2}\equiv\delta_{2}b_{2}+\beta_{2}b_{2}+\beta_{3}^{*}b_{3}+\beta_{3}b_{3}+\beta_{2}^{*}b_{2}+\delta_{3}b_{3}$

$\in{\mathbb{F}}_{2}(1_{{\cal A}},b_{2}+b_{3}),$ which implies
$\beta_{3}^{*}+\beta_{3}+\delta_{3}=\beta_{2}+\beta_{2}^{*}+\delta_{2}.$ ∎

######  Theorem 4.2.

Let ${\cal A}$ be a unital ${\mathbb{F}}_{2}$-algebra of dimension $3$. Then
$l({\cal A})=1$ if and only if there is a basis ${\cal
B}=\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ whose multiplication table
satisfies one of the following 3 conditions:

  * 1.

$a_{2}^{2}\equiv 0$, $a_{3}^{2}\equiv 0$, $a_{2}a_{3}\equiv 0$,
$a_{3}a_{2}\equiv 0$;

  * 2.

$a_{2}^{2}\equiv a_{2}$, $a_{3}^{2}\equiv a_{3}$, $a_{2}a_{3}\equiv 0$,
$a_{3}a_{2}\equiv 0$;

  * 3.

$a_{2}^{2}\equiv 0$, $a_{3}^{2}\equiv a_{3}$, $a_{2}a_{3}\equiv 0$,
$a_{3}a_{2}\equiv a_{3}$.

###### Proof.

Let $\left\\{1_{{\cal A}}=b_{1},b_{2},b_{3}\right\\}$ be an arbitrary basis of
${\cal A}$. If $l({\cal A})=1$, then $b^{2}\in{\mathbb{F}}(1_{{\cal A}},b)$
for any $b\in{\cal A}$. So $b^{2}\equiv 0$ or $b^{2}\equiv b$. (Notice that
$1_{{\cal A}}^{2}\equiv 0$ and $1_{{\cal A}}^{2}\equiv 1_{{\cal A}}$ and it is
the only nonzero element that satisfies both conditions). We will distinguish
3 cases.

Case 1: For every $a\in{\cal A}$, $a^{2}\equiv 0$.

Using Lemma 4.1 we know that there are elements $\beta_{2}$,
$\beta_{2}^{\ast}$, $\beta_{3}$, $\beta_{3}^{\ast}$ $\in{\mathbb{F}}_{2}$
(notice that $\delta_{2}=\delta_{3}=0)$ such that
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{2}^{\ast}b_{3}$,
$b_{3}b_{2}\equiv\beta_{3}b_{3}+\beta_{2}^{\ast}b_{2}$ and
$\beta_{2}+\beta_{2}^{\ast}=\beta_{3}+\beta_{3}^{\ast}$. Then
$b_{2}b_{3}+b_{3}b_{2}\equiv(\beta_{3}+\beta_{3}^{\ast})b_{3}+(\beta_{2}+\beta_{2}^{\ast})b_{2}=\beta(b_{2}+b_{3})$,
where $\beta=\beta_{2}+\beta_{2}^{\ast}=\beta_{3}+\beta_{3}^{\ast}$. But
$(b_{2}+b_{3})^{2}\equiv 0$ by our assumption, so $\beta=0$. Then
$\beta_{2}=\beta_{2}^{\ast}$, $\beta_{3}=\beta_{3}^{\ast}$ and so
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{3}b_{3}\equiv b_{3}b_{2}$ and
$b_{2}b_{3}+b_{3}b_{2}\equiv 0$.

Taking the basis ${\cal B}_{1}=\left\\{a_{1}=1_{{\cal
A}},a_{2}=\beta_{3}1_{{\cal A}}+b_{2},a_{3}=\beta_{2}1_{{\cal
A}}+b_{3}\right\\}$ the multiplication table satisfies $a_{2}^{2}\equiv 0$,
$a_{3}^{2}\equiv 0$,
$a_{2}a_{3}\equiv\beta_{3}b_{3}+\beta_{2}b_{2}+b_{2}b_{3}=0$ and
$a_{3}a_{2}\equiv 0$.

Case 2. For every $a\not\equiv 0$, $a^{2}\equiv a$.

