# On the inner structure of 3-dimensional Leibniz algebras

L.A. Kurdachenko, O.O. Pypka
Oles Honchar Dnipro National University, Dnipro, Ukraine
e-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
I.Ya. Subbotin
National University, Los Angeles, USA
e-mail<EMAIL_ADDRESS>
###### Abstract

Let $L$ be an algebra over a field $F$ with the binary operations $+$ and
$[,]$. Then $L$ is called a left Leibniz algebra if
$[[a,b],c]=[a,[b,c]]-[b,[a,c]]$ for all $a,b,c\in L$. We describe the inner
structure of left Leibniz algebras having dimension 3.

Key Words: Leibniz algebra, nilpotent Leibniz algebra, dimension.

2020 MSC: 17A32, 17A60, 17A99.

##  1 Introduction

Let $L$ be an algebra over a field $F$ with the binary operations $+$ and
$[,]$. Then, $L$ is called a Leibniz algebra (more precisely, a left Leibniz
algebra) if, for all elements $a,b,c\in L$, it satisfies the Leibniz identity:

$[[a,b],c]=[a,[b,c]]-[b,[a,c]].$

We will also use another form of this identity:

$[a,[b,c]]=[[a,b],c]+[b,[a,c]].$

Leibniz algebras first appeared in the paper of A. Blokh [2] where they were
called $D$-algebras. However, at that time, these works were not in demand.
Only after two decades was there a real rise in interest towards Leibniz
algebras. It happened thanks to the rediscovery of these algebras by J.-L.
Loday [9] (see also [8, Section 10.6]), who used the term “Leibniz algebras”
since it was Leibniz who discovered and proved the Leibniz rule for
differentiation of the product of functions. The main motivation for the
introduction of Leibniz algebras was the study of periodicity phenomena in
algebraic K-theory. The Leibniz algebras appeared to be naturally related to
several areas such as differential geometry, homological algebra, classical
algebraic topology, algebraic K-theory, loop spaces, non-commutative geometry,
physics and so on. Nowadays, the theory of Leibniz algebras is one of actively
developing areas of modern algebra.

Note that Lie algebras are a partial case of Leibniz algebras. Conversely, if
$L$ is a Leibniz algebra in which $[a,a]=0$ for every element $a\in L$, then
it is a Lie algebra. Thus, Leibniz algebras can be seen as a non-commutative
generalization of Lie algebras.

The theory of Leibniz algebras has been intensely developing in many different
directions. Some results of this theory were presented in the recent book [1].

One of the first steps in the theory of Leibniz algebras is the description of
algebras with small dimensions. Unlike Lie algebras, the situation with
Leibniz algebras of dimension 3 is very diverse. Leibniz algebras of dimension
3 are mostly described. The description of Leibniz algebras of dimensions 4
and 5 is quite complex. The list of papers devoted to these studies is quite
large and we will not give it here in full. We only note that the Section 3.1
of book [1] is devoted to study of right Leibniz algebras having dimension 3.
The investigation of Leibniz algebras having dimension 3 was carried out in
articles [3, 4, 5, 10, 11, 12]. Some of these papers use the language of the
right Leibniz algebras, whilst others use the language of left Leibniz
algebras. Basically, the description is reduced to determining the structural
constants of these algebras. However, the structural constants do not always
give an idea of the internal structure of these algebras. Elucidation of the
structure requires some additional analysis. The overall picture seems
fragmented. Thus, the articles dealt with Leibniz algebras over concrete
fields of real, complex, $p$-adic numbers, etc. Therefore, we cannot decide if
the internal structure of these algebras really contains a complete
description. For example, when passing from the field of rational numbers to
the field of real numbers, some types of algebras disappear altogether.
Furthermore, some sections describe right Leibniz algebras, while others
describe left Leibniz algebras. Moreover, only structural constants were
found. None of the articles considered the internal structure. So, under these
circumstances, in order to observe the entire scope, there are two options
that arise: analyze these articles and really make sure everything is done
there or do it yourself all over again. The second option is preferable since
it is better to complete the list within your own framework if it turns out
that not all types of algebras have been considered. More importantly, we are
interested in the description of the inner structure, not just structural
constants. Of course, you can extract some information about the structure
from structural constants, but it is more logical to obtain the structural
constants in the process of describing the inner structure.

Therefore, in the current article, we present a description of left Leibniz
algebras having dimension 3, focusing on clarifying their structure and
obtaining structural constants by passing. This consideration of the structure
of Leibniz algebras of dimension 3 is carried out over an arbitrary field $F$,
and when studying these specific types of algebras, additional natural
restrictions on the field $F$ appear. These restrictions are very significant
in some cases. Some types of algebras can exist only if sufficiently strict
restrictions are imposed. Our goal was the most detailed description of these
algebras, reflecting all the nuances of their structure.

##  2 Main results.

Let $L$ be a Leibniz algebra over a field $F$. Then $L$ is called abelian if
$[a,b]=0$ for every elements $a,b\in L$. In particular, an abelian Leibniz
algebra is a Lie algebra.

If $A,B$ are subspaces of $L$, then $[A,B]$ will denote a subspace generated
by all elements $[a,b]$ where $a\in A$, $b\in B$. As usual, a subspace $A$ of
$L$ is called a subalgebra of $L$, if $[a,b]\in A$ for every $a,b\in A$. It
follows that $[A,A]\leqslant A$. A subalgebra $A$ of $L$ is called a left
(respectively right) ideal of $L$, if $[b,a]\in A$ (respectively $[a,b]\in A$)
for every $a\in A$, $b\in L$. In other words, if $A$ is a left (respectively
right) ideal, then $[L,A]\leqslant A$ (respectively $[A,L]\leqslant A$). 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.

Every Leibniz algebra $L$ possesses the following specific ideal. Denote by
$\mathrm{\mathbf{Leib}}(L)$ the subspace generated by the elements $[a,a]$,
$a\in L$. It is not hard to prove that $\mathrm{\mathbf{Leib}}(L)$ is an ideal
of $L$. The ideal $\mathrm{\mathbf{Leib}}(L)$ is called the Leibniz kernel of
algebra $L$.

We note the following important property of the Leibniz kernel:

$[[a,a],x]=[a,[a,x]]-[a,[a,x]]=0.$

The left (respectively right) center $\zeta^{\mathrm{left}}(L)$ (respectively
$\zeta^{\mathrm{right}}(L)$) of a Leibniz algebra $L$ is defined by the rule:

$\zeta^{\mathrm{left}}(L)=\\{x\in L|\ [x,y]=0\ \mbox{for each element }y\in
L\\}$

(respectively,

$\zeta^{\mathrm{right}}(L)=\\{x\in L|\ [y,x]=0\ \mbox{for each element }y\in
L\\}).$

It is not hard to prove that the left center of $L$ is an ideal, but that is
not true for the right center. Moreover,
$\mathrm{\mathbf{Leib}}(L)\leqslant\zeta^{\mathrm{left}}(L)$ so that
$L/\zeta^{\mathrm{left}}(L)$ is a Lie algebra. The right center is a
subalgebra of $L$ and, in general, the left and right centers are distinct
(see, for example, [7]).

The center $\zeta(L)$ of $L$ is defined by the rule:

$\zeta(L)=\\{x\in L|\ [x,y]=0=[y,x]\ \mbox{for each element }y\in L\\}.$

The center is an ideal of $L$.

Now we define the upper central series

$\langle
0\rangle=\zeta_{0}(L)\leqslant\zeta_{1}(L)\leqslant\ldots\zeta_{\alpha}(L)\leqslant\zeta_{\alpha+1}(L)\leqslant\ldots\zeta_{\eta}(L)$

of a Leibniz algebra $L$ by the following rule: $\zeta_{1}(L)=\zeta(L)$ is the
center of $L$, and recursively,
$\zeta_{\alpha+1}(L)/\zeta_{\alpha}(L)=\zeta(L/\zeta_{\alpha}(L))$ for all
ordinals $\alpha$, and
$\zeta_{\lambda}(L)=\bigcup_{\mu<\lambda}\zeta_{\mu}(L)$ for the limit
ordinals $\lambda$. By definition, each term of this series is an ideal of
$L$.

Define the lower central series of $L$

$L=\gamma_{1}(L)\geqslant\gamma_{2}(L)\geqslant\ldots\gamma_{\alpha}(L)\geqslant\gamma_{\alpha+1}(L)\geqslant\ldots\gamma_{\tau}(L)$

by the rule: $\gamma_{1}(L)=L$, $\gamma_{2}(L)=[L,L]$,
$\gamma_{\alpha+1}(L)=[L,\gamma_{\alpha}(L)]$ for all ordinals $\alpha$ and
$\gamma_{\lambda}(L)=\bigcap_{\mu<\lambda}\gamma_{\mu}(L)$ for the limit
ordinals $\lambda$.

As usual, we say that a Leibniz algebra $L$ is nilpotent, if there exists a
positive integer $k$ such that $\gamma_{k}(L)=\langle 0\rangle$. More
precisely, $L$ is said to be nilpotent of nilpotency class $c$ if
$\gamma_{c+1}(L)=\langle 0\rangle$, but $\gamma_{c}(L)\neq\langle 0\rangle$.
We denote the nilpotency class of $L$ by $\mathrm{\mathbf{ncl}}(L)$.

Define the lower derived series of $L$

$L=\delta_{0}(L)\geqslant\delta_{1}(L)\geqslant\ldots\delta_{\alpha}(L)\geqslant\delta_{\alpha+1}(L)\geqslant\ldots\delta_{\nu}(L)$

by the following rule: $\delta_{0}(L)=L$, $\delta_{1}(L)=[L,L]$, and
recursively $\delta_{\alpha+1}(L)=[\delta_{\alpha}(L),\delta_{\alpha}(L)]$ for
all ordinals $\alpha$ and
$\delta_{\lambda}(L)=\bigcap_{\mu<\lambda}\delta_{\mu}(L)$ for the limit
ordinals $\lambda$. If $\delta_{n}(L)=\langle 0\rangle$ for some positive
integer $n$, then we say that $L$ is a soluble Leibniz algebra.

As usual, we say that a Leibniz algebra $L$ is finite dimensional if the
dimension of $L$ as a vector space over $F$ is finite.

If $\mathrm{\mathbf{dim}}_{F}(L)=1$, then $L$ is abelian.

If $\mathrm{\mathbf{dim}}_{F}(L)=2$, then we obtain the following types of
Leibniz algebras:

$\begin{split}\mathrm{\mathbf{Lei}}_{1}(2,F)&=Fa\oplus Fb\ \mbox{ where
}[a,a]=b,[a,b]=[b,a]=[b,b]=0;\\\ \mathrm{\mathbf{Lei}}_{2}(2,F)&=Fc\oplus Fd\
\mbox{ where }[c,c]=[c,d]=d,[d,c]=[d,d]=0\end{split}$

(see, for example, [6]).

Moving on to Leibniz algebras of dimension 3, we immediately note that we will
consider Leibniz algebras, which are not Lie algebras. This means that their
Leibniz kernel is non-zero. Then, the factor-algebra over Leibniz kernel has
dimension, at most, 2. Note that Lie algebras having dimension at most 2 are
soluble. Thus, we obtain the following

######  Proposition 2.1.

Let $L$ be a Leibniz algebra over a field $F$. Suppose that $L$ is not a Lie
algebra. If $L$ has dimension $3$, then $L$ is soluble.

For the Leibniz kernel $\mathrm{\mathbf{Leib}}(L)$ of a Leibniz but not a Lie
algebra, $L$ having dimension 3 will give us only two possibilities:
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$ and
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$.

First, we will consider the situation when
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$. Immediately, we
obtain the following two subcases:

(IA) the center of $L$ includes $\mathrm{\mathbf{Leib}}(L)$;

(IB) the Leibniz kernel of $L$ is not central.

For each of these subcases, we have the following two possibilities:

(IA1) the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is abelian;

(IA2) the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is not abelian;

and

(IB1) the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is abelian;

(IB2) the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is not abelian.

Consider these cases.

######  Theorem 2.2.

Let $L$ be a Leibniz algebra over a field $F$ having dimension $3$. Suppose
that $L$ is not a Lie algebra. If the center of $L$ includes the Leibniz
kernel, $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$ and the
factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is abelian, then $L$ is an
algebra of one of the following types.

  1. (i)

$\mathrm{\mathbf{Lei}}_{3}(3,F)=L_{3}$ is a direct sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}$. Moreover, $A$ is a nilpotent cyclic
Leibniz algebra of dimension $2$, $[A,B]=[B,A]=\langle 0\rangle$, so that
$L_{3}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{3})=[L_{3},L_{3}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{3})=\zeta(L_{3})=\zeta^{\mathrm{right}}(L_{3})=Fa_{2}\oplus
Fa_{3}$, $L_{3}$ is nilpotent and $\mathrm{\mathbf{ncl}}(L_{3})=2$.

  2. (ii)

$\mathrm{\mathbf{Lei}}_{4}(3,F)=L_{4}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=Fa_{3}$,
$[B,A]=\langle 0\rangle$, so that $L_{4}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=[a_{1},a_{2}]=a_{3}$,
$[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{4})=[L_{4},L_{4}]=\zeta^{\mathrm{right}}(L_{4})=\zeta(L_{4})=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{4})=Fa_{2}\oplus Fa_{3}$, $L_{4}$ is nilpotent and
$\mathrm{\mathbf{ncl}}(L_{4})=2$.

  3. (iii)

$\mathrm{\mathbf{Lei}}_{5}(3,F)=L_{5}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=\langle 0\rangle$,
$[B,A]=Fa_{3}$, so that $L_{5}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{2},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{5})=[L_{5},L_{5}]=\zeta^{\mathrm{left}}(L_{5})=\zeta(L_{5})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{5})=Fa_{2}\oplus Fa_{3}$, $L_{5}$ is nilpotent and
$\mathrm{\mathbf{ncl}}(L_{5})=2$.

  4. (iv)

$\mathrm{\mathbf{Lei}}_{6}(3,F)=L_{6}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=[B,A]=Fa_{3}$, so
that $L_{6}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{2},a_{1}]=a_{3}$, $[a_{1},a_{2}]=\alpha a_{3}$ $(\alpha\neq
0)$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{6})=[L_{6},L_{6}]=\zeta^{\mathrm{right}}(L_{6})=\zeta^{\mathrm{left}}(L_{6})=\zeta(L_{6})=Fa_{3}$,
$L_{6}$ is nilpotent and $\mathrm{\mathbf{ncl}}(L_{6})=2$.

  5. (v)

$\mathrm{\mathbf{Lei}}_{7}(3,F)=L_{7}$ is a sum of two nilpotent cyclic ideals
$A=Fa_{1}\oplus Fa_{3}$ and $C=Fa_{2}\oplus Fa_{3}$, $[A,C]=[C,A]=\langle
0\rangle$, so that $L_{7}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=a_{3}$, $[a_{2},a_{2}]=\beta a_{3}$ $(\beta\neq 0)$,
$[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{7})=[L_{7},L_{7}]=\zeta^{\mathrm{right}}(L_{7})=\zeta^{\mathrm{left}}(L_{7})=\zeta(L_{7})=Fa_{3}$.
Moreover, polynomial $X^{2}+\beta$ has no root in field $F$, $L_{7}$ is
nilpotent and $\mathrm{\mathbf{ncl}}(L_{7})=2$.

  6. (vi)

$\mathrm{\mathbf{Lei}}_{8}(3,F)=L_{8}$ is a sum of two nilpotent cyclic ideals
$A=Fa_{1}\oplus Fa_{3}$ and $C=Fa_{2}\oplus Fa_{3}$, $[A,C]=Fa_{3}$,
$[C,A]=\langle 0\rangle$, so that $L_{8}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$,
$[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=\alpha a_{3}$, $[a_{2},a_{2}]=\beta
a_{3}$ $(\alpha,\beta\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{8})=[L_{8},L_{8}]=\zeta^{\mathrm{right}}(L_{8})=\zeta^{\mathrm{left}}(L_{8})=\zeta(L_{8})=Fa_{3}$.
Moreover, polynomial $X^{2}+\alpha X+\beta$ has no root in field $F$, $L_{8}$
is nilpotent and $\mathrm{\mathbf{ncl}}(L_{8})=2$.

