Leibniz A-algebras

TL;DR

This paper extends the concept of A-algebras from Lie algebras to Leibniz algebras, exploring their properties and significance in mathematical physics and algebraic structures.

Contribution

It generalizes known results about A-algebras to Leibniz algebras, broadening the understanding of their structure and applications.

Findings

Generalized A-algebra properties to Leibniz algebras

Connected Leibniz A-algebras to Lie algebra results

Enhanced understanding of Leibniz algebra structures

Abstract

A finite-dimensional Lie algebra is called an A-algebra if all of its nilpotent subalgebras are abelian. These arise in the study of constant Yang-Mills potentials and have also been particularly important in relation to the problem of describing residually finite varieties. They have been studied by several authors, including Bakhturin, Dallmer, Drensky, Sheina, Premet, Semenov, Towers and Varea. In this paper we establish generalisations of many of these results to Leibniz algebras.

Equations

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

$$L^1=L,L^{k+1}=[L^k,L]and{L}^{(0)}=L,L^{(k+1)}=[L^{(k)},L^{(k)}]for\; allk=2,3,\dots$$

$$L_x^{k+1}(A)$$

$$=[L_x^k(A),x]\subseteq R_x^k(A).$$

$$x+A\in \frac{Z(L)}{A}\cap \frac{L^2+A}{A}\subseteq Z\left(\frac{L}{A}\right)\cap {\left(\frac{L}{A}\right)}^2,$$

$$[C,[L,C]]\subseteq [[C,L],C]+[C^2,L]\subseteq [L,C]$$

$$[C,R_C^{k+1}(L)]=[C,[R_C^k(L),C]]\subseteq [[C,R_C^k(L)],C]+[C^2,R_C^k(L)]\subseteq R_C^{k+1}(L).$$

$$U\ne z+U\in (Z(L)\cap L^2+U)/U\subseteq Z(L/U)\cap (L/U{)}^2.$$

$$[x,[L,x]]\subseteq [[x,L],x]+[x^2,L]\subseteq [L,x]+[x^2,M+Fx]\subseteq [L,x]$$

$$[x,R_x^{k+1}(L)]=[x,[R_x^k(L),x]]\subseteq [[x,R_x^k(L)],x]+[x^2,R_x^k(L)]\subseteq R_x^{k+1}(L),$$

$$L/D^2=(B+D^2)/D^2\dot{+}(C+D^2)/D^2$$

$$D/D^2=(B\cap D+D^2)/D^2\dot{+}(C\cap D+D^2)/D^2$$

$$D=B\cap D+C\cap D+D^2=B\cap D\dot{+}C\cap D.$$

$$\left[\frac{D+B}{B},\frac{E+B}{B}\right]+\left[\frac{E+B}{B},\frac{D+B}{B}\right]=\frac{B}{B},$$

$$[D,E]+[E,D]\subseteq [D+B,E+B]+[E+B,D+B]\subseteq Band$$

$$[D,E]+[E,D]\subseteq B\cap D=0;$$

$$[E,B]\subseteq [[E,L^{(n-1)}],L^{(n-1)}]=[[E,B+E],B+E]\subseteq [B,B+E]=[B,E],$$

$$Z(L^{(i)})=(Z(L^{(i)})\cap L^{(i+1)})+(Z(L^{(i)})\cap A_i)=Z(L^{(i)})\cap A_i\subseteq N\cap A_i.$$

$$[L,L_0\cap L^2]=[L_0+L_1,L_0\cap L^2]\subseteq (L_0\cap L^2)+L^{(2)}=L_0\cap L^2.$$

$$[L_0,L_1]\subseteq [L_0,[L^2,U]]\subseteq [[L_0,L^2],U]+[[L_0,U],L^2]\subseteq [L^2,U]=L_1,$$

$$K=[K,U]=[K,D]soK\subseteq L_{1}andU\cap L^2\subseteq L_0\cap L^2.$$

$$L_0^2\subseteq L_0\cap L^2=(U\cap L^2)+(L_0\cap K)=U\cap L^2,$$

$$U=L_0\cap U=L_0^2\oplus (E\cap U)=(U\cap L^2)\oplus (E\cap U).(\ast )$$

$$L=L_1+L_0=L^2+L_0=L^2+E=L^2\dot{+}C$$

$$C\cap U=(E\cap U)\oplus (C\cap U\cap L^2)=E\cap U,$$

$$L_1=\bigcap _{i=1}^{\infty }{R}_x^i(L)\subseteq [[L,A],A]\subseteq [A,A]\subseteq N,$$

$$[b,[b,L^2]]\subseteq [b^2,L^2]+[[b,L^2],b]\subseteq [L^2,b]=0,$$

Full text

Mathematics Subject Classification 2000: 17B05, 17B20, 17B30, 17B50.

Key Words and Phrases: Lie algebras, Leibniz algebras, $A$-algebras, Frattini ideal, solvable, nilpotent, completely solvable, metabelian, monolithic, cyclic Leibniz algebras.

1 Introduction

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

[TABLE]

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 algebra. 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 ([4]) who called them $D$-algebras, though they attracted more widespread interest, and acquired their current name, through work by Loday and Pirashvili ([9], [10]).

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

We define the following series:

[TABLE]

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 $L$.

A Lie algebra $L$ is called an $A$-algebra if all of its nilpotent subalgebras are abelian. This is analogous to the concept of an $A$-group: a finite group with the property that all of its Sylow subgroups are abelian. They have been studied and used by a number of authors, including Bakhturin and Semenov [5], Dallmer [6], Drensky [7], Sheina [15] and [16], Premet and Semenov [13], Semenov [14] and Towers and Varea [19], [20]. They arise in the study of constant Yang-Mills potentials and have also been particularly important in relation to the problem of describing residually finite varieties (see [5], [15], [16], [14] and [13]).

It would seem to be worthwhile examining this same concept for Leibniz algebras, both because there has been much interest in seeing which properties of Lie algebras generalise to Leibniz algebras, but also because Leibniz algebras can be used to define consistent generalisations of Yang-Mills functionals.

In section two we consider the non-solvable case. Here we collect together the preliminary results that we need, including the fact that for Leibniz $A$-algebras the derived series coincides with the lower nilpotent series. The main result is an analogue of the structure theorem of Premet and Semenov ([13]).

