On some Leibniz algebras, having small dimension
The first step in the study of all types of algebras is the description of such algebras having small dimensions. The structure of 3-dimensional Leibniz algebras is more complicated than 1 and 2-dimensional cases. In this paper, we consider the structure of Leibniz algebras of dimension 3 over the f...
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| Cite this: | On some Leibniz algebras, having small dimension / V.S. Yashchuk // Algebra and Discrete Mathematics. — 2019. — Vol. 27, № 2. — С. 292–308. — Бібліогр.: 16 назв. — англ. |
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| citation_txt | On some Leibniz algebras, having small dimension / V.S. Yashchuk // Algebra and Discrete Mathematics. — 2019. — Vol. 27, № 2. — С. 292–308. — Бібліогр.: 16 назв. — англ. |
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| description | The first step in the study of all types of algebras is the description of such algebras having small dimensions. The structure of 3-dimensional Leibniz algebras is more complicated than 1 and 2-dimensional cases. In this paper, we consider the structure of Leibniz algebras of dimension 3 over the finite fields. In some cases, the structure of the algebra essentially depends on the characteristic of the field, in others on the solvability of specific equations in the field, and so on.
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 27 (2019). Number 2, pp. 292–308
c© Journal “Algebra and Discrete Mathematics”
On some Leibniz algebras,
having small dimension
Viktoriia S. Yashchuk
Communicated by L. A. Kurdachenko
Abstract. The first step in the study of all types of algebras
is the description of such algebras having small dimensions. The
structure of 3-dimensional Leibniz algebras is more complicated
than 1 and 2-dimensional cases. In this paper, we consider the
structure of Leibniz algebras of dimension 3 over the finite fields.
In some cases, the structure of the algebra essentially depends on
the characteristic of the field, in others on the solvability of specific
equations in the field, and so on.
Introduction
Let L be an algebra over finite field F with the binary operations +
and [·, ·]. Then L is called a Leibniz algebra (more precisely a left Leibniz
algebra) if it satisfies the (left) Leibniz identity
[[a, b], c] = [a, [b, c]]− [b, [a, c]]
for all a, b, c ∈ L.
We will also use another form of this identity:
[a, [b, c]] = [[a, b], c] + [b, [a, c]].
The author would like to express her sincere thanks to her scientific advisor Professor
L. A. Kurdachenko.
2010 MSC: 17A32, 17A60.
Key words and phrases: Leibniz algebra, ideal, factor-algebra, Leibniz kernel,
finite dimensional Leibniz algebra, nilpotent Leibniz algebra, left (right) center, Frattini
subalgebra.
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V. S. Yashchuk 293
Leibniz algebras appeared first in the papers of A.M. Bloh [5–7], in
which he called them the D-algebras. However, in that time these works
were not in demand, and they have not been properly developed. Only
after two decades, a real interest to Leibniz algebras rose. It happened
thanks to the work of J.-L. Loday [15] (see also [16, Section 10.6]), who
“rediscovered” these algebras and used the term Leibniz algebras since it
was Gottfried Wilhelm Leibniz who discovered and proved the Leibniz
rule for differentiation of functions.
Let L be a Leibniz algebra over finite field F . If A,B are subspaces
of L, then [A,B] will denote a subspace generated by all elements [a, b]
where a ∈ A, b ∈ B. As usual, a subspace A of L is called a subalgebra of
L, if [x, y] ∈ A for every x, y ∈ A. It follows that [A,A] 6 A.
Let L be a Leibniz algebra over finite field F , M be non-empty subset
of L, then 〈M〉 denote the subalgebra of L generated by M .
A subalgebra A of L is called a left (respectively right) ideal of L,
if [y, x] ∈ A (respectively [x, y] ∈ A) for every x ∈ A, y ∈ L. In other
words, if A is a left (respectively right) ideal, then [L,A] 6 A (respectively
[A,L] 6 A).
A subalgebra A of L is called an ideal of L (more precisely, two-sided
ideal) if it is both a left ideal and a right ideal, that is [y, x], [x, y] ∈ A for
every x ∈ A, y ∈ L.
If A is an ideal of L, we can consider a factor-algebra L/A. It is not
hard to see that this factor-algebra also is a Leibniz algebra.
As usual, a Leibniz algebra L is called abelian if [x, y] = 0 for all
elements x, y ∈ L. In abelian Leibniz algebra every subspace is a subalgebra
and an ideal.
Denote by Leib(L) the subspace, generated by the elements [a, a],
a ∈ L. It is possible to prove that Leib(L) is an ideal of L such that
L/Leib(L) is a Lie algebra. Conversely, if H is an ideal of L such that
L/H is a Lie algebra, then Leib(L) 6 H.
The ideal Leib(L) is called a Leibniz kernel of algebra L.
We note a following important property of Leibniz kernel:
[[a, a], x] = [a, [a, x]]− [a, [a, x]] = 0.
This property shows that Leib(L) is an abelian subalgebra of L.
As usual, we say that a Leibniz algebra L is finite dimensional, if the
dimension L as a vector space over F is finite.
The first step in the study of all types of algebras is the description of
such algebras having small dimensions.
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294 On some Leibniz algebras, having small dimension
If dimF (L) = 1, then L is an abelian Lie algebra, that is L = Fa for
some element a and [a, a] = 0.
If dimF (L) = 2 and L is not a Lie algebra, then there are the following
two non-isomorphic Leibniz algebras:
L1 = Fa+ Fb, [a, a] = b, [b, a] = [a, b] = [b, b] = 0,
and
L2 = Fc+ Fd, [c, c] = [c, d] = d, [d, c] = [d, d] = 0.
(see, for example, a survey [12]). The structure of 3-dimensional Leibniz
algebras is more complicated. The study of Leibniz algebras, having
dimension 3, has been conducted in the papers ([1,2,8,10]) for the fields of
characteristic 0, moreover for a field C of complex numbers or algebraically
closed field of characteristic 0. In this paper, we consider the opposite
situation, where the structure of Leibniz algebras of dimension 3 over the
finite field will be described. As we will see later, the situation here is
much more diverse, in some cases the structure of the algebra essentially
depends on the characteristic of the field, in others on the solvability of
specific equations in the field, and so on. We will see that the Leibniz
algebras of dimension 3 is soluble, therefore a first natural step of our
study is a consideration of nilpotent algebras.
