WEAK C-IDEALS OF LEIBNIZ ALGEBRAS

David A. TOWERS and Zekiye CILOGLU

Department of Mathematics and Statistics

Lancaster University

Lancaster LA1 4YF

England

<EMAIL_ADDRESS>
and

Department of Mathematics

Suleyman Demirel University

$32260$, Isparta

TURKEY

<EMAIL_ADDRESS>

###### Abstract

A subalgebra $B$ of a Leibniz algebra $L$ is called a weak c-ideal of $L$ if
there is a subideal $C$ of $L$ such that $L=B+C$ and $B\cap C\subseteq B_{L}$
where $B_{L}$ is the largest ideal of $L$ contained in $B.$ This is analogous
to the concept of a weakly c-normal subgroup, which has been studied by a
number of authors. We obtain some properties of weak c-ideals and use them to
give some characterisations of solvable and supersolvable Leibniz algebras
generalising previous results for Lie algebras. We note that one-dimensional
weak c-ideals are c-ideals, and show that a result of Turner in [15, Theorem
3.2.9] classifying Leibniz algebras in which every one-dimensional subalgebra
is a c-ideal is false for general Leibniz algebras, but holds for symmetric
ones.
Mathematics Subject Classification 2010: 17B05, 17B20, 17B30, 17B

Key Words and Phrases: Weak c-ideal; Frattini ideal; Leibniz algebra;
Symmetric Leibniz algebra; Solvable; Supersolvable

##  1 Introduction

An algebra $L$ over a field $F$ is called a Leibniz algebra if, for every
$x,y,z\in L$, we have

$[x,[y,z]]=[[x,y],z]-[[x,z],y]$

In other words the right multiplication operator $R_{x}:L\rightarrow
L:y\mapsto[y,x]$ is a derivation of $L$. As a result such algebras are
sometimes called right Leibniz algebras, and there is a corresponding notion
of left Leibniz algebras, which satisfy

$[x,[y,z]]=[[x,y],z]+[y,[x,z]].$

Clearly the opposite of a right (left) Leibniz algebra is a left (right)
Leibniz algebra, so, in most situations, it does not matter which definition
we use. A symmetric Leibniz algebra $L$ is one which is both a right and left
Leibniz algebra and in which $[[x,y],[x,y]]=0$ for all $x,y\in L$. This last
identity is only needed in characteristic two, as it follows from the right
and left Leibniz identities otherwise (see [7, Lemma 1]). Symmetric Leibniz
algebras $L$ are flexible, power associative and have $x^{3}=0$ for all $x\in
L$ (see [6, Proposition 2.37]), and so, in a sense, are not far removed from
Lie algebras.

Every Lie algebra is a Leibniz algebra and every Leibniz algebra satisfying
$[x,x]=0$ for every element is a Lie algebra. They were introduced in 1965 by
Bloh ([3]) who called them $D$-algebras, though they attracted more widespread
interest, and acquired their current name, through work by Loday and
Pirashvili ([8], [9]). They have natural connections to a variety of areas,
including algebraic $K$-theory, classical algebraic topology, differential
geometry, homological algebra, loop spaces, noncommutative geometry and
physics. A number of structural results have been obtained as analogues of
corresponding results in Lie algebras.

The Leibniz kernel is the set $I=$ span$\\{x^{2}:x\in L\\}$. Then $I$ is the
smallest ideal of $L$ such that $L/I$ is a Lie algebra. Also $[L,I]=0$.

We define the following series:

$L^{1}=L,L^{k+1}=[L^{k},L](k\geq 1)\hbox{ and
}L^{(0)}=L,L^{(k+1)}=[L^{(k)},L^{(k)}](k\geq 0).$

Then $L$ is nilpotent of class n (resp. solvable of derived length n) if
$L^{n+1}=0$ but $L^{n}\neq 0$ (resp. $L^{(n)}=0$ but $L^{(n-1)}\neq 0$) for
some $n\in{\mathbb{N}}$. It is straightforward to check that $L$ is nilpotent
of class n precisely when every product of $n+1$ elements of $L$ is zero, but
some product of $n$ elements is non-zero.The nilradical, $N(L)$, (resp.
radical, $R(L)$) is the largest nilpotent (resp. solvable) ideal of $L$.

In [14] we introduced the concept of a weak c-ideal, which is analogous to the
concept of a weakly c-normal subgroup as introduced by Zhu, Guo and Shum in
[16] and which has since been further studied by a number of authors. It is a
generalisation of the concept of a c-ideal, as introduced in [13]. Here we
investigate the extent to which the results in [14] can be extended to Leibniz
algebras.