Section three contains the basic structure theorems for solvable Leibniz $A$-algebras. First they split over each term in their derived series. This leads to a decomposition of $L$ as $L=A_{n}\dot{+}A_{n-1}\dot{+}\ldots\dot{+}A_{0}$ where $A_{i}$ is an abelian subalgebra of $L$ and $L^{(i)}=A_{n}\dot{+}A_{n-1}\dot{+}\ldots\dot{+}A_{i}$ for each $0\leq i\leq n$. It is shown that the ideals of $L$ relate nicely to this decomposition: if $K$ is an ideal of $L$ then $K=(K\cap A_{n})\dot{+}(K\cap A_{n-1})\dot{+}\ldots\dot{+}(K\cap A_{0})$; moreover, if $N$ is the nilradical of $L$, $Z(L^{(i)})=N\cap A_{i}$. We also see that the result in Theorem 2.7 (iii)(a) holds when $L$ is solvable without any restrictions on the underlying field.

The fourth section looks at Leibniz $A$-algebras in which $L^{2}$ is nilpotent. These are metabelian and so the results of section three simplify. In addition we can locate the position of the maximal nilpotent subalgebras: if $U$ is a maximal nilpotent subalgebra of $L$ then $U=(U\cap L^{2})\oplus(U\cap C)$ where $C$ is a Cartan subalgebra of $L$.

Section five is devoted to Leibniz $A$-algebras having a unique minimal ideal $W$. Again some of the results of sections three and four simplify. In particular, $N=Z_{L}(W)$, and if $L$ is strongly solvable the maximal nilpotent subalgebras of $L$ are $L^{2}$ and the Cartan subalgebras of $L$ (that is, the subalgebras that are complementary to $L^{2}$.) We also give necessary and sufficient conditions for a Leibniz algebra with a unique minimal ideal to be a strongly solvable $A$-algebra.

In section six we illustrate some of the previous results by examining the subclass of cyclic Leibniz algebras.

The final section is devoted to generalising a result of Drensky ([7]). This shows that a solvable Leibniz algebra over an algebraically closed field has derived length at most three.

Throughout $L$ will denote a finite-dimensional algebra over a field $F$. Algebra direct sums will be denoted by $\oplus$, whereas vector space direct sums will be denoted by $\dot{+}$. The centre of $L$ is $Z(L)=\{z\in L\mid[z,x]=[x,z]=0$ for all $x\in L\}$. If $U$ is a subalgebra of $L$, the centraliser of $U$ in $L$ is $C_{L}(U)=\{x\in L\mid[x,U]=[U,x]=0\}$. We say that $L$ is monolithic with monolith $W$ if $W$ is the unique minimal ideal of $L$. The Frattini ideal of $L$, $\phi(L)$, is the largest ideal of $L$ contained in all maximal subalgebras of $L$; we call $L$ $\phi$-free if $\phi(L)=0$.

2 The non-solvable case

First we note that the class of Leibniz $A$-algebras is closed with respect to subalgebras, factor algebras and direct sums. Also that there is always a unique maximal abelian ideal, and it is the nilradical.

Lemma 2.1

Let $L$ be a Lie $A$-algebra and let $N$ be its nilradical. Then

• (i)

$N$* is the unique maximal abelian ideal of $L$;*

• (ii)

if $B$ and $C$ are abelian ideals of $L$, we have $[B,C]=0$.

Proof. (i) Clearly $N$ is abelian and contains every abelian ideal of $L$.

(ii) Simply note that $B,C\subseteq N$. $\Box$

Lemma 2.2

If $L$ is a Leibniz $A$-algebra over any field and $B$ is an ideal of $L$, then $L/B$ is a Leibniz $A$-algebra.

Proof. Let $U$ be a subalgebra of $L$ such that $U/B$ is nilpotent. If $B\subseteq\phi(U)$ the $U$ is nilpotent (see [1]) and hence abelian.

So suppose that $B\not\subseteq\phi(U)$. Then there is a maximal subalgebra $M$ of $U$ such that $U=B+M$. Choose $C$ to be a subalgebra of $L$ which is minimal with respect to $U=B+C$. Then $B\cap C\subseteq\phi(C)$ and $U/B\cong C/B\cap C$. It follows, by [1] again, that $C$ is nilpotent and hence abelian.

So, in either case, $U/B$ is abelian and $L/B$ is an $A$-algebra. $\Box$

Lemma 2.3

Let $B$, $C$ be ideals of the Leibniz algebra $L$.

• (i)

If $L/B$, $L/C$ are $A$-algebras, then $L/(B\cap C)$ is an $A$-algebra.

• (ii)

If $L=B\oplus C$, where $B,C$ are $A$-algebras, then $L$ is an $A$-algebra.

Proof. (i) Let $U/(B\cap C)$ be a nilpotent subalgebra of $L/(B\cap C)$. Then $(U+B)/B$ is a nilpotent subalgebra of $L/B$, which is an $A$-algebra. It follows that $U^{2}\subseteq B$. Similarly, $U^{2}\subseteq C$, whence the result.

(ii) This follows from (i). $\Box$

We define the nilpotent residual, $\gamma_{\infty}(L)$, of $L$ be the smallest ideal of $L$ such that $L/\gamma_{\infty}(L)$ is nilpotent. Clearly this is the intersection of the terms of the lower central series for $L$. Then the lower nilpotent series for $L$ is the sequence of ideals $N_{i}(L)$ of $L$ defined by $N_{0}(L)=L$, $N_{i+1}(L)=\gamma_{\infty}(N_{i}(L))$ for $i\geq 0$.

For Leibniz $A$-algebras we have the following result.

Lemma 2.4

Let $L$ be a Leibniz $A$-algebra. Then the lower nilpotent series coincides with the derived series.

Proof. Since $L/L^{(1)}$ is nilpotent we have $N_{1}(L)\subseteq L^{(1)}$. Also $L/N_{1}(L)$ is nilpotent and hence abelian, by Lemma 2.1 (ii), so $L^{(1)}\subseteq N_{1}(L)$. Repetition of this argument gives $N_{i}(L)=L^{(i)}$ for each $i\geq 0$.

If $F$ has characteristic zero, then every solvable Leibniz $A$-algebra over $F$ is metabelian, since $L^{2}$ is nilpotent. This is not the case, however, when $F$ is any field of characteristic $p>0$ (see [18, Example 2.1]).

A main problem encountered when trying to generalise results about Lie algebras to the case of Leibniz algebras is the lack of anti-symmetry, so that one-sided ideals exist in a Leibniz algebra. The following lemma is used several times in this paper to overcome this difficulty.

Lemma 2.5

Let $A$ be an abelian ideal of a Leibniz algebra $L$ and suppose that $x^{2}\in A$. Then $L_{x}^{n}(A)\subseteq R_{x}^{n-1}(A)$ for all $n\geq 1$.

