This note contains a few introductory results on strongly Lie nilpotent rings and, in particular, an analogue of a well known theorem of P. Hall on nilpotent groups

Catino, F.

Miccoli, M. M.

English

ESE - Salento University Publishing

Note di Matematica 20, n. 2, 2000/2001, 35–41.Note on strongly Lie nilpotent ringsF. Catino and M. M. MiccoliDipartimento di Matematica, Universita` di LecceVia Prov.le Lecce–Arnesano, P.O. Box 193 I - 73100 Lecce, Italyfrancesco.catino@unile.it, maddalena.miccoli@unile.itReceived: 18 February 1997; accepted: 24 November 2000.Abstract. This note contains a few introductory results on strongly Lie nilpotent rings and,in particular, an analogue of a well known theorem of P. Hall on nilpotent groups.Keywords: ring, central chains, strongly Lie nilpotent ringMSC 2000 classification: 16A221 IntroductionLet R be an associative ring. For all a, b ∈ R we set a ◦ b = ab − ba. It iswell-known that (R,+, ◦) is a Lie ring. For all A,B ⊆ R, the additive subgroupof R generated by all Lie products a ◦ b (a ∈ A, b ∈ B) is denoted by A ◦B.Now we put γ1(R) = R and for any n ∈ N, n > 1, γn(R) = γn−1(R)◦R.If there exists c ∈ N such that γc+1(R) = 0, then R is called a Lie nilpotentring.We deﬁne the Lie powers R(n)(n ∈ N) as follows: R(1) = R, and for alln ∈ N, n > 1, R(n) is the ideal of R generated by R(n−1) ◦ R. If there existsc ∈ N such that R(c+1) = 0, then R is called a strongly Lie nilpotent ring(see [7]).Clearly, γn ⊆ R(n) for all n ∈ N, thus a strongly Lie nilpotent ring is Lie-nilpotent.There are many results on strongly Lie nilpotent group rings, see for exampleBovdi’s paper [2].The 2nd section of this note contains a few developments in the spirit ofJennings’ paper [4]. In the 3rd section, an analogue of a well known theorem ofP. Hall on nilpotent groups for strongly Lie nilpotent rings is obtained.2 Central series of idealsWe recall that if I and J are ideals of a ring R and I ⊆ J , then J/Iis called a central factor if J ◦R ⊆ I or, equivalently, J/I belongs to the centreZ(R/I) of the ring R/I.36 E. BallicoA chain (J (λ)) of ideals of a ring R is called a central series of R if everyfactor J (λ+1)/J (λ) is central (see [4]).The lower central series of a ring R is the descending series whose termsR(α) are deﬁned by setting: R(1) = R and, for α > 1, R(α) =⋂β<α R(β) if αis a limit ordinal and R(α) is the ideal of R generated by R(α−1) ◦R, otherwise.Following an idea of Jennings [4], we now deﬁne the upper central series ofan arbitrary ring.If B is an additive subgroup of a ring R, then the set M := {x|x ∈ B, Rx ⊆B} is the largest left ideal of R which is contained in B. Moreover, the setF := {y|y ∈ M, yR ⊆ M} is the largest ideal of R which is contained in B.It is easy to see that F (R) = {y|y ∈ Z(R), yR ⊆ Z(R)} is the largestideal of R which is contained in the centre Z(R) of R. The ideal F (R) is calledthe strong centre of R. We remark that the annihilator of a ring R is containedin F (R).The