ON THE SUBALGEBRA LATTICE OF A RESTRICTED LIE ALGEBRA

. In this paper we study the lattice of restricted subalgebras of a restricted Lie algebra. In particular, we consider those algebras in which this lattice is dually atomistic, lower or upper semimodular, or in which every restricted subalgebra is a quasi-ideal. The fact that there are one-dimensional subalgebras which are not restricted results in some of these conditions being weaker than for the corresponding conditions in the non-restricted case.


Introduction
The relationship between the structure of a group and that of its lattice of subgroups is highly developed and has attracted the attention of many leading algebraists.According to Schmidt ([18]), the origin of the subject can be traced back to Dedekind, who studied the lattice of ideals in a ring of algebraic integers; he discovered and used the modular identity, which is also called the Dedekind law, in his calculation of ideals.Since then modularity and lattice conditions related to it have been studied in a number of contexts.The lattice of submodules of a module over a ring is modular, and hence so is the lattice of subgroups of an abelian group.The lattice of normal subgroups of a group is also modular, but the lattice of all subgroups is not in general.
The study of the subalgebra lattice of a finite-dimensional Lie algebra was popular in the 1980's and in the 90's (see, for example, [3,7,8,9,10,11,13,14,25,26,27,28,29,30]), but interest then waned.The likely reason is that most of the conditions under investigation were too strong and so few algebras satisfied them.However, the lattice of restricted subalgebras of a restricted Lie algebra is fundamentally different; for example, not every element spans a one-dimensional restricted subalgebra.Thus, one could expect more interesting results to hold for restricted algebras and, as we shall see, this is indeed the case.
In Section 2 we fix some notation and terminology and introduce some results that are needed later.In Section 3 we study restricted Lie algebras that are dually atomistic; that is, such that every restricted subalgebra is an intersection of maximal restricted subalgebras.We show that such algebras over an algebraically closed field of positive characteristic are solvable or semisimple, and then characterise the solvable ones more precisely.It turns out that they are more abundant than in the non-restricted case.We then investigate those restricted Lie algebras all of whose subalgebras (not necessarily restricted) are intersections of maximals.It is shown that if the ground field is algebraically closed then there are no such algebras that are perfect.
The objective in Section 4 is to study restricted Lie algebras L in which every restricted subalgebra is a restricted quasi-ideal; that is, such that [S, H] ⊆ S +H for all restricted subalgebras S, H of L. These are characterised over an algebraically closed field of characteristic different from 2, and the nilpotent ones more generally over a perfect field of characteristic different from 2. In this regard, we also mention that restricted Lie algebras over perfect fields all of whose restricted subalgebras are ideals were characterized by the second author in [19].Section 5 then goes on to consider J-algebras and lower semimodular restricted Lie algebras.The main result here is the same as for the corresponding situation as in the non-restricted case if the underlying field is algebraically closed, but it is pointed out that the assumption of algebraic closure cannot be omitted.
The final section is devoted to studying upper semimodular restricted Lie algebras.It is shown that, over an algebraically closed field, such an algebra is either almost abelian or nilpotent of class at most 2.This is proved by considering first upper semimodular restricted Lie algebras that are generated by their one-dimensional restricted subalgebras.It is also shown that over such fields the conditions that L is modular, L is upper semimodular and every restricted subalgebra of L is a quasi-ideal are equivalent.

Preliminaries
Here we fix some notation and terminology and introduce some results that will be needed later.Unless otherwise stated, throughout the paper all algebras are supposed finite-dimensional.Let L be a Lie algebra over a field F. As usual, the derived series for L is defined inductively by Similarly, the lower central series is defined inductively by Also, L is said to be supersolvable if it admits a complete flag made up of ideals of L, that is, there exists a chain of ideals of L such that dim L j = j for every 0 ≤ j ≤ n.The centre of L is denoted by Z(L), and C B (A) = {x ∈ B : [x, A] = 0} denotes the centraliser in a subalgebra B of another subalgebra A. Also, the ascending central series is defined inductively by is defined to be the maximal nilpotent ideal, and the solvable radical, denoted by R(L), is defined to be the maximal solvable ideal.For every x ∈ L, the adjoint map of x is defined by ad(x) : L → L, a → [x, a].If S is a subalgebra of L, then the largest ideal of L contained in S is called the core of L and is denoted by S L .The Frattini subalgebra F (L) of L is the intersection of all maximal subalgebras of L; the Frattini ideal of L is φ(L) = F (L) L .The abelian socle, Asoc(L), is the sum of the minimal abelian ideals of L. We will denote algebra direct sums by ⊕, whereas direct sums of the vector space structure alone will be written as +.