Let L be a Leibniz algebra. Define the lower central series
L = γ1(L) > γ2(L) > . . . γα(L) > γα+1(L) > . . . γδ(L)
of L by the following rule: γ1(L) = L, γ2(L) = [L,L], and recursively
γα+1(L) = [L, γα(L)] for all ordinals α and γλ(L) =
⋂
µ<λ
γµ(L) for the
limit ordinals λ.It is possible to shows that every term of this series is an
ideal of L. The last term γδ(L) is called the lower hypocenter of L. We
have γδ(L) = [L, γδ(L)].
If α = k is a positive integer, then γk(L) = [L, [L, [L, . . .] . . .]].
A Leibniz algebra L is called nilpotent if there exists a positive integer
k such that γk(L) = 〈0〉. More precisely, L is said to be nilpotent of
nilpotency class c if γc+1(L) = 〈0〉, but γc(L) 6= 〈0〉. We denote by ncl(L)
the nilpotency class of L.
The left (respectively right) center ζ left(L) (respectively ζright(L)) of
a Leibniz algebra L is defined by the rule:
ζ left(L) = {x ∈ L| [x, y] = 0 for each element y ∈ L}
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V. S. Yashchuk 295
(respectively,
ζright(L) = {x ∈ L| [y, x] = 0 for each element y ∈ L}).
It is not hard to prove that the left center of L is an ideal, but it is not true
for the right center. Moreover, Leib(L) 6 ζ left(L), so that L/ζ left(L) is a
Lie algebra. The right center is a subalgebra of L, and in general, the left
and right centers are different; they even may have different dimensions.
The corecponding examples could be find in [13].
Of course we will consider a case when L is a not Lie algebra.
1. Nilpotent Leibniz algebra of dimension 3
In this section we will suppose that L is nilpotent.
Since ncl(L) 6 dimF (L), ncl(L) 6 3.
Let L be a Leibniz algebra over field F . The intersection of the
maximal subalgebras of L is called the Frattini subalgebra of L and de-
noted by Frat(L). If L does not include maximal subalgebras, then put
L = Frat(L).
We will need the following important property of Frattini subalgebras.
Proposition 1. Let L be a finite dimensional Leibniz algebra over finite
field F . If L is nilpotent then [L,L] = Frat(L).
Indeed since L is nilpotent, every maximal subalgebra of L is an ideal
[3, Lemma 2.2], so we can apply Proposition 7 of paper [12].
Theorem 1. Let L be a nilpotent Leibniz algebra over finite field F . If L is
a not Lie algebra and ncl(L) = 3 = dimF (L), then L has a basis {a, b, c}
such that [a, a] = b, [a, b] = c, [c, a] = [a, c] = [b, a] = [c, b] = [b, c] =
[b, b] = [c, c] = 0. Moreover, Leib(L) = ζ left(L) = [L,L] = Fb ⊕ Fc,
ζright(L) = ζ(L) = γ3(L) = Fc. In particular, L is a nilpotent cyclic
Leibniz algebra.
Proof. Since γ1(L) = L 6= γ2(L) = [L,L] 6= γ3(L) 6= 〈0〉, dimF (γ2(L)) =
2, dimF (γ3(L)) = 1 = dimF (L/γ2(L)). Let a be an element of L such
that a /∈ γ2(L). Then Proposition 1 shows that L is a cyclic algebra,
generated by an element a. Put b = [a, a], then b ∈ γ2(L) = [L,L] and
b ∈ Leib(L). It follows that [b, a] = 0. If we suppose that [a, b] = 0, then
b ∈ ζ(L). But in this case γ2(L) 6 ζ(L) and γ3(L) = 〈0〉, that contradicts
to our assumption. Thus [a, b] = c 6= 0. Then c ∈ γ3(L), that follows that
γ3(L) = Fc and [c, a] = [a, c] = [c, b] = [b, c] = 0.
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296 On some Leibniz algebras, having small dimension
The case when ncl(L) = 2 break up on two subcases. First: there exists
an element d /∈ γ2(L) = [L,L] such that [d, d] = 0. Second: [d, d] 6= 0 for
each element d /∈ γ2(L). In the second case [d, d] is a non-zero element of
ζ(L). Then ζ(L) = F [d, d] 6 〈d〉. Since a factor-algebra L/ζ(L) is abelian,
a cyclic subalgebra 〈d〉 is an ideal. Then and every non-zero subalgebra of
L is an ideal.
The following two theorems consider the both these subcase separately.
Further writing L = A ⊕ B means that L is a direct sum of the
subspaces A and B or the subalgebras A and B. If L = A⊕B and A is
an ideal of L and B is a subalgebra of L, then we will say that L is a
semidirect sum of A and B and use the following symbol L = A ⊣ B.
Theorem 2. Let L be a nilpotent Leibniz algebra over finite field F .
Suppose that L is a not Lie algebra, dimF (L) = 3, ncl(L) = 2 and L has
an element b /∈ γ2(L) such that [b, b] = 0. Then L is an algebra of one of
the following types:
I. L = A ⊕ B, where A, B are the ideals, B = Fb, [b, b] = 0, A =
Fa⊕ Fc is a cyclic nilpotent subalgebra, [a, a] = c, [c, a] = [a, c] =
[c, c] = 0. Moreover, Leib(L) = [L,L] = Fc, ζ left(L) = ζright(L) =
ζ(L) = Fb⊕ Fc.
II. L = A ⊣ B, where A = Fa ⊕ Fc is a cyclic nilpotent subalgebra,
[a, a] = c, [c, a] = [a, c] = 0, B is an abelian subalgebra, B = Fb,
[b, b] = 0, and [a, b] = c, [b, a] = [b, c] = [c, b] = [c, c] = 0. Moreover,
Leib(L) = [L,L] = ζright(L) = ζ(L) = Fc, ζ left(L) = Fb⊕ Fc.