In section two we give some basic properties of weak c-ideals; in particular,
it is shown that weak c-ideals inside the Frattini subalgebra of a Leibniz
algebra $L$ are necessarily ideals of $L$. In section three we first show that
all maximal subalgebras of $L$ are weak c-ideals of $L$ if and only if $L$ is
solvable and that $L$ has a solvable maximal subalgebra that is a weak c-ideal
if and only if $L$ is solvable. Unlike the corresponding results for c-ideals,
it is necessary to restrict the underlying field to characteristic zero, as
was shown by an example in [14]. Finally we have that if all maximal nilpotent
subalgebras of $L$ are weak c-ideals, or if all Cartan subalgebras of $L$ are
weak c-ideals and $F$ has characteristic zero, then $L$ is solvable.

In section four we show that if $L$ is a solvable symmetric Lie algebra over a
general field and every maximal subalgebra of each maximal nilpotent
subalgebra of $L$ is a weak c-ideal of $L$ then $L$ is supersolvable. If each
of the maximal nilpotent subalgebras of $L$ has dimension at least two then
the assumption of solvability can be removed. Similarly if the field has
characteristic zero and $L$ is not three-dimensional simple then this
restriction can be removed.

In the final section we see that every one-dimensional subalgebra is a weak
c-ideal if and only if it is a c-ideal, and go on to study the class of
Leibniz algebras in which every one-dimensional subalgebra is a c-ideal. It is
shown that the cyclic subalgebras in this class are at most two dimensional.
There is a characterisation of all of the algebras in this class given by
Turner in [15, Theorem 3.2.9], but an example is given to show that this
result is false. A number of properties of such algebras are given, but a full
classification appears complicated. However, it is shown finally that Turner’s
result does hold for symmetric Leibniz algebras.

Throughout, the term ’Leibniz algebra’ will refer to a finite-dimensional
right Leibniz algebra over a field $F$. If $A,B$ are subalgebras of $L$ with
$A\subseteq B$, the centraliser of $A$ in $B$, $C_{B}(A)=\\{b\in
B\mid[b,A]+[A,b]=0\\}$. The normaliser of $A$ in $B$, $N_{B}(A)=\\{b\in
B\mid[b,A]+[A,b]\subseteq A\\}$. Algebra direct sums will be denoted by
$\oplus$, whereas direct sums of the vector space structure alone will be
written as $\dot{+}$. Subsets will be denoted by ‘$\subseteq$’ and proper
subsets by ‘$\subset$’.

##  2 Preliminary Results

###### Definition 1

[10] Let $L$ be a Leibniz algebra over a field $F$ and $B$ be a subalgebra of
$L$. We call $B$ a subideal of $L$ if there is a chain of subalgebras

$B=B_{t}\subset B_{t-1}\subset...\subset B_{0}=L$

such that $B_{i}$ is an ideal of $B_{i-1}$ for each $1\leq i\leq t.$

###### Definition 2

[10] Let $L$ be a Leibniz algebra and $H$ a subalgebra of $L$. Then $H$ is
called a c-ideal of $L$ if there is an ideal $K$ of $L$ such that $L=H+K$ and
$H\cap K$ is contained in the core of $H$ (with respect to $L$), denoted by
$H_{L}$, where this is the largest ideal of $L$ contained in $H$.

###### Definition 3

A subalgebra $B$ of a Leibniz algebra $L$ is a weak c-ideal of $L$ if there
exists a subideal $C$ in $L$ such that $L=B+C$ and $B\cap C\subseteq B_{L}$.

###### Definition 4

A Leibniz algebra $L$ is called weakly c-simple if $L$ does not contain any
weak c-ideals other than $L$, the trivial subalgebra $0$ and the Leibniz
kernel $I.$ It is simple if these same three subalgebras are the only ideals
of $L$.

###### Lemma 2.1

Let $L$ be a Leibniz algebra. Then the following statements are valid:

  1. 1.

Let $B$ be a subalgebra of $L$. If $B$ is a c-ideal of $L$ then $B$ is a weak
c-ideal of $L.$

  2. 2.

$L$ is weakly c-simple if and only if $L$ is simple.

  3. 3.

If $B$ is a weak c-ideal of $L$ and $K$ is a subalgebra with $B\subseteq
K\subseteq L$ then $B$ is a weak c-ideal of $K$.

  4. 4.

If $A$ is an ideal of $L$ and $A\subseteq B$ then $B$ is a weak c-ideal of $L$
if and only if $B/A$ is a weak c-ideal of $L/A.$

Proof.

  1. 1.

This is apparent from the definition.

  2. 2.

Suppose first that $L$ is simple and let $B$ be a weak c-ideal with $B\neq L$.
Then by definition of weak c-ideal

$L=B+C\hbox{ and }B\cap C\subseteq B_{L}$

where $C$ is a subideal of $L.$ But since $L$ is simple $B_{L}$ must be $0$ or
$I.$ If $B_{L}=0$ then $C=L$, since $C\neq 0$ and is a subideal of $L$, whence
$B=0$. If $B_{L}=I$ then, similarly, $C=L$ and hence $B=I$. Hence $L$ is
weakly c-simple.

Conversely, suppose $L$ is weakly c-simple. Then, since every ideal of $L$ is
a weak c-ideal, $L$ must be simple.

  3. 3.