We say that L is dually atomistic if every subalgebra of L is an intersection of maximal subalgebras of L. The Lie algebra L is said to be almost abelian if L = Fx +A, where A is an abelian ideal of L and ad(x) acts as the identity map on A. Scheiderer proved in [17] that, over a field of characteristic zero, every dually atomistic Lie algebra is abelian, almost abelian or simple.Here we establish a slightly weaker version of this which is valid over any field.Proposition 2.1.If L is a dually atomistic Lie algebra over any field then L is either abelian, almost abelian or semisimple.
Proof.Let L be dually atomistic and suppose that L is not semisimple.Then Asoc(L) = 0 and L splits over Asoc(L), by [24,Theorem 7.3].Furthermore, the minimal abelian ideals of L are one-dimensional, by [17, Lemma 1], so we can write Let M be a maximal subalgebra of L with a 1 ∈ M .We shall show that L (∞) ⊆ M .Now L = Fa 1 + M , so M has codimension one in L. It follows that L/M L is as described in [2, Theorems 3.1 and 3.2].Also, [Fa 1 , M L ] ⊆ Fa 1 ∩ M = 0. We consider the three cases given in [2, Theorem 3.1] separately.
Case (a): Case (c): 2 , where L m (Γ) 2 is simple.Put L/M L = L, and so on.
In any case we have established that, for any maximal subalgebra M of L, either a 1 ∈ M or L (∞) ⊆ M .Suppose that L (∞) = 0. Then L (∞) = Fa 1 (since (Fa 1 ) 2 = 0), so there is an element x ∈ L (∞) \ Fa 1 .Let M be a maximal subalgebra containing x + a 1 .Then either a 1 ∈ M or L (∞) ⊆ M .In each case, Fx + Fa 1 ⊆ M .It follows that F(x + a 1 ) cannot be an intersection of maximal subalgebras of L, a contradiction.Hence, L (∞) = 0 and L is solvable.The result now follows from [17, Lemma 1].
We shall need the following result which is due to Grunewald, Kunyavskii, Nikolova and Plotkin for p > 5.However, the same proof works for p > 3 by using the Corollary in page 180 of [20]. A. Premet has pointed out that the result is also valid for p = 3, but that it relies on results that have not yet been published, so we omit this case.In what follows we shall be studying the lattice of restricted subalgebras of a restricted Lie algebra.Let L be a restricted Lie algebra over a field of characteristic p > 0. For a subset S of L, we denote by S p the restricted subalgebra generated by S. We say that L is cyclic if L = x p for some x ∈ L.An element x ∈ L is said to be semisimple if x ∈ x [p]  p and toral if x [p] = x.An abelian restricted Lie algebra consisting of semisimple elements is called a torus.An element x ∈ L is said to be p-nilpotent if x [p] n = 0 for some n > 0 (in this case, the minimal n with such a property is called the order of x), and L is said to be p-nilpotent if there exists n > 0 such that x [p] n = 0 for every x ∈ L. We also introduce as "restricted analogues" of earlier concepts, F p (L), the Frattini p-subalgebra of L, to be the intersection of the maximal restricted subalgebras of L, and φ p (L), the Frattini p-ideal of L, to be the largest restricted ideal of L which is contained in F p (L).The abelian p-socle, A p soc(L), is the sum of the minimal abelian restricted ideals of L. To avoid tedious repetition we shall therefore often omit the word 'restricted'.