III. L = A ⊣ B, where A = Fa ⊕ Fc is a cyclic nilpotent subalgebra,
[a, a] = c, [c, a] = [a, c] = 0, B is an abelian subalgebra, B = Fb,
[b, b] = 0, and [a, b] = c, [b, a] = γc, γ 6= 0, [b, c] = [c, b] = [c, c] = 0.
Moreover, Leib(L) = [L,L] = ζ left(L) = ζright(L) = ζ(L) = Fc.
Proof. Since γ3(L) = 〈0〉, then γ2(L) = [L,L] 6 ζ(L). An inclusion
Leib(L) 6 γ2(L) shows that dimF (γ2(L)) > 1. If we assume that
dimF (L/γ2(L)) = 1, then Proposition 1 shows that L is a cyclic al-
gebra. But in this case it is not hard to prove that dimF (L) = 2. Hence
L/γ2(L) is not cyclic, so that dimF (L/γ2(L)) = 2. Put K = γ2(L). Since
L is not a Lie algebra, there is an element a such that [a, a] = c 6= 0. An in-
clusion K 6 ζ(L) shows that a /∈ K. Since dimF (L/γ2(L)) = 2, K = Fc.
The fact that L/K is abelian implies that [a, x], [x, a] ∈ K = Fc 6 〈a〉
for each element x ∈ L. This shows that a subalgebra A = 〈a〉 = Fa⊕Fc
is an ideal of L.
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V. S. Yashchuk 297
By the choice of an element a we obtain that b /∈ 〈a〉, so that L = A⊕Fb.
An equality [b, b] = 0 shows that a subalgebra 〈b〉 is abelian, so that
〈b〉 = Fb. Thus L is a semidirect sum of an ideal 〈a〉 and 1-dimensional
abelian subalgebra Fb. Here we will have the following possibilities for
commutators [a, b] and [b, a].
(i) [a, b] = [b, a] = 0, in this case a subalgebra 〈b〉 is an ideal and L is a
direct sum of two ideals: L = (Fa⊕Fc)⊕Fb, so that L is a Leibniz
algebra of type I.
(ii) [a, b] = αc, [b, a] = βc, where (α, β) 6= (0, 0). If α 6= 0 (respectively
β 6= 0), then we can replace an element b on b1 = α−1b (respectively
on b1 = β−1b). Clearly L = 〈a〉 ⊕ Fb1 and [a, b1] = c, [b1, b1] = 0
(respectively [b1, a] = c, [b1, b1] = 0). Consider now two situations,
which appear here:
(iia) [a, b] = c, [b, a] = 0,
(iib) [a, b] = 0, [b, a] = c.
In the second case put a1 = a, b1 = a− b, then we have:
[b1, b1] = [a− b, a− b] = [a, a]− [b, a]− [a, b] + [b, b] = c− c = 0,
[a1, b1] = [a, a− b] = [a, a]− [a, b] = c− 0 = c,
[b1, a1] = [a− b, a] = [a, a]− [b, a] = c− c = 0.
This shows that in every of situation (iia) and (iib) we come to the
same algebra. Thus we obtain the Leibniz algebra of types II and III.
In the second subcase we consider a situation when [d, d] is a non-zero
element of ζ(L) for each element d /∈ ζ(L). As we noted above, in this
case every non-zero subalgebra of L is an ideal.
Theorem 3. Let L be a nilpotent Leibniz algebra over finite field F .
Suppose that L is a not Lie algebra, dimF (L) = 3, ncl(L) = 2 and
[d, d] 6= 0 for each element d /∈ γ2(L). Then L is an algebra of one of the
following types:
I. L = A+B, where A, B are the nilpotent ideals, A = 〈a〉, B = 〈b〉,
A∩B = ζ(L) = Fc, [a, a] = [b, b] = c, [c, a] = [a, c] = [c, b] = [b, c] =
[a, b] = [b, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) =
ζright(L) = ζ(L) = Fc, char(F ) 6= 2, and an equation X2 + 1 = 0
has not a solution in F .
II. L = A+B, where A, B are the nilpotent ideals, A = 〈a〉, B = 〈b〉,
A ∩ B = ζ(L) = Fc, [a, a] = c, [b, b] = ρc, where ρ is a primitive
root of identity of degree |F | − 1, [c, a] = [a, c] = [c, b] = [b, c] =
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298 On some Leibniz algebras, having small dimension
[a, b] = [b, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) =
ζright(L) = ζ(L) = Fc, char(F ) 6= 2.
III. L = A+B, where A, B are the nilpotent ideals, A = 〈a〉, B = 〈b〉,
A ∩ B = ζ(L) = Fc, [a, a] = c = [a, b], [b, b] = ηc, [c, a] = [a, c] =
[c, b] = [b, c] = [b, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] =
ζ left(L) = ζright(L) = ζ(L) = Fc and a polynomial X2 +X + η has
no roots in a field F .
Proof. As in Theorem 2 γ2(L) = [L,L] 6 ζ(L), Leib(L) 6 γ2(L),
dimF (γ2(L)) = 1, so that dimF (L/γ2(L)) = 2. Put K = γ2(L). Let
a /∈ K, then [a, a] = c 6= 0. Since dimF (γ2(L)) = 1, K = Fc. The fact
that L/K is abelian implies that [a, x], [x, a] ∈ K = Fc 6 〈a〉 for each
element x ∈ L. This shows that a subalgebra A = 〈a〉 = Fa ⊕ Fc is an
ideal of L.
Choose an element b /∈ 〈a〉. Then the subset {c, a, b} is a basis of L.
By our condition [b, b] = ηc and η 6= 0. Consider the elements [a, b] and
[b, a]. Suppose first that [a, b] = [b, a] = 0. Since F is finite, G = F\{0} is
a cyclic group by multiplication of order q = |F |−1. Suppose that η ∈ G2.
Then η = κ2 for some element 0 6= κ ∈ F . Put b1 = κ−1b, then
[b1, b1] = [κ−1b, κ−1b] = κ−2[b, b] = κ−2ηc = κ−2κ2c = c.