If $B$ is a weak c-ideal of $L$, then there exist a subideal $C$ of $L$ such
that

$L=B+C\text{ and }B\cap C\subseteq B_{L}$

Now $K=K\cap L=K\cap\left(B+C\right)=B+\left(K\cap C\right).$ Since $C$ is a
subideal of $L$ there exists a chain of subalgebras

$C=C_{n}\subset C_{n-1}\subset...\subset C_{0}=L$

where $C_{j}$ is an ideal of $C_{j-1}$ for each $1\leq j\leq n.$ If we
intersect each term in this chain with $K$ we get

$C\cap K=C_{n}\cap K\subseteq C_{n-1}\cap K\subseteq\ ...\subseteq C_{0}\cap
K=L\cap K=K$

and obviously $C_{j}\cap K$ is an ideal of $C_{j-1}\cap K$ for each $0\leq
j\leq n.$ Thus $C\cap K$ is a subideal of $K.$ Also,

$B\cap\left(C\cap K\right)\subseteq B_{K}$

and so that $B$ is a $weak\ c$-$ideal$ of $K.$

(4) Suppose first that $B/A$ is a $weak\ c$-$ideal$ of $L/A.$ Then there
exists a subideal $C/A$ of $L/A$ such that

$\frac{L}{A}=\frac{B}{A}+\frac{C}{A}\text{ and
}\frac{B}{A}\cap\frac{C}{A}\subseteq\left(\frac{B}{A}\right)_{L/A}=\frac{B_{L}}{A}$

It follows that $L=B+C$ and $B\cap C\subseteq B_{L}$ where $C$ is a subideal
of $L.$

Suppose conversely that $A$ is an ideal of $L$ with $A\subseteq B$ and $B$ is
a $weak\ c$-$ideal$ of $L.$ Then there exists a $C$ subideal of $L$ such that

$L=B+C\text{ and }B\cap C\subseteq B_{L}.$

Since $A$ is an ideal and $A\leq B$ the factor algebra

$\frac{L}{A}=\frac{B+C}{A}=\frac{B}{A}+\frac{C+A}{A}$

where $(C+A)/A$ is a subideal of $L/A$ and

$\frac{B}{A}\cap\frac{C+A}{A}=\frac{B\cap(C+A)}{A}=\frac{A+B\cap
C}{A}\subseteq\frac{B_{L}}{A}=\left(\frac{B}{A}\right)_{L/A}$

so $B/A$ is a $weak\ c$-$ideal$ of $L/A.$

$\Box$

The Frattini subalgebra $F(L)$ of a Leibniz algebra $L$ is the intersection of
all of the maximal subalgebras of $L.$ The Frattini ideal, $F(L)_{L}$, of $L$
is denoted by $\phi(L).$

The next result is a generalisation of [13, Proposition 2.2] and the same
proof works, but we will include it for completeness.

###### Proposition 2.2

Let $B$, $C$ be subalgebras of $L$ with $B\subseteq F(C).$ If $B$ is a weak
c-ideal of $L,$ then $B$ is an ideal of $L$ and $B\subseteq\varphi(L).$

Proof. Suppose that $L=B+K$ where $K$ is a subideal of $L$ and $B\cap
K\subseteq B_{L}.$ Then

$C=C\cap L=C\cap(B+K)=B+C\cap K=C\cap K$

since $B\subseteq F(C).$ Hence, $B\subseteq C\subseteq K,$ giving $B=B\cap
K\subseteq B_{L}$ and $B$ is an ideal of $L.$ It then follows from [12, Lemma
4.1] that $B\subseteq\varphi(L).$ $\Box$

An ideal $A$ is complemented in $L$ if there is a subalgebra $U$ of $L$ such
that $L=A+U$ and $A\cap U=0.$ We adapt this to define a subideal complement as
follows:

###### Definition 5

Let $L$ be a Leibniz algebra and let $B$ be a subalgebra of $L.$ Then $B$ has
a subideal complement in $L$ if there is a subideal $C$ of $L$ such that
$L=B+C$ and $B\cap C=0.$

Then we have the following lemma:

###### Lemma 2.3

If $B$ is a weak c-ideal of a Leibniz algebra $L,$ then $B/B_{L}$ has a
subideal complement in $L/B_{L}.$ Conversely, if $B$ is a subalgebra of $L$
such that $B/B_{L}$ has a subideal complement in $L/B_{L},$ then $B$ is a weak
c-ideal of $L.$

Proof. Let $B$ be a weak c-ideal of $L$. Then there exists a subideal $C$ of
$L$ such that $B+C=L$ and $B\cap C\subseteq B_{L}.$ If $B_{L}=0$ then $B\cap
C=0$ and $C$ is a subideal complement of $B$ in $L.$ So, assume that
$B_{L}\neq 0,$ then we can construct the factor algebras $B/B_{L}$ and
$(C+B_{L})/B_{L}.$ If we intersect these two factor algebras;