Again Lemma 4.1 implies that there are elements
$\beta_{2},\beta_{2}^{\ast},\beta_{3},\beta_{3}^{\ast}\in{\mathbb{F}}_{2}$
(notice that $\delta_{2}=\delta_{3}=1)$ such that
$\beta_{2}+\beta_{2}^{\ast}=\beta_{3}+\beta_{3}^{\ast}$, $b_{2}^{2}\equiv
b_{2}$, $b_{3}^{2}\equiv b_{3}$,
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{3}^{\ast}b_{3},$
$b_{3}b_{2}\equiv\beta_{3}b_{3}+\beta_{2}^{\ast}b_{2}$.

So
$b_{2}b_{3}+b_{3}b_{2}\equiv(\beta_{2}+\beta_{2}^{\ast})b_{2}+(\beta_{3}+\beta_{3}^{\ast})b_{3}=\beta(b_{2}+b_{3})$,
where $\beta=\beta_{2}+\beta_{2}^{\ast}=\beta_{3}+\beta_{3}^{\ast}$. But our
assumption implies that $(b_{2}+b_{3})^{2}\equiv b_{2}+b_{3}$ and
$(b_{2}+b_{3})^{2}=b_{2}^{2}+b_{2}b_{3}+b_{3}b_{2}+b_{3}^{2}\equiv
b_{2}+b_{3}+\beta(b_{2}+b_{3})$.

So $\beta=0$ and $\beta_{2}=\beta_{2}^{\ast}$, $\beta_{3}=\beta_{3}^{\ast}$.

That is, the multiplication table is
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{3}b_{3}\equiv b_{3}b_{2}$.

Take $a_{2}=\beta_{3}1_{{\cal A}}+b_{2},$ $a_{3}=\beta_{2}1_{{\cal A}}+b_{3}$
as before. Then ${\cal B}_{2}=\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$
is a basis of ${\cal A}$,
$a_{2}^{2}\equiv\beta_{3}b_{2}+\beta_{3}b_{2}+b_{2}^{2}\equiv b_{2}\equiv
a_{2}$, $a_{3}^{2}\equiv a_{3}$,
$a_{2}a_{3}=a_{3}a_{2}\equiv\beta_{3}b_{3}+\beta_{2}b_{2}+b_{2}b_{3}\equiv 0$.

Case 3. There are elements $b_{2}\not\equiv 0$, $b_{3}\not\equiv 0$ such that
$b_{2}^{2}\equiv 0$, $b_{3}^{2}\equiv b_{3}$. Then $\left\\{1_{{\cal
A}},b_{2},b_{3}\right\\}$ is a basis of ${\cal A}$. Again, using Lemma 4.1 we
know that there are elements
$\beta_{2},\beta_{2}^{\ast},\beta_{3},\beta_{3}^{\ast}\in{\mathbb{F}}_{2}$
$(\delta_{2}=0,\delta_{3}=1)$ such that
$b_{2}b_{3}\equiv\beta_{2}b_{2}+\beta_{3}^{\ast}b_{3}$,
$b_{3}b_{2}=\beta_{3}b_{3}+\beta_{2}^{\ast}b_{2}$,
$\beta_{2}+\beta_{2}^{\ast}=\beta_{3}+\beta_{3}^{\ast}+1$. If
$\beta_{2}+\beta_{2}^{\ast}=0$, then $\beta_{2}=\beta_{2}^{\ast}$,
$\beta_{3}=\beta_{3}^{\ast}+1$. So
$b_{2}b_{3}+b_{3}b_{2}=\beta_{2}b_{2}+(\beta_{3}+1)b_{3}+\beta_{3}b_{3}+\beta_{2}b_{2}=b_{3}$.

Taking $a_{2}=(\beta_{3}+1)1_{{\cal A}}+b_{2}$, $a_{3}=\beta_{2}1_{{\cal
A}}+b_{3}$, ${\cal B}_{3}=\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ is
a basis of ${\cal A}$, $a_{2}^{2}\equiv b_{2}^{2}\equiv 0$, $a_{3}^{2}\equiv
b_{3}^{2}\equiv b_{3}\equiv a_{3}$,

$a_{2}a_{3}=(\beta_{3}+1)b_{3}+\beta_{2}b_{2}+\beta_{2}b_{2}+(\beta_{3}+1)b_{3}\equiv
0,$
$a_{3}a_{2}\equiv\beta_{2}b_{2}+(\beta_{3}+1)b_{3}+\beta_{3}b_{3}+\beta_{2}b_{2}=b_{3}\equiv
a_{3}.$

Clearly algebras satisfying two different conditions are not isomorphic.