Proof. Clearly $[x,A]\subseteq A$ so the result holds for $n=1$. Suppose that it holds for $n\leq k$ where $k\geq 1$. Then

[TABLE]

The result follows by induction. $\Box$

Finally in this section we generalise a structure theorem of Premet and Semenov (see [13]) to Leibniz algebras. We will need the following easy lemma.

Lemma 2.6

Let $L$ be a Leibniz algebra over a field of characteristic different from $2$ such that $L/Z(L)$ is a simple three-dimensional Lie algebra. Then $L=L^{2}\dot{+}Z(L)$.

Proof. By [8, page 13], $L/Z(L)$ has a basis $e_{1}+Z(L),e_{2}+Z(L),e_{3}+Z(L)$ with products $[e_{2},e_{3}]+Z(L)=e_{1}+Z(L)$,$[e_{3},e_{1}]+Z(L)=\alpha e_{2}+Z(L)$, $[e_{1},e_{2}]+Z(L)=\beta e_{3}+Z(L)$ for some $\alpha,\beta\in F\setminus{0}$. Then it is easy to see that the subspace $S$ spanned by $[e_{1},e_{2}]$,$[e_{3},e_{1}]$,$[e_{3},e_{2}]$ is a three dimensional simple subalgebra of $L$. It follows that $Z(L)\cap S=0$ and $S=L^{2}$. Hence $L=L^{2}\dot{+}Z(L)$. $\Box$

If $K$ is an extension field of $F$, denote $K\otimes_{F}L$ by $L_{K}$.

Theorem 2.7

Let $L$ be a Leibniz $A$-algebra over a field $F$. If $F$ has characteristic $\neq 2,3$ and cohomological dimension $\leq 1$ (this means that the Brauer group of any algebraic extension of the underlying field is trivial), then

• (i)

$L^{2}\cap Z(L)=0$; and

• (ii)

$L$* has a Levi decomposition and every Levi subalgebra is representable as a direct sum of simple ideals, each one of which splits over some finite extension of the ground field into a direct sum of ideals isomorphic to $sl(2)$.*

Proof.

• (i)

Let $L$ be a minimal counter-example, so there is a non-zero element $x\in Z(L)\cap L^{2}$. Clearly Leib$(L)\neq 0$ by [13, Proposition 2]. Let $A$ be a subspace complementary to $Fx$ in $Z(L)$, so $Z(L)=Fx\dot{+}A$. Then

[TABLE]

so we have that $A=0$ and $\dim Z(L)=1$. If $B$ is a non-trivial ideal of $L$ we have $Z(L)\subseteq B$, since otherwise $L/B$ would be a counter-example of smaller dimension. It follows that $L$ is monolithic with monolith $Z(L)$. Let $M$ be a maximal ideal of $L$. Then $M^{2}\cap Z(M)=0$ and so $Z(L)\not\subseteq M^{2}$, whence $M^{2}=0$. But now either $L$ is nilpotent or there is a unique maximal ideal which is abelian and is the radical. If $L$ is nilpotent, it is abelian, which yields a contradiction.

So suppose that $L$ has a unique maximal ideal $M$ which is abelian and is the radical. Then $L/M={\mathcal{L}}$ is simple. It follows from [13, Corollary 1 and Lemma 2] that ${\mathcal{L}}$ is a Lie $p$-algebra Moreover, our assumption on the field $F$ implies that${\mathcal{L}}$ has a non-zero nilpotent element (see [11] and [12]). Hence there exists an element $u\in L\setminus M$ such that $R_{u}^{p^{m}}(L)\subseteq M$. Let $\overline{u}$ be the image of $u$ under the canonical homomorphism from $L$ to $\mathcal{L}$. The element $\overline{u}^{p^{m}}$ lies in the centre of the universal enveloping algebra $U({\mathcal{L}})$, and so in any indecomposable $L$-module $W$ the set $\lambda_{1}(W),\ldots,\lambda_{r}(W)$ of eigenvalues of $\overline{u}^{p^{m}}$ consists of elements of $\overline{F}$ that are conjugate under the Galois group $Gal(\overline{F}/F)$. The right module $M$ is indecomposable and contains $Z(L)$, and so $\lambda_{k}(M)=0$ for some $1\leq k\leq r$. It follows that $u$ acts nilpotently on the right in $L$. But now $u^{2}\in$ Leib$(L)\subseteq M$, so, using Lemma 2.5, $Fu+M$ is a nilpotent subalgebra of $L$ and thus abelian. This yields that $u\in C_{L}(M)$, and so $C_{L}(M)=L$ and $M=Z(L)$.

Now there is a finite extension $K$ of $F$ over which $(L/Z(L))_{K}$ splits as a direct sum of ideals $S_{1}/Z(L)\oplus\ldots\oplus S_{n}/Z(L)$ isomorphic to $sl(2)$, by [13, Proposition 2(ii)] again. Let $\theta:L\rightarrow L/Z(L)$ be the canonical homomorphism with $\ker(\theta)=Z(L)$ and let $\theta_{K}:L_{K}\rightarrow(L/Z(L))_{K}$ be the natural extension of $\theta$ to the corresponding algebras over the extension field. Then $\theta_{K}$ is a surjective homomorphism with $\ker(\theta_{K})=(\ker(\theta))_{K}$ (see, for example, [8]), so $(L/Z(L))_{K}\cong L_{K}/Z(L)_{K}$. Using Lemma 2.6 we thus see that $L_{K}=(L_{K})^{2}\dot{+}Z(L)_{K}$. But now $L=L^{2}\dot{+}Z(L)$, a contradiction from which the result follows.

• (ii)

We have that $L/$Leib$(L)=S/$Leib$(L)\dot{+}R/$Leib$(L)$ where $R$ is the radical of $L$ and there is a finite extension $K$ of $F$ over which $S/$Leib$(L)$ splits as a direct sum of ideals $S_{1}/$Leib$(L)\oplus\ldots\oplus S_{n}/$Leib$(L)$ isomorphic to $sl(2)$, by [13, Proposition 2(ii)]. Arguing as in the final two paragraphs of (i) we have that $L_{K}=S_{K}^{2}\dot{+}R_{K}$, from which $L=S^{2}\dot{+}R$ giving the claimed result.

$\Box$

3 Decomposition results for Solvable Leibniz $A$-algebras

Here we have the basic structure theorems for solvable Leibniz $A$-algebras. First we see that such an algebra splits over the terms in its derived series.

