Lie algebras with nilpotent length greater than that of each of their subalgebras

The main purpose of this paper is to study the finite-dimensional solvable Lie algebras described in its title, which we call {\em minimal non-${\mathcal N}$}. To facilitate this we investigate solvable Lie algebras of nilpotent length $k$, and of nilpotent length $\leq k$, and {\em extreme} Lie algebras, which have the property that their nilpotent length is equal to the number of conjugacy classes of maximal subalgebras. We characterise the minimal non-${\mathcal N}$ Lie algebras in which every nilpotent subalgebra is abelian, and those of solvability index $\leq 3$.


Introduction
Let L be a Lie algebra, and let L (0) = L, L (i+1) = [L (i) , L (i) ] be its derived series. Recall that L is solvable if there exists r such that L (r) = 0; the smallest such r is called the derived length of L. Similarly, L 1 = L, L i+1 = [L i , L] is the lower central series of L; L is nilpotent of nilpotency index r if L r+1 = 0 but L r = 0. Throughout, L will denote a finite-dimensional solvable Lie algebra over a field F . The symbol '⊕' will denote an algebra direct sum, whilst '+' will denote a direct sum of the underlying vector space structure alone. If U is a subalgebra of L we define U L , the core (with respect to L) of U , to be the largest ideal of L contained in U . We say that U is core-free in L if U L = 0.
We denote the nilradical of L by N (L). We define the upper nilpotent series of L by The nilpotent length, n(L), of L is the smallest integer n such that N n (L) = L.
We define the nilpotent residual, γ ∞ (L), of L be the smallest ideal of L such that L/γ ∞ (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 Γ i (L) of L defined by Γ 0 (L) = L, Γ i+1 (L) = γ ∞ (Γ i (L)) for i ≥ 0. First we note that this series has the same length as that of the upper nilpotent series. Proof.
Suppose that Γ s−k (L) ⊆ N k (L) for some k ≥ 1. then is a nilpotent ideal of L/N k (L) and so is contained in N k+1 (L)/N k (L). Hence Γ s−k−1 (L) ⊆ N k+1 (L), and the result follows.
Although the upper and lower nilpotent series have equal lengths, n say, we do not necessarily have that Γ n−i (L) = N i (L) for i = 0, . . . , n, as the following example shows. From now on we choose to work with the upper nilpotent series. In section 2 we investigate properties of this series, particularly its relationship to maximal subalgebras, and of Lie algebras with nilpotent length k or ≤ k. In considering factor algebras, a complication arises because, unlike the situation in group theory, there are solvable Lie algebras L in which, for an ideal I of L, N (I) may not be contained in N (L). To overcome this obstacle we introduce the notions of nilregular and strongly nilregular subalgebras of L; in particular, it is shown that if a maximal subalgebra of L has a strongly nilregular core, its nilpotent length is at most one less than that of L. The section concludes with a fundamental decomposition theorem for Lie algebras with a given nilpotent length.
In section 3 we introduce the class of extreme Lie algebras in which N i (L)/φ i (L) is a chief factor for each i = 1, . . . , n(L). These are characteriseded in relation to the decomposition result from the previous section and described explicitly in two special cases.
The final section then considers the algebras in the title of the paper. By considering their relationship to extreme Lie algebras the minimal non-N Lie algebras in which every nilpotent subalgebra is abelian, and those of solvability index ≤ 3, are characterised. The last result is that a homomorphic image of a minimal non-N Lie algebra is minimal non-N if it has a complemented minimal ideal.
Much of this is inspired by corresponding work in group theory in [5] and [4], but there are significant differences encountered in the Lie case.

Properties of the upper nilpotent series
The Frattini subalgebra of L, φ(L), is the intersection of the maximal subalgebras of L. Since L is solvable, this is an ideal of L ([2, Lemma 3.4]). The Frattini series of L is given by which yields that φ k+1 (L)/B = φ k+1 (L/B).