upper central series of a ring R is the ascending series whose termsF (α)(R) are deﬁned by setting F (0)(R) = {0} and, for alpha > 0, F (α)(R) =⋃β<α F(β)(R) if α is a limit ordinal and F (α+1)(R)/F (α)(R) = F (R/F (α)(R))otherwise. In particular, F (1)(R) is the strong centre of R.Moreover, for any positive integer kF (k)(R) = {x|x ∈ R, ∀r, s ∈ R x(1 + r) ◦ s ∈ F (k−1)(R)} (1)The following result gives some relationship between the lower central seriesand the upper central series of arbitrary ring R.Proposition 1. Let R be a ring, and let k and l be positive integers. (1)R(k) ·R(l) ⊆ R(k+l−1)(2) R(k) ◦R(l) ⊆ R(k+l)(3) (R(k))(l) ⊆ R(kl)(4) R(k) · F (l)(R) ⊆ F (l−k+1)(R) se k ≤ l(5) F (l)(R) ·R(k) ⊆ F (l−k+1)(R) se k ≤ l(6) R(k) ◦ F (l)(R) ⊆ F (l−k)(R) se k ≤ l(7) F (k)(R/F (l)(R)) = F (k+l)(R)/F (l)(R)Proof. For (1), (2) see [4], Theorem 3.3 e Theorem 3.4. We prove ourassertions by induction. First, (3) is trivial for l = 1. If l > 1, then, by (2), wehave(R(k))(l−1) ◦R(k) ⊆ Rk(l−1) ◦R(k) ⊆ R(k(l−1)+k) = R(kl)for all positive integer k. Hence (R(k))(l) ⊆ R(kl).Space curves not contained in low degree surfaces in positive characteristic 37(4): If k = 1, then, for all l ∈ NR(k)F (l)(R) = R(1)F (l)(R) ⊆ F (l)(R) ⊆ F (l−k+1)(R)Now let k > 1. For all a ∈ R(k−1), b ∈ R and c ∈ F (l)(R), the inductivehypothesis implies that(a ◦ b)c = ac ◦ b− a(c ◦ b) ∈ F (l−k+1)(R),as desired.(5): Analogously to (4).(6): If k = 1, then, for all l ∈ NR(k) ◦ F (l)(R) = R ◦ F (l)(R) ⊆ F (l−1)(R) = F (l−k)(R)Now let k > 1. For all a ∈ R(k−1), b ∈ R, r ∈ R and c ∈ F (l)(R), inductivelywe have(a ◦ b) ◦ c = b ◦ (c ◦ a) + a ◦ (b ◦ c) ∈ F (l−k)(R)Hence, by (5), we have(a ◦ b)r ◦ c = (a ◦ b) ◦ rc + r ◦ (c(a ◦ b)) ∈ F (l−k)(R)(7): If k = 1, then, for all l ∈ NF (k)(R/F (l)(R)) = F (R/F (l)(R)) = F (l+1)(R)/F (l)(R) = F (k+l)(R)/F (l)(R)Now let k > 1. For all l ∈ N and for all y ∈ R we havey + F (l)(R) ∈ F (k)(R/F (l)(R)) ⇐⇒⇐⇒ ∀a, b ∈ R (y(1 + a) ◦ b)+ F (l)(R) ∈ F (k−1)(R/F (l)(R))⇐⇒ ∀a, b ∈ R (y(1 + a) ◦ b)+ F (l)(R) ∈ F (k−1+l)(R)/F (l)(R)⇐⇒ ∀a, b ∈ R y(1 + a) ◦ b ∈ F (k−1+l)(R) ⇐⇒ y ∈ F (k+l)(R)which completes the proof. QEDCorollary 1. If R is a ring and k is a positive integer, thencharR/F (k)(R) = charR(k+1).38 E. BallicoProof. Let k ∈ N and let m := charR/F (k)(R) = 0. For all a ∈ R(k), r ∈R, we havem(a ◦ r) = a ◦mr ∈ R(k) ◦ F (k)(R) = 0,by Prop. 1 (6). Since R(k+1) is the ideal of R generated by R(k) ◦ R, it followsthat charR(k+1) divides m.Now let n := charR(k+1) = 0. For each r, r1, . . . rk, s1, . . . s1, . . . sk ∈ R wehave(· · · ((((nr(1 + r1) ◦ s1)(1 + r2) ◦ s2) · · · )(1 + rk) ◦ sk == n((· · · ((((r(1 + r1) ◦ s1)(1 + r2) ◦ s2) · · · )(1 + rk) ◦ sk)= 0By (1), it follows that nr ∈ F (k)(R). Hence charR/F (k)(R) divides nIt follows immediately that charR/F (k)(R) = 0 if and only if charR(k+1) =0. QEDThe following proposition gives a relation between the characteristic of thefactors of the upper central series of a ring and that of its strong