Lemma 3.1

Let $L$ be any solvable Leibniz algebra with nilradical $N$. Then $Z_{L}(N)\subseteq N$

Proof. Suppose that $Z_{L}(N)\not\subseteq N$. Then there is a non-trivial abelian ideal $A/(N\cap Z_{L}(N)$ of $L/(N\cap Z_{L}(N)$ inside $Z_{L}(N)/(N\cap Z_{L}(N)$. But now $A^{3}\subseteq[N,A]=0$, so $A$ is a nilpotent ideal of $L$. It follows that $A\subseteq N\cap Z_{L}(N)$, a contradiction. $\Box$

Theorem 3.2

Let $L$ be a solvable Leibniz $A$-algebra. Then $L$ splits over each term in its derived series. Moreover, the Cartan subalgebras of $L^{(i)}/L^{(i+2)}$ are precisely the subalgebras that are complementary to $L^{(i+1)}/L^{(i+2)}$ for $i\geq 0$.

Proof. Suppose that $L^{(n+1)}=0$ but $L^{(n)}\neq 0$. First we show that $L$ splits over $L^{(n)}$. Clearly we can assume that $n\geq 1$. Let $C$ be a Cartan subalgebra of $L^{(n-1)}$ (this exists in any solvable Leibniz algebra: the proof is the same as that for Lie algebras in [21, Corollary 4.4.1.2]) and let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{C}$. Then $L_{1}=\cap_{k=1}^{\infty}R_{C}^{k}(L)\subseteq L^{(n)}$, and so $L_{1}$ is an abelian right ideal of $L$. Also $L^{(n-1)}=L_{1}\dot{+}L_{0}\cap L^{(n-1)}$ and $L_{0}\cap L^{(n-1)}=(L^{(n-1)})_{0}=C$, which is abelian.

Now

[TABLE]

Suppose that $[C,R_{C}^{k}(L)]\subseteq R_{C}^{k}(L)$ for $k\geq 1$. Then

[TABLE]

It follows that $[C,L_{1}]\subseteq L_{1}$ and thus that $L_{1}$ is an ideal of $L^{(n-1)}$. But $L^{(n-1)}/L_{1}$ is abelian, whence $L^{(n)}\subseteq L_{1}$ and $L=L_{0}\dot{+}L^{(n)}$.

So we have that $L=L^{(n)}\dot{+}B$ where $B=L_{0}$ is a subalgebra of $L$. Clearly $B^{(n)}=0$, so, by the above argument, $B$ splits over $B^{(n-1)}$, say $B=B^{(n-1)}\dot{+}D$. But then $L=L^{(n)}\dot{+}(B^{(n-1)}\dot{+}D)=L^{(n-1)}\dot{+}D$. Continuing in this way gives the desired result. $\Box$

This gives us the following fundamental decomposition result.

Corollary 3.3

Let $L$ be a solvable Lie $A$-algebra of derived length $n+1$. Then

• (i)

$L=A_{n}\dot{+}A_{n-1}\dot{+}\ldots\dot{+}A_{0}$* where $A_{i}$ is an abelian subalgebra of $L$ for each $0\leq i\leq n$; and*

• (ii)

$L^{(i)}=A_{n}\dot{+}A_{n-1}\dot{+}\ldots\dot{+}A_{i}$* for each $0\leq i\leq n$*

Proof. (i) By Theorem 3.2 there is a subalgebra $B_{n}$ of $L$ such that $L=L^{(n)}\dot{+}B_{n}$. Put $A_{n}=L^{(n)}$. Similarly $B_{n}=A_{n-1}\dot{+}B_{n-1}$ where $A_{n-1}=(B_{n})^{(n-1)}$. Continuing in this way we get the claimed result. Note, in particular, that it is apparent from the construction that $A_{k}\cap(A_{k-1}+...+A_{0})=0$ for each $1\leq k\leq n$, and that it is easy to see from this that the sum is a vector space direct sum.

(ii) We have that $L^{(n)}=A_{n}$. Suppose that $L^{(k)}=A_{n}\dot{+}\ldots\dot{+}A_{k}$ for some $1\leq k\leq n$. Then $L=L^{(k)}\dot{+}B_{k}$ and $A_{k-1}=B_{k}^{(k-1)}$ by the construction in (i). But now $L^{(k-1)}\subseteq L^{(k)}+B_{k}^{(k-1)}\subseteq L^{(k-1)}$, whence $L^{(k-1)}=A_{n}\dot{+}\ldots\dot{+}A_{k-1}$ and the result follows by induction. $\Box$

Now we show that the result in Theorem 2.7 (iii)(a) holds when $L$ is solvable without any restrictions on the underlying field.

Theorem 3.4

Let $L$ be a solvable Leibniz $A$-algebra. Then $Z(L)\cap L^{2}=0$.

Proof. Let $L$ be a minimal counter-example and let $z\in Z(L)\cap L^{2}$. Put $Z(L)=U\dot{+}Fz$. Then $U$ is an ideal of $L$ and

[TABLE]

The minimality of $L$ implies that $U=0$, so $Z(L)=Fz$. But now if $K$ is an ideal of $L$ which does not contain $Z(L)$, then $K\neq z+K\in Z(L/K)\cap(L/K)^{2}$ similarly, contradicting the minimality of $L$. It follows that $L$ is monolithic with monolith $Z(L)$.

Now let $M$ be a maximal ideal of $L$. Then $Z(M)\cap M^{2}=0$ by the minimality of $L$, so $Z(L)\not\subseteq M^{2}$, whence $M^{2}=0$. It follows that $L=M\dot{+}Fx$ for some $x\in L$ and $M$ is abelian. Let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{x}$. Then $L_{1}=\cap_{i=1}^{\infty}R_{x}^{i}(L)\subseteq M$, and $[L_{1},L_{0}]\subseteq L_{1}$, so $L_{1}$ is a right ideal of $L$.

Now

[TABLE]

since $x^{2}\in I\subseteq M$, so $[x^{2},M]=0$. Suppose that $[x,R_{x}^{k}(L)]\subseteq R_{x}^{k}(L)$. Then

[TABLE]

since $R_{x}^{k}(L)\subseteq[L,x]=[M+Fx,x]\subseteq M$, whence $[x^{2},R_{x}^{k}(L)]=0$. It follows that $[L,L_{1}]=[x,L_{1}]\subseteq L_{1}$ and $L_{1}$ is an ideal of $L$.

If $L_{1}\neq 0$ then $Z(L)\subseteq L_{1}\cap L_{0}=0$, a contradiction. Hence $L_{1}=0$ and $R_{x}$ is nilpotent. But then $L=M+Fx$ is nilpotent and hence abelian, and the result follows. $\Box$

Next we aim to show the relationship between ideals of $L$ and the decomposition given in Corollary 3.3. First we need the following lemma.