In particular, if char(F ) = 2, then G2 = G. We take an arbitrary
element d = λa+ µb1, where λ 6= 0, µ 6= 0 and find [d, d]. By such choice
d /∈ ζ(L), so that [d, d] 6= 0. We have
[d, d] = [λa+ µb1, λa+ µb1]
= λ2[a, a] + λµ[b1, a] + λµ[a, b1] + µ2[b1, b1]
= λ2c+ µ2c.
Suppose that λ2 + µ2 = 0. Since µ 6= 0, we obtain y2 + 1 = 0 where
y = λµ−1. Thus we see that a polynomial X2 + 1 has no roots in a field
F . Since in a field F such that char(F ) = 2 a polynomial X2 + 1 has
root 1, the characteristic of F must be not 2. Thus we obtain the Leibniz
algebras of type I.
For example, if F = F5, F13, an equation y2 + 1 = 0 has a solution,
if F = F3, F7, F11, an equation y2 + 1 = 0 has not a solution. Thus we
can see that the algebras have the same defining relation, but have the
different properties.
Consider the case when η /∈ G2. Since G is cyclic, then G/G2 is a
cyclic group of order 2, so that G = G2 ∪ ρG2, where ρ is a primitive
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V. S. Yashchuk 299
root of identity of degree q. Then η = ρκ2 for some element κ ∈ F . Put
b1 = κ−1b, then
[b1, b1] = [κ−1b, κ−1b] = κ−2[b, b] = κ−2ηc = κ−2ρκ2c = ρc.
As we noted above in this case char(F ) 6= 2. Let again d = λa+ µb1,
where λ 6= 0, µ 6= 0. We have
[d, d] = [λa+ µb1, λa+ µb1] = λ2[a, a] + µ2[b1, b1]
= λ2c+ µ2ρc = (λ2 + µ2ρ)c.
Suppose that λ2 + µ2ρ = 0. Since µ 6= 0, we obtain y2 + ρ = 0, where
y = λµ−1. By the choice ρ is a generator of a multiplicative cyclic group
G = F\{0}, so that ρ /∈ G2. Since 〈ρ〉 6 〈−ρ〉, −ρ /∈ G2. It follows that
an equation y2 + ρ = 0 has not a solution in F . Then [d, d] 6= 0 for every
element d /∈ Fc. As we have seen above in this case every subalgebra of L
is an ideal. Thus we come to a Leibniz algebra of type II.
Suppose that [b, a] = αc 6= 0. Then α 6= 0. Put b1 = αa − b, then
[b1, a] = [αa− b, a] = α[a, a]− [b, a] = αc− αc = 0. Therefore further we
will suppose that [b, a] = 0.
We have [a, b] = βc for some element β ∈ F . The case when β = 0
we have already considered. Assume therefore that [a, b] = βc 6= 0. Put
b1 = β−1b, then [a, b1] = [a, β−1b] = β−1[a, b] = β−1βc = c. Let again
d = λa+ µb where λ 6= 0, µ 6= 0. We have
[d, d] = [λa+ µb1, λa+ µb1] == λ2[a, a] + λµ[a, b1] + µ2[b1, b1]
= λ2c+ λµc+ µ2ηc = (λ2 + λµ+ µ2η)c, where η = [b1, b1].
Suppose that λ2 + λµ+ µ2η = 0. Since µ 6= 0, we obtain an equation
y2+y+η = 0 where y = λµ−1. Hence we see that a polynomial X2+X+η
has no roots in a field F . Thus we obtain the Leibniz algebra of type III.
As we can see, in this last case the properties of the algebra depend
on whether the polynomial above has roots in the field F , and this also
depends on the choice of the element η. The difference appears even over
fields of the same characteristic. So, if F = F2, then for η only one value
is possible η = 1. But an equation y2 + y+1 = 0 has no solution in a field
F = F2. Hence the respectively algebra has no element d /∈ Fc such that
[d, d] = 0. It follows that every subalgebra of L is an ideal. If F = F4 and
η = 1, then an equation y2 + y + 1 = 0 has solutions in a field F4. In this
case the respectively algebra L has one dimensional subalgebra, which is
no ideal.
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300 On some Leibniz algebras, having small dimension
2. Non-nilpotent Leibniz algebras of dimension 3 with
1-dimensional Leibniz kernel over finite fields
The next step is a consideration of a case when L is non-nilpotent. We
will consider a Leibniz algebras of dimension 3 over finite fields, which
are not Lie algebras. It follows that Leib(L) 6= 〈0〉. Since Leib(L) is
an abelian ideal, L 6= Leib(L). Hence for Leib(L) we have only two
possibility: dimF (Leib(L)) = 1, dimF (Leib(L)) = 2.
In this section we consider the case when dimF (Leib(L)) = 1, so that
dimF (L/Leib(L)) = 2.
Theorem 4. Let L be a non-nilpotent Leibniz algebra over finite field F .
Suppose that L is a not Lie algebra, dimF (L) = 3 and dimF (Leib(L)) =
1. Then L is an algebra of one of the following types:
I. L = A⊕B, where A, B are the ideals, B = Fb, [b, b] = 0, A is a cyclic
subalgebra, A = Fa ⊕ Fc, where [a, a] = c = [a, c], [c, a] = [c, b] =
[b, c] = [a, b] = [b, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] = Fc,
ζ left(L) = Fb⊕ Fc, ζright(L) = ζ(L) = Fb.
II. L = A ⊣ B, where B = Fb, [b, b] = 0, A = Fa ⊕ Fc is a cyclic
subalgebra, [a, a] = c = [a, c], [a, b] = c, [c, a] = [c, b] = [b, c] =
[b, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] = Fc, ζ left(L) =
Fb⊕ Fc, ζ(L) = ζright(L) = 〈0〉.
III. L = A ⊣ B, where B = Fb, [b, b] = 0, A = Fa ⊕ Fc is a cyclic
subalgebra, [a, a] = c = [a, c], [b, a] = [b, c] = c, [c, a] = [c, b] =
[a, b] = [c, c] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) = Fc,
ζright(L) = Fb, ζ(L) = 〈0〉.