For the converse notice that ${\mathbb{F}}_{2}$-algebras satisfying one of the
3 conditions above satisfy also ${\rm(C2)}$.

So it suffices to prove that if an algebra ${\cal A}$ over ${\mathbb{F}}_{2}$
has a basis that satisfies condition ${\rm(C2)}$ then the algebra has length
1.

Let $\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ be a basis of ${\cal A}$
with the indicated multiplication table. In order to prove that $l({\cal
A})=1$ we only need to prove that for arbitrary elements
$a=\lambda_{2}a_{2}+\lambda_{3}a_{3}$, $b=\mu_{2}a_{2}+\mu_{3}a_{3}$ in
${\mathbb{F}}_{2}(a_{2},a_{3})$, the product $ab\in{\mathbb{F}}_{2}(1_{{\cal
A}},a,b)$.

The result is clear if at least one of the elements ($a$ or $b$) is equal to
$0$. So we can assume that $a,b\in\\{a_{2},a_{3},a_{2}+a_{3}\\}$.

Again the result is clear if $a=b$ since we have proved in the first
implication that $(a_{2}+a_{3})^{2}\in{\mathbb{F}}_{2}(1_{{\cal
A}},a_{2}+a_{3})$ if and only if
$\beta_{3}^{*}+\beta_{3}+\delta_{3}=\beta_{2}+\beta_{2}^{*}+\delta_{2}.$

The case $a=a_{2}$, $b=a_{3}$ is given directly by the conditions and the case
$a=a_{2}$, $b=a_{2}+a_{3}$ (similarly $a=a_{3}$, $b=a_{2}+a_{3}$) is clear,
since

$ab=a_{2}(a_{2}+a_{3})\equiv\delta_{2}a_{2}+\beta_{2}a_{2}+\beta_{3}^{*}a_{3}=(\delta_{2}+\beta_{2}+\beta_{3}^{*})a_{2}+\beta_{3}^{*}(a_{2}+a_{3}),$

that is, $a_{2}(a_{2}+a_{3})\in{\mathbb{F}}_{2}(1_{{\cal
A}},a_{2},a_{2}+a_{3})$. ∎

######  Remark 4.3.

Notice that the algebra
${\mathbb{F}}_{2}\oplus{\mathbb{F}}_{2}\oplus{\mathbb{F}}_{2}$ that appears
mentioned in Theorem 1.5 as one associative algebra of length 1 appears here
as one of the algebras of type 2.

######  Theorem 4.4.

Let ${\mathbb{F}}$ be a proper extension of ${\mathbb{F}}_{2}$ and ${\cal A}$
a unital ${\mathbb{F}}$-algebra of dimension $3$. Then $l({\cal A})=1$ if and
only if there is a basis ${\cal B}=\left\\{a_{1}=1_{{\cal
A}},a_{2},a_{3}\right\\}$ of ${\cal A}$ whose multiplication table satisfies
the following conditions:

  * (i)

$a_{2}^{2}\equiv\delta_{2}a_{2}$, $a_{3}^{2}\equiv\delta_{3}a_{3}$,
$\delta_{2},\delta_{3}\in{\mathbb{F}}_{2}$;

  * (ii)

$a_{2}a_{3}\equiv\beta_{2}a_{2}+\beta_{3}^{\ast}a_{3}$,
$a_{3}a_{2}\equiv\beta_{3}a_{3}+\beta_{2}^{\ast}a_{2}$, $\beta_{2}$,
$\beta_{2}^{\ast}$, $\beta_{3}$, $\beta_{3}^{\ast}\in{\mathbb{F}}$ and
$\beta_{2}+\beta_{2}^{\ast}+\delta_{3}=0=\beta_{3}+\beta_{3}^{\ast}+\delta_{2}$.