Lemma 3.5

Let $L$ be a solvable Leibniz $A$-algebra of derived length $\leq n+1$, and suppose that $L=B\dot{+}C$ where $B=L^{(n)}$ and $C$ is a subalgebra of $L$. If $D$ is an ideal of $L$ then $D=(B\cap D)\dot{+}(C\cap D)$.

Proof. Let $L$ be a counter-example for which dim$L$ + dim$D$ is minimal. Suppose first that $D^{2}\neq 0$. Then $D^{2}=(B\cap D^{2})\dot{+}(C\cap D^{2})$ by the minimality of $L$. Moreover, since

[TABLE]

we have

[TABLE]

whence

[TABLE]

We therefore have that $D^{2}=0$. Similarly, by considering $L/B\cap D$, we have that $B\cap D=0$.

Put $E=C^{(n-1)}$. Then $(D+B)/B$ and $(E+B)/B$ are abelian ideals of the Leibniz $A$-algebra $L/B$, and so

[TABLE]

by Lemma 2.1 (ii), whence

[TABLE]

[TABLE]

that is, $D\subseteq Z_{L}(E)$. But $Z_{L}(E)=Z_{B}(E)+Z_{C}(E)$. For, suppose that $x=b+c\in Z_{L}(E)$, where $b\in B$, $c\in C$. Then $0=[x,E]=[b,E]+[c,E]$, so $[b,E]=-[c,E]\in B\cap C=0$. Similarly, $[E,b]=-[E,c]=0$, so that $Z_{L}(E)\subseteq Z_{B}(E)+Z_{C}(E)$. But the reverse inclusion is clear, so equality follows.

Now $L^{(n-1)}\subseteq B+E\subseteq L^{(n-1)}$, so $B=L^{(n)}=(B+E)^{2}=[B,E]+[E,B]$. But

[TABLE]

so $B=[B,E]$. Let $L^{(n-1)}=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L^{(n-1)}$ relative to $R_{E}$. Then $B\subseteq L_{1}$ so that $Z_{B}(E)\subseteq L_{0}\cap L_{1}=0$, whence $D\subseteq Z_{L}(E)=Z_{C}(E)\subseteq C$ and the result follows. $\Box$

Theorem 3.6

Let $L$ be a solvable Leibniz $A$-algebra of derived length $n+1$ with nilradical $N$, and let $K$ be an ideal of $L$ and $A$ a minimal ideal of $L$. Then, with the same notation as Corollary 3.3,

• (i)

$K=(K\cap A_{n})\dot{+}(K\cap A_{n-1})\dot{+}\ldots\dot{+}(K\cap A_{0})$;

• (ii)

$N=A_{n}\oplus(N\cap A_{n-1})\oplus\ldots\oplus(N\cap A_{0})$;

• (iii)

$Z(L^{(i)})=N\cap A_{i}$* for each $0\leq i\leq n$; and*

• (iv)

$A\subseteq N\cap A_{i}$* for some $0\leq i\leq n$.*

Proof. (i) We have that $L=A_{n}\dot{+}B_{n}$ where $A_{n}=L^{(n)}$ from the proof of Corollary 3.3. It follows from Lemma 3.5 that $K=(K\cap A_{n})+(K\cap B_{n})$. But now $K\cap B_{n}$ is an ideal of $B_{n}$ and $B_{n}=A_{n-1}\dot{+}B_{n-1}$. Applying Lemma 3.5 again gives $K\cap B_{n}=(K\cap A_{n-1})\dot{+}(K\cap B_{n-1})$. Continuing in this way gives the required result.

(ii) This is clear from (i), since $A_{n}=L^{(n)}=N\cap A_{n}$.

(iii) We have that $L^{(i)}=L^{(i+1)}\dot{+}A_{i}$ from Corollary 3.3, and also that $Z(L^{(i)})\cap L^{(i+1)}=0$ from Theorem 3.4. Thus, using Lemma 3.5,

[TABLE]

It remains to show that $N\cap A_{i}\subseteq Z(L^{(i)})$; that is, $[N\cap A_{i},L^{(i)}]+[L^{(i)},N\cap A_{i}]=0$. We use induction on the derived length of $L$. If $L$ has derived length one the result is clear. So suppose it holds for Leibniz algebras of derived length $\leq k$, and let $L$ have derived length $k+1$. Then $B=A_{k-1}+\dots+A_{0}$ is a solvable Leibniz $A$-algebra of derived length $k$, and, if $N$ is the nilradical of $L$, then $N\cap A_{i}$ is inside the nilradical of $B$ for each $0\leq i\leq k-1$, so $[N\cap A_{i},B^{(i)}]+[B^{(i)},N\cap A_{i}]=0$ for $0\leq i\leq k-1$, by the inductive hypothesis. But $[N\cap A_{i},A_{k}]=[N\cap A_{i},L^{(k)}]\subseteq[N,N]=0$, for $0\leq i\leq k$, whence $[N\cap A_{i},L^{(i)}]=[N\cap A_{i},A_{k}+B^{(i)}]=0$ for $0\leq i\leq k$. Similarly, $[L^{(i)},N\cap A_{i}]=0$.

(iv) We have $A\subseteq L^{(i)}$, $A\not\subseteq L^{(i+1)}$ for some $0\leq i\leq n$. Now $[L^{(i)},A]\subseteq[L^{(i)},L^{(i)}]=L^{(i+1)}$, so $[L^{(i)},A]\neq A$. It follows that $[L^{(i)},A]=0$. Similarly, $[A,L^{(i)}]=0$, whence $A\subseteq Z(L^{(i)})=N\cap A_{i}$, by (ii). $\Box$

The final result in this section shows when two ideals of a Leibniz $A$-algebra centralise each other.

Proposition 3.7

Let $L$ be a Leibniz $A$-algebra and let $B,D$ be ideals of $L$. Then $B\subseteq Z_{L}(D)$ if and only if $B\cap D\subseteq Z(B)\cap Z(D)$.

Proof. Suppose first that $B\subseteq Z_{L}(D)$. Then $[B\cap D,D]+[D,B\cap D]=0=[B\cap D,B]+[B,B\cap D]$, whence $B\cap D\subseteq Z(B)\cap Z(D)$.

Conversely, suppose that $B\cap D\subseteq Z(B)\cap Z(D)$. Then $[B,D]+[D,B]\subseteq B\cap D\subseteq Z(B+D)$ which yields that $[B,D]+[D,B]\subseteq(B+D)^{2}\cap Z(B+D)=0$, by Theorem 3.4. Hence $B\subseteq Z_{L}(D)$. $\Box$

4 Completely solvable Leibniz $A$-algebras

A Leibniz algebra $L$ is called completely solvable if $L^{2}$ is nilpotent. Over a field of characteristic zero every solvable Leibniz algebra is completely solvable. Clearly completely solvable Leibniz $A$-algebras are metabelian so we would expect stronger results to hold for this class of algebras. First the decomposition theorem takes on a simpler form.