IV. L = A ⊣ B,, where B = Fb, [b, b] = 0, A = Fa ⊕ Fc is a cyclic
subalgebra, [a, a] = c = [a, c], [a, b] = a = −[b, a], [b, c] = −2c,
[c, a] = [c, b] = [c, c] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) =
Fc, ζright(L) = ζ(L) = 〈0〉.
V. L = A ⊣ B, where B = Fb, [b, b] = 0, A = Fa ⊕ Fc is a cyclic
subalgebra, [a, a] = c, [a, c] = 0, [a, b] = a+γc, γ ∈ F , [b, a] = −a+γc,
[b, c] = −2c, [c, a] = [c, b] = 0. Moreover, Leib(L) = [L,L] =
ζ left(L) = Fc, ζright(L) = ζ(L) = 〈0〉 whenever char(F ) 6= 2 and
ζright(L) = ζ(L) = Fc whenever char(F ) = 2.
Proof. Since L is not Lie algebra, there is an element a such that [a, a] =
c 6= 0. Then c ∈ K = Leib(L), so that K = Fc. An inclusion Leib(L) 6
ζ left(L) implies that [c, y] = 0 for each element y ∈ L, in particular,
[c, a] = 0. A subalgebra 〈a〉 has dimension 2, therefore either [a, c] = 0 or
[a, c] = c (see, for example, a survey [12]). Suppose that a Lie algebra L/K
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V. S. Yashchuk 301
is abelian and let first [a, c] = 0. Also assume that there is an element
b /∈ 〈a〉 such that [b, b] = 0. The fact that L/K is abelian implies that
[a, b], [b, a] ∈ Fc. If [a, b] = 0 = [b, a], then a subalgebra 〈b〉 = Fb is an
ideal and L is a direct sum of two subalgebras: L = (Fa⊕Fc)⊕Fb. Since
L is non-nilpotent, then for [a, c] only one variant remains: [a, c] = c. Thus
we obtain the Leibniz algebra of type I.
Let now [a, b] = αc, [b, a] = βc, where (α, β) 6= (0, 0). If α 6= 0
(respectively β 6= 0), then we can replace an element b on b1 = α−1b
(respectively on b1 = β−1b). Clearly L = 〈a〉 ⊕ Fb1 and [a, b1] = c,
[b1, b1] = 0 (respectively [b1, a] = c, [b1, b1] = 0). Consider further all three
situation, which appear here.
(a) Suppose that [a, b] = c, [b, a] = 0. Then
[b, c] = [b, [a, a]] = [[b, a], a] + [a, [b, a]] = 0.
If we assume that [a, c] = 0, then K 6 ζ(L), which follows that an
algebra L is nilpotent. This situation was considered early. Thus we obtain
the Leibniz algebra of type II.
(b) Suppose that [a, b] = 0, [b, a] = c. Then
[b, c] = [b, [a, a]] = [[b, a], a] + [a, [b, a]] = [c, a] + [a, c] = [a, c].
If we assume that [a, c] = 0, then K 6 ζ(L), so we again come to
nilpotent case. Thus we obtain the Leibniz algebra of type III.
(c) Suppose that [a, b] = c, [b, a] = βc, where β 6= 0. Then
[b, c] = [b, [a, a]] = [[b, a], a] + [a, [b, a]] = β[c, a] + β[a, c] = β[a, c].
On the other hand,
[b, c] = [b, [a, b]] = [[b, a], b] + [a, [b, b]] = β[c, b] + [a, 0] = 0.
If we assume that [a, c] = 0, then K 6 ζ(L), so we again come to nilpotent
case. Then an equality [b, c] = β[a, c] implies that β = 0. Thus we obtain
a contradiction, which shows that the last case is impossible.
Suppose now that [d, d] 6= 0 for each element d /∈ K. It follows that
[d, d] is a non-zero element of K. Then K = F [d, d] 6 〈d〉. Since a factor-
algebra L/K is abelian, a cyclic subalgebra 〈d〉 is an ideal. If 0 6= d ∈ K,
then 〈d〉 = K, in particular, 〈d〉 again is an ideal. Then and every non-zero
subalgebra of L is an ideal. But in this case an algebra L must be nilpotent
([14, Theorem A]). This case was considered above.
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302 On some Leibniz algebras, having small dimension
Assume now that dimF (Leib(L)) = 1 and a Lie algebra L/K is
non-abelian. Again there exists an element a such that [a, a] 6= 0. Since
L/K is a Lie algebra, [a, a] ∈ K. If we suppose that a + K ∈ ζ(L/K),
then an equality dimF (L/K) = 2 implies that L/K is abelian, and we
obtain a contradiction. Thus we can find the element y such that K 6=
[a+K, y +K]. The fact that L/K is a Lie algebra of dimension 2 implies
that L/K 6= [L/K,L/K] (see, for example, [11, Chapter 1, Section 4]).
Then dimF ([L/K,L/K]) = 1. Put d = [a, y], then [L/K,L/K] = 〈d +
K〉 = D/K. As above d +K /∈ ζ(L/K). We have one of two following
possibilities: 〈d+K〉 = 〈a+K〉 or the intersection 〈d+K〉 ∩ 〈a+K〉 is
zero.
Consider a first possibility, let 〈d+K〉 = 〈a+K〉. Suppose that there
exists an element b /∈ D such that [b, b] = 0. Then L = D ⊕ 〈b〉, where a
subalgebra 〈b〉 has a dimension one, that is 〈b〉 = Fb. Since c = [a, a] ∈ K,
D = Fa ⊕ Fc. Since a + K /∈ ζ(L/K), [a + K, b + K] 6= K, that is
[a+K, b+K] = δ(a+K) for some non-zero element δ ∈ F . Put b1 = δ−1b,
then
[a+K, b1 +K] = [a+K, δ−1b+K] = δ−1[a+K, b+K]
= δ−1(d+K) = δ−1δ(a+K) = a+K.
Since [b1, b1] = δ−2[b, b] = 0, we can assume further that [a+K, b+K] =
a+K. It follows that [a, b] = a+ γc for some element γ ∈ F . Since L/K
is a Lie algebra, [b + K, a + K] = −[a + K, b + K] = −a + K, so that
[b, a] = −a + γ1c for some element γ1 ∈ F . An inclusion c ∈ Leib(L)
implies that [c, x] = 0 for all x ∈ L, in particular, [c, a] = [c, b] = 0. For
element [b, c] we obtain:
[b, c] = [b, [a, a]] = [[b, a], a] + [a, [b, a]] = [−a+ γ1c, a] + [a,−a+ γ1c]
= [−a, a] + [a,−a] = −2c.