###### Proof.

Assume that $l({\cal A})=1$. Take $\left\\{b_{1}=1_{{\cal
A}},b_{2},b_{3}\right\\}$ a basis of ${\cal A}$. By Corollary 2.2 we know that
$b_{i}^{2}\equiv\gamma_{i}b_{i}$, $i=2,3$, for some
$\gamma_{i}\in{\mathbb{F}}$. If $\gamma_{i}\neq 0$, taking
$a_{i}=\gamma_{i}^{-1}b_{i}$, we get a basis ${\cal B}=\left\\{1_{{\cal
A}},a_{2},a_{3}\right\\}$ of ${\cal A}$ satisfying that
$a_{2}^{2}\equiv\delta_{2}a_{2}$ and $a_{3}^{2}\equiv\delta_{3}a_{3}$,
$\delta_{i}=0$ or $1$, that is, $\delta_{2},\delta_{3}\in{\mathbb{F}}_{2}$.

Furthermore, $a_{2}a_{3}\equiv\beta_{2}a_{2}+\beta_{3}^{*}a_{3}$, and
$a_{3}a_{2}\equiv\beta_{3}a_{3}+\beta_{2}^{*}a_{2}$ for some
$\beta_{2},\beta_{3},\beta_{2}^{*},\beta_{3}^{*}\in{\mathbb{F}}$, by
Proposition 2.1.

Corollary 2.2 implies that, for all $\lambda\in{\mathbb{F}}$,

$(a_{2}+\lambda
a_{3})^{2}\equiv\delta_{2}a_{2}+\lambda(\beta_{2}a_{2}+\beta_{3}^{*}a_{3}+\beta_{3}a_{3}+\beta_{2}^{*}a_{2})+\lambda^{2}\delta_{3}a_{3}\in{\mathbb{F}}(1_{{\cal
A}},a_{2}+\lambda a_{3}).$

This implies that
$\lambda(\delta_{2}+\lambda\beta_{2}+\lambda\beta_{2}^{*})=\lambda\beta_{3}^{*}+\lambda\beta_{3}+\lambda^{2}\delta_{3}$,
that is,
$\lambda(\beta_{3}^{*}+\beta_{3}+\delta_{2})=\lambda^{2}(\beta_{2}+\beta_{2}^{*}+\delta_{3}).$

Since $|{\mathbb{F}}|\geq 4$, we can conclude that
$\beta_{3}^{*}+\beta_{3}+\delta_{2}=0=\beta_{2}+\beta_{2}^{*}+\delta_{3}$.

Conversely, let us assume the existence of a basis whose multiplication table
satisfies the conditions (i) and (ii), as described in the theorem. By
Proposition 2.1 we only need to prove that for arbitrary elements $a,b$ of
${\cal A}$, $ab\in{\mathbb{F}}(1_{{\cal A}},a,b)$. Since this claim is
obviously true when $\\{1_{{\cal A}},a,b\\}$ is a basis, we only have to
consider the case in which the 3 elements are linearly dependent, what reduces
to prove that, for an arbitrary element $a\in{\cal A}$ the element
$a^{2}\in{\mathbb{F}}(1_{{\cal A}},a)$. Since every element $a\in{\cal A}$ can
be expressed as $a=\mu 1_{{\cal A}}+a^{\prime}$, where
$a^{\prime}\in{\mathbb{F}}(a_{2},a_{3})$, and $a^{2}=\mu^{2}1_{{\cal
A}}+a^{\prime 2}$, the claim is true for any $a\in{\mathbb{F}}(1_{{\cal
A}},a_{2},a_{3})$ if and only if it is true for any
$a^{\prime}\in{\mathbb{F}}(a_{2},a_{3})$. So, let us assume that
$a=\lambda_{2}a_{2}+\lambda_{3}a_{3}$. First of all notice that
$a_{2}a_{3}+a_{3}a_{2}\equiv\delta_{3}a_{2}+\delta_{2}a_{3}$. Then

$a^{2}=\lambda_{2}^{2}a_{2}^{2}+\lambda_{2}\lambda_{3}(a_{2}a_{3}+a_{3}a_{2})+\lambda_{3}^{2}a_{3}^{2}\equiv\lambda_{2}^{2}\delta_{2}a_{2}+\lambda_{2}\lambda_{3}(\delta_{2}a_{3}+\delta_{3}a_{2})+\lambda_{3}^{2}\delta_{3}a_{3}.$