Theorem 4.1

Let $L$ be a completely solvable Leibniz $A$-algebra with nilradical $N$. Then $L=L^{2}\dot{+}B$, where $L^{2}$ is abelian and $B$ is an abelian subalgebra of $L$, and $N=L^{2}\oplus Z(L)$.

Proof. We have that $L=L^{2}\dot{+}B$, where $B$ is an abelian subalgebra of $L$, by Theorem 3.2. Also, $L^{2}$ is nilpotent and so abelian. Moreover, $N=L^{2}+N\cap B$ and $N\cap B=Z(L)$, by Theorem 3.6.

Next we see that the minimal ideals are easy to locate.

Theorem 4.2

Let $L=L^{2}\dot{+}B$ be a completely solvable Leibniz $A$-algebra and let $A$ be a minimal ideal of $L$. Then

• (i)

$A\subseteq L^{2}$* or $A\subseteq B$;*

• (ii)

$A\subseteq B$* if and only if $A\subseteq Z(L)$ (in which case dim $A=1$); and*

• (iii)

$A\subseteq L^{2}$* if and only if $[A,L]=A$.*

Proof. (i) and (ii) follow from Theorem 3.6 (iii) and (iv).

(iii) Suppose that $A\subseteq L^{2}$. Then $[A,L]+[L,A]\neq 0$ from (ii), so $[A,L]+[L,A]=A$. But $[L,A]=0$ or $[x,a]=-[a,x]$ for all $x\in L,a\in A$, by [1, Lemma 1.9]. Hence $[A,L]=A]$.

The converse is clear. $\Box$

Corollary 4.3

Let $L$ be a completely solvable Leibniz $A$-algebra. Then $L$ is $\phi$-free if and only if $L^{2}\subseteq$ Asoc$L$.

Proof. Suppose first that $L$ is $\phi$-free. Then $L^{2}\subseteq N=$ Asoc$L$, by [2, Theorem 2.4].

So suppose now that $L^{2}\subseteq$ Asoc$L$. Then $L$ splits over Asoc$L$ by Theorem 3.2. But now $L$ is $\phi$-free by [2, Proposition 3.1]. $\Box$

Finally we can identify the maximal nilpotent subalgebras of $L$. First we need the following lemma.

Lemma 4.4

Let $L$ be a metabelian Leibniz algebra, and let $U$ be a maximal nilpotent subalgebra of $L$. Then $U\cap L^{2}$ is an abelian ideal of $L$ and $L^{2}=(U\cap L^{2})\oplus K$ where $K$ is an ideal of $L$ and $[K,U]=K$.

Proof. Let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{U}$. Then $L_{1}=\cap_{i=1}^{\infty}L({\rm ad}\,U)^{i}\subseteq L^{2}$, and so $L^{2}=(L_{0}\cap L^{2})\dot{+}L_{1}$. Now

[TABLE]

Similarly, $[L_{0}\cap L^{2},L]\subseteq L_{0}\cap L^{2}$ so $L_{0}\cap L^{2}$ is an ideal of $L$. Also, $U^{2}\subseteq L_{0}\cap L^{2}$ and an induction argument similar to that in Lemma 2.5 shows that $L_{U}^{k}(L_{0}\cap L^{2})\subseteq R_{U}^{k-1}(L_{0}\cap L^{2})$ for $k\geq 1$. It follows that $U+(L_{0}\cap L^{2})$ is a nilpotent subalgebra of $L$, and so $L_{0}\cap L^{2}\subseteq U\cap L^{2}$. The reverse inclusion is clear.

Next, $[L^{2},L_{1}]\subseteq L^{(2)}=0$, so $[L^{2},U]=[L_{1},U]=L_{1}$. But now,

[TABLE]

so $L_{1}$ is an ideal of $L$. Hence we can put $K=L^{2}$. $\Box$

Theorem 4.5

Let $L$ be a completely solvable Leibniz $A$-algebra, and let $U$ be a maximal nilpotent subalgebra of $L$. Then $U=(U\cap L^{2})\oplus(U\cap C)$ where $C$ is a Cartan subalgebra of $L$.

Proof. Put $U=(U\cap L^{2})\oplus D$, so $D$ is an abelian subalgebra of $L$. Let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{D}$. As in Lemma 4.4, $L_{1}$ is an abelian right ideal of $L$.

Now put $L^{2}=(U\cap L^{2})\oplus K$ as given by Lemma 4.4. Then

[TABLE]

Hence

[TABLE]

since $L_{0}\cap K\subseteq L_{0}\cap L_{1}=0$.

Next put $L_{0}=L_{0}^{2}\dot{+}E$ where $E$ is an abelian subalgebra of $L_{0}$. Then

[TABLE]

Finally put $E=(E\cap L^{2})\oplus C$ where $E\cap U\subseteq C$. Then

[TABLE]

so $C$ is a Cartan subalgebra of $L$, by Theorem 3.2. Moreover, $E\cap U\subseteq C\cap U$, so (*) implies that

[TABLE]

since $C\cap L^{2}=0$. But now (*) becomes $U=(U\cap L^{2})\oplus(U\cap C)$ where $C$ is a Cartan subalgebra of $L$, as claimed. $\Box$

5 Monolithic solvable Leibniz $A$-algebras

Monolithic Lie algebras play a part in the application of Lie $A$-algebras to the study of residually finite varieties, so it seems worthwhile to investigate whether the extra properties they have are inherited by their Leibniz counterparts

Theorem 5.1

Let $L$ be a monolithic solvable Leibniz $A$-algebra of derived length $n+1$ with monolith $W$. Then, with the same notation as Corollary 3.3,

• (i)

$W$* is abelian;*

• (ii)

$Z(L)=0$* and either $[L,W]=W$ or $[W,L]=W$;*

• (iii)

$N=A_{n}=L^{(n)}$;

• (iv)

$N=Z_{L}(W)$; and

• (v)

$L$* is $\phi$-free if and only if $W=N$.*

Proof. (i) Clearly $W\subseteq L^{(n)}$, which is abelian.

(ii) If $Z(L)\neq 0$ then $W\subseteq Z(L)\cap L^{2}=0$, by Theorem 3.4, a contradiction. Hence $Z(L)=0$. It follows from this that $[L,W]+[W,L]\neq 0$. But $[L,W]$ is an ideal of $L$, so either $[L,W]=W$ or $[L,W]=0$, in which case $[W,L]=0$..