centre.Proposition 2. If R is a ring such that charF (R) = 0, then the charac-teristic of F (k+1)(R)/F (k)(R) divides the characteristic of F (R), for each non-negative integer k.Proof. Let n := charF (R) = 0. We show by induction on k that nx ∈F (k)(R), for all x ∈ F (k+1)(R) and k ∈ N0.For k = 0, there is nothing to prove. Let k ≥ 1 and assume that ny ∈F (k−1)(R) for each y ∈ F (k)(R). Let x ∈ F (k−1)(R). For all r, s ∈ R we havex(1+r)◦s ∈ F (k)(R). Inductively, n(x(1+r)◦s) ∈ F (k−1)(R). Hence (nx)(1+r)◦sbelongs to F (k−1)(R) and nx ∈ F (k)(R), by (1). QED3 Analogue of a theorem of P. HallIn [4], Jennings proves that a ring is strongly Lie nilpotent if and only if ithas a ﬁnite central series. Moreover, we haveProposition 3. Let R be a ring. If c ∈ N and 0 = I0 ⊂ . . . ⊂ Ic = R is acentral series of R, thenR(c−k+1) ⊆ Ik ⊆ F (k)(R)for each k ∈ {0, 1, . . . c}Proof. The ﬁrst inclusion holds by [4] (Theorem 2.1). We prove, by induc-tion on k, that Ik ⊆ F (k)(R). For k = 0, there is nothing to prove. Let k ≥ 1Space curves not contained in low degree surfaces in positive characteristic 39and assume that Ik−1 ⊆ F (k−1)(R). Let z ∈ Ik. Since Ik/Ik−1 is a central factor,we have inductivelyz(1 + r) ◦ s ∈ Ik−1 ⊆ F (k−1)(R)for all r, s ∈ R. Hence z ∈ F (k)(R), by (1). QEDThe proposition shows that the lower and upper central series of any stronglyLie nilpotent ring R have the same length c. This length c is called the stronglyLie nilpotent class of RThe following result is analogous to one obtained for nilpotent rings (see [5],1.2.6).Proposition 4. If R is a strongly Lie nilpotent ring, then charR = 0 if andonly if charF (R) = 0.Proof. If charF (R) = 0, then clearly charR = 0. Conversely, let charR =0 and assume that charF (R) = m = 0. If c is the strongly nilpotent class of R,then R(c) ⊆ F (R). Hence charR(c) = 0. Let i := min{j|j ∈ N, charR(j) = 0}and let n := charR(i). Then there is an element x ∈ R(i−1) such that mnx = 0.For all y, z ∈ R, we havenx(1 + y) ◦ z = n(x(1 + y) ◦ z) = n(x ◦ z + xy ◦ z) = 0By (1), nx ∈ F (R), therefore mnx = 0, a contradiction to the choice of x.QEDThe results above are examples of a strong analogy between the theories ofnilpotent groups and strongly Lie nilpotent rings.In particular, we recall the well-known theorem of P. Hall for nilpotentgroups: if N is a normal subgroups of a group G and N,G/N ′ are nilpotent, thenG is nilpotent (see [6]). A version of this theorem for Lie algebras is contained,for example, in [1].We give a version of the theorem of P. Hall for strongly Lie nilpotent rings.Lemma 1. Let R be a ring, I an ideal of R such that its strong centre F (I)is an ideal of R and M the largest ideal of R contained in I ◦ I.If there is a ﬁnite central series of R between F (I) and I, then there is aﬁnite central series of R between 0 and M .Proof. Let t ∈ N andF (I) = I0 ⊂ I1 ⊂ · · · ⊂ It = I (2)a ﬁnite central series of R between F (I) and I.For each i ∈ N, i ≤ 2t, let Bi the additive