$\displaystyle\frac{B}{B_{L}}\cap\frac{C+B_{L}}{B_{L}}$ $\displaystyle=$
$\displaystyle\frac{B\cap(C+B_{L})}{B_{L}}$ $\displaystyle=$
$\displaystyle\frac{B_{L}+(B\cap C)}{B_{L}}$ $\displaystyle\subseteq$
$\displaystyle\frac{B_{L}}{B_{L}}=0$

Hence, $(C+B_{L})/B_{L}$ is a subideal complement of $B/B_{L}$ in $L/B_{L}.$
Conversely, if $K$ is a subideal of $L$ such that $K/B_{L}$ is a subideal
complement of $B/B_{L}$ in $L/B_{L}$ then we have that

$\frac{L}{B_{L}}=\frac{B}{B_{L}}+\frac{K}{B_{L}}\text{ and
}\frac{B}{B_{L}}\cap\frac{K}{B_{L}}=0$

Then $L=B+K$ and $B\cap K=0\subseteq B_{L}$, whence $B$ is a weak c-ideal of
$L.$ $\Box$

##  3 Some characterisations of solvable algebras

###### Theorem 3.1

Let $L$ be a Leibniz algebra over a field $F$ of characteristic zero and let
$B$ be an ideal of $L$. Then $B$ is solvable if and only if every maximal
subalgebra of $L$ not containing $B$ is a weak $c$-ideal of $L.$

Proof. Suppose first that $B$ is solvable and let $M$ be a maximal subalgebra
of $L$ not containing $B$. Then there exists $k\in{\mathbb{N}}$ such that
$B^{(k+1)}\subseteq M$, but $B^{(k)}\not\subseteq M$. Clearly $L=M+B^{(k)}$
and $B^{(k)}\cap M$ is an ideal of $L$, so $B^{(k)}\cap M\subseteq M_{L}$. It
follows that $M$ is a $c$-ideal and hence a weak $c$-ideal of $L$.

Conversely, suppose every maximal subalgebra of $L$ not containing $B$ is a
weak $c$-ideal of $L$. Let $M/I$ be a maximal subalgebra of $L/I$ not
containing $(B+I)/I$. Then $M$ is a maximal subalgebra of $L$ not containing
$B$, and so is a weak $c$-ideal of $L$. Hence $M/I$ is a weak $c$-ideal of
$L/I$, by Lemma 2.1. Since $L/I$ is a Lie algebra, it follows from [14,
Theorem 3.2] that $(B+I)/I$ is solvable. It follows that $B+I$, and hence $B$,
is solvable. $\Box$

###### Corollary 3.2

Let $L$ be a Leibniz algebra over a field $F$ of characteristic zero. Then $L$
is solvable if and only if every maximal subalgebra of $L$ is a weak $c$-ideal
of $L$.

###### Lemma 3.3

Let $L=U+C$ be a Leibniz algebra, where $U$ is a solvable subalgebra of $L$
and $C$ is a subideal of $L$. Then there exists $n_{0}\in{\mathbb{N}}$ such
that $L^{(n_{0})}\subseteq C$.

Proof. This is the same as for [14, Lemma 3.5]. $\Box$

###### Theorem 3.4

Let $L$ be a Leibniz algebra over a field $F$ of characteristic zero. Then $L$
has a solvable maximal subalgebra that is a weak c-ideal of $L$ if and only if
$L$ is solvable.

Proof. Suppose first that $L$ has a solvable maximal subalgebra $M$ that is a
weak c-ideal of $L$. We show that $L$ is solvable. Let $L$ be a minimal
counter-example. Then there is a subideal $K$ of $L$ such that $L=M+K$ and
$M\cap K\subseteq M_{L}$. If $M_{L}\neq 0$ then $L/M_{L}$ is solvable, by the
minimality assumption, and $M_{L}$ is solvable, whence $L$ is solvable, a
contradiction. It follows that $M_{L}=0$ and $L=M\dot{+}K$. If $R$ is the
solvable radical of $L$ then $R\subseteq M_{L}=0$, so $L$ is a semisimple Lie
algebra, by [2, Theorem 1]. But now, for all $n\geq 1$, $L=L^{(n)}\subseteq
K\neq L$, by Lemma 3.3, a contradiction. The result follows.

The converse follows from Corollary 3.2. $\Box$

###### Theorem 3.5

Let $L$ be a Leibniz algebra over a field of characteristic zero such that all
maximal nilpotent subalgebras are weak $c$-ideals of $L$. Then $L$ is
solvable.