Dually atomistic Lie algebras
We say that a restricted Lie algebra L is dually atomistic if every restricted subalgebra of L is an intersection of maximal restricted subalgebras of L. It is easy to see that if L is dually atomistic then so is every factor algebra of L, and if L is dually atomistic then it is φ p -free.Lemma 3.1.Let L be a dually atomistic restricted Lie algebra.Then: [24,Theorem 6.5] and [15,Theorem 3.5]. (ii (iii) By (i), every subspace of N (L) is a subalgebra of L. Let S be any subspace of N (L).Then where M is the set consisting of all maximal restricted subalgebras of L containing S p .Therefore, S p is an intersection of restricted ideals of L, by (ii), and so is itself a restricted ideal of L. Proposition 3.2.Let L be a dually atomistic restricted Lie algebra over an algebraically closed field F. Then L is solvable or semisimple.[15,Theorems 3.4 and 4.2].If B = 0, then L is nilpotent and we are done.Assume therefore that B = 0.
Suppose first that , and let M be a maximal restricted subalgebra containing z + b p .Then z, b ∈ M , so we must have which implies that z is semisimple, from the second sum.This must hold for every choice of z ∈ Z(L), so Z(L) is a torus of L, by [23, Chapter 2, Theorem 3.10].A similar argument shows that if z is not semisimple then every b must be, in which case B 2 p is a torus of L. Hence, either Z(L) or B 2 p is a torus.In the latter case, B 2 p is abelian, contradicting the fact that B is semisimple.In the former case, both Z(L) and B 2 p have a toral element: z and b, say.But then Therefore suppose that N (L) = Z(L).Then there is a minimal restricted ideal A with A ⊆ N (L) and A∩Z(L) = 0.Moreover, if a ∈ A, we have that a [p] ∈ A∩Z(L), so A = Fa with a [p] = 0, by Lemma 3.1(iii).Let M be a maximal restricted subalgebra of L such that a / ∈ M .We have L = M +A, by [15, Lemma 2.1], so M has codimension one in L, and, as in Proposition 2.1, p is abelian.But this means that L is solvable.For a field F of characteristic p > 0, we will denote by F[t, σ] the skew polynomial ring over F in the indeterminate t with respect to the Frobenius endomorphism σ of F. We recall F[t, σ] is the ring consisting of all polynomials f = i≥0 α i t i with respect to the usual sum and multiplication defined by the condition t • α = α p t for every α ∈ F. Proposition 3.3.Let L be a solvable restricted Lie algebra over any field F. If L is dually atomistic then Proof.The nilradical N (L) of L is non-zero and abelian by Lemma 3.1(i).As L is φ p -free, L = N (L) +B for some restricted subalgebra B of L, and We can write , where we allow that r could be 0. Since B acts faithfully on A we cannot have r = n.Then We introduce another piece of notation before presenting the following results.We say that a Lie algebra is restricted dually atomistic if it is restricted and every subalgebra is an intersection of maximal subalgebras.Proposition 3.4.Let L be a perfect restricted dually atomistic Lie algebra over any field F of characteristic p > 0. Then every subalgebra of L is restricted.
Proof.Arguing as in [15,Lemma 3.7], it is immediate to prove that every maximal subalgebra of L is self-idealising.It follows from [15, Lemma 3.9] that every maximal subalgebra of L is restricted.The result now follows from the fact that L is dually atomistic.
Theorem 3.5.There are no perfect restricted dually atomistic Lie algebras over an algebraically closed field.