  7. (vii)

$\mathrm{\mathbf{Lei}}_{9}(3,F)=L_{9}$ is a sum of two nilpotent cyclic ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$ such that $[A,B]=\langle
0\rangle$, $[B,A]=Fa_{3}$, so that $L_{9}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=a_{3}$, $[a_{2},a_{1}]=a_{3}$, $[a_{2},a_{2}]=\sigma
a_{3}\ (\sigma\neq 0)$,
$[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{9})=[L_{9},L_{9}]=\zeta^{\mathrm{right}}(L_{9})=\zeta(L_{9})=\zeta^{\mathrm{left}}(L_{9})=Fa_{3}$.
Moreover, polynomial $X^{2}+X+\sigma$ has no root in field $F$, $L_{9}$ is
nilpotent and $\mathrm{\mathbf{ncl}}(L_{9})=2$.

  8. (viii)

$\mathrm{\mathbf{Lei}}_{10}(3,F)=L_{10}$ is a sum of two nilpotent cyclic
ideals $A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$ such that
$[A,B]=[B,A]=Fa_{3}$, so that $L_{10}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=a_{3}$, $[a_{2},a_{1}]=a_{3}$, $[a_{1},a_{2}]=\tau a_{3}\
(\tau\neq 0)$, $[a_{2},a_{2}]=\sigma a_{3}\ (\sigma\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{10})=[L_{10},L_{10}]=\zeta^{\mathrm{right}}(L_{10})=\zeta^{\mathrm{left}}(L_{10})=\zeta(L_{10})=Fa_{3}$.
Moreover, polynomial $X^{2}+(\tau+1)X+\sigma$ has no root in field $F$,
$L_{10}$ is nilpotent and $\mathrm{\mathbf{ncl}}(L_{10})=2$.

Proof. We note that the center $\zeta(L)$ has dimension at most 2. Suppose
first that $\mathrm{\mathbf{dim}}_{F}(\zeta(L))=2$. Since $L$ is not a Lie
algebra, there is an element $a_{1}$ such that $[a_{1},a_{1}]=a_{3}\neq 0$. We
note that $a_{3}\in\zeta(L)$. It follows that
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{3}]=0$. Being an abelian algebra of
dimension 2, $\zeta(L)$ has a direct decomposition $\zeta(L)=Fa_{2}\oplus
Fa_{3}$ for some element $a_{2}$. Put $A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}$,
then $B\leqslant\zeta(L)$, so that $B$ is an ideal of $L$. Clearly,
$L=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}=A\oplus B$ and $A$ is also an ideal of
$L$. Moreover, $A$ is a nilpotent cyclic Leibniz algebra of dimension 2. Thus,
we come to the following type of nilpotent Leibniz algebras:

$\displaystyle L_{3}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{1},a_{2}]=$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that $\mathrm{\mathbf{Leib}}(L_{3})=[L_{3},L_{3}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{3})=\zeta^{\mathrm{right}}(L_{3})=\zeta(L_{3})=Fa_{2}\oplus
Fa_{3}$, $\mathrm{\mathbf{ncl}}(L_{3})=2$.

Suppose now that the center of $L$ has dimension 1. In this case,
$\zeta(L)=\mathrm{\mathbf{Leib}}(L)$. Since $L$ is not a Lie algebra, there is
an element $a_{1}$ such that $[a_{1},a_{1}]=a_{3}\neq 0$. We note that
$a_{3}\in\zeta(L)$. It follows that
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{3}]=0$. Then $\zeta(L)=Fa_{3}$. Since
$L/\mathrm{\mathbf{Leib}}(L)$ is abelian, for every element $x\in L$ we have
$[a_{1},x],[x,a_{1}]\in\zeta(L)\leqslant\langle a_{1}\rangle=Fa_{1}\oplus
Fa_{3}$. It follows that subalgebra $\langle a_{1}\rangle$ is an ideal of $L$.
Since $\mathrm{\mathbf{dim}}_{F}(\langle a_{1}\rangle)=2$, $\langle
a_{1}\rangle\neq L$.

Suppose first that there exists an element $b$ such that $b\not\in\langle
a_{1}\rangle$ and $[b,b]=0$. We have $[b,a_{1}]=\gamma a_{3}$ for some
$\gamma\in F$. The following two cases appear here: $\gamma=0$ and $\gamma\neq
0$. Let $\gamma=0$. Then $[a_{1},b]=\alpha a_{3}$ for some $\alpha\in F$. If
we suppose that $\alpha=0$, then $b\in\zeta(L)$. But in this case,
$\mathrm{\mathbf{dim}}_{F}(\zeta(L))=2$, and we obtain a contradiction, which
shows that $\alpha\neq 0$. Put $a_{2}=\alpha^{-1}b$, then
$[a_{2},a_{2}]=[a_{2},a_{1}]=0$, $[a_{1},a_{2}]=a_{3}$, and we come to the
following nilpotent Leibniz algebra:

$\displaystyle L_{4}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{2}]=a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that $\mathrm{\mathbf{Leib}}(L_{4})=[L_{4},L_{4}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{4})=Fa_{2}\oplus Fa_{3}$,
$\zeta(L_{4})=\zeta^{\mathrm{right}}(L_{4})=Fa_{3}$,
$\mathrm{\mathbf{ncl}}(L_{4})=2$.

Let $\gamma\neq 0$. Put $a_{2}=\gamma^{-1}b$, then $[a_{2},a_{2}]=0$,
$[a_{2},a_{1}]=a_{3}$. We have $[a_{1},a_{2}]=\alpha a_{3}$ for some element
$\alpha\in F$. If $\alpha=0$, then $a_{2}\in\zeta^{\mathrm{right}}(L)$, and we
come to the following nilpotent Leibniz algebra:

$\displaystyle L_{5}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{2},a_{1}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that $\mathrm{\mathbf{Leib}}(L_{5})=[L_{5},L_{5}]=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{5})=Fa_{2}\oplus Fa_{3}$,
$\zeta(L_{5})=\zeta^{\mathrm{left}}(L_{5})=Fa_{3}$,
$\mathrm{\mathbf{ncl}}(L_{5})=2$.

Suppose that $\alpha\neq 0$. Then, we come to the following type of nilpotent
Leibniz algebras:

$\displaystyle L_{6}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{2},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=\alpha a_{3}\
(\alpha\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{6})=[L_{6},L_{6}]=\zeta^{\mathrm{right}}(L_{6})=\zeta^{\mathrm{left}}(L_{6})=\zeta(L_{6})=Fa_{3}$,
$\mathrm{\mathbf{ncl}}(L_{6})=2$.

Suppose now that $[b,b]\neq 0$ for every element $b\not\in\langle
a_{1}\rangle$. In particular, it follows that $b\not\in\zeta(L)$. Put
$[b,b]=\beta a_{3}$ where $\beta\in F$. We have $[b,a_{1}]=\gamma a_{3}$ and
$[a_{1},b]=\alpha a_{3}$ for some elements $\alpha,\gamma\in F$. If
$\alpha=\gamma=0$, then put $a_{2}=b$ and denote by $C$ the subalgebra
generates by $a_{2}$. Then $C$ is a cyclic nilpotent ideal such that
$[A,C]=[C,A]=\langle 0\rangle$. Furthermore, let $u=\lambda a_{1}+\mu
a_{2}+\nu a_{3}$ be the arbitrary element of $L$. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=\lambda^{2}[a_{1},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}+\beta\mu^{2}a_{3}=(\lambda^{2}+\beta\mu^{2})a_{3}.$

If $u\not\in\langle a_{1}\rangle$, then $(\lambda,\mu)\neq(0,0)$. It follows
that polynomial $X^{2}+\beta$ has no root in field $F$. Thus, we come to the
following type of nilpotent Leibniz algebras:

$\displaystyle L_{7}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{2},a_{2}]=\beta a_{3}\ (\beta\neq 0),$
$\displaystyle[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{7})=[L_{7},L_{7}]=\zeta^{\mathrm{right}}(L_{7})=\zeta^{\mathrm{left}}(L_{7})=\zeta(L_{7})=Fa_{3}$.
Moreover, polynomial $X^{2}+\beta$ has no root in field $F$,
$\mathrm{\mathbf{ncl}}(L_{7})=2$.

Suppose now that $\gamma=0$ and $\alpha\neq 0$. Put $a_{2}=b$. Then,
$[a_{2},a_{1}]=0$, $[a_{1},a_{2}]=\alpha a_{3}$, $[a_{2},a_{2}]=\beta a_{3}$.
Let $\lambda a_{1}+\mu a_{2}+\nu a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},a_{2}]+\lambda\mu[a_{2},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu\alpha a_{3}+\mu^{2}\beta
a_{3}=(\lambda^{2}+\lambda\mu\alpha+\mu^{2}\beta)a_{3}.$

As above, $\lambda\neq 0$, $\mu\neq 0$. It follows that polynomial
$X^{2}+\alpha X+\beta$ has no root in field $F$. Thus, we come to the
following type of nilpotent Leibniz algebras:

$\displaystyle L_{8}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=\alpha a_{3}\ (\alpha\neq
0),[a_{2},a_{2}]=\beta a_{3}\ (\beta\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{8})=[L_{8},L_{8}]=\zeta^{\mathrm{right}}(L_{8})=\zeta^{\mathrm{left}}(L_{8})=\zeta(L_{8})=Fa_{3}$.
Moreover, polynomial $X^{2}+\alpha X+\beta$ has no root in field $F$,
$\mathrm{\mathbf{ncl}}(L_{8})=2$.

Suppose now that $\gamma\neq 0$ and $\alpha=0$. Put $a_{2}=\gamma^{-1}b$.
Then, $[a_{2},a_{1}]=a_{3}$, $[a_{1},a_{2}]=0$,
$[a_{2},a_{2}]=\gamma^{-2}\beta a_{3}=\sigma a_{3}$. Let $\lambda a_{1}+\mu
a_{2}+\nu a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{2},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu a_{3}+\mu^{2}\sigma
a_{3}=(\lambda^{2}+\lambda\mu+\mu^{2}\sigma)a_{3}.$

As above, $\lambda\neq 0$, $\mu\neq 0$. It follows that polynomial
$X^{2}+X+\sigma$ has no root in field $F$. Thus, we come to the following type
of nilpotent Leibniz algebras:

$\displaystyle L_{9}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{2},a_{1}]=a_{3},$ $\displaystyle[a_{2},a_{2}]=\sigma
a_{3}\ (\sigma\neq 0),$
$\displaystyle[a_{1},a_{2}]=[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{9})=[L_{9},L_{9}]=\zeta^{\mathrm{right}}(L_{9})=\zeta^{\mathrm{left}}(L_{9})=\zeta(L_{9})=Fa_{3}$.
Moreover, polynomial $X^{2}+X+\sigma$ has no root in field $F$,
$\mathrm{\mathbf{ncl}}(L_{9})=2$.

Suppose now that $\gamma\neq 0$ and $\alpha\neq 0$. Put $a_{2}=\gamma^{-1}b$.
Then, $[a_{2},a_{1}]=a_{3}$, $[a_{1},a_{2}]=\gamma^{-1}\alpha a_{3}=\tau
a_{3}$, $[a_{2},a_{2}]=\gamma^{-2}\beta a_{3}=\sigma a_{3}$. Let $\lambda
a_{1}+\mu a_{2}+\nu a_{3}$ be an arbitrary element of L. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},a_{2}]+\lambda\mu[a_{2},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu\tau a_{3}+\lambda\mu
a_{3}+\mu^{2}\sigma
a_{3}=(\lambda^{2}+\lambda\mu(\tau+1)+\mu^{2}\sigma)a_{3}.$

As above, $\lambda\neq 0$, $\mu\neq 0$. It follows that polynomial
$X^{2}+(\tau+1)X+\sigma$ has no root in field $F$. Thus, we come to the
following type of nilpotent Leibniz algebras:

$\displaystyle L_{10}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{2},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=\tau
a_{3}\ (\tau\neq 0),[a_{2},a_{2}]=\sigma a_{3}\ (\sigma\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{10})=[L_{10},L_{10}]=\zeta^{\mathrm{right}}(L_{10})=\zeta^{\mathrm{left}}(L_{10})=\zeta(L_{10})=Fa_{3}$.
Moreover, polynomial $X^{2}+(\tau+1)X+\sigma$ has no root in field $F$,
$\mathrm{\mathbf{ncl}}(L_{10})=2$. $\Box$

######  Theorem 2.3.

Let $L$ be a Leibniz algebra of dimension $3$ over a field $F$. Suppose that
$L$ is not a Lie algebra. If the center of $L$ includes the Leibniz kernel,
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$ and factor-algebra
$L/\mathrm{\mathbf{Leib}}(L)$ is non-abelian, then $L$ is an algebra of one of
the following types.

  1. (i)

$\mathrm{\mathbf{Lei}}_{11}(3,F)=L_{11}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=[B,A]=Fa_{1}$, so
that $L_{11}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{11})=\zeta^{\mathrm{left}}(L_{11})=\zeta^{\mathrm{right}}(L_{11})=\zeta(L_{11})=Fa_{3}$,
$[L_{11},L_{11}]=Fa_{1}\oplus Fa_{3}$, $L_{11}$ is non-nilpotent.

  2. (ii)

$\mathrm{\mathbf{Lei}}_{12}(3,F)=L_{12}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, so that $L_{12}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}-\alpha a_{3}$,
$[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{12})=\zeta^{\mathrm{left}}(L_{12})=\zeta^{\mathrm{right}}(L_{12})=\zeta(L_{12})=Fa_{3}$,
$[L_{12},L_{12}]=Fa_{1}\oplus Fa_{3}$, $L_{12}$ is non-nilpotent.

  3. (iii)

$\mathrm{\mathbf{Lei}}_{13}(3,F)=L_{13}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A$, $B$ are
nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}$, so
that $L_{13}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}$, $[a_{2},a_{2}]=\gamma a_{3}$
$(\gamma\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{13})=\zeta^{\mathrm{left}}(L_{13})=\zeta^{\mathrm{right}}(L_{13})=\zeta(L_{13})=Fa_{3}$,
$[L_{13},L_{13}]=Fa_{1}\oplus Fa_{3}$. Moreover, polynomial $X^{2}+\gamma$ has
no root in field $F$, $L_{13}$ is non-nilpotent.

  4. (iv)

$\mathrm{\mathbf{Lei}}_{14}(3,F)=L_{14}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A,B$ are nilpotent
cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus Fa_{3}$,
so that $L_{14}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}-\alpha a_{3}$, $[a_{2},a_{1}]=a_{1}+\alpha a_{3}\
(\alpha\neq 0)$, $[a_{2},a_{2}]=\gamma a_{3}\ (\gamma\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{14})=\zeta^{\mathrm{left}}(L_{14})=\zeta^{\mathrm{right}}(L_{14})=\zeta(L_{14})=Fa_{3}$,
$[L_{14},L_{14}]=Fa_{1}\oplus Fa_{3}$. Moreover, polynomial $X^{2}+\gamma$ has
no root in field $F$, $L_{14}$ is non-nilpotent.

  5. (v)

$\mathrm{\mathbf{Lei}}_{15}(3,F)=L_{15}$ is a direct sum of ideal $B=Fa_{2}$
and a subalgebra $A=Fa_{1}\oplus Fa_{3}$. Moreover, $A$ is a nilpotent cyclic
Leibniz algebra of dimension $2$, $[A,B]=[B,A]=Fa_{2}$, so that
$L_{15}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=a_{2}$, $[a_{2},a_{1}]=-a_{2}$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{15})=\zeta^{\mathrm{left}}(L_{15})=\zeta^{\mathrm{right}}(L_{15})=\zeta(L_{15})=Fa_{3}$,
$[L_{15},L_{15}]=Fa_{2}\oplus Fa_{3}$, $L_{15}$ is non-nilpotent.