Proof. Suppose that $L$ is not solvable but that all maximal nilpotent
subalgebras of $L$ are weak $c$-ideals of $L$. Let $L=R\dot{+}S$ be the Levi
decomposition of $L$, where $S\neq 0$ is a semisimple Lie algebra ([2, Theorem
1]). Let $B$ be a maximal nilpotent subalgebra of $S$ and $U$ be a maximal
nilpotent subalgebra of $L$ containing it. Then there is a subideal $C$ of $L$
such that $L=U+C$ and $U\cap C\subseteq U_{L}$. It follows from Lemma 3.3 that
$S=S^{(n_{0})}\subseteq L^{(n_{0})}\subseteq C$, and so $B\subseteq U\cap
C\subseteq U_{L}$, whence $S\cap U_{L}\neq 0$. But $S\cap U_{L}$ is an ideal
of $S$ and so is semisimple. Since $U$ is nilpotent this is a contradiction.
$\Box$

###### Definition 6

A Cartan subalgebra of a Leibniz algebra $L$ is a nilpotent subalgebra $C$
such that $C=N_{L}(C)$. Over a field of characteristic zero such subalgebras
certainly exist (see [1, Section 6]).

The following result is a generalisation of a result of Dixmier in [5].

###### Lemma 3.6

Let $L$ be a Leibniz algebra over a field of characteristic zero with non-zero
Levi factor $S$. If $H$ is a Cartan subalgebra of $S$ and $B$ is a Cartan
subalgebra of its centralizer in the the solvable radical of $L$, then $H+B$
is a Cartan subalgebra of $L$.

Proof. Let $R$ be the solvable radical of $L$. Then $(H+B)^{r}=H^{r}+B^{r}$
for all $r\geq 1$, so $H+B$ is a nilpotent subalgebra of $L$. Let $x\in
N_{L}(H+B)$ and put $x=s+r$ where $s\in S$ and $r\in R$. Then

$[x,H]+[H,x]=[s,H]+[H,s]+[r,H]+[H,r]\subseteq H+B,$

so $[H,s]+[s,H]\subseteq H$, whence $s\in N_{S}(H)=H$. Moreover,

$[x,B]+[B,x]=[s,B]+[B,s]+[r,B]+[B,r]\subseteq H+B,$

so $[B,r]+[r,B]\subseteq B$, since $[s,B]+[B,s]=0$, whence $r\in
N_{R}(B)=B$.Thus, $N_{L}(H+B)=H+B$ and $H+B$ is a Cartan subalgebra of $L$.
$\Box$

###### Theorem 3.7

Let $L$ be a Leibniz algebra over a field of characteristic zero in which
every Cartan subalgebra of $L$ is a weak $c$-ideal of $L$. Then $L$ is
solvable.

Proof. Suppose that every Cartan subalgebra of $L$ is a weak $c$-ideal of $L$
and that $L$ has a non-zero Levi factor $S.$ Let $H$ be a Cartan subalgebra of
$S$ and let $B$ be a Cartan subalgebra of its centralizer in the solvable
radical of $L.$ Then $C=H+B$ is a Cartan subalgera of $L$, by Lemma 3.6, and
there is a subideal $K$ of $L$ such that $L=C+K$ and $C\cap K\subseteq C_{L}$.
Now there is an $n_{0}\geq 2$ such that $L^{(n_{0})}\subseteq K$ by Lemma 3.3.
But $S\subseteq L^{(n_{0})}\subseteq K$ and so $C\cap S\subseteq C\cap
K\subseteq C_{L}$ giving $C\cap S\subseteq C_{L}\cap S=0,$ a contradiction. It
follows that $S=0$ and, hence, that $L$ is solvable. $\Box$

##  4 Some characterisations of supersolvable algebras

In this section we will restrict attention to symmetric Leibniz algebras. We
know of no examples of a Leibniz algebra which is not symmetric and for which
the results are false, but have been unable to establish them in that more
general case. First we need the following lemma which holds in the general
case.

###### Lemma 4.1

[15, Lemma 5.1.2] Let $L$ be a Leibniz algebra over any field, let $A$ be an
ideal of $L$ and $U/A$ be a maximal nlpotent subalgebra of $L/A$. Then,
$U=C+A$ where $C$ is a maximal nilpotent subalgebra of $L.$

###### Theorem 4.2

Let $L$ be a solvable symmetric Leibniz algebra over any field $F$ in which
every maximal subalgebra of each maximal nilpotent subalgebra of $L$ is a weak
c-ideal of $L.$ Then $L$ is supersolvable.

Proof. Let $L$ be a minimal counter-example. If $I=0$ the result follows from
[14, Theroem 4.5], so suppose that $I\neq 0$ and let $A$ be a minimal ideal of
$L$ contained in $I$. Since $[L,I]=[I,L]=0$, $\dim A=1$. Let $U/A$ be a
maximal nilpotent subalgebra of $L/A$ and let $B/A$ be a maximal subalgebra of
$U/A$. Then $U=C+A$ where $C$ is a maximal nilpotent subalgebra of $L$, by
Lemma 4.1. But $C+A$ is nilpotent, so $A\subseteq C=U$. Hence, $B$ is a
maximal subalgebra of $C$ and there is a subideal $K$ of $L$ such that $L=B+K$
and $B\cap K\subseteq B_{L}$. Now

$\frac{L}{A}=\frac{B}{A}+\frac{K+A}{A}\hbox{ and }\frac{K+A}{A}\hbox{ is a
subideal of }\frac{L}{A}.$

Moreover,

$\frac{B}{A}\cap\frac{K+A}{A}=\frac{A+B\cap
K}{A}\subseteq\frac{A+B_{L}}{A}\subseteq\left(\frac{B}{A}\right)_{L/A}.$

It follows that $L/A$ satisfies the same hypothesis as $L$, and so $L/A$ is
supersolvable, by the minimality of $L$. Hence, $L$ is supersolvable. $\Box$

If $L$ has no one-dimensional maximal nilpotent subalgebras, we can remove the
solvability assumption from the above result provided that $F$ has
characteristic zero.