Further,
[[a, b], b] = [a, [b, b]]− [b, [a, b]] = −[b, [a, b]] = −[b, a+ γc]
= −[b, a]− γ[b, c] = a− γ1c+ 2γc = a+ (2γ − γ1)c,
and [[a, b], b] = [a+γc, b] = [a, b] = a+γc, so that a+(2γ−γ1)c = a+γc,
which implies that γ1 = γ.
A subalgebra 〈a〉 has dimension 2, therefore we have two cases:
(i) [a, c] = 0 and (ii) [a, c] = c (see, for example, a survey [12]).
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V. S. Yashchuk 303
If [a, c] = c, then
[[b, a], a] = [b, [a, a]]− [a, [b, a]] = [b, c]− [a,−a+ γ1c]
= −2c+ [a, a]− γ1[a, c] = −2c+ c− γ1[a, c],
and [[b, a], a] = [−a+ γ1c, a] = −c, which implies that −2c+ c− γ1[a, c] =
−c, so that γ1 = 0. Hence if [a, c] = c, then we obtain the Leibniz algebra
of type IV.
If [a, c] = 0, then we obtain the Leibniz algebra of type V.
Assume now that [b, b] 6= 0 for each element b /∈ D. As above we can
choose an element b /∈ D such that [a+K, b+K] = a+K. It follows that
[a, b] = a+ γc for some element γ ∈ F . Since L/K is a Lie algebra,
[b+K, a+K] = −[a+K, b+K] = −a+K,
so that [b, a] = −a + γ1c for some element γ1 ∈ F . Since [b, b] 6= 0,
[b, b] = ηc for some non-zero element η ∈ F . An inclusion c ∈ Leib(L)
implies that [c, x] = 0 for all x ∈ L, in particular, [c, a] = [c, b] = 0. For
element [b, c] we obtain again
[b, c] = [b, [a, a]] = [[b, a], a] + [a, [b, a]] = [−a+ γ1c, a] + [a,−a+ γ1c]
= [−a, a] + [a,−a] = −2c.
Further,
[[a, b], b] = [a, [b, b]]− [b, [a, b]] = [a, ηc]− [b, [a, b]] = η[a, c]− [b, a+ γc]
= η[a, c]− [b, a]− γ[b, c] = η[a, c] + a− γ1c+ 2γc
= η[a, c] + a+ (2γ − γ1)c.
If [a, c] = 0, then [[a, b], b] = a+ (2γ − γ1)c, if [a, c] = c, then [[a, b], b] =
a+(η+2γ−γ1)c. On the other hand, [[a, b], b] = [a+γc, b] = [a, b] = a+γc,
so that a+ (2γ − γ1)c = a+ γc or a+ (η + 2γ − γ1)c = a+ γc. Thus if
[a, c] = 0, then γ1 = γ, if [a, c] = c, then γ1 = η + γ.
Furthermore,
[[b, a], a] = [b, [a, a]]− [a, [b, a]] = [b, c]− [a,−a+ γ1c]
= −2c+ [a, a]− γ1[a, c] = −2c+ c− γ1[a, c] = −c− γ1[a, c].
and [[b, a], a] = [−a + γ1c, a] = −c, which implies that γ1[a, c] = 0. In
particular, if [a, c] = c then γ1 = 0. As we have seen above, in this case
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304 On some Leibniz algebras, having small dimension
γ1 = η + γ, therefore η = −γ. Suppose that [a, c] = c and consider an
element a+ b+ c. We have
[a+ b+ c, a+ b+ c] = [a, a] + [a, b] + [a, c] + [b, a] + [b, b] + [b, c]
= c+ a+ γc+ c− a− γc− 2c = 0.
Since a+ b+ c /∈ D, we obtain a contradiction, which shows that a variant
when [a, c] = c is impossible.
Hence must be [a, c] = 0. Let x = λa+µb+νc be an arbitrary element
of L such that x /∈ D, where λ, µ, ν are some elements of F . We have
[λa+ µb+ νc, λa+ µb+ νc]
= λ2[a, a] + λµ[a, b] + λµ[b, a] + µ2[b, b] + µν[b, c]
= λ2c+ λµ(a+ γc) + λµ(−a+ γc) + µ2ηc− 2µνc
= (λ2 + 2λµγ + µ2η − 2µν)c.
Consider an equation λ2 + 2λµγ + µ2η − 2µν = 0. If F is a finite field
of characteristic 2, then we obtain λ2 + µ2η = 0. Put µ = 1, by such
choice x /∈ D. An equation λ2 + η = 0 has a solution in a field F , because
F = F2. Thus if F is a finite field of characteristic 2, then there exists an
element x /∈ D such that [x, x] = 0, and we obtain a contradiction with our
assumption. Suppose now that char(F ) 6= 2. The fact that char(F ) 6= 2
implies that an equation 2x = a has a solution 1
2
a for each element a ∈ F .
Put λ = µ = 1, so we come to equation 1 + 2γ + η − 2ν = 0. Thus if we
put λ = µ = 1, ν = 1
2
(1 + 2γ + η), then x /∈ D and [x, x] = 0, and we
again obtain a contradiction.
3. Non-nilpotent cyclic Leibniz algebras of dimension 3
over finite fields
The next step is a consideration of a case when L is non-nilpotent or
dimF (Leib(L)) = 2. Here there appear two variants: L is a cyclic algebra
and L is a non-cyclic algebra. In this section we will consider a case when
Leibniz algebra of dimension 3 is cyclic.