Proof.Suppose that L is a counterexample of minimal dimension.By Proposition 3.4, L is simple as a Lie algebra, and hence its absolute toral rank is just the dimension of a maximal torus T (cf.[22, §1.2]).Given two linearly independent elements x, y ∈ T , Proposition 3.4 forces 0 = (x + λy) [p] ∈ F(x + λy) for all λ ∈ F, but this cannot happen since F is algebraically closed.Hence, L has absolute toral rank 1.Now, if F has characteristic p = 2, 3, then [21, Theorem 6.5] yields that L is solvable or isomorphic to sl(2, F) or to psl(3, F).Otherwise, L has a restricted subalgebra with a quotient isomorphic to sl(2, F), by Lemma 2.2 and Proposition 3. As well as the three-dimensional non-split simple Lie algebra, which is dually atomistic in the characteristic zero case, there exist other perfect dually atomistic simple restricted Lie algebras over a perfect field which is not algebraically closed.For example, let L be the seven-dimensional simple Lie algebra over a perfect field of characteristic 3 constructed by Gein in [8,Example 2].This algebra L can be endowed with a [p]-mapping such that every element is semisimple.Any two linearly independent elements of L generate a three-dimensional non-split restricted subalgebra which is maximal in L. Any second-maximal restricted subalgebra is then one-dimensional, and every one-dimensional restricted subalgebra S is inside more than one maximal restricted subalgebra whose intersection is S.
We finish this section by studying the so-called atomistic restricted Lie algebras, those in which every restricted subalgebra is generated by minimal restricted subalgebras.
Proposition 3.6.Let F be an algebraically closed field of characteristic p > 0. A restricted Lie algebra L over F is atomistic if and only if every p-nilpotent cyclic restricted subalgebra is one-dimensional.
Proof.Note that L is atomistic if and only if all its cyclic restricted subalgebras are atomistic.Consider the cyclic restricted subalgebra C, whose semisimple elements form a torus T , and whose p-nilpotent elements form a p-nilpotent restricted subalgebra P .By [23, Chapter 2, Theorem 3.6], T is atomistic.From [23, Chapter 2, Theorem 3.5], it follows that C = T ⊕ P , so C is atomistic if and only if P is atomistic.But this is equivalent to requiring that dim P = 1 ([16, Theorem 3.8]).The result follows.

Restricted quasi-ideals
A restricted subalgebra S of L is called a restricted quasi-ideal of L if [S, H] ⊆ S + H for all restricted subalgebras H of L. Clearly, every restricted subalgebra that is a quasi-ideal is also a restricted quasi-ideal.
Denote by L [p] the restricted subalgebra generated by all the elements x [p] , with is a quasi-ideal.Suppose that L [p] is not an ideal of L, and factor out (L [p] ) L , so we can assume that L [p] is core-free.Then, by [1, Theorem 3.6], there are three possibilities which we will consider in turn.
Suppose first that L [p] has codimension 1 in L. Define (L [p] ) i as in [2, (5)].Then every element x ∈ L can be written as x = x s + x n , where x s is semisimple and x n is p-nilpotent, by [23,Theorem 3.5].Moreover, all semisimple elements belong to L [p] , so L = L [p] + Fx for some p-nilpotent element x.Suppose that x [p] k = 0. Now (L [p] , by [2, Lemma 2.1], contradicting the fact that L [p] is not an ideal of L. On the other hand, [1, Theorem 3.6(c)] cannot hold, as the three-dimensional simple Lie algebra W (1, 2) 2 over a field of characteristic 2 is not restrictable.To see this simply note that the derivation ad 2 (x) is not inner.
Finally, suppose that [1, Theorem 3.6(d)] holds.Then L = L 2 + Fy where ad(y) acts as the identity map on L 2 and L [p] = Fy.Let x ∈ L 2 .We have ad p (y) = ad(y) and ad p (x) = 0 for every x ∈ L 2 .Therefore, as L is centerless, the p-mapping of L is determined by the conditions y [p] = y and x [p] = 0.This implies The converse is straightforward.
Proposition 4.3.Let L be a restricted Lie algebra such that every restricted subalgebra of L is a restricted quasi-ideal.Then L 2 ⊆ L [p] .It follows that L 3 = L p+1 ; in particular, if L is nilpotent, then L has nilpotency class at most 2.
Lemma 4.4.Let L be a restricted Lie algebra over an algebraically closed field of characteristic p > 0 in which every restricted subalgebra is a restricted quasi-ideal.
If H is a Cartan subalgebra of L, then L has root space decomposition where Φ is the set of roots α for which −α is also a root, and Ψ is the remaining set of roots.