  6. (vi)

$\mathrm{\mathbf{Lei}}_{16}(3,F)=L_{16}$ is a sum of abelian ideal
$B=Fa_{2}\oplus Fa_{3}$ and a subalgebra $A=Fa_{1}\oplus Fa_{3}$. Moreover,
$A$ is a nilpotent cyclic Leibniz algebra of dimension $2$,
$[A,B]=[B,A]=Fa_{2}\oplus Fa_{3}$, so that $L_{16}=Fa_{1}\oplus Fa_{2}\oplus
Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=a_{2}+\alpha a_{3}$,
$[a_{2},a_{1}]=-a_{2}-\alpha a_{3}\ (\alpha\neq 0)$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{16})=\zeta^{\mathrm{left}}(L_{16})=\zeta^{\mathrm{right}}(L_{16})=\zeta(L_{16})=Fa_{3}$,
$[L_{16},L_{16}]=Fa_{2}\oplus Fa_{3}$, $L_{16}$ is non-nilpotent.

Proof. We note that the center $\zeta(L)$ has dimension at most 2. Suppose
first that $\mathrm{\mathbf{dim}}_{F}(\zeta(L))=2$. Since $L$ is a not Lie
algebra, there is an element $a_{1}$ such that $[a_{1},a_{1}]=a_{3}\neq 0$. We
note that $a_{3}\in\zeta(L)$. It follows that
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{3}]=0$. Being an abelian algebra of
dimension 2, $\zeta(L)$ has a direct decomposition $\zeta(L)=Fa_{2}\oplus
Fa_{3}$ for some element $a_{2}$. We have $[a_{1},a_{2}]=[a_{2},a_{1}]=0$.
But, in this case, the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is
abelian, and we obtain a contradiction. This contradiction shows that
$\zeta(L)$ has dimension 1 and hence, $\zeta(L)=\mathrm{\mathbf{Leib}}(L)$.

As noted above, $L/\mathrm{\mathbf{Leib}}(L)$ has an ideal
$C/\mathrm{\mathbf{Leib}}(L)$ of dimension 1 (i.e.,
$C=Fc\oplus\mathrm{\mathbf{Leib}}(L)$ for some element $c$). If $[c,c]\neq 0$
without loss of generality, we can put $c=a_{1}$. The ideal $\langle
a_{1}\rangle=Fa_{1}\oplus Fa_{3}=A$ is nilpotent and has codimension 1. Let
$b$ be an element such that $L=A\oplus Fb$. We have $[b,b]=\gamma a_{3}$ for
some element $\gamma\in F$. As noted above, in this case, $[b,a_{1}]\in
a_{1}+Fa_{3}$ so that $[b,a_{1}]=a_{1}+\alpha a_{3}$ for some element
$\alpha\in F$. We have also $[a_{1},b]=-a_{1}+\beta a_{3}$ for some element
$\beta\in F$. Using the equality

$[b,[a_{1},b]]=[[b,a_{1}],b]+[a_{1},[b,b]]=[[b,a_{1}],b],$

we obtain $[b,-a_{1}+\beta a_{3}]=[a_{1}+\alpha a_{3},b]$. It follows that
$-[b,a_{1}]=[a_{1},b]$, so that $[a_{1},b]=-a_{1}-\alpha a_{3}$.

Suppose that $\gamma=\alpha=0$. Put $a_{2}=b$. Then, $[a_{2},a_{2}]=0$,
$[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}$. Thus, we come to the following
type of Leibniz algebras:

$\displaystyle L_{11}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{11})=\zeta^{\mathrm{left}}(L_{11})=\zeta^{\mathrm{right}}(L_{11})=\zeta(L_{11})=Fa_{3}$,
$[L_{11},L_{11}]=Fa_{1}\oplus Fa_{3}$, $L_{11}$ is non-nilpotent.

Suppose now that $\gamma=0$ and $\alpha\neq 0$. Put again $a_{2}=b$. Then,
$[a_{2},a_{2}]=0$, $[a_{1},a_{2}]=-a_{1}-\alpha a_{3}$,
$[a_{2},a_{1}]=a_{1}+\alpha a_{3}$. Then, we come to the following type of
Leibniz algebras:

$\displaystyle L_{12}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}-\alpha
a_{3},[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{12})=\zeta^{\mathrm{left}}(L_{12})=\zeta^{\mathrm{right}}(L_{12})=\zeta(L_{12})=Fa_{3}$,
$[L_{12},L_{12}]=Fa_{1}\oplus Fa_{3}$, $L_{12}$ is non-nilpotent.

Suppose that $\gamma\neq 0$ and $\alpha=0$. Put again $a_{2}=b$. Then,
$[a_{2},a_{2}]=\gamma a_{3}$, $[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}$.
Let $\lambda a_{1}+\mu a_{2}+\nu a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},a_{2}]+\lambda\mu[a_{2},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\mu a_{1}+\mu^{2}\gamma
a_{3}=(\lambda^{2}+\mu^{2}\gamma)a_{3}.$

As above, $\lambda\neq 0$, $\mu\neq 0$. It follows that polynomial
$X^{2}+\gamma$ has no root in field $F$. Then, we come to the following type
of Leibniz algebras:

$\displaystyle L_{13}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1},$
$\displaystyle[a_{2},a_{2}]=\gamma a_{3}\ (\gamma\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{13})=\zeta^{\mathrm{left}}(L_{13})=\zeta^{\mathrm{right}}(L_{13})=\zeta(L_{13})=Fa_{3}$,
$[L_{13},L_{13}]=Fa_{1}\oplus Fa_{3}$. Moreover, polynomial $X^{2}+\gamma$ has
no root in field $F$, $L_{13}$ is non-nilpotent.

Suppose now that $\gamma\neq 0$ and $\alpha\neq 0$. Put again $a_{2}=b$. Then,
$[a_{2},a_{2}]=\gamma a_{3}$, $[a_{1},a_{2}]=-a_{1}-\alpha a_{3}$,
$[a_{2},a_{1}]=a_{1}+\alpha a_{3}$. Let $\lambda a_{1}+\mu a_{2}+\nu a_{3}$ be
an arbitrary element of $L$. Then,

$\displaystyle[\lambda a_{1}+\mu a_{2}+\nu a_{3},\lambda a_{1}+\mu a_{2}+\nu
a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},a_{2}]+\lambda\mu[a_{2},a_{1}]+\mu^{2}[a_{2},a_{2}]=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu(-a_{1}-\alpha
a_{3})+\lambda\mu(a_{1}+\alpha a_{3})+\mu^{2}\gamma
a_{3}=(\lambda^{2}+\mu^{2}\gamma)a_{3}.$

As above, $\lambda\neq 0$, $\mu\neq 0$. It follows that polynomial
$X^{2}+\gamma$ has no root in field $F$. Then, we come to the following type
of Leibniz algebras:

$\displaystyle L_{14}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}-\alpha
a_{3},[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0),$
$\displaystyle[a_{2},a_{2}]=\gamma a_{3}\ (\gamma\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{14})=\zeta^{\mathrm{left}}(L_{14})=\zeta^{\mathrm{right}}(L_{14})=\zeta(L_{14})=Fa_{3}$,
$[L_{14},L_{14}]=Fa_{1}\oplus Fa_{3}$. Moreover, polynomial $X^{2}+\gamma$ has
no root in field $F$, $L_{14}$ is non-nilpotent.

Suppose now that $[c,c]=0$. Put again $\mathrm{\mathbf{Leib}}(L)=Fa_{3}$.
Since $\mathrm{\mathbf{Leib}}(L)=\zeta(L)$, the ideal
$C=Fc\oplus\mathrm{\mathbf{Leib}}(L)$ is abelian. Suppose that there exists an
element $b\not\in C$ such that $[b,b]=0$. Using the above arguments without
loss of generality, we can assume that $[b,c]\in c+Fa_{3}$,
$[c,b]\in-c+Fa_{3}$, so that $[b,c]=c+\alpha a_{3}$, $[c,b]=-c+\beta a_{3}$
for some elements $\alpha,\beta\in F$. Using the equality

$[[b,c],b]=[b,[c,b]]-[c,[b,b]]=[b,[c,b]],$

we obtain $[c+\alpha a_{3},b]=[b,-c+\beta a_{3}]$. It follows that
$[c,b]=-[b,c]$, so that $[c,b]=-c-\alpha a_{3}$. Let $u=\lambda c+\mu b+\nu
a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\lambda c+\mu b+\nu a_{3},\lambda c+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[c,c]+\lambda\mu[c,b]+\lambda\mu[b,c]+\mu^{2}[b,b]=\lambda\mu[c,b]+\lambda\mu[b,c]=0.$

Thus, we obtain a contradiction with the fact that $L$ is not a Lie algebra.
This contradiction shows that $[b,b]\neq 0$ for every element $b\not\in C$.
Hence, $[b,b]=\gamma a_{3}$ where $\gamma$ is a non-zero element of field $F$.
Without loss of generality, we can assume that $[b,b]=a_{3}$. Since
$[b,b]\in\mathrm{\mathbf{Leib}}(L)=\zeta(L)$, we obtain that $[c,b]=-[b,c]$,
so that $[c,b]=-c-\alpha a_{3}$.

If $\alpha=0$, then $[b,c]=c$, $[c,b]=-c$. Put $b=a_{1}$, $c=a_{2}$, then
$Fa_{2}$ is an ideal of $L$, and we come to the following type of Leibniz
algebras:

$\displaystyle L_{15}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{1},a_{2}]=a_{2},[a_{2},a_{1}]=-a_{2},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{15})=\zeta^{\mathrm{left}}(L_{15})=\zeta^{\mathrm{right}}(L_{15})=\zeta(L_{15})=Fa_{3}$,
$[L_{15},L_{15}]=Fa_{2}\oplus Fa_{3}$, $L_{15}$ is non-nilpotent.

If $\alpha\neq 0$, then $[b,c]=c+\alpha a_{3}$, $[c,b]=-c-\alpha a_{3}$, and
we come to the following type of Leibniz algebras:

$\displaystyle L_{16}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=a_{2}+\alpha
a_{3},[a_{2},a_{1}]=-a_{2}-\alpha a_{3}\ (\alpha\neq 0),$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{16})=\zeta^{\mathrm{left}}(L_{16})=\zeta^{\mathrm{right}}(L_{16})=\zeta(L_{16})=Fa_{3}$,
$[L_{16},L_{16}]=Fa_{2}\oplus Fa_{3}$, $L_{16}$ is non-nilpotent. $\Box$

######  Theorem 2.4.

Let $L$ be a Leibniz algebra of dimension $3$ over a field $F$. Suppose that
$L$ is not a Lie algebra. If the center of $L$ does not include the Leibniz
kernel, $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$ and the
factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is abelian, then $L$ is an
algebra of one of the following types.

  1. (i)

$\mathrm{\mathbf{Lei}}_{17}(3,F)=L_{17}$ is a direct sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}$. Moreover, $A$ is a non-nilpotent
cyclic Leibniz algebra of dimension $2$ and $B=\zeta(L)$, $[A,B]=[B,A]=\langle
0\rangle$, so that $L_{17}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{17})=[L_{17},L_{17}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{17})=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{17})=\zeta(L_{17})=Fa_{2}$, $L_{17}$ is non-
nilpotent.

  2. (ii)

$\mathrm{\mathbf{Lei}}_{18}(3,F)=L_{18}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and a subalgebra $B=Fa_{2}$. Moreover, $A$ is a non-
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=Fa_{3}$,
$[B,A]=\langle 0\rangle$, so that $L_{18}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=[a_{1},a_{2}]=[a_{1},a_{3}]=a_{3}$,
$[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{18})=[L_{18},L_{18}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{18})=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{18})=\zeta(L_{18})=\langle 0\rangle$, $L_{18}$ is
non-nilpotent.

  3. (iii)

$\mathrm{\mathbf{Lei}}_{19}(3,F)=L_{19}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and a subalgebra $B=Fa_{2}$. Moreover, $A$ is a non-
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=\langle 0\rangle$,
$[B,A]=Fa_{3}$, so that $L_{19}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{19})=[L_{19},L_{19}]=\zeta^{\mathrm{left}}(L_{19})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{19})=Fa_{2}$, $\zeta(L_{19})=\langle 0\rangle$,
$L_{19}$ is non-nilpotent.

  4. (iv)

$\mathrm{\mathbf{Lei}}_{20}(3,F)=L_{20}$ is a sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A$ is a
non-nilpotent cyclic Leibniz algebra of dimension $2$, $B$ is a nilpotent
cyclic Leibniz algebra of dimension $2$, $[A,B]=Fa_{3}$, $[B,A]=\langle
0\rangle$, so that $L_{20}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{2},a_{2}]=\sigma a_{3}\ (\sigma\neq
0)$,
$[a_{1},a_{2}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{20})=[L_{20},L_{20}]=\zeta^{\mathrm{left}}(L_{20})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{20})=\zeta(L_{20})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\gamma+\beta^{2}\sigma\neq 0$, $L_{20}$ is non-nilpotent.

  5. (v)

$\mathrm{\mathbf{Lei}}_{21}(3,F)=L_{21}$ is a sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A$ is a
non-nilpotent cyclic Leibniz algebra of dimension $2$, $B$ is a nilpotent
cyclic Leibniz algebra of dimension $2$, $[A,B]=Fa_{3}$, $[B,A]=\langle
0\rangle$, so that $L_{21}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{2}]=[a_{1},a_{3}]=a_{3}$, $[a_{2},a_{2}]=\tau a_{3}\
(\tau\neq 0)$,
$[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{21})=[L_{21},L_{21}]=\zeta^{\mathrm{left}}(L_{21})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{21})=\zeta(L_{21})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\beta+\alpha\gamma+\beta^{2}\tau\neq 0$, $L_{21}$ is non-
nilpotent.

  6. (vi)

$\mathrm{\mathbf{Lei}}_{22}(3,F)=L_{22}$ is a sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A,B$ are
non-nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{3}$,
so that $L_{22}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3}$,
$[a_{2},a_{2}]=\tau a_{3}\ (\tau\neq 0)$,
$[a_{1},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{22})=[L_{22},L_{22}]=\zeta^{\mathrm{left}}(L_{22})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{22})=\zeta(L_{22})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq 0$,
$L_{22}$ is non-nilpotent.

  7. (vii)

$\mathrm{\mathbf{Lei}}_{23}(3,F)=L_{23}$ is a sum of two ideals
$A=Fa_{1}\oplus Fa_{3}$ and $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A,B$ are
non-nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{3}$,
so that $L_{23}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=\delta a_{3}\ (\delta\neq 0)$, $[a_{2},a_{2}]=\tau a_{3}\
(\tau\neq 0)$, $[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{23})=[L_{23},L_{23}]=\zeta^{\mathrm{left}}(L_{23})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{23})=\zeta(L_{23})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\beta\delta+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq
0$, $L_{23}$ is non-nilpotent.

Proof. Since $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$,
$\mathrm{\mathbf{Leib}}(L)\cap\zeta(L)=\langle 0\rangle$. If we suppose that
the center $\zeta(L)$ has dimension 2, then
$L=\mathrm{\mathbf{Leib}}(L)\oplus\zeta(L)$. But, in this case, $L$ is abelian
and we obtain a contradiction. Suppose now that
$\mathrm{\mathbf{dim}}_{F}(\zeta(L))=1$.