###### Corollary 4.3

Let $L$ be a symmetric Leibniz algebra over a field $F$ of characteristic zero
in which every maximal nilpotent subalgebra has dimension at least two. If
every maximal subalgebra of each maximal nilpotent subalgebra of $L$ is a weak
c-ideal of $L,$ then $L$ is supersolvable.

Proof. Let $N$ be the nilradical of $L,$ and let $x\notin N.$ Then $x\in C$
for some maximal nilpotent subalgebra $C$ of $L.$ Since $\dim C>1$, there is a
maximal subalgebra $B$ of $C$ with $x\in B.$ Then there is a subideal $K$ of
$L$ such that $L=B+K$ and $B\cap K\subseteq B_{L}\leq C_{L}\leq N$. Clearly,
$x\notin K,$ since otherwise $x\in B\cap K\leq N.$

Now $L/I=(B+I)/I+(K+I)/I$ where $(B+I)/I$ is nilpotent and $(K+I)/I$ is a
subideal of $L/I$. It follows from [14, Lemma 4.2] that $L^{r}\subseteq K+I$
for some $r\in{\mathbb{N}}$. Hence $L^{r+1}\subseteq K$. We have shown that if
$x\notin N$ there is a subideal $K$ of $L$ with $x\notin K$ and
$L^{r+1}\subseteq K$. Suppose that $L$ is not solvable. Then there is a
semisimple Levi factor $S$ of $L$. Choose $x\in S$. Then $x\in
S=S^{r+1}\subseteq K$, a contradiction. Thus $L$ is solvable and the result
follows from Theorem 4.2. $\Box$

If $L$ has a one-dimensional maximal nilpotent subalgebra, then we can also
remove the solvability assumption from Theorem 4.4., provided that underlying
field $F$ has again characteristic zero and $L$ is not three-dimensional
simple.

###### Corollary 4.4

Let $L$ be a symmetric Leibniz algebra over a field $F$ of characteristic
zero. If every maximal subalgebra of each maximal nilpotent subalgebra of $L$
is a weak $c$-ideal of $L,$ then $L$ is supersolvable or three dimensional
simple.

Proof. If every maximal nilpotent subalgebra of $L$ has dimension at least
two, then $L$ is supersolvable by Corollary 4.3. So we need only consider the
case where $L$ has a one-dimensional maximal nilpotent subalgebra say $Fx$.
Suppose first that $I=0$l. Then $L$ is a Lie algebra and the result follows
from [14, Corollary 4.7].

So now let $I\neq 0$ and let $L$ be a minimal-counter-example. Then $L$ has a
minimal ideal $A\subseteq I.$ As in the proof of Theorem 4.2, $L/A$ satisfies
the same hypothesis as $L$ and so is supersolvable or three-dimensional
simple. In the former case, $L$ is solvable and so is supersolvable, by
Theorem 4.2. In the latter case, $L=A\oplus S$ where $S$ is three-dimensional
simple, by Levi’s Theorem. But now $L$ is a Lie algebra and the result follows
again from [14, Corollary 4.7]. $\Box$

##  5 Leibniz algebras in which every one-dimensional subalgebra is a weak
c-ideal

###### Proposition 5.1

For a one-dimensional subalgebra $Fx$ of a Leibniz algebra $L$ the following
are equivalent:

  * (i)

$Fx$ is a weak c-ideal of $L$;

  * (ii)

$Fx$ is a c-ideal of $L$; and

  * (iii)

either $Fx$ is an ideal of $L$, or there is an ideal $B$ of $L$ such that
$L=B\dot{+}Fx$ and $x\notin L^{2}$.

Proof. (i) and (ii) are equivalent since a subideal of codimension one in $L$
is an ideal.

If (ii) holds, then there is an ideal $B$ in $L$ such that $L=Fx+B$, and
$Fx\cap B\subseteq(Fx)_{L}=0$ or $Fx$. The former imples that $L=B\dot{+}Fx$
and $x\notin L^{2}\subseteq B$; the latter implies that $Fx$ is an ideal of
$L$. Hence (iii) holds. The converse is clear. $\Box$

###### Definition 7

Put $J=\\{x\in L\mid x^{2}=0\\}$. Note that $I\subseteq J$.

###### Corollary 5.2

Let $L$ be a Leibniz algebra over any field. Then every one-dimensional
subalgebra is a $c$-ideal if and only if $L^{2}\cap J\subseteq Asoc(L)$.