If $\delta_{2}=\delta_{3}=\delta$, we get
$a^{2}\equiv(\lambda_{2}^{2}+\lambda_{2}\lambda_{3})\delta
a_{2}+(\lambda_{2}\lambda_{3}+\lambda_{3}^{2})\delta a_{3},$ that is,

$a^{2}\equiv(\lambda_{2}+\lambda_{3})\delta(\lambda_{2}a_{2}+\lambda_{3}a_{3}),$

what proves the result in this case.

When $\delta_{2}\neq\delta_{3}$, we can assume, without loss of generality,
that $\delta_{2}=1$ and $\delta_{3}=0$. So
$a^{2}\equiv\lambda_{2}^{2}a_{2}+\lambda_{2}\lambda_{3}a_{3}=\lambda_{2}a,$
what finishes the proof of the theorem. ∎

######  Corollary 4.5.

Let ${\mathbb{F}}$ be a proper extension of ${\mathbb{F}}_{2}$ and ${\cal A}$
a unital ${\mathbb{F}}$-algebra of dimension 3. Then $l({\cal A})=1$ if and
only if there is a basis ${\cal B}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},a_{3}^{\ast}\right\\}$ whose multiplication table satisfies
one of the following 3 conditions:

  * 1.

${a_{2}^{\ast}}^{2}\equiv 0,{a_{3}^{\ast}}^{2}\equiv
0,a_{2}^{\ast}a_{3}^{\ast}\equiv 0$, $a_{3}^{\ast}a_{2}^{\ast}\equiv 0$;

  * 2.

${a_{2}^{\ast}}^{2}\equiv a_{2}^{\ast}$, ${a_{3}^{\ast}}^{2}\equiv
a_{3}^{\ast}$, $a_{2}^{\ast}a_{3}^{\ast}\equiv
a_{3}^{\ast},a_{3}^{\ast}a_{2}^{\ast}\equiv a_{2}^{\ast}$;

  * 3.

${a_{2}^{\ast}}^{2}\equiv 0,{a_{3}^{\ast}}^{2}\equiv a_{3}^{\ast}$,
$a_{2}^{\ast}a_{3}^{\ast}\equiv 0$, $a_{3}^{\ast}a_{2}^{\ast}\equiv
a_{3}^{\ast}$.

###### Proof.

By Theorem 4.4 we know that if ${\cal A}$ is an algebra of length 1 then there
is a basis ${\cal B}=\left\\{a_{1}=1_{{\cal A}},a_{2},a_{3}\right\\}$ with a
multiplication table given by

$\displaystyle a_{2}^{2}$ $\displaystyle\equiv\delta_{2}a_{2},$ $\displaystyle
a_{3}^{2}$ $\displaystyle\equiv\delta_{3}a_{3}$ $\displaystyle a_{2}a_{3}$
$\displaystyle\equiv\beta_{2}a_{2}+\beta_{3}^{\ast}a_{3},$ $\displaystyle
a_{3}a_{2}\equiv$ $\displaystyle\beta_{3}a_{3}+\beta_{2}^{\ast}a_{2},$

where
$\delta_{2},\beta_{2},\beta_{2}^{\ast},\delta_{3},\beta_{3},\beta_{3}^{\ast}\in{\mathbb{F}}$,
and
$\beta_{2}+\beta_{2}^{\ast}+\delta_{3}=0=\beta_{3}+\beta_{3}^{\ast}+\delta_{2}.$

We will distinguish 3 cases, when both $\delta_{2}$ and $\delta_{3}$ are
either 0 or 1 and when one of them is 0 and the other is 1.

Case 1. If $\delta_{2}=\delta_{3}=0$, $b_{2}^{2}\equiv 0$, $b_{3}^{2}\equiv
0$, $\beta_{2}+\beta_{2}^{\ast}=0=\beta_{3}+\beta_{3}^{\ast}$, then
$\beta_{2}=\beta_{2}^{\ast}$, $\beta_{3}=\beta_{3}^{\ast}$, then
$a_{2}a_{3}\equiv\beta_{2}a_{2}+\beta_{3}a_{3}\equiv a_{3}a_{2}$, then
$a_{2}a_{3}+a_{3}a_{2}\equiv 0$.