Since $L$ is not a Lie algebra, there is an element $a_{1}$ such that
$[a_{1},a_{1}]=a_{3}\neq 0$. Then, $\mathrm{\mathbf{Leib}}(L)=Fa_{3}$. Then,
$A=\langle a_{1}\rangle=Fa_{1}\oplus Fa_{3}$ is a subalgebra of $L$. It is
obvious that $A\cap\zeta(L)=\langle 0\rangle$, so that $L=A\oplus\zeta(L)$,
$A$ is an ideal of $L$ and $B=\zeta(L)=Fa_{2}$. If we suppose that $A$ is
nilpotent (that is, $[a_{1},a_{3}]=0$), then
$\mathrm{\mathbf{Leib}}(L)=\mathrm{\mathbf{Leib}}(A)=\zeta(A)\leqslant\zeta(L)$,
and we obtain a contradiction. Thus, $A$ is not nilpotent. As we have seen
above, we can choose an element $a_{1}$ such that $[a_{1},a_{3}]=a_{3}$. Thus,
we come to the following type of Leibniz algebras:

$\displaystyle L_{17}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that $\mathrm{\mathbf{Leib}}(L_{17})=[L_{17},L_{17}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{17})=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{17})=\zeta(L_{17})=Fa_{2}$, $L_{17}$ is non-
nilpotent.

Suppose now that $\zeta(L)$ is zero. Since $L$ is not a Lie algebra, there is
an element $a_{1}$ such that $[a_{1},a_{1}]=a_{3}\neq 0$. Then,
$\mathrm{\mathbf{Leib}}(L)=Fa_{3}$. Since $L/\mathrm{\mathbf{Leib}}(L)$ is
abelian, a subalgebra $A=\langle a_{1}\rangle=Fa_{1}\oplus Fa_{3}$ is an ideal
of $L$. Suppose that $A$ is nilpotent (that is, $[a_{1},a_{3}]=0$). Let $b$ be
an element of $L$ such that $b\not\in A$. We have:

$[b,a_{3}]=[b,[a_{1},a_{1}]]=[[b,a_{1}],a_{1}]+[a_{1},[b,a_{1}]]=[\lambda
a_{3},a_{1}]+[a_{1},\lambda a_{3}]=0.$

Since $[a_{3},b]=0$, we obtain that $a_{3}\in\zeta(L)$, and we obtain a
contradiction. This contradiction shows that subalgebra $A=\langle
a_{1}\rangle=Fa_{1}\oplus Fa_{3}$ is not nilpotent. As we have seen above, we
can choose an element $a_{1}$ such that $[a_{1},a_{3}]=a_{3}$.

Suppose that $L$ contains an element $b$ such that $b\not\in A$ and $[b,b]=0$.
Since $L/\mathrm{\mathbf{Leib}}(L)$ is abelian, $[b,a_{1}]=\lambda a_{3}$,
$[a_{1},b]=\mu a_{3}$ for some elements $\lambda,\mu\in F$. As above,

$[b,a_{3}]=[a_{1},[b,a_{1}]]=[a_{1},\lambda
a_{3}]=\lambda[a_{1},a_{3}]=\lambda a_{3}.$

If $\lambda=\mu=0$, then $b\in\zeta(L)$, and we obtain a contradiction with
our assumption concerning the inclusion of $\zeta(L)$.

Suppose, now, that $\lambda=0$, $\mu\neq 0$. Put $a_{2}=\mu^{-1}b$, then
$[a_{2},a_{1}]=0$, $[a_{1},a_{2}]=a_{3}$, $[a_{2},a_{2}]=0$. As seen above, we
can observe that $[a_{2},a_{3}]=0$, and we obtain that $[a_{2},A]=\langle
0\rangle$. It follows that $a_{2}\in\zeta^{\mathrm{left}}(L)$. Thus, we come
to the following type of Leibniz algebras:

$\displaystyle L_{18}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{2}]=[a_{1},a_{3}]=a_{3},$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that $\mathrm{\mathbf{Leib}}(L_{18})=[L_{18},L_{18}]=Fa_{3}$,
$\zeta^{\mathrm{left}}(L_{18})=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{18})=\zeta(L_{14})=\langle 0\rangle$, $L_{18}$ is
non-nilpotent.

Suppose now that $\lambda\neq 0$, $\mu=0$. Put $a_{2}=\lambda^{-1}b$, then
$[a_{2},a_{1}]=a_{3}$, $[a_{1},a_{2}]=0$, $[a_{2},a_{2}]=0$. Since
$[a_{3},a_{2}]=0$, $[A,a_{2}]=\langle 0\rangle$. It follows that
$a_{2}\in\zeta^{\mathrm{right}}(L)$. Thus, we come to the following type of
Leibniz algebras:

$\displaystyle L_{19}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where}$
$\displaystyle[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{19})=[L_{19},L_{19}]=\zeta^{\mathrm{left}}(L_{19})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{19})=Fa_{2}$, $\zeta(L_{19})=\langle 0\rangle$,
$L_{19}$ is non-nilpotent.

Suppose now that $\mu\neq 0$, $\lambda\neq 0$. We have:

$\displaystyle 0=[a_{1},0]=[a_{1},[b,b]]=[[a_{1},b],b]+[b,[a_{1},b]]=$
$\displaystyle[b,[a_{1},b]]=[b,\mu a_{3}]=\mu[b,a_{3}]=\mu\lambda a_{3}.$

It follows that $\mu\lambda=0$, and we obtain a contradiction.

Suppose that $[b,b]\neq 0$ for every element $b$ such that $b\not\in A$. Since
$L/\mathrm{\mathbf{Leib}}(L)$ is abelian, $[b,a_{1}]=\lambda a_{3}$,
$[a_{1},b]=\mu a_{3}$ for some elements $\lambda,\mu\in F$, and $[b,b]=\sigma
a_{3}$ for some non-zero element $\sigma\in F$.

As above,

$[b,a_{3}]=[a_{1},[b,a_{1}]]=[a_{1},\lambda
a_{3}]=\lambda[a_{1},a_{3}]=\lambda a_{3}.$

If $\lambda=\mu=0$, then $[b,a_{1}]=[a_{1},b]=[b,a_{3}]=0$. Put $a_{2}=b$. Let
$u=\alpha a_{1}+\beta a_{2}+\gamma a_{3}$ be an arbitrary element of $L$.
Then,

$\displaystyle[u,u]=[\alpha a_{1}+\beta a_{2}+\gamma a_{3},\alpha a_{1}+\beta
a_{2}+\gamma a_{3}]=$
$\displaystyle\alpha^{2}[a_{1},a_{1}]+\alpha\gamma[a_{1},a_{3}]+\beta^{2}[a_{2},a_{2}]=$
$\displaystyle\alpha^{2}a_{3}+\alpha\gamma
a_{3}+\beta^{2}a_{3}=(\alpha^{2}+\alpha\gamma+\beta^{2}\sigma)a_{3}.$

If $u\not\in A$, then $[u,u]\neq 0$. It follows that
$\alpha^{2}+\alpha\gamma+\beta^{2}\sigma\neq 0$. Thus, we come to the
following type of Leibniz algebras:

$\displaystyle L_{20}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$ $\displaystyle[a_{2},a_{2}]=\sigma a_{3}\
(\sigma\neq 0),$
$\displaystyle[a_{1},a_{2}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{20})=[L_{20},L_{20}]=\zeta^{\mathrm{left}}(L_{20})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{20})=\zeta(L_{20})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\gamma+\beta^{2}\sigma\neq 0$, $L_{20}$ is non-nilpotent.

Suppose now that $\lambda=0$, $\mu\neq 0$. Put $a_{2}=\mu^{-1}b$. Then
$[a_{2},a_{1}]=[a_{2},a_{3}]=0$, $[a_{1},a_{2}]=a_{3}$,
$[a_{2},a_{2}]=\mu^{-2}[b,b]=\mu^{-2}\sigma a_{3}=\tau a_{3}$. Let $\alpha
a_{1}+\beta a_{2}+\gamma a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\alpha a_{1}+\beta a_{2}+\gamma a_{3},\alpha a_{1}+\beta
a_{2}+\gamma a_{3}]=$
$\displaystyle\alpha^{2}[a_{1},a_{1}]+\alpha\beta[a_{1},a_{2}]+\alpha\gamma[a_{1},a_{3}]+\beta^{2}[a_{2},a_{2}]=$
$\displaystyle\alpha^{2}a_{3}+\alpha\beta a_{3}+\alpha\gamma
a_{3}+\beta^{2}\tau
a_{3}=(\alpha^{2}+\alpha\beta+\alpha\gamma+\beta^{2}\tau)a_{3}.$

As above, $\alpha^{2}+\alpha\beta+\alpha\gamma+\beta^{2}\tau\neq 0$. Thus, we
come to the following type of Leibniz algebras:

$\displaystyle L_{21}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{2}]=[a_{1},a_{3}]=a_{3},$
$\displaystyle[a_{2},a_{2}]=\tau a_{3}\ (\tau\neq 0),$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{21})=[L_{21},L_{21}]=\zeta^{\mathrm{left}}(L_{21})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{21})=\zeta(L_{21})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\beta+\alpha\gamma+\beta^{2}\tau\neq 0$, $L_{21}$ is non-
nilpotent.

Suppose now that $\lambda\neq 0$, $\mu=0$. Put $a_{2}=\lambda^{-1}b$. Then
$[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3}$, $[a_{1},a_{2}]=0$,
$[a_{2},a_{2}]=\lambda^{-2}[b,b]=\lambda^{-2}\sigma a_{3}=\tau a_{3}$. Let
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\alpha a_{1}+\beta a_{2}+\gamma a_{3},\alpha a_{1}+\beta
a_{2}+\gamma a_{3}]=$
$\displaystyle\alpha^{2}[a_{1},a_{1}]+\alpha\gamma[a_{1},a_{3}]+\alpha\beta[a_{2},a_{1}]+\beta^{2}[a_{2},a_{2}]+\beta\gamma[a_{2},a_{3}]=$
$\displaystyle\alpha^{2}a_{3}+\alpha\gamma a_{3}+\alpha\beta
a_{3}+\beta^{2}\tau a_{3}+\beta\gamma a_{3}=$
$\displaystyle(\alpha^{2}+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma)a_{3}.$

As above, $\alpha^{2}+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq
0$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{22}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where}$
$\displaystyle[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3},$
$\displaystyle[a_{2},a_{2}]=\tau a_{3}\ (\tau\neq 0),$
$\displaystyle[a_{1},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{22})=[L_{22},L_{22}]=\zeta^{\mathrm{left}}(L_{22})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{22})=\zeta(L_{22})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not inA$, then
$\alpha^{2}+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq 0$,
$L_{22}$ is non-nilpotent.

Suppose now that $\lambda\neq 0$, $\mu\neq 0$. Put $a_{2}=\lambda^{-1}b$. Then
$[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=\lambda^{-1}[a_{1},b]=\lambda^{-1}\mu a_{3}=\delta a_{3}$,
$[a_{2},a_{2}]=\lambda^{-2}[b,b]=\lambda^{-2}\sigma a_{3}=\tau a_{3}$. Let
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\alpha a_{1}+\beta a_{2}+\gamma a_{3},\alpha a_{1}+\beta
a_{2}+\gamma a_{3}]=$
$\displaystyle\alpha^{2}[a_{1},a_{1}]+\alpha\beta[a_{1},a_{2}]+\alpha\gamma[a_{1},a_{3}]+$
$\displaystyle\alpha\beta[a_{2},a_{1}]+\beta^{2}[a_{2},a_{2}]+\beta\gamma[a_{2},a_{3}]=$
$\displaystyle\alpha^{2}a_{3}+\alpha\beta\delta a_{3}+\alpha\gamma
a_{3}+\alpha\beta a_{3}+\beta^{2}\tau a_{3}+\beta\gamma a_{3}=$
$\displaystyle(\alpha^{2}+\alpha\beta\delta+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma)a_{3}.$

As above,
$\alpha^{2}+\alpha\beta\delta+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq
0$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{23}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where}$
$\displaystyle[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=\delta a_{3}\ (\delta\neq 0),[a_{2},a_{2}]=\tau
a_{3}\ (\tau\neq 0),$
$\displaystyle[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{23})=[L_{23},L_{23}]=\zeta^{\mathrm{left}}(L_{23})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{23})=\zeta(L_{23})=\langle 0\rangle$. Moreover, if
$\alpha a_{1}+\beta a_{2}+\gamma a_{3}\not\in A$, then
$\alpha^{2}+\alpha\beta\delta+\alpha\gamma+\alpha\beta+\beta^{2}\tau+\beta\gamma\neq
0$, $L_{23}$ is non-nilpotent. $\Box$

Let $L$ be a Leibniz algebra over a field $F$, $M$ be non-empty subset of $L$
and $H$ be a subalgebra of $L$. Put

$\displaystyle\mathrm{\mathbf{Ann}}_{H}^{\mathrm{left}}(M)=\\{a\in H|\
[a,M]=\langle 0\rangle\\},$
$\displaystyle\mathrm{\mathbf{Ann}}_{H}^{\mathrm{right}}(M)=\\{a\in H|\
[M,a]=\langle 0\rangle\\}.$

The subset $\mathrm{\mathbf{Ann}}_{H}^{\mathrm{left}}(M)$ is called the left
annihilator of $M$ in subalgebra $H$. The subset
$\mathrm{\mathbf{Ann}}_{H}^{\mathrm{right}}(M)$ is called the right
annihilator of $M$ in subalgebra $H$. The intersection

$\mathrm{\mathbf{Ann}}_{H}(M)=\mathrm{\mathbf{Ann}}_{H}^{\mathrm{left}}(M)\cap\mathrm{\mathbf{Ann}}_{H}^{\mathrm{right}}(M)$

is called the annihilator of $M$ in subalgebra $H$.

It is not hard to see that all of these subsets are subalgebras of $L$.
Moreover, if $M$ is an ideal of $L$, then $\mathrm{\mathbf{Ann}}_{H}(M)$ is an
ideal of $L$ (see, for example, [7]).

######  Theorem 2.5.

Let $L$ be a Leibniz algebra over a field $F$ having dimension $3$. Suppose
that $L$ is not a Lie algebra. Suppose that the center of $L$ does not include
the Leibniz kernel, $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$,
and the factor-algebra $L/\mathrm{\mathbf{Leib}}(L)$ is non-abelian. Then, $L$
is an algebra of one of the following types.

  1. (i)

$\mathrm{\mathbf{Lei}}_{24}(3,F)=L_{24}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$,
$\mathrm{\mathbf{char}}(F)\neq 2$, $[A,B]=Fa_{1}$, $[B,A]=Fa_{1}\oplus
Fa_{3}$, so that $L_{24}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}$,
$[a_{2},a_{3}]=2a_{3}$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{24})=\zeta^{\mathrm{left}}(L_{24})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{24})=\zeta(L_{24})=\langle 0\rangle$,
$[L_{24},L_{24}]=Fa_{1}\oplus Fa_{3}$, $L_{24}$ is non-nilpotent.

  2. (ii)

$\mathrm{\mathbf{Lei}}_{25}(3,F)=L_{25}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$,
$\mathrm{\mathbf{char}}(F)\neq 2$, $[A,B]=[B,A]=Fa_{1}\oplus Fa_{3}$, so that
$L_{25}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}+\alpha a_{3}$, $[a_{2},a_{1}]=a_{1}+\alpha a_{3}\
(\alpha\neq 0)$, $[a_{2},a_{3}]=2a_{3}$,
$[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{25})=\zeta^{\mathrm{left}}(L_{25})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{25})=\zeta(L_{25})=\langle 0\rangle$,
$[L_{25},L_{25}]=Fa_{1}\oplus Fa_{3}$, $L_{25}$ is non-nilpotent.