Proof. Clearly, if $Fx$ is a one-dimensional subalgebra of $L$, then $x\in J$.
It follows from Proposition 5.1 that if every one-dimensional subalgebra of
$L$ is a $c$-ideal, then $L^{2}\cap J\subseteq Asoc(L)$.

So suppose that $L^{2}\cap J\subseteq Asoc(L)$, and let $Fx$ be a one-
dimensional subalgebra of $L$. If $x\in L^{2}$, then $Fx$ is an ideal of $L$.
If $x\notin L^{2}$, let $B$ be a subspace containing $L^{2}$ which is
complementary to $Fx$. Then $B$ is an ideal of $L$ and $L=Fx\dot{+}B$. hence,
$Fx$ is a $c$-ideal, by Proposition 5.1 (iii). $\Box$

###### Proposition 5.3

Let $L=\langle x\rangle$ be a cyclic Leibniz algebra. Then every one-
dimensional subalgebra of $L$ is a c-ideal if and only if $\dim L\leq 2$.

Proof. Suppose that every one-dimensional subalgebra of $L$ is a c-ideal and
that $\dim L>1$. If $Fx^{2}$ is not an ideal of $L$, there is an ideal $B$ of
$L$ such that $L=Fx^{2}+B$. But then $x+\lambda x^{2}\in B$ for some
$\lambda\in F$, whence $x^{2}=[x,x+\lambda x^{2}]\in B$, a contradiction. Thus
$Fx^{2}$ is an ideal of $L$, $x^{3}=\lambda x^{2}$ for some $\lambda\in F$ and
$L=Fx+Fx^{2}$.

Suppose conversely that $\dim L=2$. Then $L=Fx+Fx^{2}$, where $x^{3}=0$ or
$x^{3}=x^{2}$. In the former case, the only one-dimensional subalgebra is
$Fx^{2}$ and that is an ideal of $L$. In the latter case, the one-dimensional
subalgebras are $Fx^{2}$, which is an ideal, and $F(x-x^{2})$, which is
complemented by $Fx^{2}$. $\Box$

In [15] the following result appears.

###### Theorem 5.4

Let $L$ be a Leibniz algebra over any field $F$ . Then all one-dimensional
subalgebras of $L$ are $c$-ideals of L if and only if:

  * (i)

$L^{3}=0$; or

  * (ii)

$L=A\oplus B$, where $A$ is an abelian ideal of $L$ and $B$ is an almost
abelian ideal of $L$.

Proof. See [15, Theorem 3.2.9, page 26]. $\Box$

Turner defines a subalgebra $B$ of a Leibniz algebra $L$ to be almost abelian
if $B=Fx\dot{+}D$ where, $D$ is abelian and $[d,x]\subseteq D$ for all $d\in
D$. (She actually has the products the other way around as she is dealing with
left Leibniz algebras, whereas, here we are concerned with right Leibniz
algebras.) However, this definition is problematic as it appears to be assumed
in the proof that $[d,x]=d$ for all $d\in D$, and that does not follow from
the definition. Also, nothing is said about $[x,d]$. Elsewhere in the
literature there have been defined two types of almost abelian Leibniz
algebras: $B=Fx\dot{+}D$ is called an almost abelian Lie algebra if
$[d,x]=-[x,d]=d$ for all $d\in D$, and is an almost abelian non-Lie Leibniz
algebra if $[d,x]=d$ for all $d\in D$, all other products being zero in each
case.

Moreover, the result is false, as the following example shows.

###### Example 5.1

Let $L$ be the three-dimensional Leibniz algebra over a field of
characteristic different from $2$ with basis $a,b,x$ and non-zero products
$a^{2}=b$, $[a,x]=-[x,a]=\frac{1}{2}a$, $[b,x]=b$. Then $(\alpha a+\beta
b+\gamma x)^{2}=0$ if and only if $\alpha^{2}=-\beta\gamma$. If $\alpha=0$
then, either $\beta-0$, in which case $Fx$ is complemented by the ideal
$Fa+Fb$, or $\gamma=0$, in which case $Fb$ is an ideal of $L$. If $\alpha\neq
0$, then $F(\alpha a+\beta b+\frac{\alpha^{2}}{\beta}x)$ is complemented by
the ideal $Fa+Fb$. It follows that every one-dimensional subalgebra is a
$c$-ideal. However, $L$ is not of the form given in Theorem 5.4.

In fact, the structure of Leibniz algebras in which all one-dimensional
subalgebras are $c$-ideals can be more complicated than is claimed by Theorem
5.4. The best that we can achieve currently is the following.

###### Lemma 5.5

Let $L$ be a Leibniz algebra in which every one-dimensional subalgebra is a
c-ideal. Then

  * (i)

all minimal abelian ideals are one dimensional;

  * (ii)

if $Asoc(L)=Fa_{1}\oplus\ldots\oplus Fa_{r}$ and $x\in L$, then $a_{i}\in
Z(L)$ or $[a_{i},x]=\lambda_{x}a_{i}$ for some $0\neq\lambda_{x}\in F$, $1\leq
i\leq r$.