Proof.Let T be a maximal torus, H = C L (T ) and let L = H +α∈Π L α be the corresponding root space decomposition.Then 0 also, giving the root space decomposition claimed.Suppose that every restricted subalgebra of L is a restricted quasi-ideal.Let S be the subspace spanned by the semisimple elements of L and let P be the subspace spanned by the p-nilpotent elements of L. Then S and P are subalgebras of L, since [x, y] ∈ x p + y p , and, if F is perfect, L = S + P .Moreover, both are restricted, since and x [p] , y [p] are semisimple (respectively, p-nilpotent) if so are x, y, and s i (x, y) ∈ x, y p .Proposition 4.5.Let L be a nilpotent restricted Lie algebra over a perfect field of characteristic different from 2. Then every restricted subalgebra of L is a restricted quasi-ideal of L if and only if L = S ⊕ P , where S is a toral ideal and P is a p-nilpotent ideal in which every restricted subalgebra is a restricted quasi-ideal.
Proof.Suppose that every restricted subalgebra of L is a restricted quasi-ideal of L .By Proposition 4.3, L 3 = 0 and L [p] ⊆ Z(L).Then, for all x, y ∈ L, (x + y) [p] = x [p] + y [p] , so S, P are just the sets of semisimple and p-nilpotent elements of L respectively.Then S ∩ P = 0 and S ⊆ Z(L).It follows that L = S ⊕ P and that S is toral.
The converse is straightforward.
Corollary 4.6.Let L be a restricted Lie algebra over an algebraically closed field of characteristic different from 2 in which every restricted subalgebra of L is a restricted quasi-ideal of L. Then L has a Cartan subalgebra H such that H = S ⊕ P where S is a torus and P is the set of p-nilpotent elements in H, and L = S +N where N is an ideal, N 3 = 0 and N [p] ⊆ Z(H).

J-algebras and lower semimodular restricted Lie algebras
For this section, it will be useful to handle the following result.
Lemma 5.1.Let L be a restricted Lie algebra over an algebraically closed field of characteristic p > 0. If L is supersolvable, then L admits a complete flag made up of restricted ideals of L.
Proof.Plainly, it is enough to show that L has a one-dimensional restricted ideal, from which the conclusion will follow by induction.Suppose dim L > 1, the claim being trivial otherwise.Consider a complete flag of ideals of L. If the ideal L 1 is restricted, then we are done.Thus we can suppose that there exists x ∈ L 1 such that x [p] / ∈ L 1 .As L 1 is an abelian ideal, the restricted subalgebra H generated by x [p] is contained in the centre of L. Since the ground field is algebraically closed, by [23,Chapter 2,Theorem 3.6] we see that H contains a toral element t.We conclude that I = Ft is a one-dimensional restricted ideal of L, as desired.
Note that the assumption that the ground field is algebraically closed is essential for the validity of Lemma 5.1.In fact, over arbitrary fields of positive characteristic, there can be cyclic restricted Lie algebras of arbitrary dimension with no non-zero proper restricted subalgebras (cf.[16,Proposition 3.1]).
Let L be a restricted Lie algebra.A restricted subalgebra U of L is called lower semimodular in L if U ∩ B is maximal in B for every restricted subalgebra B of L such that U is maximal in U, B p .We say that L is lower semimodular if every restricted subalgebra of L is lower semimodular in L.
If U , V are restricted subalgebras of L with U ⊆ V , a J-series (or Jordan-Dedekind series) for (U, V ) is a series of restricted subalgebras such that U i is a maximal subalgebra of U i+1 for 0 ≤ i ≤ r − 1.This series has length equal to r.We shall call L a J-algebra if, whenever U and V are restricted subalgebras of L with U ⊆ V , all J-series for (U, V ) have the same finite length, d(U, V ).Put d(L) = d(0, L).Proposition 5.2.For a solvable restricted Lie algebra L over an algebraically closed field of characteristic p > 0 the following are equivalent: (i) L is lower semimodular; (ii) L is a J-algebra; and (iii) L is supersolvable.