  3. (iii)

$\mathrm{\mathbf{Lei}}_{26}(3,F)=L_{26}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Furthermore, $A$ is a non-
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=Fa_{1}$,
$[B,A]=Fa_{1}\oplus Fa_{3}$, so that $L_{26}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}$,
$[a_{2},a_{1}]=a_{1}$, $[a_{2},a_{3}]=2a_{3}$,
$[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{26})=\zeta^{\mathrm{left}}(L_{26})=Fa_{3}$,
$[L_{26},L_{26}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{26})=\zeta(L_{26})=\langle 0\rangle$. Moreover,
$\mathrm{\mathbf{char}}(F)\neq 2$, $L_{26}$ is non-nilpotent.

  4. (iv)

$\mathrm{\mathbf{Lei}}_{27}(3,F)=L_{27}$ is a direct sum of ideal
$A=Fa_{1}\oplus Fa_{3}$ and subalgebra $B=Fa_{2}$. Moreover, $A$ is a non-
nilpotent cyclic Leibniz algebra of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, so that $L_{27}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}+\beta a_{3}$
$(\beta=\alpha(1+\alpha)^{-1})$, $[a_{2},a_{1}]=a_{1}+\alpha a_{3}$,
$[a_{2},a_{3}]=(2+\alpha)a_{3}$ $(\alpha\neq 0,\alpha\neq-1,\alpha\neq-2)$,
$[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{27})=\zeta^{\mathrm{left}}(L_{27})=Fa_{3}$,
$[L_{27},L_{27}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{27})=\zeta(L_{27})=\langle 0\rangle$, $L_{27}$ is
non-nilpotent.

  5. (v)

$\mathrm{\mathbf{Lei}}_{28}(3,F)=L_{28}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $B$ is a non-nilpotent
cyclic Leibniz algebra of dimension $2$, $\mathrm{\mathbf{char}}(F)\neq 2$,
$[A,B]=Fa_{1}$, $[B,A]=Fa_{1}\oplus Fa_{3}$, so that $L_{28}=Fa_{1}\oplus
Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}$,
$[a_{2},a_{1}]=a_{1}$, $[a_{2},a_{2}]=\gamma a_{3}$ $(\gamma\neq 0)$,
$[a_{2},a_{3}]=2a_{3}$,
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{28})=\zeta^{\mathrm{left}}(L_{28})=Fa_{3}$,
$[L_{28},L_{28}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{28})=\zeta(L_{28})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\mu^{2}\gamma+2\mu\nu\neq 0$, $L_{28}$ is non-nilpotent.

  6. (vi)

$\mathrm{\mathbf{Lei}}_{29}(3,F)=L_{29}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Furthermore, $A$ is a
nilpotent cyclic Leibniz algebra of dimension $2$, $B$ is a non-nilpotent
cyclic Leibniz algebra of dimension $2$, $\mathrm{\mathbf{char}}(F)\neq 2$,
$[A,B]=[B,A]=Fa_{1}\oplus Fa_{3}$, so that $L_{29}=Fa_{1}\oplus Fa_{2}\oplus
Fa_{3}$ where $[a_{1},a_{1}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}+\alpha a_{3}$,
$[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0)$, $[a_{2},a_{2}]=\gamma
a_{3}\ (\gamma\neq 0)$, $[a_{2},a_{3}]=2a_{3}$,
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{29})=\zeta^{\mathrm{left}}(L_{29})=Fa_{3}$,
$[L_{29},L_{29}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{29})=\zeta(L_{29})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+2\lambda\mu\alpha+\mu^{2}\gamma+2\mu\nu\neq 0$, $L_{29}$ is non-
nilpotent.

  7. (vii)

$\mathrm{\mathbf{Lei}}_{30}(3,F)=L_{30}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A$, $B$ are non-
nilpotent cyclic Leibniz algebras of dimension $2$,
$\mathrm{\mathbf{char}}(F)\neq 2$, $[A,B]=[B,A]=Fa_{1}\oplus Fa_{3}$, so that
$L_{30}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}+\gamma a_{3}$
$(\gamma\neq 0)$, $[a_{2},a_{1}]=a_{1}$, $[a_{2},a_{2}]=\gamma a_{3}$,
$[a_{2},a_{3}]=2a_{3}$, $[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{30})=\zeta^{\mathrm{left}}(L_{30})=Fa_{3}$,
$[L_{30},L_{30}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{30})=\zeta(L_{30})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\gamma+\lambda\nu+\mu^{2}\gamma+2\mu\nu\neq 0$,
$L_{30}$ is non-nilpotent.

  8. (viii)

$\mathrm{\mathbf{Lei}}_{31}(3,F)=L_{31}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A,B$ are non-
nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, so that $L_{31}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}$, $[a_{2},a_{1}]=a_{1}-a_{3}$,
$[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{31})=\zeta^{\mathrm{left}}(L_{31})=Fa_{3}$,
$[L_{31},L_{31}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{31})=\zeta(L_{31})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$, $L_{31}$ is non-
nilpotent.

  9. (ix)

$\mathrm{\mathbf{Lei}}_{32}(3,F)=L_{31}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A,B$ are non-
nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, so that $L_{32}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=a_{3}$,
$[a_{1},a_{2}]=-a_{1}+\beta a_{3}\ (\beta\neq 0)$,
$[a_{2},a_{1}]=a_{1}-a_{3}$, $[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{32})=\zeta^{\mathrm{left}}(L_{32})=Fa_{3}$,
$[L_{32},L_{32}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{32})=\zeta(L_{32})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\beta+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$,
$L_{32}$ is non-nilpotent.

  10. (x)

$\mathrm{\mathbf{Lei}}_{33}(3,F)=L_{31}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A,B$ are non-
nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, $L_{33}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}$,
$[a_{2},a_{1}]=a_{1}-\gamma a_{3}\ (\gamma\neq 0,\gamma\neq 1,\gamma\neq 2)$,
$[a_{2},a_{2}]=\gamma a_{3}$, $[a_{2},a_{3}]=(2-\gamma)a_{3}$,
$[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{33})=\zeta^{\mathrm{left}}(L_{33})=Fa_{3}$,
$[L_{33},L_{33}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{33})=\zeta(L_{33})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\nu-\lambda\mu\gamma+\mu^{2}\gamma+\mu\nu(2-\gamma)\neq
0$, $L_{33}$ is non-nilpotent.

  11. (xi)

$\mathrm{\mathbf{Lei}}_{34}(3,F)=L_{34}$ is a sum of ideal $A=Fa_{1}\oplus
Fa_{3}$ and subalgebra $B=Fa_{2}\oplus Fa_{3}$. Moreover, $A,B$ are non-
nilpotent cyclic Leibniz algebras of dimension $2$, $[A,B]=[B,A]=Fa_{1}\oplus
Fa_{3}$, $L_{34}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where
$[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3}$, $[a_{1},a_{2}]=-a_{1}+\beta a_{3}\
(\beta=(\alpha+\gamma)(1+\alpha)^{-1})$, $[a_{2},a_{1}]=a_{1}+\alpha a_{3}\
(\alpha\neq 0,\alpha\neq-1,\alpha\neq-2)$, $[a_{2},a_{2}]=\gamma a_{3}\
(\gamma\neq 0)$, $[a_{2},a_{3}]=(2+\alpha)a_{3}$,
$[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{34})=\zeta^{\mathrm{left}}(L_{34})=Fa_{3}$,
$[L_{34},L_{34}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{34})=\zeta(L_{34})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\beta+\lambda\nu+\lambda\mu\alpha+\mu^{2}\gamma+\mu\nu(2+\alpha)\neq
0$, $L_{34}$ is non-nilpotent.

  12. (xii)

$\mathrm{\mathbf{Lei}}_{35}(3,F)=L_{35}$ is a direct sum of ideal $B=Fa_{2}$
and a cyclic non-nilpotent subalgebra $A=Fa_{1}\oplus Fa_{3}$ of dimension
$2$, $[A,B]=[B,A]=Fa_{2}$, so that $L_{35}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=\gamma a_{3}\ (\gamma\neq 0)$, $[a_{1},a_{2}]=a_{2}$,
$[a_{1},a_{3}]=a_{3}$, $[a_{2},a_{1}]=-a_{2}$,
$[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{35})=\zeta^{\mathrm{left}}(L_{35})=Fa_{3}$,
$[L_{35},L_{35}]=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{35})=\zeta(L_{35})=\langle 0\rangle$, $L_{35}$ is
non-nilpotent.

  13. (xiii)

$\mathrm{\mathbf{Lei}}_{36}(3,F)=L_{36}$ is a sum of abelian ideal
$B=Fa_{2}\oplus Fa_{3}$ and a cyclic non-nilpotent subalgebra $A=Fa_{1}\oplus
Fa_{3}$ of dimension $2$, $[A,B]=[B,A]=Fa_{2}\oplus Fa_{3}$, so that
$L_{36}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=\gamma a_{3}\
(\gamma\neq 0)$, $[a_{1},a_{2}]=a_{2}$, $[a_{1},a_{3}]=a_{3}$,
$[a_{2},a_{1}]=-a_{2}+\beta a_{3}\ (\beta\neq 0)$,
$[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{36})=\zeta^{\mathrm{left}}(L_{36})=Fa_{3}$,
$[L_{36},L_{36}]=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{36})=\zeta(L_{36})=\langle 0\rangle$, $L_{36}$ is
non-nilpotent.

Proof. As in the previous theorem, we can see that
$\mathrm{\mathbf{dim}}_{F}(\zeta(L))\leqslant 1$. Suppose that
$\mathrm{\mathbf{dim}}_{F}(\zeta(L))=1$. Since $L$ is not a Lie algebra, there
is an element $a_{1}$ such that $[a_{1},a_{1}]=a_{3}\neq 0$. Then,
$\mathrm{\mathbf{Leib}}(L)=Fa_{3}$. Let $A=\langle a_{1}\rangle=Fa_{1}\oplus
Fa_{3}$. An equality $\mathrm{\mathbf{Leib}}(L)\cap\zeta(L)=\langle 0\rangle$
implies that $A\cap\zeta(L)=\langle 0\rangle$, so that $L=A\oplus\zeta(L)$ and
$A$ is an ideal of $L$. But, in this case, the factor-algebra
$L/\mathrm{\mathbf{Leib}}(L)$ is abelian, and we obtain a contradiction. This
contradiction shows that $\zeta(L)=\langle 0\rangle$.

By what is noted above, $L/\mathrm{\mathbf{Leib}}(L)$ has an ideal
$C/\mathrm{\mathbf{Leib}}(L)$ of dimension 1 (i.e.,
$C=Fc\oplus\mathrm{\mathbf{Leib}}(L)$ for some element $c$). If $[c,c]\neq 0$
without loss of generality, we can put $c=a_{1}$. Then, subalgebra $\langle
a_{1}\rangle=Fa_{1}\oplus Fa_{3}=A$ is an ideal which has codimension 1. Then,
for every element $b$ such that $b\not\in A$, we have $L=A\oplus Fb$. By what
is noted above, in this case, $[b,a_{1}]\in a_{1}+Fa_{3}$, so that
$[b,a_{1}]=a_{1}+\alpha a_{3}$ for some element $\alpha\in F$. We have also
$[a_{1},b]=-a_{1}+\beta a_{3}$ for some element $\beta\in F$. Since
$[b,b]\in\mathrm{\mathbf{Leib}}(L)$, $[b,b]=\gamma a_{3}$ for some element
$\gamma\in F$.

Suppose first that $\gamma=0$. In other words, we suppose that there exists an
element $b$ such that $b\not\in A$ and $[b,b]=0$. Consider first the case when
$A$ is nilpotent. We have:

$\displaystyle[b,a_{3}]=[b,[a_{1},a_{1}]]=[[b,a_{1}],a_{1}]+[a_{1},[b,a_{1}]]=$
$\displaystyle[a_{1}+\alpha a_{3},a_{1}]+[a_{1},a_{1}+\alpha
a_{3}]=[a_{1},a_{1}]+[a_{1},a_{1}]=2a_{3}.$

In particular, if we suppose that $\mathrm{\mathbf{char}}(F)=2$, then
$[b,a_{3}]=0$. Since $[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},b]=0$,
$\mathrm{\mathbf{Leib}}(L)=Fa_{3}\leqslant\zeta(L)$, and we obtain a
contradiction. This contradiction shows that $\mathrm{\mathbf{char}}(F)\neq
2$. Further,

$[b,[a_{1},b]]=[[b,a_{1}],b]+[a_{1},[b,b]]=[a_{1}+\alpha
a_{3},b]=[a_{1},b]=-a_{1}+\beta a_{3}.$

On the other hand,

$\displaystyle[b,[a_{1},b]]=[b,-a_{1}+\beta a_{3}]=-[b,a_{1}]+\beta[b,a_{3}]=$
$\displaystyle-(a_{1}+\alpha a_{3})+2\beta a_{3}=-a_{1}+(2\beta-\alpha)a_{3}.$

It follows that $2\beta-\alpha=\beta$ or $\beta=\alpha$ and
$[a_{1},b]=-a_{1}+\alpha a_{3}$.

Suppose that $\alpha=0$. Put $a_{2}=b$. Then $[a_{1},a_{2}]=-a_{1}$,
$[a_{2},a_{1}]=a_{1}$, $[a_{2},a_{2}]=0$, $[a_{2},a_{3}]=2a_{3}$. Thus, we
come to the following type of Leibniz algebras:

$\displaystyle L_{24}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1},[a_{2},a_{3}]=2a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{24})=\zeta^{\mathrm{left}}(L_{24})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{24})=\zeta(L_{24})=\langle 0\rangle$,
$[L_{24},L_{24}]=Fa_{1}\oplus Fa_{3}$, $\mathrm{\mathbf{char}}(F)\neq 2$,
$L_{24}$ is non-nilpotent.

Suppose now that $\alpha\neq 0$. Put again $a_{2}=b$. Thus, we come to the
following type of Leibniz algebras:

$\displaystyle L_{25}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}+\alpha
a_{3},[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0),$
$\displaystyle[a_{2},a_{3}]=2a_{3},[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{25})=\zeta^{\mathrm{left}}(L_{25})=Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{25})=\zeta(L_{25})=\langle 0\rangle$,
$[L_{25},L_{25}]=Fa_{1}\oplus Fa_{3}$, $\mathrm{\mathbf{char}}(F)\neq 2$,
$L_{25}$ is non-nilpotent.

Consider now the case when $A$ is not nilpotent. As we have seen above, we can
choose an element $a_{1}$ such that $[a_{1},a_{3}]=a_{3}$. Then,

$\displaystyle[b,a_{3}]=[b,[a_{1},a_{1}]]=[[b,a_{1}],a_{1}]+[a_{1},[b,a_{1}]]=$
$\displaystyle[a_{1}+\alpha a_{3},a_{1}]+[a_{1},a_{1}+\alpha a_{3}]=$
$\displaystyle[a_{1},a_{1}]+[a_{1},a_{1}]+\alpha[a_{1},a_{3}]=(2+\alpha)a_{3}.$

Further,

$\displaystyle[b,[a_{1},b]]=[[b,a_{1}],b]+[a_{1},[b,b]]=$
$\displaystyle[a_{1}+\alpha a_{3},b]=[a_{1},b]=-a_{1}+\beta a_{3}.$

On the other hand,

$\displaystyle[b,[a_{1},b]]=[b,-a_{1}+\beta a_{3}]=-[b,a_{1}]+\beta[b,a_{3}]=$
$\displaystyle-(a_{1}+\alpha
a_{3})+\beta(2+\alpha)a_{3}=-a_{1}+(2\beta+\beta\alpha-\alpha)a_{3}.$

It follows that $2\beta+\beta\alpha-\alpha=\beta$ or
$\beta+\beta\alpha-\alpha=0$. It follows that $\beta(1+\alpha)=\alpha$. We
consider separately the case when $\alpha=0$. Then, $\beta=0$. If we suppose
that $\mathrm{\mathbf{char}}(F)=2$, then $[b,a_{3}]=0$. Since $[a_{3},b]=0$,
$b\in\mathrm{\mathbf{Ann}}_{L}(Fa_{3})=\mathrm{\mathbf{Ann}}_{L}(\mathrm{\mathbf{Leib}}(L))$.
But, $\mathrm{\mathbf{Ann}}_{L}(\mathrm{\mathbf{Leib}}(L))$ is an ideal of
$L$. Then, $[b,a_{1}]=a_{1}\in Fb\oplus Fa_{3}$, and we obtain a
contradiction. This contradiction shows that $\mathrm{\mathbf{char}}(F)\neq
2$. Put $a_{2}=b$. We come to the following type of Leibniz algebras:

$\displaystyle L_{26}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},[a_{1},a_{2}]=-a_{1},$
$\displaystyle[a_{2},a_{1}]=a_{1},[a_{2},a_{3}]=2a_{3},[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{26})=\zeta^{\mathrm{left}}(L_{26})=Fa_{3}$,
$[L_{26},L_{26}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{26})=\zeta(L_{26})=\langle 0\rangle$,
$\mathrm{\mathbf{char}}(F)\neq 2$, $L_{26}$ is non-nilpotent.