Theorem 5. Let L be a non-nilpotent cyclic Leibniz algebra of dimension 3
over finite field F . Then L is an algebra of one of the following types:
I. L = D ⊣ A, where D = Fd, [d, d] = 0, A = Fa ⊕ Fc is a cyclic
nilpotent subalgebra, [a, a] = c, [a, c] = 0, [a, d] = δd, 0 6= δ ∈ F ,
[c, a] = [c, d] = [c, c] = [d, c] = [d, a] = 0. Moreover, Leib(L) =
[L,L] = ζ left(L) = Fd⊕ Fc, ζ(L) = ζright(L) = Fc.
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V. S. Yashchuk 305
II. L = D ⊣ B, where B = Fb, [b, b] = 0, D = Fd ⊕ Fc is an
abelian subalgebra, [d, d] = [d, c] = [c, d] = [c, c] = 0, [b, c] = d,
[b, d] = γd + δd, 0 6= γ, δ ∈ F , [c, b] = [d, b] = 0. Moreover,
Leib(L) = [L,L] = ζ left(L) = Fd⊕Fc, ζright(L) = Fb, ζ(L) = 〈0〉.
Proof. Since L is cyclic and has dimension 3, L can not be a Lie algebra.
Let a be an element of L such that L = 〈a〉. Since L is a not Lie algebra,
[a, a] = c 6= 0. Moreover, c ∈ Leib(L) 6 ζ left(L), that implies [c, x] = 0
for all x ∈ L, in particular, [c, a] = 0. Since L/Leib(L) is a cyclic Lie
algebra, that dimF (L/Leib(L)) = 1, so that dimF (Leib(L)) = 2. If we
assume that [a, c] ∈ Fc, then 〈a〉 = Fa⊕Fc, so that dimF (L) = 2, and we
obtain a contradiction. Thus d = [a, c] /∈ Fc, which follows that {c, d} is a
basis of K = Leib(L). Again [d, a] = 0. Since K is an abelian subalgebra,
[c, c] = [c, d] = [d, d] = [d, c] = 0. Since K is an ideal, [a, d] ∈ K, so that
[a, d] = γc+ δd. If γ = 0 = δ, then L is a cyclic nilpotent algebra. This
situation was considered early. Suppose now that γ = 0, δ 6= 0, then
[a, d] ∈ Fd, which follows that D = Fd is an ideal of L. Put a1 = a−δ−1c,
then
[a1, a1] = [a− δ−1c, a− δ−1c] = [a, a]− δ−1[a, c] = c− δ−1d = a2,
[a1, a2] = [a− δ−11c, c− δ−1d] = [a, c]− δ−1[a, d] = d− δ−1δd = 0,
[a1, d] = [a− δ−1c, d] = [a, d] = δd.
Thus we can see that a subalgebra A = 〈a1〉 is nilpotent and has dimension
2, D ∩ A = 〈0〉, so that L is semidirect sum of an ideal D of dimension
1 and a nilpotent subalgebra A of dimension 2. Moreover, D 6 ζ left(L),
[L,L] = ζ left(L) = Leib(L) = D ⊕ [A,A]. Thus we obtain the Leibniz
algebra of type I.
Suppose now that if γ 6= 0, δ 6= 0. Thus [a, d] = γ[a, a] + δ[a, c] and
0 = [a, γa+ δc− d]. Put b = a+ γ−1δc− γ−1d, then [a, b] = 0. Further
[b, b] = [a+ γ−1δc− γ−1d, b] = [a, b] + [γ−1δc− γ−1d, b] = 0,
so that 〈b〉 = Fb. Since [c, b] = [d, b] = 0, Fb = ζright(L). Furthermore
[b, c] = [a+ γ−1δc− γ−1d, c] = [a, c] = d,
[b, d] = [a+ γ−1δc− γ−1d, d] = [a, d] = γc+ δd.
We can see that L is semidirect sum of an abelian ideal Fc⊕Fd of dimen-
sion 2 and abelian subalgebra Fb of dimension 1. Moreover, ζ left(L) =
Fc⊕ Fd = Leib(L) = [L,L], ζright(L) = Fb. Thus we obtain the Leibniz
algebra of type II.
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306 On some Leibniz algebras, having small dimension
4. Non-nilpotent Leibniz algebras of dimension 3 with
2-dimensional Leibniz kernel over finite fields
The last case of our consideration is the case when L is non-nilpotent,
non-cyclic and dimF (Leib(L)) = 2. Then dimF (L/Leib(L)) = 1. In
particular, L/Leib(L) is abelian.
Theorem 6. Let L be a non-nilpotent non-cyclic Leibniz algebra of di-
mension 3 over finite field F . Suppose that L is a not Lie algebra and
dimF (Leib(L)) = 2. Then L is an algebra of one of the following types:
I. L = A ⊣ D, where D = Fd, [d, d] = 0, A = Fa ⊕ Fc is a cyclic
subalgebra, [a, a] = c = [a, c], [a, d] = d, [c, a] = [c, d] = [d, c] =
[d, a] = [c, c] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) = Fd⊕Fc,
ζ(L) = ζright(L) = 〈0〉.
II. char(F ) 6= 2, L = A ⊣ D, where D = Fd, [d, d] = 0, A = Fa⊕Fc is
a cyclic subalgebra, [a, a] = c = [a, c], [a, d] = c+ 2d, [c, a] = [c, d] =
[d, c] = [d, a] = 0. Moreover, Leib(L) = [L,L] = ζ left(L) = Fd⊕Fc,
ζ(L) = ζright(L) = 〈0〉.
Proof. Put K = Leib(L). Since L is a not Lie algebra, there is an element
a such that [a, a] = c 6= 0. Then a /∈ K, so that L = K ⊕ Fa. An
inclusion Leib(L) 6 ζ left(L) implies that [c, y] = 0 for each element
y ∈ L, in particular, [c, a] = 0. For an element [a, c] we have the following
possibilities: [a, c] = 0, [a, c] ∈ Fc, [a, c] /∈ Fc.
Consider the first situation. Choose an element d ∈ K such that
K = Fc ⊕ Fd. Since K is abelian, [d, d] = 0 = [c, d] = [d, c]. It follows
that c ∈ ζ(L). We have [d, a] = 0 and [a, d] = γc + δd. If γ = 0, then
a subalgebra 〈d〉 = Fd is an ideal. In this case we obtain that L = D⊕A,
where D = Fd is an abelian ideal, D 6 ζ left(L), A is a nilpotent cyclic
subalgebra, A = Fa⊕Fc, where [a, a] = c, [L,L] = D⊕ [A,A] = ζ left(L),
is an abelian ideal. Thus we obtain the Leibniz algebra of above considered
Type I from Theorem 5.