(ii)⇒(iii): We first show by induction on dim L that there exists a series of restricted subalgebras from 0 to L having length dim L. Suppose L = 0.As L is solvable, it holds that L (1) p = L; otherwise, L (1) = L (1) (1) p = L (2) = 0, a contradiction.Then the inductive hypothesis ensures the existence of a series of restricted subalgebras Moreover, as L/ L ′ p is abelian, Lemma 5.1 yields the claim.Now, by hypothesis, all J-series of restricted subalgebras from 0 to L have length dim L, and consequently all maximal restricted subalgebras have codimension one in L. On the other hand, if H is a maximal subalgebra of L which is not restricted, then pick an element x of H such that x [p] / ∈ H. Then H + Fx [p] is a subalgebra of L properly containing H, so H + Fx [p] = L by the maximality of H. Therefore, every maximal subalgebra has codimension one in L, which allows to conclude that L is supersolvable, by [5,Theorem 7].
Note that the assumption of solvability is actually needed in the previous result.In fact, consider the restricted Lie algebra L = sl(2, F) over an algebraically closed field F of characteristic p > 2. Then all J-series of restricted subalgebras of L have length 3, despite the fact that L is simple.

Upper semimodular restricted Lie algebras
Let L be a restricted Lie algebra.We say that a restricted subalgebra S of L is upper semimodular in L if S is maximal in S, T p for every restricted subalgebra T of L such that S ∩ T is maximal in T .The restricted Lie algebra L is called upper semimodular if all of its restricted subalgebras are upper semimodular in L.
This section is devoted to study the structure of upper semimodular restricted Lie algebras over algebraically closed fields.In particular, our main aim of this section is to prove the following result: Theorem 6.1.Let L be a restricted Lie algebra over an algebraically closed field.The following conditions are equivalent: (i) L is upper semimodular; (ii) L is modular; (iii) every restricted subalgebra of L is a restricted quasi-ideal.Moreover, if one of the previous statements holds, then L is either almost abelian or nilpotent of class at most 2.
We start with some preliminary results.Let L be an almost abelian Lie algebra over a field F of characteristic p > 0. Then L = Fx +A, where A is an abelian ideal and ad(x) acts as the identity map on A. It is immediate to check that L is restrictable and also centerless, so it admits a unique p-mapping by [23, Chapter 2, Corollary 2.2].Explicitly, this p-mapping is given by a [p] = 0 for all a ∈ A and x [p] = x.Lemma 6.2.Let L be an upper semimodular restricted Lie algebra over an algebraically closed field of characteristic p > 0. If L is generated by two distinct one-dimensional restricted subalgebras X and Y , then L is two-dimensional.
Proof.Let Z be a nonzero proper restricted subalgebra of L. Assume first that X ⊆ Z, Y ⊆ Z.As X ∩ Y = 0 is maximal in Y , X must be maximal in L, yielding Z = X.Assume now that X, Y ⊆ Z and take a one-dimensional restricted subalgebra Z ′ of Z.By the previous case, X, Z ′ p = L. Since X ∩ Z ′ = 0 is maximal in X, Z ′ is maximal in L and Z = Z ′ .Thus, all nonzero proper restricted subalgebras of L are one-dimensional, and it follows from [32,Lemma 1.6] that L is two-dimensional.Lemma 6.3.Let F be an algebraically closed field of characteristic p > 0. Let L be a non-abelian upper semimodular restricted Lie algebra over F generated by three one-dimensional restricted subalgebras.Then, L is centerless.
Proof.Let Fx, Fy, Fz be three distinct one-dimensional restricted subalgebras generating L and suppose, by contradiction, that Z(L) = 0. Note that we can take x to be either toral or such that x [p] = 0.By Lemma 6.2 and without loss of generality, we may also assume x ∈ Z(L) and that y, z p is almost abelian, with [y, z] = z, y [p] = y and z [p] = 0.If x [p] = 0, then x + z p ∩ Fy = 0 is maximal in x + z p , but Fy is not maximal in x + z, y p = L, a contradiction.On the other hand, if x is toral, then x ∈ x − z p ⊆ x + y, y + z p , so x + y, y + z p = L. Now x + y p ∩ y + z p = 0 is maximal in x + y p , but y + z p is not maximal in L, a contradiction.Proposition 6.4.Let F be an algebraically closed field of characteristic p > 0. Any upper semimodular restricted Lie algebra L over F generated by its one-dimensional restricted subalgebras is either abelian or almost abelian.