Taking $a_{2}^{\ast}=\beta_{3}1_{{\cal A}}+a_{2}$,
$a_{3}^{\ast}=\beta_{2}1_{{\cal A}}+a_{3}$, the basis ${\cal
B}_{1}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},a_{3}^{\ast}\right\\}$ satisfies condition 1.

Case 2. If $\delta_{2}=\delta_{3}=1$, $a_{2}^{2}\equiv a_{2}$,
$a_{3}^{2}\equiv a_{3}$, $\beta_{2}^{\ast}=\beta_{2}+1$,
$\beta_{3}^{\ast}=\beta_{3}+1$, then
$a_{2}a_{3}\equiv\beta_{2}a_{2}+(\beta_{3}+1)a_{3}$,
$a_{3}a_{2}\equiv(\beta_{2}+1)a_{2}+\beta_{3}a_{3}$.

Consider the basis ${\cal B}_{2}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast}=a_{2}+\beta_{3}1_{{\cal
A}},a_{3}^{\ast}=a_{3}+\beta_{2}1_{{\cal A}}\right\\}$. It satisfies condition
2.

Case 3. If $\delta_{2}=0,\delta_{3}=1$, then $\beta_{2}^{\ast}=\beta_{2}+1,$
$\beta_{3}^{\ast}=\beta_{3}$. So $a_{2}^{2}\equiv 0$, $a_{3}^{2}\equiv 1$,
$a_{2}a_{3}=\beta_{2}a_{2}+\beta_{3}a_{3}$,
$a_{3}a_{2}=\beta_{2}a_{2}+(\beta_{3}+1)a_{3}$. Take
$a_{2}^{\ast}=a_{2}+\beta_{3}1_{{\cal A}}$ and
$a_{3}^{\ast}=a_{3}+\beta_{2}1_{{\cal A}}$. Then the basis ${\cal
B}_{3}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},a_{3}^{\ast}\right\\}$ satisfies condition 3. ∎

Finally we will address the case of dimension greater than 3.

######  Theorem 4.6.

Let ${\mathbb{F}}$ be a field of characteristic 2 and ${\cal A}$ an
${\mathbb{F}}$-algebra with ${\rm dim}\ {\cal A}\geq 4$. Then $l({\cal A})=1$
if and only if there is a basis ${\cal B}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},\ldots,a_{n}^{\ast}\right\\}$ of ${\cal A}$ whose
multiplication table satisfies one of the following two conditions:

  * (I)

${a_{i}^{\ast}}^{2}\equiv 0$, for $i=2,\ldots,n$,
$a_{i}^{\ast}a_{j}^{\ast}\equiv
a_{j}^{\ast}a_{i}^{\ast}\equiv\beta_{ij}a_{i}^{\ast}+\beta_{ji}a_{j}^{\ast}$,

  * (II)

${a_{i}^{\ast}}^{2}\equiv a_{i}^{\ast}$, for $i=2,\ldots,n$,
$a_{i}^{\ast}a_{j}^{\ast}\equiv\beta_{ij}a_{i}^{\ast}+(1+\beta_{ji})a_{j}^{\ast}$.

Notice that in the last case
$a_{i}^{\ast}a_{j}^{\ast}+a_{j}^{\ast}a_{i}^{\ast}\equiv
a_{i}^{\ast}+a_{j}^{\ast}$.

###### Proof.

Assume $l({\cal A})=1$ and consider $\left\\{1_{{\cal
A}},b_{2},\ldots,b_{n}\right\\}$ a basis of ${\cal A}$. Then
$b_{i}^{2}\equiv\delta_{i}b_{i}$, for some $\delta_{i}\in{\mathbb{F}}$, $2\leq
i\leq n$, by Corollary 2.2.

Arguing as in Theorem 4.4, we can assume that
$\delta_{i}\in\left\\{0,1\right\\}$, for all $2\leq i\leq n$.