Assume now that γ 6= 0. Put d1 = γ−1d, then
[a, d1] = γ−1[a, d] = γ−1(γc+ δd) = c+ δd1.
If we suppose that δ = 0, then again L/ζ(L) is abelian and L is nilpotent.
This case has been considered above. Therefore we assume that δ 6= 0. We
have
[a+ d1, a+ d1] = [a, a] + [a, d1] = c+ c+ δd1 = 2c+ δd1.
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V. S. Yashchuk 307
If char(F ) = 2, then Fd1 6 〈a+ d1〉, which follows that a ∈ 〈a+ d1〉 and
c = [a, a] ∈ 〈a+d1〉. Thus we can see, that in this case L is a cyclic algebra.
Suppose that char(F ) 6= 2. Then from δ(a+d1)−[a+d1, a+d1] ∈ 〈a+d1〉,
we obtain δa−2c ∈ 〈a+d1〉. It follows that δ2c = [δa−2c, δa−2c] ∈ 〈a+d1〉,
Since δ2 6= 0, Fc 6 〈a+ d1〉. In turn out it follows that Fd1 6 〈a+ d1〉,
and again we obtain that L is a cyclic.
Suppose now that 0 6= [a, c] ∈ Fc. Then a subalgebra 〈a〉 has dimension
2 and is non-nilpotent. Therefore we can suppose that [a, c] = c (see, for
example, a survey [12]). Choose an element d ∈ K such that K = Fc⊕Fd.
Since K is abelian, [d, d] = 0 = [c, d] = [d, c]. We have [d, a] = 0 and
[a, d] = γc + δd. If γ = 0, then a subalgebra 〈d〉 = Fd is an ideal. In
this case we obtain that L = D ⊕ A, where D = Fd is an abelian ideal,
D 6 ζ left(L), A is a non-nilpotent cyclic subalgebra of dimension 2,
A = Fa⊕ Fc, where [a, a] = c = [a, c], [L,L] = D ⊕ [A,A] = ζ left(L), is
an abelian ideal. Thus we obtain the Leibniz algebra of type I.
Assume now that γ 6= 0. Put d1 = γ−1d, then
[a, d1] = γ−1[a, d] = γ−1(γc+ δd) = c+ δd1.
If we suppose that δ = 0, then [a, d1] ∈ Fc, which implies that L/Fc
is abelian. In turn out it follows that Leib(L) 6 Fc. But in this case
dimF (Leib(L)) = 1, and we obtain a contradiction. This contradiction
shows δ 6= 0. We have
[a+ d1, a+ d1] = [a, a] + [a, d1] = c+ c+ δd1 = 2c+ δd1.
If char(F ) = 2, then Fd1 6 〈a + d1〉, which follows that a ∈ 〈a + d1〉
and c = [a, a] ∈ 〈a+ d1〉. Thus we can see, that in this case L is a cyclic
algebra.
Suppose that char(F ) 6= 2. Then from δ(a+ d1)− [a+ d1, a+ d1] ∈
〈a+ d1〉, we obtain δa− 2c ∈ 〈a+ d1〉. It follows that
(δ2 − 2δ)c = δ2c− 2δc = [δa− 2c, δa− 2c] ∈ 〈a+ d1〉.
By above δ 6= 0. If δ 6= 2, then Fc 6 〈a+ d1〉. In turn out it follows
that Fd1 6 〈a + d1〉, and again we obtain that L is a cyclic. Suppose
that δ = 2, that is [a, d1] = c+ 2d1. Hence, if L is not cyclic, then L is a
Leibniz algebra of a type II.
Suppose now that [a, c] /∈ Fc. In this case [a, c] = αc + βd besides
β 6= 0. We can see that a Leibniz algebra L is cyclic. This case has been
considered above.
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308 On some Leibniz algebras, having small dimension
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Contact information
V. S. Yashchuk Department of Geometry and Algebra, Faculty
of Mechanics and Mathematics, Oles Honchar
Dnipro National University, Gagarin ave., 72,
Dnipro, 49010,Ukraine
E-Mail(s): Viktoriia.S.Yashchuk@gmail.com
Received by the editors: 28.02.2018
and in final form 22.03.2018.
|
| id | nasplib_isofts_kiev_ua-123456789-188439 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T17:47:25Z |
| publishDate | 2019 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Yashchuk, V.S. 2023-03-01T15:54:47Z 2023-03-01T15:54:47Z 2019 On some Leibniz algebras, having small dimension / V.S. Yashchuk // Algebra and Discrete Mathematics. — 2019. — Vol. 27, № 2. — С. 292–308. — Бібліогр.: 16 назв. — англ. 1726-3255 2010 MSC: 17A32, 17A60 https://nasplib.isofts.kiev.ua/handle/123456789/188439 The first step in the study of all types of algebras is the description of such algebras having small dimensions. The structure of 3-dimensional Leibniz algebras is more complicated than 1 and 2-dimensional cases. In this paper, we consider the structure of Leibniz algebras of dimension 3 over the finite fields. In some cases, the structure of the algebra essentially depends on the characteristic of the field, in others on the solvability of specific equations in the field, and so on. The author would like to express her sincere thanks to her scientific advisor Professor L. A. Kurdachenko. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On some Leibniz algebras, having small dimension Article published earlier |
| spellingShingle | On some Leibniz algebras, having small dimension Yashchuk, V.S. |
| title | On some Leibniz algebras, having small dimension |
| title_full | On some Leibniz algebras, having small dimension |
| title_fullStr | On some Leibniz algebras, having small dimension |
| title_full_unstemmed | On some Leibniz algebras, having small dimension |
| title_short | On some Leibniz algebras, having small dimension |
| title_sort | on some leibniz algebras, having small dimension |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/188439 |
| work_keys_str_mv | AT yashchukvs onsomeleibnizalgebrashavingsmalldimension |