Suppose that $\alpha\neq 0$. The equality $\beta(1+\alpha)=\alpha$ shows that
$\alpha\neq-1$. In this case, $\beta=\alpha(1+\alpha)^{-1}$. If we suppose
that $\alpha=-2$, then $[b,a_{3}]=0$. Using the above arguments, we obtain a
contradiction. This contradiction shows that $\alpha\neq-2$. Put $a_{2}=b$. We
come to the following type of Leibniz algebras:

$\displaystyle L_{27}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}+\beta
a_{3}\ (\beta=\alpha(1+\alpha)^{-1}),[a_{2},a_{1}]=a_{1}+\alpha a_{3},$
$\displaystyle[a_{2},a_{3}]=(2+\alpha)a_{3}\ (\alpha\neq
0,\alpha\neq-1,\alpha\neq-2),$
$\displaystyle[a_{2},a_{2}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{27})=\zeta^{\mathrm{left}}(L_{27})=Fa_{3}$,
$[L_{27},L_{27}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{27})=\zeta(L_{27})=\langle 0\rangle$, $L_{27}$ is
non-nilpotent.

Suppose now that $[b,b]\neq 0$ for every element $b$ such that $b\not\in A$,
so that $[b,b]=\gamma a_{3}$ and $\gamma\neq 0$. Consider first the case when
$A$ is nilpotent. We have:

$\displaystyle[b,a_{3}]=[b,[a_{1},a_{1}]]=[[b,a_{1}],a_{1}]+[a_{1},[b,a_{1}]]=$
$\displaystyle[a_{1}+\alpha a_{3},a_{1}]+[a_{1},a_{1}+\alpha
a_{3}]=[a_{1},a_{1}]+[a_{1},a_{1}]=2a_{3}.$

Again, we obtain that $\mathrm{\mathbf{char}}(F)\neq 2$. Using the arguments
above, we can get that $[a_{1},b]=-a_{1}+\alpha a_{3}$ again. We consider
separately the case when $\alpha=0$ (that is, $[a_{1},b]=-a_{1}$,
$[b,a_{1}]=a_{1}$). Let $\lambda a_{1}+\mu b+\nu a_{3}$ be the arbitrary
element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\mu a_{1}+\mu^{2}\gamma
a_{3}+2\mu\nu a_{3}=$ $\displaystyle(\lambda^{2}+\mu^{2}\gamma+2\mu\nu)a_{3}.$

Then, a condition $[b,b]\neq 0$ for every element $b\not\in A$ yields that
$\lambda^{2}+\mu^{2}\gamma+2\mu\nu\neq 0$. Put $a_{2}=b$. Thus, we come to the
following type of Leibniz algebras:

$\displaystyle L_{28}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1},$
$\displaystyle[a_{2},a_{2}]=\gamma a_{3}\ (\gamma\neq
0),[a_{2},a_{3}]=2a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{28})=\zeta^{\mathrm{left}}(L_{28})=Fa_{3}$,
$[L_{28},L_{28}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{28})=\zeta(L_{28})=\langle 0\rangle$,
$\mathrm{\mathbf{char}}(F)\neq 2$. Moreover, if $\lambda a_{1}+\mu a_{2}+\nu
a_{3}\not\in A$, then $\lambda^{2}+\mu^{2}\gamma+2\mu\nu\neq 0$, $L_{28}$ is
non-nilpotent.

Suppose that $\alpha\neq 0$. Let $\lambda a_{1}+\mu b+\nu a_{3}$ be the
arbitrary element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu(-a_{1}+\alpha
a_{3})+\lambda\mu(a_{1}+\alpha a_{3})+\mu^{2}\gamma a_{3}+2\mu\nu a_{3}=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\mu\alpha
a_{3}+\lambda\mu a_{1}+\lambda\mu\alpha a_{3}+\mu^{2}\gamma a_{3}+2\mu\nu
a_{3}=$
$\displaystyle(\lambda^{2}+2\lambda\mu\alpha+\mu^{2}\gamma+2\mu\nu)a_{3}.$

As above, $\lambda^{2}+2\lambda\mu\alpha+\mu^{2}\gamma+2\mu\nu\neq 0$. Put
again $a_{2}=b$. Then, we come to the following type of Leibniz algebras:

$\displaystyle L_{29}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}+\alpha
a_{3},[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq 0),$
$\displaystyle[a_{2},a_{2}]=\gamma a_{3}\ (\gamma\neq
0),[a_{2},a_{3}]=2a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{29})=\zeta^{\mathrm{left}}(L_{29})=Fa_{3}$,
$[L_{29},L_{29}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{29})=\zeta(L_{29})=\langle 0\rangle$,
$\mathrm{\mathbf{char}}(F)\neq 2$. Moreover, if $\lambda a_{1}+\mu a_{2}+\nu
a_{3}\not\in A$, then $\lambda^{2}+2\lambda\mu\alpha+\mu^{2}\gamma+2\mu\nu\neq
0$, $L_{29}$ is non-nilpotent.

Suppose, now, that $[b,b]\neq 0$ for every element $b$ such that $b\not\in A$
and a subalgebra $A$ is not nilpotent. As we have seen above, we can choose an
element $a_{1}$ such that $[a_{1},a_{3}]=a_{3}$. Then,

$\displaystyle[b,a_{3}]=[b,[a_{1},a_{1}]]=[[b,a_{1}],a_{1}]+[a_{1},[b,a_{1}]]=$
$\displaystyle[a_{1}+\alpha a_{3},a_{1}]+[a_{1},a_{1}+\alpha a_{3}]=$
$\displaystyle[a_{1},a_{1}]+[a_{1},a_{1}]+\alpha[a_{1},a_{3}]=(2+\alpha)a_{3}.$

If we suppose that $\alpha=-2$, then $[b,a_{3}]=0$. Since $[a_{3},b]=0$,
$b\in\mathrm{\mathbf{Ann}}_{L}(Fa_{3})=\mathrm{\mathbf{Ann}}_{L}(\mathrm{\mathbf{Leib}}(L))$.
But, $\mathrm{\mathbf{Ann}}_{L}(\mathrm{\mathbf{Leib}}(L))$ is an ideal of
$L$. Then, $[b,a_{1}]\in Fb\oplus Fa_{3}$, and we obtain a contradiction. This
contradiction shows that $\alpha\neq-2$. Further,

$\displaystyle[b,[a_{1},b]]=[[b,a_{1}],b]+[a_{1},[b,b]]=[a_{1}+\alpha
a_{3},b]+[a_{1},\gamma a_{3}]=$
$\displaystyle[a_{1},b]+\gamma[a_{1},a_{3}]=-a_{1}+\beta a_{3}+\gamma
a_{3}=-a_{1}+(\beta+\gamma)a_{3}.$

On the other hand,

$\displaystyle[b,[a_{1},b]]=[b,-a_{1}+\beta a_{3}]=-[b,a_{1}]+\beta[b,a_{3}]=$
$\displaystyle-(a_{1}+\alpha
a_{3})+\beta(2+\alpha)a_{3}=-a_{1}+(2\beta+\beta\alpha-\alpha)a_{3}.$

It follows that $2\beta+\beta\alpha-\alpha=\beta+\gamma$ or
$\beta(1+\alpha)=\alpha+\gamma$. We consider separately the case when
$\alpha=0$ and $\alpha=-1$.

Let $\alpha=0$, then $\beta=\gamma$ and $[b,a_{3}]=2a_{3}$. Using the
arguments given above, we can obtain that $\mathrm{\mathbf{char}}(F)\neq 2$.
In this case, $[b,a_{1}]=a_{1}$ and $[a_{1},b]=-a_{1}+\gamma a_{3}$. Let
$\lambda a_{1}+\mu b+\nu a_{3}$ be the arbitrary element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\nu[a_{1},a_{3}]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu(-a_{1}+\gamma a_{3})+\lambda\mu
a_{3}+\lambda\mu a_{1}+\mu^{2}\gamma a_{3}+2\mu\nu a_{3}=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\mu\gamma
a_{3}+\lambda\nu a_{3}+\lambda\mu a_{1}+\mu^{2}\gamma a_{3}+2\mu\nu a_{3}=$
$\displaystyle(\lambda^{2}+\lambda\mu\gamma+\lambda\nu+\mu^{2}\gamma+2\mu\nu)a_{3}.$

Then, the condition $[b,b]\neq 0$ for every element $b\not\in A$ yields that
$\lambda^{2}+\lambda\mu\gamma+\lambda\nu+\mu^{2}\gamma+2\mu\nu\neq 0$. Put
$a_{2}=b$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{30}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}+\gamma
a_{3}\ (\gamma\neq 0),[a_{2},a_{1}]=a_{1},[a_{2},a_{2}]=\gamma
a_{3},[a_{2},a_{3}]=2a_{3},$
$\displaystyle[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{30})=\zeta^{\mathrm{left}}(L_{30})=Fa_{3}$,
$[L_{30},L_{30}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{30})=\zeta(L_{30})=\langle 0\rangle$,
$\mathrm{\mathbf{char}}(F)\neq 2$. Moreover, if $\lambda a_{1}+\mu a_{2}+\nu
a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\gamma+\lambda\nu+\mu^{2}\gamma+2\mu\nu\neq 0$,
$L_{30}$ is non-nilpotent.

Let now $\alpha=-1$. Then $\gamma=1$, $[b,b]=[b,a_{3}]=a_{3}$,
$[b,a_{1}]=a_{1}-a_{3}$, $[a_{1},b]=-a_{1}+\beta a_{3}$. If $\beta=0$, then
$[a_{1},b]=-a_{1}$. Let $\lambda a_{1}+\mu b+\nu a_{3}$ be the arbitrary
element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\nu[a_{1},a_{3}]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\nu
a_{3}+\lambda\mu(a_{1}-a_{3})+\mu^{2}a_{3}+\mu\nu a_{3}=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\nu a_{3}+\lambda\mu
a_{1}-\lambda\mu a_{3}+\mu^{2}a_{3}+\mu\nu a_{3}=$
$\displaystyle(\lambda^{2}+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu)a_{3}.$

As above, $\lambda^{2}+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$. Put
$a_{2}=b$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{31}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where}$
$\displaystyle[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1}-a_{3},$
$\displaystyle[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{31})=\zeta^{\mathrm{left}}(L_{31})=Fa_{3}$,
$[L_{31},L_{31}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{31})=\zeta(L_{31})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$, $L_{31}$ is non-
nilpotent.

Suppose that $\beta\neq 0$. Let $\lambda a_{1}+\mu b+\nu a_{3}$ be the
arbitrary element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\nu[a_{1},a_{3}]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu(-a_{1}+\beta a_{3})+\lambda\nu
a_{3}+\lambda\mu(a_{1}-a_{3})+\mu^{2}a_{3}+\mu\nu a_{3}=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\mu\beta
a_{3}+\lambda\nu a_{3}+\lambda\mu a_{1}-\lambda\mu a_{3}+\mu^{2}a_{3}+\mu\nu
a_{3}=$
$\displaystyle(\lambda^{2}+\lambda\mu\beta+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu)a_{3}.$

As above,
$\lambda^{2}+\lambda\mu\beta+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$. Put
$a_{2}=b$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{32}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where}$
$\displaystyle[a_{1},a_{1}]=[a_{1},a_{3}]=[a_{2},a_{2}]=[a_{2},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=-a_{1}+\beta a_{3}\ (\beta\neq
0),[a_{2},a_{1}]=a_{1}-a_{3},$
$\displaystyle[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{32})=\zeta^{\mathrm{left}}(L_{32})=Fa_{3}$,
$[L_{32},L_{32}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{32})=\zeta(L_{32})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\beta+\lambda\nu-\lambda\mu+\mu^{2}+\mu\nu\neq 0$,
$L_{32}$ is non-nilpotent.

Suppose now that $\alpha\neq 0$ and $\alpha\neq-1$. As we have noted above,
$\alpha\neq-2$. We obtain $\beta=(\alpha+\gamma)(1+\alpha)^{-1}$. If
$\beta=0$, then $\alpha=-\gamma$, $[a_{1},b]=-a_{1}$, $[b,a_{1}]=a_{1}-\gamma
a_{3}$, $[b,a_{3}]=(2-\gamma)a_{3}$. Let $\lambda a_{1}+\mu b+\nu a_{3}$ be
the arbitrary element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\nu[a_{1},a_{3}]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}-\lambda\mu a_{1}+\lambda\nu
a_{3}+\lambda\mu(a_{1}-\gamma a_{3})+\mu^{2}\gamma
a_{3}+\mu\nu(2-\gamma)a_{3}=$ $\displaystyle\lambda^{2}a_{3}-\lambda\mu
a_{1}+\lambda\nu a_{3}+\lambda\mu a_{1}-\lambda\mu\gamma a_{3}+\mu^{2}\gamma
a_{3}+\mu\nu(2-\gamma)a_{3}=$
$\displaystyle(\lambda^{2}+\lambda\nu-\lambda\mu\gamma+\mu^{2}\gamma+\mu\nu(2-\gamma))a_{3}.$

As above,
$\lambda^{2}+\lambda\nu-\lambda\mu\gamma+\mu^{2}\gamma+\mu\nu(2-\gamma)\neq
0$. Put $a_{2}=b$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{33}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$
$\displaystyle[a_{1},a_{2}]=-a_{1},[a_{2},a_{1}]=a_{1}-\gamma a_{3}\
(\gamma\neq 0,\gamma\neq 1,\gamma\neq 2),$ $\displaystyle[a_{2},a_{2}]=\gamma
a_{3},[a_{2},a_{3}]=(2-\gamma)a_{3},[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{33})=\zeta^{\mathrm{left}}(L_{33})=Fa_{3}$,
$[L_{33},L_{33}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{33})=\zeta(L_{33})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\nu-\lambda\mu\gamma+\mu^{2}\gamma+\mu\nu(2-\gamma)\neq
0$, $L_{33}$ is non-nilpotent.