subgroup of R generated by⋃h+k=i Ih ◦ Ik, and let Bi be the ideal R generated by Bi.40 E. BallicoEvidently0 = B1 ⊆ B2 ⊆ · · · ⊆ B2t (3)We show that (3) is a central series of R.It is suﬃcient to prove that, for all a ∈ Ih, b ∈ Ik such that h + k = i andfor all r, s, v ∈ R we havea ◦ b ◦ v ∈ Bi−1(a ◦ b)r ◦ v ∈ Bi−1r(a ◦ b) ◦ v ∈ Bi−1r(a ◦ b)s ◦ v ∈ Bi−1Since (2) is a central series, by the Jacobi identity, we havea ◦ b ◦ v = a ◦ v ◦ b + v ◦ b ◦ a ∈ Bi−1 ⊆ Bi−1Moreover (cfr. [3], Lemma 2)(a ◦ b)(r ◦ v) = v(a ◦ r) ◦ b− r(a ◦ b ◦ v)−a ◦ r ◦ bv + a ◦ br ◦ v−a ◦ b ◦ v ◦ r + a ◦ r ◦ b ◦ v ∈ Bi−1.Hence(a ◦ b)r ◦ v = (a ◦ b ◦ v)r + (a ◦ b)(r ◦ v) ∈ Bi−1.It follows thatr(a ◦ b) ◦ v = −(a ◦ b ◦ r) ◦ v + (a ◦ b)r ◦ v ∈ Bi−1.Finally,s(a ◦ b)r ◦ v = s((a ◦ b)r ◦ v) + (s ◦ v)(a ◦ b)r ∈ Bi−1.Hence for all i ∈ N, 1 < i ≤ 2t we have(Bi ∩M) ◦R ⊆ (Bi ◦R) ∩ (M ◦R) ⊆ Bi−1 ∩M.Therefore0 = B1 ∩M ⊆ · · · ⊆ B2t ∩M = Mis a ﬁnite central series of R between 0 and M . QEDTheorem 1. Let R be a ring, I an ideal of R such that its strong centreF (I) is an ideal of R, and let M be the largest ideal of R contained in I ◦ I.If I and R/M are strongly Lie nilpotent rings, then R is strongly Lie nilpo-tent.Space curves not contained in low degree surfaces in positive characteristic 41Proof. We proceed by induction on the strongly Lie nilpotent class c ofI. If c = 1, then I = F (I) and I ◦ I = 0. It follows that M = 0. Hence R isstrongly Lie nilpotent.If c = 2, then I ◦ I ⊆ F (I). Hence M ⊆ F (I). As R/M is strongly nilpotent,it follows that R/F (I) is strongly Lie nilpotent. Now, I/F (I) is an ideal ofR/F (I), and therefore there is a ﬁnite central series of R between F (I) and I.By 1, there is a ﬁnite central series of R between 0 and M . It follows that R isstrongly Lie nilpotent.If c > 3 and M is the largest ideal of R/F (I) contained in I/F (I) ◦ I/F (I),then F (I) ⊆ M and M = M/F (I). Since (R/F (I))/M ∼= R/M , we have that(R/F (I))/M is strongly Lie nilpotent. Now I/F (I) is strongly Lie nilpotent ofclass c− 1 and, inductively R/F (I) is strongly Lie nilpotent.Proceeding as in the case of c = 2, we complete our proof. QEDReferences[1] Yu. A. Bahturin: Identical Relations in Lie Algebra, VNU Science Press, Utrecht, 1987[2] A. A. Bovdi: The group of units of a group algebra of characteristic p, Publ. Math.Debrecen 52 (1998), 193–244.[3] H. Laue, On the associated Lie ring and the adjoint group of a radical ring, Canad. Math.Bull. 27 (1984), 215–222[4] S. A. Jennings, Central chains of ideals in an associative ring, Duke Math. J. 9 (1942),341–355.[5] R. L. Kruse, D. T. Price, Nilpotent Rings, Gordon and Breach, New York, 1969[6] D. J. S. Robinson, A Course in the Theory of Groups, Springer, New York, 1982[7] S. K. Sehgal, Topics in group ring, Marcel Dekker, New York, 1978

Note on strongly Lie nilpotent rings

CATINO, Francesco

MICCOLI, Maria Maddalena

Archivio Istituzionale della Ricerca- Università del Salento

Note on strongly Lie nilpotent rings.