(iii) We have $N=A_{n}\oplus N\cap A_{n-1}\oplus\ldots\oplus N\cap A_{0}$ by Theorem 3.6(i). Moreover, $N\cap A_{i}$ is an ideal of $L$ for each $0\leq i\leq n-1$, by Theorem 3.6(iii). But if $N\cap A_{i}\neq 0$ then $W\subseteq A_{n}\cap N\cap A_{i}=0$ if $i\neq n$. This contradiction yields the result.

(iv) We have that $L=N\dot{+}B$ for some subalgebra $B$ of $L$, by Theorem 3.2 and (iii). Put $C=Z_{L}(W)$ and note that $N\subseteq C$. Suppose that $N\neq C$. Then $C=N\dot{+}B\cap C$. Choose $A/N$ to be a minimal ideal of $L/N$, so that $A^{2}\subseteq N$. Pick $x\in A\setminus N$ and let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{x}$. Then

[TABLE]

which is abelian. Hence $N=L_{1}\dot{+}N\cap L_{0}$. Now $N\cap L_{0}$ is an ideal of $L$, since $[L_{1},N\cap L_{0}]+[N\cap L_{0},L_{1}]\subseteq N^{2}=0$ and it is clearly invariant under $L_{0}$. Moreover, $Fx+N\cap L_{0}$ is a nilpotent subalgebra of $L$, since $x^{2}\in$ Leib$(L)\subseteq N$, $x^{2}\in L_{0}$ and using Lemma 2.5. Hence it is abelian, and so $[N\cap L_{0},x]=0$ and $[N,x]=[L_{1},x]=L_{1}$. It follows that $L_{1}=R_{x}^{k}(N)$ for all $k\geq 1$. But now, a straightforward induction proof shows that $[L_{0},L_{1}]\subseteq[L_{0},R_{x}^{k}(N)]\subseteq L_{1}+[R_{x}^{k}(L_{0}),N]$ for all $k\geq 1$. Since $R_{x}^{k}(L_{0})=0$ for some $k$ this yields that $[L_{0},L_{1}]\subseteq L_{1}$. Thus $L_{1}$ is an abelian ideal of $L$, and so $L_{1}=0$, as, otherwise, $W\subseteq L_{1}\cap L_{0}=0$.This yields that $Fx+N$ is nilpotent and thus abelian, whence $A\subseteq Z_{L}(N)\subseteq N$, by Lemma 3.1. This contradiction implies that $N=C$.

(v) Clearly $W=$ Asoc$L$. Suppose first that $L$ is $\phi$-free. Then $W=$ Asoc$L=N$, by [17, Theorem 7.4]. So suppose now that Asoc$L=W=N$. Then $L$ splits over Asoc$L$ by Theorem 3.2 and (iii). But now $L$ is $\phi$-free by [17, Theorem 7.3]. $\Box$

It is shown in [18] that monolithic solvable Lie $A$-algebras are not necessarily metabelian. However, when a Leibniz $A$-algebra is strongly solvable the situation is more straightforward.

Theorem 5.2

Let $L$ be a monolithic strongly solvable Leibniz $A$-algebra. Then the maximal nilpotent subalgebras of $L$ are $L^{2}$ and the Cartan subalgebras of $L$ (that is, the subalgebras that are complementary to $L^{2}$.)

Proof. Let $U$ be a maximal nilpotent subalgebra of $L$ and let $W$ be the monolith of $L$. Then $L^{2}=(U\cap L^{2})\oplus K$ where $U\cap L^{2},K$ are ideals of $L$ and $[U,K]=K$, by Lemma 4.4. Either $W\subseteq U\cap L^{2}$ and $K=0$ or else $W\subseteq K$ and $U\cap L^{2}=0$.

In the former case $N=L^{2}\subseteq U$, by Theorem 5.1. But then $U\subseteq Z_{L}(N)\subseteq N$, by Lemma 3.1, so $U=L^{2}$. In the latter case $U$ is a Cartan subalgebra of $L$, by Theorem 4.5. $\Box$

Finally we give necessary and sufficient conditions for a monolithic algebra to be a strongly solvable Leibniz $A$-algebra.

Lemma 5.3

Let $L=L^{2}\dot{+}B$ be a metabelian Leibniz algebra, where $B$ is a subalgebra of $L$, and suppose that $[L^{2},b]=L^{2}$ for all $b\in B$. Then $L$ is a strongly solvable $A$-algebra.

Proof. Let $U$ be a maximal nilpotent subalgebra of $L$. We have $L^{2}=(U\cap L^{2})\oplus K$ where $K$ is an ideal of $L$ and $[U,K]=K$, by Lemma 4.4. Let $u=x+b\in U$, where $x\in L^{2}$, $b\in B$. Then $L^{2}=[L^{2},b]=[L^{2},u]$, so $L^{2}=R_{u}^{i}(L^{2})$ for all $i\geq 1$. It follows that $L^{2}=K$ from which $U^{2}\subseteq U\cap L^{2}=0$ and $L$ is an $A$-algebra. $\Box$

Theorem 5.4

Let $L$ be a monolithic Leibniz algebra. Then $L$ is a strongly solvable $A$-algebra if and only if $L=L^{2}\dot{+}B$ is metabelian, where $B$ is a subalgebra of $L$ and $[L^{2},b]=L^{2}$ for all $b\in B$ (or, equivalently, $R_{b}$ acts invertibly on $L^{2}$).

Proof. Suppose first that $L$ is a strongly solvable $A$-algebra. Then $L=L^{2}\dot{+}B$ is metabelian, where $B$ is a subalgebra of $L$, by Theorem 3.2. Let $b\in B$ and let $L=L_{0}\dot{+}L_{1}$ be the Fitting decomposition of $L$ relative to $R_{b}$. It is easy to see, as in Lemma 4.4, that $L^{2}=(L^{2}\cap L_{0})\dot{+}L_{1}$ and $L^{2}\cap L_{0}$ and $L_{1}$ are ideals of $L$, so $L^{2}=L^{2}\cap L_{0}$ or $L^{2}=L_{1}$ as $L$ is monolithic. The former implies that $[L^{2},b]=0$. But then

[TABLE]

so $L^{2}+Fb$ is a nilpotent subalgebra of $L$ and hence is abelian. This yields that $L^{2}$ and $Fb$ are ideals of $L$, which is impossible. It follows that $L^{2}=L_{1}$, whence $[L^{2},b]=L^{2}$. If $\theta=R_{b}|_{L^{2}}$ then $L^{2}={\rm Ker}\,\theta\dot{+}{\rm Im}\,\theta$, so ${\rm Ker}\,\theta=\{0\}$ and $\theta$ is invertible.