Suppose that $\beta\neq 0$. Then $[a_{1},b]=-a_{1}+\beta a_{3}\
(\beta=(\alpha+\gamma)(1+\alpha)^{-1})$, $[b,a_{1}]=a_{1}+\alpha a_{3}$,
$[b,a_{3}]=(2+\alpha)a_{3}$. Let $\lambda a_{1}+\mu b+\nu a_{3}$ be the
arbitrary element of $L$. We have:

$\displaystyle[\lambda a_{1}+\mu b+\nu a_{3},\lambda a_{1}+\mu b+\nu a_{3}]=$
$\displaystyle\lambda^{2}[a_{1},a_{1}]+\lambda\mu[a_{1},b]+\lambda\nu[a_{1},a_{3}]+\lambda\mu[b,a_{1}]+\mu^{2}[b,b]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda^{2}a_{3}+\lambda\mu(-a_{1}+\beta a_{3})+\lambda\nu
a_{3}+\lambda\mu(a_{1}+\alpha a_{3})+\mu^{2}\gamma
a_{3}+\mu\nu(2+\alpha)a_{3}=$ $\displaystyle\lambda^{2}a_{3}-\lambda\mu
a_{1}+\lambda\mu\beta a_{3}+\lambda\nu a_{3}+\lambda\mu a_{1}+\lambda\mu\alpha
a_{3}+\mu^{2}\gamma a_{3}+\mu\nu(2+\alpha)a_{3}=$
$\displaystyle(\lambda^{2}+\lambda\mu\beta+\lambda\nu+\lambda\mu\alpha+\mu^{2}\gamma+\mu\nu(2+\alpha))a_{3}.$

As above,
$\lambda^{2}+\lambda\mu\beta+\lambda\nu+\lambda\mu\alpha+\mu^{2}\gamma+\mu\nu(2+\alpha)\neq
0$. Put $a_{2}=b$. Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{34}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{3}]=a_{3},$ $\displaystyle[a_{1},a_{2}]=-a_{1}+\beta
a_{3}\ (\beta=(\alpha+\gamma)(1+\alpha)^{-1}),$
$\displaystyle[a_{2},a_{1}]=a_{1}+\alpha a_{3}\ (\alpha\neq
0,\alpha\neq-1,\alpha\neq-2),$ $\displaystyle[a_{2},a_{2}]=\gamma a_{3}\
(\gamma\neq 0),[a_{2},a_{3}]=(2+\alpha)a_{3},$
$\displaystyle[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{34})=\zeta^{\mathrm{left}}(L_{34})=Fa_{3}$,
$[L_{34},L_{34}]=Fa_{1}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{34})=\zeta(L_{34})=\langle 0\rangle$. Moreover, if
$\lambda a_{1}+\mu a_{2}+\nu a_{3}\not\in A$, then
$\lambda^{2}+\lambda\mu\beta+\lambda\nu+\lambda\mu\alpha+\mu^{2}\gamma+\mu\nu(2+\alpha)\neq
0$, $L_{34}$ is non-nilpotent.

Suppose now that $[c,c]=0$. Put again $\mathrm{\mathbf{Leib}}(L)=Fa_{3}$. Let
$b$ be an element such that $b\not\in C$.

Suppose that a subalgebra $C=Fc\oplus\mathrm{\mathbf{Leib}}(L)$ is not
abelian. As we have seen above, we can choose an element $a_{3}$ such that
$[c,a_{3}]=a_{3}$. Since $\mathrm{\mathbf{Leib}}(L)=Fa_{3}$ is an ideal,
$[b,a_{3}]=\eta a_{3}$ for some element $\eta\in F$. Using the above arguments
without loss of generality we can assume that $[b,c]\in c+Fa_{3}$,
$[c,b]\in-c+Fa_{3}$, so that $[b,c]=c+\alpha a_{3}$, $[c,b]=-c+\beta a_{3}$
for some elements $\alpha,\beta\in F$. Then

$\displaystyle a_{3}=[c,a_{3}]=[c+\alpha
a_{3},a_{3}]=[[b,c],a_{3}]=[b,[c,a_{3}]]-[c,[b,a_{3}]]=$
$\displaystyle[b,a_{3}]-[c,\eta a_{3}]=\eta a_{3}-\eta[c,a_{3}]=\eta
a_{3}-\eta a_{3}=0.$

This contradiction shows that a subalgebra $C$ is abelian.

Note that $\eta\neq 0$. In fact, otherwise $a_{3}\in\zeta(L)$, and we obtain a
contradiction. This contradiction shows that $[b,a_{3}]\neq 0$ for every
element $b\not\in C$. As we have seen above, we can choose an element $b$ such
that $[b,a_{3}]=a_{3}$.

Suppose first that there exists an element $b\not\in C$ such that $[b,b]\neq
0$. It follows that $[b,b]=\gamma a_{3}$ where $\gamma$ is a non-zero element
of $F$. Then, subalgebra $B=Fb\oplus Fa_{3}$ is non-abelian. We have now

$\displaystyle[c,a_{3}]=[c,\gamma^{-1}[b,b]]=\gamma^{-1}[c,[b,b]]=\gamma^{-1}([[c,b],b]+[b,[c,b]])=$
$\displaystyle\gamma^{-1}([-c+\beta a_{3},b]+[b,-c+\beta
a_{3}])=\gamma^{-1}(-[c,b]-[b,c]+\beta[b,a_{3}])=$
$\displaystyle\gamma^{-1}(c-\beta a_{3}-c-\alpha a_{3}+\beta
a_{3})=-\alpha\gamma^{-1}a_{3}.$

On the other hand, we proved above that $[c,a_{3}]=0$. Since $\gamma\neq 0$,
it follows that $\alpha=0$. Hence, $[b,c]=c$.

If $\beta=0$, then put $a_{1}=b$, $a_{2}=c$. A subalgebra $Fc$ is an ideal of
$L$ and we come to the following type of Leibniz algebras:

$\displaystyle L_{35}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=\gamma a_{3}\ (\gamma\neq 0),$
$\displaystyle[a_{1},a_{2}]=a_{2},[a_{1},a_{3}]=a_{3},[a_{2},a_{1}]=-a_{2},$
$\displaystyle[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{35})=\zeta^{\mathrm{left}}(L_{35})=Fa_{3}$,
$[L_{35},L_{35}]=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{35})=\zeta(L_{35})=\langle 0\rangle$, $L_{35}$ is
non-nilpotent.

If $\beta\neq 0$, then put $a_{1}=b$, $a_{2}=c$. A subalgebra $Fc$ is not an
ideal of $L$ and we come to the following type of Leibniz algebras:

$\displaystyle L_{36}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=\gamma a_{3}\ (\gamma\neq 0),$
$\displaystyle[a_{1},a_{2}]=a_{2},[a_{1},a_{3}]=a_{3},[a_{2},a_{1}]=-a_{2}+\beta
a_{3}\ (\beta\neq 0),$
$\displaystyle[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{36})=\zeta^{\mathrm{left}}(L_{36})=Fa_{3}$,
$[L_{36},L_{36}]=Fa_{2}\oplus Fa_{3}$,
$\zeta^{\mathrm{right}}(L_{36})=\zeta(L_{36})=\langle 0\rangle$, $L_{36}$ is
non-nilpotent.

Suppose now that $[b,b]=0$ for each element $b\not\in C$. Let $u=\lambda c+\mu
b+\nu a_{3}$ be an arbitrary element of $L$. Then,

$\displaystyle[\lambda c+\mu b+\nu a_{3},\lambda c+\mu b+\nu
a_{3}]=\lambda\mu[c,b]+\lambda\mu[b,c]+\mu\nu[b,a_{3}]=$
$\displaystyle\lambda\mu(-c+\beta a_{3})+\lambda\mu(c+\alpha a_{3})+\mu\nu
a_{3}=$ $\displaystyle(\lambda\mu\beta+\lambda\mu\alpha+\mu\nu)a_{3}.$

If $u\not\in C$, then $\mu\neq 0$. If $\lambda=0$, $\mu=\nu=1$, then
$[u,u]=a_{3}\neq 0$, and we obtain a contradiction. This contradiction shows
that this situation is not possible. $\Box$

The following natural situation appears when
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$. Immediately, we
obtain the following two subcases:

(IIA) the intersection $\zeta(L)\cap\mathrm{\mathbf{Leib}}(L)$ is not trivial;

(IIB) $\zeta(L)\cap\mathrm{\mathbf{Leib}}(L)=\langle 0\rangle$.

######  Theorem 2.6.

Let $L$ be a nilpotent Leibniz algebra over a field $F$ having dimension $3$,
which is not a Lie algebra. Suppose that
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$. Then, $L$ is an
algebra of the following type:

  1. $\mathrm{\mathbf{Lei}}_{37}(3,F)=L_{37}$ is a cyclic nilpotent Leibniz algebra, so that $L_{37}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{2}$, $[a_{1},a_{2}]=a_{3}$, $[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$, $\mathrm{\mathbf{Leib}}(L_{37})=[L_{37},L_{37}]=\zeta^{\mathrm{left}}(L_{37})=Fa_{2}\oplus Fa_{3}$, $\zeta^{\mathrm{right}}(L_{37})=\zeta(L_{37})=Fa_{3}$, $\mathrm{\mathbf{ncl}}(L_{37})=3$.

Proof. Since $\mathrm{\mathbf{dim}}_{F}(L)=3$,
$\mathrm{\mathbf{ncl}}(L)\leqslant 3$. Suppose first that
$\mathrm{\mathbf{ncl}}(L)=3$. Then, $L$ has an upper central series of a
length 3:

$\langle 0\rangle=C_{0}\leqslant C_{1}\leqslant C_{2}\leqslant C_{3}=L.$

Every factor of this series must be non-trivial. Therefore every factor of
this series has dimension 1. Let $a_{1}$ be an element of $L$ such that
$a_{1}\not\in C_{2}$. The fact that $L/C_{2}$ is abelian implies that
$a_{2}=[a_{1},a_{1}]\in C_{2}$. Suppose that $a_{2}\in C_{1}$. Since
$\mathrm{\mathbf{dim}}_{F}(C_{1})=1$, $C_{1}=Fa_{2}$. Choose an element $b\in
C_{2}$ such that $b\not\in C_{1}$. Then, $[a_{1},b],[b,a_{1}],[b,b]\in C_{1}$,
so that $[a_{1},b]=\alpha a_{2}$, $[b,a_{1}]=\beta a_{2}$, $[b,b]=\gamma
a_{2}$ for some elements $\alpha,\beta,\gamma\in F$. It is not hard to see
that the elements $a_{1},b,a_{2}$ generate $L$. Let
$\lambda_{1}a_{1}+\lambda_{2}b+\lambda_{3}a_{2}$,
$\mu_{1}a_{1}+\mu_{2}b+\mu_{3}a_{2}$ be two arbitrary elements of $L$. We
have:

$\displaystyle[\lambda_{1}a_{1}+\lambda_{2}b+\lambda_{3}a_{2},\mu_{1}a_{1}+\mu_{2}b+\mu_{3}a_{2}]=$
$\displaystyle\lambda_{1}\mu_{1}[a_{1},a_{1}]+\lambda_{1}\mu_{2}[a_{1},b]+\lambda_{1}\mu_{3}[a_{1},a_{2}]+$
$\displaystyle\lambda_{2}\mu_{1}[b,a_{1}]+\lambda_{2}\mu_{2}[b,b]+\lambda_{2}\mu_{3}[b,a_{2}]=$
$\displaystyle\lambda_{1}\mu_{1}a_{2}+\lambda_{1}\mu_{2}\alpha
a_{2}+\lambda_{2}\mu_{1}\beta a_{2}+\lambda_{2}\mu_{2}\gamma a_{2}\in C_{1}.$

It follows that $[L,L]\leqslant C_{1}=\zeta(L)$ and hence,
$\mathrm{\mathbf{ncl}}(L)=2$. This contradiction shows that $a_{2}\\\ not\in
C_{1}$. Then, $a_{3}=[a_{1},a_{2}]\neq 0$. It follows that $C_{2}=Fa_{2}\oplus
Fa_{3}$. We have $[a_{2},a_{3}]=[a_{3},a_{2}]=0$, so that $C_{2}$ is an
abelian subalgebra, $[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{2},a_{1}]=0$. It follows
that

$\displaystyle[a_{1}+a_{2},a_{1}+a_{2}]=[a_{1},a_{1}]+[a_{1},a_{2}]+[a_{2},a_{1}]+[a_{2},a_{2}]=$
$\displaystyle a_{2}+a_{3}\in\mathrm{\mathbf{Leib}}(L),$

and hence, $\mathrm{\mathbf{Leib}}(L)=C_{2}$. Thus, we come to the following
type of Leibniz algebras:

$\displaystyle L_{37}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{2},[a_{1},a_{2}]=a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{37})=[L_{37},L_{37}]=\zeta^{\mathrm{left}}(L_{37})=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{37})=\zeta^{\mathrm{right}}(L_{37})=Fa_{3}$,
$\mathrm{\mathbf{ncl}}(L_{37})=3$.

Suppose now that $\mathrm{\mathbf{ncl}}(L)=2$. Then, $L$ has an upper central
series of a length 2:

$\langle 0\rangle=C_{0}\leqslant C_{1}\leqslant C_{2}=L.$

Here, we have two possibilities: $\mathrm{\mathbf{dim}}_{F}(C_{1})=1$ or
$\mathrm{\mathbf{dim}}_{F}(C_{1})=2$. Since $L/C_{1}$ is abelian,
$\mathrm{\mathbf{Leib}}(L)\leqslant C_{1}$. Then, the fact that
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$ implies that
$\mathrm{\mathbf{dim}}_{F}(C_{1})=2$.

Since $L$ is a not Lie algebra, there is an element $a_{1}$ such that
$[a_{1},a_{1}]=a_{3}\neq 0$. Then, $a_{1}\not\in\zeta(L)$,
$a_{3}\in\mathrm{\mathbf{Leib}}(L)\leqslant\zeta(L)$. It follows that
$[a_{1},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{3}]=0$. We can see that $A=\langle
a_{1}\rangle=Fa_{1}\oplus Fa_{3}$. In particular, $A\cap C_{1}=Fa_{3}$. We
have $C_{1}=Fa_{2}\oplus Fa_{3}$ for some element $a_{2}$. Then, $L=A\oplus
Fa_{2}$. Since $a_{2}\in\zeta(L)$, $Fa_{2}$ is an ideal of $L$. The choice of
$a_{2}$ implies that
$[a_{1},a_{2}]=[a_{2},a_{1}]=[a_{2},a_{3}]=[a_{3},a_{2}]=[a_{2},a_{2}]=0$. It
follows that factor-algebra $L/Fa_{3}$ is abelian. Then,
$\mathrm{\mathbf{Leib}}(L)\leqslant Fa_{3}$, in particular,
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$, and we obtain a
contradiction. This contradiction shows that the case
$\mathrm{\mathbf{ncl}}(L)=2$ is not possible. $\Box$

######  Theorem 2.7.

Let $L$ be a non-nilpotent Leibniz algebra over a field $F$ having dimension
$3$, which is not a Lie algebra. Suppose that $\zeta(L)\neq\langle 0\rangle$
and that $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$. Then, $L$
is an algebra of the following type:

  1. $\mathrm{\mathbf{Lei}}_{38}(3,F)=L_{38}$ is a cyclic Leibniz algebra, $L_{24}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{2}$, $[a_{1},a_{2}]=a_{2}+a_{3}$, $[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$, $\mathrm{\mathbf{Leib}}(L_{38})=\zeta^{\mathrm{left}}(L_{38})=[L_{38},L_{38}]=Fa_{2}\oplus Fa_{3}$, $\zeta^{\mathrm{right}}(L_{38})=\zeta(L_{38})=Fa_{3}$.