Università del Salento: ESE - Salento University Publishing

Note di Matematica 20, n. 2, 2000/2001, 35–41.Note on strongly Lie nilpotent ringsF. Catino and M. M. MiccoliDipartimento di Matematica, Universita` di LecceVia Prov.le Lecce–Arnesano, P.O. Box 193 I - 73100 Lecce, Italyfrancesco.catino@unile.it, maddalena.miccoli@unile.itReceived: 18 February 1997; accepted: 24 November 2000.Abstract. This note contains a few introductory results on strongly Lie nilpotent rings and,in particular, an analogue of a well known theorem of P. Hall on nilpotent groups.Keywords: ring, central chains, strongly Lie nilpotent ringMSC 2000 classification: 16A221 IntroductionLet R be an associative ring. For all a, b ∈ R we set a ◦ b = ab − ba. It iswell-known that (R,+, ◦) is a Lie ring. For all A,B ⊆ R, the additive subgroupof R generated by all Lie products a ◦ b (a ∈ A, b ∈ B) is denoted by A ◦B.Now we put γ1(R) = R and for any n ∈ N, n > 1, γn(R) = γn−1(R)◦R.If there exists c ∈ N such that γc+1(R) = 0, then R is called a Lie nilpotentring.We define the Lie powers R(n)(n ∈ N) as follows: R(1) = R, and for alln ∈ N, n > 1, R(n) is the ideal of R generated by R(n−1) ◦ R. If there existsc ∈ N such that R(c+1) = 0, then R is called a strongly Lie nilpotent ring(see [7]).Clearly, γn ⊆ R(n) for all n ∈ N, thus a strongly Lie nilpotent ring is Lie-nilpotent.There are many results on strongly Lie nilpotent group rings, see for exampleBovdi’s paper [2].The 2nd section of this note contains a few developments in the spirit ofJennings’ paper [4]. In the 3rd section, an analogue of a well known theorem ofP. Hall on nilpotent groups for strongly Lie nilpotent rings is obtained.2 Central series of idealsWe recall that if I and J are ideals of a ring R and I ⊆ J , then J/Iis called a central factor if J ◦R ⊆ I or, equivalently, J/I belongs to the centreZ(R/I) of the ring R/I.36 E. BallicoA chain (J (λ)) of ideals of a ring R is called a central series of R if everyfactor J (λ+1)/J (λ) is central (see [4]).The lower central series of a ring R is the descending series whose termsR(α) are defined by setting: R(1) = R and, for α > 1, R(α) =⋂β<α R(β) if αis a limit ordinal and R(α) is the ideal of R generated by R(α−1) ◦R, otherwise.Following an idea of Jennings [4], we now define the upper central series ofan arbitrary ring.If B is an additive subgroup of a ring R, then the set M := {x|x ∈ B, Rx ⊆B} is the largest left ideal of R which is contained in B. Moreover, the setF := {y|y ∈ M, yR ⊆ M} is the largest ideal of R which is contained in B.It is easy to see that F (R) = {y|y ∈ Z(R), yR ⊆ Z(R)} is the largestideal of R which is contained in the centre Z(R) of R. The ideal F (R) is calledthe strong centre of R. We remark that the annihilator of a ring R is containedin F (R).The upper central series of a ring R is the ascending series whose termsF (α)(R) are defined by setting F (0)(R) = {0} and, for alpha > 0, F (α)(R) =⋃β<α F(β)(R) if α is a limit ordinal and F (α+1)(R)/F (α)(R) = F (R/F (α)(R))otherwise. In particular, F (1)(R) is the strong centre of R.Moreover, for any positive