Lemma 2.2 If
A is an ideal of L with N r−1 (L) ⊆ A ⊆ N r (L), then n(L/A) = n(L) − r or n(L) − r + 1.
Then it is easy to see that K i /K i−1 = N (L/K i−1 ), and a straightforward induction argument shows that If K n(L)−r = N n(L) (L) we have n(L/A) = n(L) − r; otherwise, n(L/A) = n(L) − r + 1.

Lemma 2.3
Let M be a maximal subalgebra of the solvable Lie algebra L. Then Proof.
(i) This is a straightforward induction proof.
(ii) It is easy to see that this holds for i = 1. So suppose it holds for (iii) This follows from (i) and (ii).
(iv) Suppose that k is the smallest positive integer such that N k (L) ⊆ M .  In general, over a field of characteristic p > 0, N (L) is not a characteristic ideal of L. The best known example is due to Jacobson and first appeared in [7]; however, it is not solvable. We shall see next that N (L) is characteristic in L whenever φ(L) is. First we need some lemmas. Proof. This is easy to check.
Proposition 2.6 Let L be a φ-free Lie algebra over a field of characteristic p > 2. Then N (L) is a characteristic ideal of L.  Proof. This follows easily from Lemma 2.5 and Proposition 2.6.
However, there are solvable Lie algebras L in which N (L) is not a characteristic ideal, as the following example shows. If we form the split extension X = F d+L, where [d, x] = D(x) for all x ∈ L, then L is an ideal of X, but N (L) is not.
Consequently, we shall need the following result from [12]. Proposition 2.8 Let I be a nilpotent subideal of a Lie algebra L over a field F . If F has characteristic zero, or has characteristic p and L has no subideal with nilpotency class greater than or equal to p − 1, then I ⊆ N , where N is the nilradical of L.
Let L be a Lie algebra over a field F and let U be a subalgebra of L. We call the largest integer r such that N r (L) ⊆ U the compatibility index of U . As in [13], if F has characteristic p > 0, we will call U nilregular if the nilradical of U has nilpotency class less than p − 1. If U has compatibility index r, we say that U is strongly nilregular if N k (U )/N k−1 (U ) has nilpotency class less than p − 1 for k = 1, . . . , r. If F has characteristic zero we regard every subalgebra of L as being nilregular. Then we have the following result.
We shall call L primitive if it has a core-free maximal subalgebra. It is said to be primitive of type 1 if it has a unique minimal ideal that is abelian; since L is solvable this is the only type that can occur here (see [11]). If B is an ideal of L and U/B is a subalgebra of L/B, the centraliser of But then an easy induction The result follows by induction.
Note that the above result is not true for all maximal subalgebras, as is shown in the next example.
and L are both maximal subalgebras of X of compatibility index 1. However, Let N (k), N (≤ k) denote the classes of Lie algebras of nilpotent length k and of nilpotent length ≤ k respectively. Of course, over a field of characteristic zero, every Lie algebra L ∈ N (≤ 2). However, over a field of characteristic p > 0 it is easy to construct Lie algebras L ∈ N (k) for any k ∈ N.
A class H of finite-dimensional solvable Lie algebras is called a homomorph if H contains, along with an algebra L, all epimorphic images of Proposition 2.11 The class N (≤ k) is saturated formation for each k ≥ 1.
Proof. It is shown that N (1) is a saturated formation in [2,Lemma 3.7]. Suppose that it holds for k = r. Then N (≤ r + 1) is clearly a homomorph. Suppose that L/A, L/B ∈ N (≤ r + 1). Let S/A = N (L/A) and T /B = N (L/B). Then Corollary 2.12 Let L ∈ N (k) have more than one minimal ideal. Then there is at least one minimal ideal A of L such that L/A ∈ N (k).
Proof. If A 1 , . . . , A n are minimal ideals of L, where n > 1, and L/A i ∈ N (≤ k − 1) for all 1 ≤ k ≤ n, then L ∈ N (≤ k − 1), by Proposition 2.11. Proposition 2.13 Let L be a solvable Lie algebra over a field F . If M is a maximal subalgebra of L for which M L is strongly nilregular, then Proof. Let r be the compatibility index of M . Then L = M + N r+1 (L) and Next we have a fundamental decomposition result.
Proof. Let L ∈ N (n), U i be as in Lemma 2.14 and put The converse is clear.