  * (iii)

Let $Asoc(L)=Z(L)\oplus D$ where $[a,x]=\lambda_{x}a$ for all $a\in D$, $x\in
L$. Then, either $D=0$ or $L=(Z(L)\oplus D)\dot{+}C\dot{+}Fx$ where
$[D,C]=[C,D]=0$, $(Z(L)\oplus D)\dot{+}C$ is an ideal of $L$ and $[a,x]=a$ for
all $a\in D$.

Proof.

  * (i)

Let $A$ be a minimal abelian ideal of $L$ and let $a\in A$. If $A\neq Fa$ then
there is an ideal $K$ of $L$ such that $L=Fa\dot{+}K$. But $A\cap K=0$ so
$A=Fa$, a contradiction.

  * (ii)

Suppose that $[a_{i},x]=\lambda a_{i}$ and $[a_{j},x]=\mu a_{j}$ where
$\lambda\neq\mu$. Then $F(a_{i}+a_{j})$ is not an ideal of $L$ and so there is
an ideal $M$ of $L$ such that $L=F(a_{i}+a_{j})\dot{+}M$. Clearly, one of
$a_{i}$ and $a_{j}$ does not belong to $M$. Suppose that $a_{i}\notin M$, so
$L=Fa_{i}\oplus M$ and $a_{i}\in Z(L)$.

  * (iii)

Let $\Lambda:L\rightarrow F$ be given by $\Lambda(x)=\lambda_{x}$. This is a
linear transformation. Hence, either Im $\Lambda=0$, in which case $D=0$, or
else $L=$ Ker $\Lambda\dot{+}Fx$ and $\lambda_{x}=1$. Put $L=(Z\oplus
D)\dot{+}C\dot{+}Fx$, where $C\subseteq$ Ker $\Lambda$. Let $d\in D$, $c\in C$
with $[c,d]=\lambda d$. Then

$\lambda^{2}d=\lambda[c,d]=[c[c,d]]=[c^{2},d]-[[c,d],c]=0,$

so $\lambda=0$. Hence $[D,C]=[C,D]=0$. It is straightforward to check that Ker
$\Lambda$ is an ideal of $L$.

$\Box$

However, we can retrieve Theorem 5.4 for symmetric Leibniz algebras.

###### Theorem 5.6

Let $L$ be a symmetric Leibniz algebra over any field $F$ . Then all one-
dimensional subalgebras of $L$ are $c$-ideals of $L$ if and only if:

  * (i)

$L^{3}=0$; or

  * (ii)

$L=A\oplus B$, where $A$ is an abelian ideal of $L$ and $B$ is an almost
abelian Lie ideal of $L$.

Proof. Suppose that all one-dimensional subalgebras of $L$ are $c$-ideals of
L. First note that, if $x,y\in L$, then $[x,y]^{2}\in L^{2}\cap J$, so
$L^{2}\subseteq Asoc(L)$, by Corollary 5.2. Also, $L$ must have the structure
given in Lemma 5.5 (iii). If $D=0$ then $L^{2}\subseteq Z(L)$ and (i) holds.
So suppose that $D\neq 0$.

Let $0\neq a\in D$. If $z\in Z$, then $F(z+a)$ is not an ideal of $L$ and so
$z+a\not\in L^{2}$. But $a=[a,x]\in L^{2}$, so $z\not\in L^{2}$ and $Z\cap
L^{2}=0$. It follows that $L^{2}=D$. Now, if $c\in C$, we have that $c^{2}\in
L^{2}=D$, so $c^{3}=c^{2}$. But $c^{3}=0$, by [6, Proposition 2.17]. If
$[c,x]=0$, then $[x,c]\in L^{2}=D$ and so $[x,c]=[[x,c],x]=[x,[c,x]]=0$, since
$L$ is also a flexible algebra, by [6, Proposition 2.17] again. Hence $Fc$ is
an ideal of $L$. But then $c\in L^{2}\cap C=D\cap C=0$.

So suppose that $[c,x]\neq 0$. Then $[c-[c,x],x]=0$ and
$[x,c-[c,x]]=[x,c]-[x,[c,x]]=[x,c]-[[x,c],x]=0$, using the flexible law again.
Also, $[c-[c,x],c-[c,x]]=-[c,[c,x]]-[[c,x],c]=-[c,[c,x]]-[c,[x,c]]=0$, since
$[c,x]+[x,c]\in I$. It follows that $F(c-[c,x])$ is an ideal of $L$ and hence
is inside $L^{2}=D$. Thus, $c\in C\cap D=0$, so $C=0$.

Finally, for all $a\in D$, $[a,x]=a$. Also, $[a,x]+[x,a]\in I$, so
$0=[[a,x],x]+[[x,a],x]=[a,x]+[x,a]$ and $[x,a]=-a$. Thus, $L$ is as described
in (ii). $\Box$

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