The converse follows from Lemma 5.3. $\Box$

6 Cyclic Leibniz algebras

Cyclic Leibniz algebras, $L$, are generated by a single element. In this case $L$ has a basis $a,a^{2},\ldots,a^{n}(n>1)$ and product $[a^{n},a]=\alpha_{2}a^{2}+\ldots+\alpha_{n}a^{n}$. Let $T$ be the matrix for $R_{a}$ with respect to the above basis. Then $T$ is the companion matrix for $p(x)=x^{n}-\alpha_{n}x^{n-1}-\ldots-\alpha_{2}x=p_{1}(x)^{n_{1}}\ldots p_{r}(x)^{n_{r}}$, where the $p_{j}$ are the distinct irreducible factors of $p(x)$. Then we have the following result.

Theorem 6.1

$L$* is a cyclic Leibniz $A$-algebra if and only if $\alpha_{2}\neq 0$, and then $L=L^{2}\dot{+}F(a^{n}-\alpha_{n}a^{n-1}-\dots-\alpha_{2}a)$ and we can take $p_{1}(x)^{n_{1}}=x$.*

Proof. If $\alpha_{2}=0$ we have that $L$ is nilpotent but not abelian, so $L$ is not an $A$-algebra. If $\alpha_{2}\neq 0$, it is easy to check that $Fb=F(a^{n}-\alpha_{n}a^{n-1}-\dots-\alpha_{2}a)$ is a subalgebra of $L$ which complements $L^{2}$, and $[L^{2},b]=L^{2}$. It follows from Lemma 5.3 that $L$ is an $A$-algebra. Moreover, $p(x)$ is divisible by $x$ only once. $\Box$

Theorem 6.2

The cyclic Leibniz $A$-algebra $L$ is monolithic if and only if $p(x)$ has exactly two irreducible factors (one of which is $x$).

Proof. This follows easily from [3, Corollary 4.5]. $\Box$

Corollary 6.3

The cyclic Leibniz $A$-algebra $L$ is monolithic and $\phi$-free if and only if $p(x)=xp_{2}(x)$

Proof. Theorem 6.2 and [3, Corollary 4.2]. $\Box$

Corollary 6.4

If the underlying field is algebraically closed, then the cyclic Leibniz $A$-algebra $L$ is monolithic and $\phi$-free if and only if it is two dimensional with $[a^{2},a]=a^{2}$.

Proof. Clearly $p(x)$ is quadratic, so $L$ is two dimensional, and replacing $a$ by $(1/\sqrt{\alpha_{2}})a$ gives the claimed multiplication. $\Box$

7 Solvable Leibniz $A$-algebras over an algebraically closed field

The following result was proved for Lie algebras by Drensky in [7].

Theorem 7.1

Let $L$ be a solvable Leibniz $A$-algebra over an algebraically closed field $F$. Then the derived length of $L$ is at most $3$.

Proof. First note that we can assume that the ground field is of characteristic $p>0$, since otherwise $L$ is strongly solvable and so of derived length at most $2$. Suppose that $L$ is a minimal counter-example, so the derived length of $L$ is four.

Let $A$ be a minimal ideal of $L$ contained in Leib$(L)$, and put $N=L^{(2)}$. We have that $L^{(3)}=A$. Put $\bar{L}=L/$Leib$(L)$ and for each $x\in L$ write $\bar{x}=x+$ Leib$(L)$. Then $A$ is an irreducible right $\bar{L}$-module, and hence an irreducible right $U$-module, where $U$ is the universal enveloping algebra of $\bar{L}$. Let $\phi$ be the corresponding representation of $U$ and let $\bar{x}\in\bar{L}$, $n\in N$. Then $[[\bar{x},\bar{n}],\bar{n}]=\bar{0}$, whence $[\bar{x},\bar{n}^{p}]=0$ and so $\bar{n}^{p}\in Z=Z(U)$.

Let $n_{1},n_{2}\in N$. Then $\bar{n}_{1}^{p},\bar{n}_{2}^{p}\in Z$, so $\alpha_{1}\bar{n}_{1}^{p}+\alpha_{2}\bar{n}_{2}^{p}\in\hbox{ker}(\phi)$, for some $\alpha_{1},\alpha_{2}\in F$, since dim $\phi(Z)\leq 1$, by Schur’s Lemma. Since $F$ is algebraically closed, there are $\beta_{1},\beta_{2}\in F$ such that $\alpha_{1}=\beta_{1}^{p},\alpha_{2}=\beta_{2}^{p}$, so $(\beta_{1}\bar{n}_{1}+\beta_{2}\bar{n}_{2})^{p}=\beta_{1}^{p}\bar{n}_{1}^{p}+\beta_{2}^{p}\bar{n}_{2}^{p}\in\hbox{ker}(\phi)$, since $[\bar{n}_{1},\bar{n}_{2}]=\bar{0}$. It follows from this together with Lemma 2.5 that $A+F(\beta_{1}n_{1}+\beta_{2}n_{2})$ is a nilpotent subalgebra of $L$ and hence abelian. Thus $\beta_{1}\bar{n}_{1}+\beta_{2}\bar{n}_{2}\in\hbox{ker}(\phi)$ and so dim $\phi(\bar{N})\leq 1$. Hence $Z_{N}(A)$ has codimension at most $1$ in $N$.

Then $\dim N/Z_{N}(A))\leq 1$. Suppose that $\dim N/Z_{N}(A))=1$. Put $S=L/Z_{N}(A)$. Then dim$(S^{(2)})=1$. It follows that $S/Z_{L}(S^{(2)})\subseteq R_{S}(S^{(2)})$ and so has dimension at most one, giving $[S^{(1)},S^{(2)}]+[S^{(2)},S^{(1)}]=0$. But now $S^{(1)}$ is nilpotent but not abelian. As $S$ must be an $A$-algebra, this is a contradiction. We therefore have that dim $(L^{(2)}/Z_{L^{(2)}}(A))=0$, whence $[A,L^{(2)}]=0$.

Now we can include $L^{(3)}$ in a chief series for $L$. So let $0=A_{0}\subset A_{1}\subset\ldots\subset A_{r}=L^{(3)}$ be a chain of ideals of $L$ each maximal in the next. By the above we have $[A_{i},L^{(2)}]\subseteq A_{i-1}$ for each $1\leq i\leq r$. It follows that $L^{(2)}$ is a nilpotent subalgebra of $L$ and hence abelian. We infer that $L^{(3)}=0$, a contradiction. The result follows. $\Box$