Furthermore, by Proposition 2.1, $b_{i}b_{j}\in{\mathbb{F}}(1_{{\cal
A}},b_{i},b_{j})$, for all $2\leq i\neq j\leq n$. So
$b_{i}b_{j}\equiv\beta_{ij}b_{i}+\beta_{ji}^{*}b_{j}$, for some
$\beta_{ij},\beta_{ji}^{*}\in{\mathbb{F}}$. Using again Proposition 2.1, we
know that for all distinct $i,j,k\in\left\\{2,\ldots,n\right\\}$, the
following conditions hold:

  * (i)

$b_{i}(b_{j}+b_{k})\equiv\beta_{ij}b_{i}+\beta_{ji}^{*}b_{j}+\beta_{ik}b_{i}+\beta_{ki}^{*}b_{k}\in{\mathbb{F}}(1_{{\cal
A}},b_{i},b_{j}+b_{k})$, which implies $\beta_{ji}^{*}=\beta_{ki}^{*}$. Let us
define $\beta_{i}^{*}=\beta_{ji}^{*}$ for any $j\neq
i\in\left\\{2,\ldots,n\right\\}$;

  * (ii)

$(b_{j}+b_{k})b_{i}\equiv\beta_{ji}b_{j}+\beta_{ij}^{*}b_{i}+\beta_{ki}b_{k}+\beta_{ik}^{*}b_{i}\in{\mathbb{F}}(1_{{\cal
A}},b_{j}+b_{k},b_{i})$, which implies $\beta_{ji}=\beta_{ki}$. Let us define
$\beta_{i}=\beta_{ji}$ for any $j\neq i\in\left\\{2,\ldots,n\right\\}$;

  * (iii)

$(b_{i}+b_{j})(b_{i}+b_{k})\equiv\delta_{i}b_{i}+\beta_{k}b_{i}+\beta_{i}^{*}b_{k}+\beta_{i}b_{j}+\beta_{j}^{*}b_{i}+\beta_{k}b_{j}+\beta_{j}^{*}b_{k}\in{\mathbb{F}}(1_{{\cal
A}},b_{i}+b_{j},b_{i}+b_{k})$ which implies
$\delta_{i}+\beta_{k}+\beta_{j}^{*}=\beta_{i}+\beta_{k}+\beta_{i}^{*}+\beta_{j}^{*}$.
Then, by (i) and (ii), we can conclude that
$\beta_{i}+\beta_{i}^{*}=\delta_{i}$.

Therefore, $b_{i}b_{j}\equiv\beta_{j}b_{i}+(\beta_{i}+\delta_{i})b_{j}$ and
$b_{j}b_{i}\equiv\beta_{i}b_{j}+(\beta_{j}+\delta_{j})b_{i}$, for all
$i,j\in\left\\{2,\ldots,n\right\\}$ such that $i\neq j$.

Claim. We can assume that $\delta_{2}=\cdots=\delta_{n}$.

Indeed, if either all scalars are 0 or all scalars are 1, there is nothing to
prove. Suppose that $r-1$ scalars are 0 and $n-r$ are 1. Without loss of
generality (reordering elements), we can assume that
$\delta_{2}=\cdots=\delta_{r}=0$ and $\delta_{r+1}=\cdots=\delta_{n}=1$.

Observe that

$(b_{s}+b_{r+1})^{2}\equiv\delta_{s}b_{s}+\beta_{r+1}b_{s}+\beta_{s}^{*}b_{r+1}+\beta_{s}b_{r+1}+\beta_{r+1}^{*}b_{s}+\delta_{r+1}b_{r+1}$
$=(\beta_{r+1}+\beta_{r+1}^{*})b_{s}+(\beta_{s}+\beta_{s}^{*}+1)b_{r+1}=\delta_{r+1}b_{s}+(\delta_{s}+1)b_{r+1}=b_{s}+b_{r+1},$

for all $2\leq s\leq r$.

Thus, ${\cal B}=\left\\{1_{{\cal
A}},a_{2}=b_{2}+b_{r+1},\ldots,a_{r}=b_{r}+b_{r+1},a_{r+1}=b_{r+1},\ldots,a_{n}=b_{n}\right\\}$
is a basis of ${\cal A}$ such that $a_{i}^{2}\equiv a_{i}$.