Proof.By Lemma 6.2, all the restricted subalgebras of L generated by two onedimensional restricted subalgebras are abelian or almost abelian.Suppose that y, x 1 p is almost abelian, where Fy, Fx 1 are restricted subalgebras of L with [y, x 1 ] = x 1 , y [p] = y and x [p] 1 = 0. Write L = y, x 1 , . . ., x s p , where y, x 1 , . . ., x s are linearly independent.We claim that y, x i p is almost abelian for i = 2, . . ., s. Suppose otherwise that [y, x i ] = 0 for some i = 1.By Lemma 6.3, we must have [x 1 , x i ] = 0. Then x 1 , x i p would be almost abelian and [x 1 , x i ] = λx 1 for some λ ∈ F, λ = 0.But then y + λ −1 x i ∈ Z( y, x 1 , x i p ) = 0 by Lemma 6.3, a contradiction.Note also that [y, x i ] / ∈ Fy, as otherwise y [p] = 0. Therefore, we can clearly assume that [y, Therefore, L = x 1 , . . ., x s p +Fy is an almost abelian restricted Lie algebra of dimension s + 1, as desired.
Note that the hypothesis of F being algebraically closed is essential for our results.Indeed, the Lie algebra L over a perfect field of characteristic 3 given by Gein in [8, Example 2], with the p-mapping indicated in Section 3, is upper semimodular, generated by its minimal restricted subalgebras and semisimple.The reader could ask if, ruling out the hypothesis of F being algebraically closed, any upper semimodular restricted Lie algebra generated by its minimal restricted subalgebras would be abelian, almost abelian or semisimple, in a way somehow similar to the situation in the ordinary Lie algebra setting (see [10]).However, this is not the case either: the restricted Lie algebra Fx ⊕ L, with x [p] = 0, is generated by its minimal restricted subalgebras and it is upper semimodular, but it is neither abelian, nor almost abelian, nor semisimple.Furthermore, it is even possible to pick a modular restricted subalgebra of Fx ⊕ L which does not lie in any of these three cases.Proposition 6.5.Let F be an algebraically closed field of characteristic p > 0, and let L be an upper semimodular restricted Lie algebra over F. Let B be the restricted subalgebra generated by the one-dimensional restricted subalgebras of L. If B is almost abelian, then L = B.
Proof.Assume L = B.By Proposition 3.6, there exists a p-nilpotent element x ∈ L of order 2. Write B = A +Fy, where A is a strongly abelian restricted ideal of B, and y is a toral element which acts as the identity map on A. Since x [p] ∈ A, we have ad p (x)(y) = [x [p] , y] = −x [p] .Set w = ad p−1 (x)(y), and note that [x, w] = −x [p]  and [x [p] , w] = [x, w [p] ] = 0.
As x p ∩ x [p] , y p = Fx [p] is maximal in x [p] , y p = Fx [p] + Fy, one has that x p must be maximal in x, x [p] , y p = x, y p .We have x p x, w p ⊆ x, y p .
It follows that y ∈ x, w p = x p + w p , from which [x, y] = λ[x, w] = −λx [p] , for some λ ∈ F. But then −x [p] = ad p (x)(y) = −λad p−1 (x)(x [p] ) = 0, a contradiction.Therefore, L = B and L is almost abelian.Theorem 6.6.Let F be an algebraically closed field of characteristic p > 0. Any upper semimodular restricted Lie algebra L over F is either abelian, almost abelian or of the form L = x 1 , . . ., x r , B p , where x i is p-nilpotent of order n i > 1 for all i = 1, . . ., r, B is an abelian restricted subalgebra and [L, L] ⊆ x 1 , . . ., x r p .