Extreme Lie Algebras
The Lie algebra L is monolithic if it has a unique minimal ideal A, the monolith of L. If B is an ideal of L and A/B is a minimal ideal of L/B we say that A/B is a chief factor of L. The series  We say that the chief factor A/B is complemented if there is a maximal subalgebra M of L such that L = A + M and A ∩ M = B. We define c(L) to be the number of complemented chief factors in a chief series for L. This is independent of the particular chief series chosen, by [11,Theorem 2.3].
Let x ∈ L and let ad x be the corresponding inner derivation of L. If F has characteristic zero suppose that (ad x) n = 0 for some n; if F has characteristic p suppose that x ∈ I where I is a nilpotent ideal of L of class less than p. Put Then exp(ad x) is an automorphism of L. We call the group I(L) generated by all such automorphisms the group of inner automorphisms of L. Two subsets U, V are conjugate in L if U = α(V ) for some α ∈ I(L).
Then we have the following characterisation of extreme Lie algebras. Proof. (i) ⇒ (ii) : Let L be extreme and consider the series There is a unique conjugacy class of maximal subalgebras of L complementing the chief factor N i (L)/φ i (L) for each i = 1, 2, . . . , n(L), by [1]. But each maximal subalgebra of L must complement one of the complemented chief factors in the above series, and must, therefore, belong to one of these n(L) conjugacy classes. Hence n(L) = m(L).   Conversely, suppose that L satisfies (iv) and consider L/φ i (L). Since N i (L)/φ i (L) is the direct sum of complemented minimal ideals of L/φ i (L), as above, it follows that N i (L)/φ i (L) is a chief factor of L/φ i (L). Hence L is extreme.
Lemma 3.4 Let L be an extreme Lie algebra.
(i) If L is nilpotent, then dim L = 1.
(ii) This follows from Lemma 3.2 and (i). Proof. We have that Also dim B n = 1 by Lemma 3.4 (ii).
If S is a subalgebra of L, we will denote by S the image of S under the canonical homomorphism from L onto L/φ(L). We have the following characterisation of those Lie algebras L ∈ N (≤ 2) that are extreme, which includes all extreme Lie algebras over a field of characteristic zero. Proof. This is just the cases n = 1, 2 in Theorem 3.5. Conversely, if dim(L/φ(L)) ≤ 2 then L/φ(L) is supersolvable, and so L is supersolvable, by [1,Theorem 6]. Clearly L is also extreme in each of cases (i) and (ii).
Example 3.1 It is easy to check that every three-dimensional Lie algebra as described in Corollary 3.7 has a basis x, y, z with non-zero products [x, y] = y + z, [x, z] = αz for some 0 = α ∈ F . Moreover, no two of these with different values of α are isomorphic.

Minimal non-N algebras
If X is a class of Lie algebras, we say that L is minimal non-X if every proper subalgebra of L, but not L itself, belongs to X . We say that L is minimal non-N if it is minimal non-N (≤ k) for some k; in other words, if its nilpotent length is greater than that of any of its proper subalgebras. Over a field of characteristic zero a Lie algebra can only be minimal non-N (1) and these are described in [9].  In group theory, every minimal non-N (k) group is extreme, and so a natural question is whether this holds for Lie algebras. We show next that this is 'usually' the case for Lie algebras. We call a class H of Lie algebras a semi-homomorph if, for all L ∈ H, which has nilpotency class < p − 1 for i = 1, . . . , r − 1, since M L is strongly nilregular. Hence M/N (L) has a strongly nilregular core.
The reverse inclusion is clear and the claim is established.