In any case, we can consider a basis ${\cal
B}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},\ldots,a_{n}^{\ast}\right\\}$ of ${\cal A}$ such that
${a_{i}^{\ast}}^{2}\equiv\delta a_{i}^{\ast}$, for all $2\leq i\leq n$, where
$\delta\in\left\\{0,1\right\\}$. Clearly, $\delta=0$ gives statement (1), and
$\delta=1$ gives statement (2).

Conversely, suppose that there is a basis ${\cal
B}^{\ast}=\left\\{a_{1}^{\ast}=1_{{\cal
A}},a_{2}^{\ast},\ldots,a_{n}^{\ast}\right\\}$ that satisfies either (I) or
(II). Let $a=\lambda_{1}1_{{\cal A}}+\cdots+\lambda_{n}{a_{n}}^{\ast}$ and
$b=\gamma_{1}1_{{\cal A}}+\cdots+\gamma_{n}a_{n}^{\ast}$ be arbitrary elements
of ${\cal A}$. Then

Case 1: If ${a_{i}^{\ast}}^{2}\equiv 0$, for $i=2,\ldots,n$,
$a_{i}^{\ast}a_{j}^{\ast}\equiv
a_{j}^{\ast}a_{i}^{\ast}\equiv\beta_{j}a_{i}^{\ast}+\beta_{i}a_{j}^{\ast}$,
for all $i\neq j=2,\ldots,n$, then

$ab\equiv(\gamma_{2}\beta_{2}+\cdots+\gamma_{n}\beta_{n})a+(\lambda_{2}\beta_{2}+\cdots+\lambda_{n}\beta_{n})b\in{\mathbb{F}}(1_{{\cal
A}},a,b).$

Case 2: If ${a_{i}^{\ast}}^{2}\equiv a_{i}^{\ast}$, for $i=2,\ldots,n$,
$a_{i}^{\ast}a_{j}^{\ast}\equiv\beta_{j}a_{i}+(1+\beta_{i})a_{j}$, for all
$i\neq j=2,\ldots,n$, then

$ab\equiv(\gamma_{2}\beta_{2}+\cdots+\gamma_{n}\beta_{n})a+(\lambda_{2}+\cdots+\lambda_{n}+\lambda_{2}\beta_{2}+\cdots+\lambda_{n}\beta_{n})b\in{\mathbb{F}}(1_{{\cal
A}},a,b).$

In both cases, we can conclude that $l({\cal A})=1$, by Proposition 2.1, which
completes our proof. ∎

######  Remark 4.7.

If $a=\lambda_{1}1_{{\cal A}}+\lambda_{2}a_{2}+\cdots+\lambda_{n}a_{n}$ is an
arbitrary element of an algebra ${\cal A}$ satisfying Theorem 4.6 (I) then
$a^{2}\equiv 0$. Similarly, if ${\cal A}$ satisfies Theorem 4.6 (II) then
$a^{2}\equiv(\lambda_{2}+\cdots+\lambda_{n})a.$ So Theorem 4.6 is a kind of
classification theorem in the line of Theorem 1.5. The same applies to Theorem
4.2 and Corollary 4.5.

## Conclusion

We have given a complete characterization of algebras of length 1, without
additional conditions neither on the algebra nor on the characteristic of the
field.

In case of characteristic 2 we have obtained a kind of classification similar
to the one obtained in the associative case (see Theorem 1.5).

In case of ${\rm char}({\mathbb{F}})\neq 2$, our result characterizes algebras
of length one through the existence of a particular basis (an special basis).
Of course there are classification problems that remain opened and that
probably only could be addressed under additional conditions on the algebra.

However our result allows the construction of an easy algorithm to decide if a
given algebra, with some known basis and the corresponding multiplication
table, has length one.

## Acknowledgements

Authors acknowledge the referee for many valuable comments that have helped to
improve the paper.

The investigations of the first author are supported by Russian Science
Foundation, Project 17-11-01124. The second-named author have been partially
supported by Spanish Government, Project MTM2017-83506-C2-2-P and by
Principado de Asturias, Project FC – GRUPIN – IDI/2018/000193. The third
author thanks Oviedo University, where part of this study took place, for the
hospitality.

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