Proof.Let B be the restricted subalgebra generated by the one-dimensional restricted subalgebras of L. By Proposition 6.4, B is either abelian or almost abelian.If L = B, then B is abelian by Proposition 6.5, and every x i / ∈ B is p-nilpotent of order n i > 1 by Proposition 3.6.
To prove that [L, L] ⊆ x 1 , . . ., x r p , it suffices to see that [x i , b] ∈ x i p , for i = 1, . . ., r and b ∈ B such that b p is one-dimensional.Take such a b ∈ B. If b ∈ x i p , then we are done.Otherwise, x i p ∩ b p = 0 is maximal in b p = Fb, and then x i p must be maximal in x i , b p .Write w = ad r−1 (x i )(b) = 0, where r is such that ad r (x i )(b) = 0. We have the following chain of inclusions Then, w ∈ x i p .Assume now that ad r−k (x i )(b) ∈ x i p for some k > 1, and set w ′ = ad r−k−1 (x i )(b).Again, it is clear that x i p ⊆ x i , w ′ p ⊆ x i , b p , where one inclusion has to be an equality.By assumption, if b ∈ x i , w ′ p = x i p + w ′ p , then [x i , b] ∈ x i p .Therefore w ′ ∈ x i p , and by induction we have that [x i , b] ∈ x i p .
Note that, although any abelian or almost abelian restricted Lie algebra is upper semimodular, the converse of Theorem 6.6 does not hold, as the following example shows.
Example 6.7.Let L = x, y, z p with x [p] 2 = y [p] = z [p] = 0 and [x, y] = z as the only non-zero product.Then the restricted subalgebra B = Fx [p] ⊕ Fy ⊕ Fz generated by all the one-dimensional restricted subalgebras is abelian.However, L is not upper semimodular, as x p ∩ Fy = 0 is maximal in Fy, but x p is not maximal in x, y p = L.
Proposition Let F be an algebraically closed field of characteristic p > 0, and let L be an upper semimodular restricted Lie algebra over F.Then, L is almost abelian or nilpotent.
Proof.Assume that L is not almost abelian.Let T be a torus of L. By [23, Chapter 2, Theorem 3.6], T has a basis consisting of toral elements and therefore T ⊆ B, in the notation of Theorem 6.6.Consequently, every semisimple element of L belongs to B, and the restricted subalgebra T formed by the semisimple elements of L is the unique maximal torus of L. Suppose, by contradiction, that L is not nilpotent.Consider the Cartan subalgebra H = C L (T) and the associated root space decomposition L = H +( α∈Φ L α ).Then there exists α ∈ Φ and a toral element t ∈ T such that α(t) = 0. Let x ∈ L α , x = 0.By [23,  x [p] n for every n > 0. Now, by [23, Chapter 2, Theorem 3.4] we see that (t+x) [p] n ∈ T for some sufficiently large n, and so we deduce from (1) that x ∈ H, a contradiction.Corollary 6.9.Let F be an algebraically closed field of characteristic p > 0, and let L be an upper semimodular restricted Lie algebra over F.Then, L is also lower semimodular and a J-algebra.
Proof.It follows from Proposition 6.8 and Proposition 5.2.
Proposition 6.10.Let F be an algebraically closed field of characteristic p > 0, and let L be an upper semimodular restricted Lie algebra over F.Then, every restricted subalgebra of L is a restricted quasi-ideal.

Lemma 2 . 2 .
Every simple Lie algebra L over an algebraically closed field F of characteristic p > 3 contains a subalgebra S with a quotient isomorphic to sl(2, F).Proof.The proof is the same as for [12, Lemma 3.2] with the reference to [31, Part II, Corollary 1.4] replaced by [20, page 180, Corollary].
by [15, Theorems 3.4 and 4.2].Let a ∈ A. Then C B (A) is a restricted ideal of L and C B (A) ∩ A = 0, so C B (A) = 0 and B acts faithfully on A. Also ad 2 (a) = 0 and so ad(a [p] ) = 0, whence a [p] ∈ Z(L) for all a ∈ A.