Proof. We note that a Leibniz algebra of dimension 1 is abelian. Then, by our
conditions, we obtain that $\mathrm{\mathbf{dim}}_{F}(\zeta(L))=1$. If we
suppose that $\zeta(L)\cap\mathrm{\mathbf{Leib}}(L)=\langle 0\rangle$, then
$L=\zeta(L)\oplus\mathrm{\mathbf{Leib}}(L)$, so that $L$ is abelian, and we
obtain a contradiction. This contradiction shows that
$\zeta(L)\leqslant\mathrm{\mathbf{Leib}}(L)$. Put $C=\zeta(L)$ and let $c$ be
an element such that $C=Fc$. Since $C\neq\mathrm{\mathbf{Leib}}(L)$, the
factor-algebra $L/C$ is not a Lie algebra. Using the information above about
the structure of the Leibniz algebras of dimension 2, we obtain that
$L/C=F(b+C)\oplus F(d+C)$ where

$\displaystyle[d+C,d+C]=b+C,[d+C,b+C]=b+C,$
$\displaystyle[b+C,d+C]=[b+C,b+C]=C.$

Without loss of generality, we may assume that $[d,d]=b$. Then, we have
$[d,b]=b+\alpha c$, $[b,d]=0$ for some element $\alpha\in F$. The fact that
$\mathrm{\mathbf{Leib}}(L)$ is abelian implies that $[b,b]=[b,c]=[c,b]=0$. Let
$\lambda_{1}d+\lambda_{2}b+\lambda_{3}c$ be an arbitrary element of $L$. We
have:

$\displaystyle[\lambda_{1}d+\lambda_{2}b+\lambda_{3}c,\lambda_{1}d+\lambda_{2}b+\lambda_{3}c]=\lambda_{1}^{2}b+\lambda_{1}\lambda_{2}(b+\alpha
c)=$
$\displaystyle(\lambda_{1}^{2}+\lambda_{1}\lambda_{2})b+\lambda_{1}\lambda_{2}\alpha
c.$

Thus, we can see that if $\alpha=0$, then $\mathrm{\mathbf{Leib}}(L)=Fb$. In
particular, we have $\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=1$,
and we obtain a contradiction. This contradiction shows that $\alpha\neq 0$.
Put $a_{1}=d$, $a_{2}=b$, $a_{3}=\alpha c$, then

$[a_{1}-a_{2},a_{1}-a_{2}]=[a_{1},a_{1}]-[a_{2},a_{1}]-[a_{1},a_{2}]+[a_{2},a_{2}]=a_{2}-a_{2}-a_{3}=-a_{3}.$

Thus, $\mathrm{\mathbf{Leib}}(L)=Fa_{2}\oplus Fa_{3}$, and we come to the
following type of Leibniz algebras:

$\displaystyle L_{38}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{2},[a_{1},a_{2}]=a_{2}+a_{3},$
$\displaystyle[a_{1},a_{3}]=[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{38})=\zeta^{\mathrm{left}}(L_{38})=[L_{38},L_{38}]=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{38})=\zeta^{\mathrm{right}}(L_{38})=Fa_{3}$. $\Box$

######  Theorem 2.8.

Let $L$ be a Leibniz algebra over a field $F$ having dimension $3$ and $L$ not
be a Lie algebra. Suppose that $\zeta(L)=\langle 0\rangle$ and that
$\mathrm{\mathbf{dim}}_{F}(\mathrm{\mathbf{Leib}}(L))=2$. Then, $L$ is an
algebra of one of the following types.

  1. (i)

$\mathrm{\mathbf{Lei}}_{39}(3,F)=L_{39}$ is a direct sum of ideal $B=Fa_{3}$
and cyclic non-nilpotent subalgebra $A=Fa_{1}\oplus Fa_{2}$, $[A,B]=Fa_{3}$,
$[B,A]=\langle 0\rangle$, so that $L_{39}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$
where $[a_{1},a_{1}]=[a_{1},a_{2}]=a_{2}$, $[a_{1},a_{3}]=\beta a_{3}\
(\beta\neq 0)$,
$[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{39})=\zeta^{\mathrm{left}}(L_{39})=[L_{39},L_{39}]=Fa_{2}\oplus
Fa_{3}$,$\zeta(L_{39})=\zeta^{\mathrm{right}}(L_{39})=\langle 0\rangle$.

  2. (ii)

$\mathrm{\mathbf{Lei}}_{40}(3,F)=L_{40}$ is a cyclic Leibniz algebra, so that
$L_{40}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{2}$,
$[a_{1},a_{2}]=a_{2}+\gamma a_{3}$ $(\gamma\neq 0)$, $[a_{1},a_{3}]=\beta
a_{3}\ (\beta\neq 0)$,
$[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{40})=\zeta^{\mathrm{left}}(L_{40})=[L_{40},L_{40}]=Fa_{2}\oplus
Fa_{3}$, $\zeta^{\mathrm{right}}(L_{40})=\zeta(L_{40})=\langle 0\rangle$.

  3. (iii)

$\mathrm{\mathbf{Lei}}_{41}(3,F)=L_{41}$ is a cyclic Leibniz algebra, so that
$L_{41}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{2}$,
$[a_{1},a_{2}]=\gamma a_{3}\ (\gamma\neq 0)$, $[a_{1},a_{3}]=a_{3}$,
$[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{41})=\zeta^{\mathrm{left}}(L_{41})=[L_{41},L_{41}]=Fa_{2}\oplus
Fa_{3}$, $\zeta^{\mathrm{right}}(L_{41})=\zeta(L_{41})=\langle 0\rangle$.

  4. (iv)

$\mathrm{\mathbf{Lei}}_{42}(3,F)=L_{42}$ is a cyclic Leibniz algebra, so that
$L_{42}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}$ where $[a_{1},a_{1}]=a_{2}$,
$[a_{1},a_{2}]=a_{3}$, $[a_{1},a_{3}]=\beta a_{2}+\gamma a_{3}$,
$[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0$,
$\mathrm{\mathbf{Leib}}(L_{42})=\zeta^{\mathrm{left}}(L_{42})=[L_{42},L_{42}]=Fa_{2}\oplus
Fa_{3}$, $\zeta^{\mathrm{right}}(L_{42})=\zeta(L_{42})=\langle 0\rangle$.
Moreover, polynomial $X^{2}-\gamma X-\beta$ is irreducible over field $F$.

Proof. Suppose first that $\mathrm{\mathbf{Leib}}(L)$ includes an ideal $K$ of
dimension 1. Let $c$ be an element such that $K=Fc$. Since
$K\neq\mathrm{\mathbf{Leib}}(L)$, the factor-algebra $L/K$ is not a Lie
algebra. Using the above information about the structure of the Leibniz
algebras of dimension 2, we obtain that $L/K=F(b+K)\oplus F(d+K)$ where

$\displaystyle[d+K,d+K]=b+K,[d+K,b+K]=b+K,$
$\displaystyle[b+K,d+K]=[b+K,b+K]=K$

or

$\displaystyle[d+K,d+K]=b+K,$ $\displaystyle[d+K,b+K]=[b+K,d+K]=[b+K,b+K]=K.$

Consider the first situation. Without loss of generality, we may assume that
$[d,d]=b$. Then, we have $[b,d]=0$, $[d,b]=b+\alpha c$ for some element
$\alpha\in F$. The fact that $\mathrm{\mathbf{Leib}}(L)$ is abelian implies
that $[b,b]=[b,c]=[c,b]=0$. Since $\zeta(L)=\langle 0\rangle$, $[d,c]=\beta c$
for some non-zero element $\beta\in F$. Put $a_{3}=\beta c$, then $K=Fa_{3}$
and $[b,a_{3}]=[a_{3},b]=[a_{3},d]=[a_{3},a_{3}]=0$, $[d,b]=b+\gamma a_{3}$
where $\gamma=\alpha\beta^{-1}$. Let
$u=\lambda_{1}d+\lambda_{2}b+\lambda_{3}a_{3}$ be an arbitrary element of $L$.
We have:

$\displaystyle[u,u]=[\lambda_{1}d+\lambda_{2}b+\lambda_{3}a_{3},\lambda_{1}d+\lambda_{2}b+\lambda_{3}a_{3}]=$
$\displaystyle\lambda_{1}^{2}[d,d]+\lambda_{1}\lambda_{2}[d,b]+\lambda_{1}\lambda_{3}[d,a_{3}]=$
$\displaystyle\lambda_{1}^{2}b+\lambda_{1}\lambda_{2}(b+\gamma
a_{3})+\lambda_{1}\lambda_{3}\beta a_{3}=$
$\displaystyle(\lambda_{1}^{2}+\lambda_{1}\lambda_{2})b+(\gamma\lambda_{1}\lambda_{2}+\lambda_{1}\lambda_{3}\beta)a_{3}.$

If we put $\lambda_{1}=1$, $\lambda_{2}=-1$,
$\lambda_{3}=\beta^{-1}(1+\gamma)$, then we obtain $[u,u]=a_{3}$. Put
$a_{1}=d$, $a_{2}=b$.

If $\alpha=0$ and hence, $\gamma=0$, then a subalgebra $Fa_{2}$ is an ideal of
$L$ and $A=\langle a_{1}\rangle=Fa_{1}\oplus Fa_{2}$ is a cyclic subalgebra.
Thus, we come to the following type of Leibniz algebras:

$\displaystyle L_{39}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=[a_{1},a_{2}]=a_{2},$ $\displaystyle[a_{1},a_{3}]=\beta a_{3}\
(\beta\neq 0),$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{39})=\zeta^{\mathrm{left}}(L_{39})=[L_{39},L_{39}]=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{39})=\zeta^{\mathrm{right}}(L_{39})=\langle 0\rangle$.

If $\alpha\neq 0$, then $a_{3}=\gamma^{-1}([a_{1},a_{2}]-a_{2})$. It follows
that $L$ is a cyclic algebra. Thus, we come to the following type of Leibniz
algebras:

$\displaystyle L_{40}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{2},$ $\displaystyle[a_{1},a_{2}]=a_{2}+\gamma a_{3}\
(\gamma\neq 0),[a_{1},a_{3}]=\beta a_{3}\ (\beta\neq 0),$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{40})=\zeta^{\mathrm{left}}(L_{40})=[L_{40},L_{40}]=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{40})=\zeta^{\mathrm{right}}(L_{40})=\langle 0\rangle$.

Consider now a situation when $L/K=F(b+K)\oplus F(d+K)$ where

$\displaystyle[d+K,d+K]=b+K,$ $\displaystyle[d+K,b+K]=[b+K,d+K]=[b+K,b+K]=K.$

Without loss of generality we may assume that $[d,d]=b$. Then we have
$[b,d]=0$, $[d,b]=\alpha c$ for some element $\alpha\in F$. The fact that
$\mathrm{\mathbf{Leib}}(L)$ is abelian implies that $[b,b]=[b,c]=[c,b]=0$.
Since $\zeta(L)=\langle 0\rangle$, $[d,c]=\beta c$ for some non-zero element
$\beta\in F$. Put $a_{3}=c$, then $K=Fa_{3}$ and
$[b,a_{3}]=[a_{3},b]=[a_{3},d]=[a_{3},a_{3}]=0$, $[d,b]=\alpha a_{3}$.

If $\alpha=0$, then $Fb$ lies in the center of $L$, and we obtain a
contradiction.

Suppose that $\alpha\neq 0$. Put $a_{1}=\beta^{-1}d$, then $[a_{1},c]=c$.
Further $[a_{1},a_{1}]=[\beta^{-1}d,\beta^{-1}d]=\beta^{-2}b=a_{2}$. We have
$[a_{2},a_{2}]=[a_{2},a_{1}]=[a_{2},c]=[c,a_{2}]=0$,
$[a_{1},a_{2}]=[\beta^{-1}d,\beta^{-2}b]=\beta^{-3}\alpha c=\gamma c$. Thus we
come to the following type of Leibniz algebra:

$\displaystyle L_{41}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{2},$ $\displaystyle[a_{1},a_{2}]=\gamma a_{3}\ (\gamma\neq
0),[a_{1},a_{3}]=a_{3},$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{41})=\zeta^{\mathrm{left}}(L_{41})=[L_{41},L_{41}]=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{41})=\zeta^{\mathrm{right}}(L_{41})=\langle 0\rangle$.

Suppose now that $\mathrm{\mathbf{Leib}}(L)$ does not include proper non-zero
ideals. Since $L$ is a non-Lie algebra, there is an element $d$ such that
$[d,d]=b\neq 0$. Then, $d\not\in\mathrm{\mathbf{Leib}}(L)$. Put
$K=\mathrm{\mathbf{Leib}}(L)$. By our assumption, $[d,b]=c\not\in Fb$. The
fact that $\mathrm{\mathbf{dim}}_{F}(K)=2$ implies that $K=Fb\oplus Fc$. Then,
$[d,c]=\beta b+\gamma c$ for some elements $\beta,\gamma\in F$. The mapping
$x\rightarrow[d,x]$, $x\in K$ is linear. Our conditions imply that a
polynomial $X^{2}-\gamma X-\beta$ is irreducible over a field $F$. Put
$a_{1}=d$, $a_{2}=b$, $a_{3}=c$. Thus, we come to the following type of
Leibniz algebras:

$\displaystyle L_{42}=Fa_{1}\oplus Fa_{2}\oplus Fa_{3}\ \mbox{where
}[a_{1},a_{1}]=a_{2},[a_{1},a_{2}]=a_{3},$ $\displaystyle[a_{1},a_{3}]=\beta
a_{2}+\gamma a_{3},$
$\displaystyle[a_{2},a_{1}]=[a_{2},a_{2}]=[a_{2},a_{3}]=[a_{3},a_{1}]=[a_{3},a_{2}]=[a_{3},a_{3}]=0.$

Note also that
$\mathrm{\mathbf{Leib}}(L_{42})=\zeta^{\mathrm{left}}(L_{42})=[L_{42},L_{42}]=Fa_{2}\oplus
Fa_{3}$, $\zeta(L_{42})=\zeta^{\mathrm{right}}(L_{42})=\langle 0\rangle$,
polynomial $X^{2}-\gamma X-\beta$ is irreducible over field $F$. $\Box$

## References

  * [1] Sh. Ayupov, B. Omirov, I. Rakhimov, Leibniz Algebras: Structure and Classification, CRC Press, Taylor & Francis Group, (2020).
  * [2] A. Blokh, On a generalization of the concept of Lie algebra, Dokl. Akad. Nauk SSSR, 165(3), 471-473 (1965) (in Russian).
  * [3] J.M. Casas, M.A. Insua, M. Ladra, S. Ladra, An algorithm for the classification of $3$-dimensional complex Leibniz algebras, Linear Algebra Appl., 436(9), 3747-3756 (2012); DOI:10.1016/j.laa.2011.11.039.
  * [4] I. Demir, K.C. Misra, E. Stitzinger, On some structures of Leibniz algebras, Recent Advances in Representation Theory, Quantum Groups, Algebraic Geometry, and Related Topics, Contemporary Mathematics, 623, 41-54 (2014). DOI:10.1090/conm/623/12456.
  * [5] A.Kh. Khudoyberdiyev, T.K. Kurbanbaev, B.A. Omirov, Classification of three-dimensional solvable $p$-adic Leibniz algebras, $p$-Adic Numbers Ultrametric Anal. Appl., 2(3), 207-221 (2010); DOI:10.1134/ S2070046610030039.
  * [6] V.V. Kirichenko, L.A. Kurdachenko, A.A. Pypka, I.Ya. Subbotin, Some aspects of Leibniz algebra theory, Algebra Discrete Math., 24(1), 1-33 (2017).
  * [7] L.A. Kurdachenko, J. Otal, A.A. Pypka, Relationships between the factors of the canonical central series of Leibniz algebras, Eur. J. Math., 2(2), 565-577 (2016); DOI:10.1007/s40879-016-0093-5.
  * [8] J.-L. Loday, Cyclic homology, Grundlehren der Mathematischen Wissenschaften, 301, Springer Verlag, (1992); DOI:10.1007/978-3-662-11389-9.
  * [9] J.-L. Loday, Une version non commutative des algèbres de Lie; les algèbras de Leibniz, Enseign. Math., 39, 269-293 (1993).
  * [10] I.S. Rakhimov, I.M. Rikhsiboev, M.A. Mohammed, An algorithm for classifications of three-dimensional Leibniz algebras over arbitrary fields, JP Journal of Algebra, Number Theory and Applications, 40(2), 181-198 (2018); DOI:10.17654/NT040020181.
  * [11] I.M. Rikhsiboev, I.S. Rakhimov, Classification of three dimensional complex Leibniz algebras, AIP Conference Proceedings, 1450(1), 358-362 (2012); DOI:10.1063/1.4724168.
  * [12] V.S. Yashchuk, On some Leibniz algebras, having small dimension, Algebra Discrete Math., 27(2), 292-308 (2019).