Now we have that
which has nilpotency class < p−1 for i = 1, . . . , r, since M L is strongly nilregular. Hence M/N (L) has a strongly nilregular core.
A solvable primitive algebra has a unique minimal, self-centralising, ideal A such that L = A+U (see [11]). We shall say that a class of Lie algebras H has the primitive quotient property if, for every primitive algebra L in H with minimal ideal A, L/A is minimal non-N . A Lie algebra L is called an A-algebra if all of its nilpotent subalgebras are abelian. These arise in the study of constant YangMills potentials and in relation to the problem of describing residually finite varieties. The structure of solvable Lie A-algebras was studied in some detail in [10]. In the case of an A-algebra the lower nilpotent series and the derived series coincide ([10, Lemma 2.3]), and so the terms "derived length" and "nilpotent length" are identical.  Proof. Let H be the class of Lie algebras of solvability index ≤ 3. Then H is clearly a semi-homomorph. Let L ∈ H be primitive with minimal ideal A = N (L) and let M be a maximal subalgebra containing A. If L has solvability index ≤ 2 it is clear that L/A is minimal non-N , so assume that L has index 3.
We  Next we seek to characterise the algebras considered in Corollaries 4.5 and 4.6. We can characterise the A-algebras that are also minimal non-N as follows.
Theorem 4.7 Let L be a Lie A-algebra of derived length n + 1 over a field F . Then L is minimal non-N if and only if the following hold.
(iv) A i is an irreducible L/L (i+1) -module for each 0 ≤ i ≤ n; and Proof. Suppose first that L is minimal non-N . Since L is an A-algebra of derived length n+1, L = A n+ A n−1 . . .+A 1+ A 0 and L (i) = A n+ A n−1 . . .+A i , where A i is an abelian subalgebra of L for each 0 ≤ i ≤ n, by [10, Corollary 3.2]. But dim A 0 = 1, by Lemma 4.1 (i), so A 0 = F x for some x ∈ L. This gives (i) and (ii). The decomposition in (i) follows from the splitting of a Lie A-algebra over each term in its derived series ([10, Theorem 3.1]), so A straightforward induction argument shows that A i ⊆ N n−i−1 for 0 ≤ i ≤ n. We now establish (iv) and (v) by induction on n. Then (iv) clearly holds for i = 0, and (v) holds for i = 0 by Lemma 4.1 (i), so suppose that they hold for all i ≥ k (k ≥ 0). Then N n−k+1 (L) = L (k) = A n+ . . .+A k and L (k+1) = A n+ . . .+A k+1 ⊆ N n−k (L). It follows that N n−k (L) = L (k+1) by the irreducibility of A k and the fact that N n−k (L) = N n−k+1 (L) (since L has nilpotent length n + 1).
Conversely, suppose that (i)-(v) hold and let M be a maximal subalgebra of L. Clearly L ∈ N (n+1). Let i be the smallest integer such that L (i) ⊆ M .
Recall the following result from [3].  In either case M is nilpotent-by-abelian. Hence L is minimal non-(nilpotentby-abelian). It is of type I, since otherwise L ∈ N (2).
Conversely, suppose that L is minimal non-(nilpotent-by-abelian) of type I. Then L has solvability index 3 and the maximal subalgebras of L are nilpotent-by-abelian, as in the paragraph above. Clearly L itself is not nilpotent-by-abelian.
Lie algebras as described in Theorem 4.9 do exist over every field of characteristic p > 0, as is shown in [3]; over an algebraically closed field they are minimal non-supersolvable ([3, Theorem 5]). Finally we show that a homomorphic image of a minimal non-N Lie algebra is minimal non-N if it has a complemented minimal ideal. Proof. Suppose there is a chief series of L through A and B in which A/B is the kth complemented factor, where 1 ≤ k ≤ c(L). If k = 1, then B ⊆ φ(L) and so n(L/B) = n(L), in which case L/B is minimal non-N because L is.
So let k > 1 and assume that the theorem holds for the (k − 1)th complemented factor C/D. Without loss of generality we may assume that D = 0. Then A/B is the second complemented factor in some chief series of L. It follows from Theorem 4.3 that L and L/N (L) are extreme, and so every chief series of L has only one complemented chief factor below N (L), by