integer kF (k)(R) = {x|x ∈ R, ∀r, s ∈ R x(1 + r) ◦ s ∈ F (k−1)(R)} (1)The following result gives some relationship between the lower central seriesand the upper central series of arbitrary ring R.Proposition 1. Let R be a ring, and let k and l be positive integers. (1)R(k) ·R(l) ⊆ R(k+l−1)(2) R(k) ◦R(l) ⊆ R(k+l)(3) (R(k))(l) ⊆ R(kl)(4) R(k) · F (l)(R) ⊆ F (l−k+1)(R) se k ≤ l(5) F (l)(R) ·R(k) ⊆ F (l−k+1)(R) se k ≤ l(6) R(k) ◦ F (l)(R) ⊆ F (l−k)(R) se k ≤ l(7) F (k)(R/F (l)(R)) = F (k+l)(R)/F (l)(R)Proof. For (1), (2) see [4], Theorem 3.3 e Theorem 3.4. We prove ourassertions by induction. First, (3) is trivial for l = 1. If l > 1, then, by (2), wehave(R(k))(l−1) ◦R(k) ⊆ Rk(l−1) ◦R(k) ⊆ R(k(l−1)+k) = R(kl)for all positive integer k. Hence (R(k))(l) ⊆ R(kl).Space curves not contained in low degree surfaces in positive characteristic 37(4): If k = 1, then, for all l ∈ NR(k)F (l)(R) = R(1)F (l)(R) ⊆ F (l)(R) ⊆ F (l−k+1)(R)Now let k > 1. For all a ∈ R(k−1), b ∈ R and c ∈ F (l)(R), the inductivehypothesis implies that(a ◦ b)c = ac ◦ b− a(c ◦ b) ∈ F (l−k+1)(R),as desired.(5): Analogously to (4).(6): If k = 1, then, for all l ∈ NR(k) ◦ F (l)(R) = R ◦ F (l)(R) ⊆ F (l−1)(R) = F (l−k)(R)Now let k > 1. For all a ∈ R(k−1), b ∈ R, r ∈ R and c ∈ F (l)(R), inductivelywe have(a ◦ b) ◦ c = b ◦ (c ◦ a) + a ◦ (b ◦ c) ∈ F (l−k)(R)Hence, by (5), we have(a ◦ b)r ◦ c = (a ◦ b) ◦ rc + r ◦ (c(a ◦ b)) ∈ F (l−k)(R)(7): If k = 1, then, for all l ∈ NF (k)(R/F (l)(R)) = F (R/F (l)(R)) = F (l+1)(R)/F (l)(R) = F (k+l)(R)/F (l)(R)Now let k > 1. For all l ∈ N and for all y ∈ R we havey + F (l)(R) ∈ F (k)(R/F (l)(R)) ⇐⇒⇐⇒ ∀a, b ∈ R (y(1 + a) ◦ b)+ F (l)(R) ∈ F (k−1)(R/F (l)(R))⇐⇒ ∀a, b ∈ R (y(1 + a) ◦ b)+ F (l)(R) ∈ F (k−1+l)(R)/F (l)(R)⇐⇒ ∀a, b ∈ R y(1 + a) ◦ b ∈ F (k−1+l)(R) ⇐⇒ y ∈ F (k+l)(R)which completes the proof. QEDCorollary 1. If R is a ring and k is a positive integer, thencharR/F (k)(R) = charR(k+1).38 E. BallicoProof. Let k ∈ N and let m := charR/F (k)(R) = 0. For all a ∈ R(k), r ∈R, we havem(a ◦ r) = a ◦mr ∈ R(k) ◦ F (k)(R) = 0,by Prop. 1 (6). Since R(k+1) is the ideal of R generated by R(k) ◦ R, it followsthat charR(k+1) divides m.Now let n := charR(k+1) = 0. For each r, r1, . . . rk, s1, . . . s1, . . . sk ∈ R wehave(· · · ((((nr(1 + r1) ◦ s1)(1 + r2) ◦ s2) · · · )(1 + rk) ◦ sk == n((· · · ((((r(1 + r1) ◦ s1)(1 + r2) ◦ s2) · · · )(1 + rk) ◦ sk)= 0By (1), it follows that nr ∈ F (k)(R). Hence charR/F (k)(R) divides nIt follows immediately that charR/F (k)(R) = 0 if and only if charR(k+1) =0. QEDThe following proposition gives a relation between the characteristic of thefactors of the upper central series of a ring and that of its strong centre.Proposition 2. If R is a ring such that charF (R) = 0, then the charac-teristic of F (k+1)(R)/F (k)(R) divides the characteristic of F (R), for each non-negative integer k.Proof. Let n := charF (R) = 0. We show by induction on k that nx ∈F (k)(R), for all x ∈ F (k+1)(R) and k ∈ N0.For k = 0, there is nothing to prove. Let k ≥ 1 and assume that ny ∈F (k−1)(R) for each y ∈ F (k)(R). Let x ∈ F (k−1)(R). For all r, s ∈ R we havex(1+r)◦s ∈ F (k)(R). Inductively, n(x(1+r)◦s) ∈ F (k−1)(R). Hence (nx)(1+r)◦sbelongs to F (k−1)(R) and nx ∈ F (k)(R), by (1). QED3 Analogue of a theorem of P. HallIn [4], Jennings proves that a ring is strongly Lie nilpotent if and only if ithas a finite central series. Moreover, we haveProposition 3. Let R be a ring. If c ∈ N and 0 = I0 ⊂ . . . ⊂ Ic = R is acentral series of R, thenR(c−k+1) ⊆ Ik ⊆ F (k)(R)for each k ∈ {0, 1, . . . c}Proof. The first inclusion holds by [4] (Theorem 2.1). We prove, by induc-tion on k, that Ik ⊆ F (k)(R). For k = 0, there is nothing to prove. Let k ≥ 1Space curves not contained in low degree surfaces in positive characteristic 39and assume that Ik−1 ⊆ F (k−1)(R). Let z ∈ Ik. Since Ik/Ik−1 is a central factor,we have inductivelyz(1 + r) ◦ s ∈ Ik−1 ⊆ F (k−1)(R)for all r, s ∈ R. Hence z ∈ F (k)(R), by (1). QEDThe proposition shows that the lower and upper central series of any stronglyLie nilpotent ring R have the same length c. This length c is called the stronglyLie nilpotent class of RThe following result is analogous to one obtained for nilpotent rings (see [5],1.2.6).Proposition 4. If R is a strongly Lie nilpotent ring, then charR = 0 if andonly if charF (R) = 0.Proof. If charF (R) = 0, then clearly charR = 0. Conversely, let charR =0 and assume that charF (R) = m = 0. If c is the strongly nilpotent class of R,then R(c) ⊆ F (R). Hence charR(c) = 0. Let i := min{j|j ∈ N, charR(j) = 0}and let n := charR(i). Then there is an element x ∈ R(i−1) such that mnx = 0.For all y, z ∈ R, we havenx(1 + y) ◦ z = n(x(1 + y) ◦ z) = n(x ◦ z + xy ◦ z) = 0By (1), nx ∈ F (R), therefore mnx = 0, a contradiction to the choice of x.QEDThe results above are examples of a strong analogy between the theories ofnilpotent groups and strongly Lie nilpotent rings.In particular, we recall the well-known theorem of P. Hall for nilpotentgroups: if N is a normal subgroups of a group G and N,G/N ′ are nilpotent, thenG is nilpotent (see [6]). A version of this theorem for Lie algebras is contained,for example, in [1].We give a version of the theorem of P. Hall for strongly Lie nilpotent rings.Lemma 1. Let R be a ring, I an ideal of R such that its strong centre F (I)is an ideal of R and M the largest ideal of R contained in I ◦ I.If there is a finite central series of R between F (I) and I, then there is afinite central series of R between 0 and M .Proof. Let t ∈ N andF (I) = I0 ⊂ I1 ⊂ · · · ⊂ It = I (2)a finite central series of R between F (I) and I.For each i ∈ N, i ≤ 2t, let Bi the additive subgroup of R generated by⋃h+k=i Ih ◦ Ik, and let Bi be the ideal R generated by Bi.40 E. BallicoEvidently0 = B1 ⊆ B2 ⊆ · · · ⊆ B2t (3)We show that (3) is a central series of R.It is sufficient to prove that, for all a ∈ Ih, b ∈ Ik such that h + k = i andfor all r, s, v ∈ R we havea ◦ b ◦ v ∈ Bi−1(a ◦ b)r ◦ v ∈ Bi−1r(a ◦ b) ◦ v ∈ Bi−1r(a ◦ b)s ◦ v ∈ Bi−1Since (2) is a central series, by the Jacobi identity, we havea ◦ b ◦ v = a ◦ v ◦ b + v ◦ b ◦ a ∈ Bi−1 ⊆ Bi−1Moreover (cfr. [3], Lemma 2)(a ◦ b)(r ◦ v) = v(a ◦ r) ◦ b− r(a ◦ b ◦ v)−a ◦ r ◦ bv + a ◦ br ◦ v−a ◦ b ◦ v ◦ r + a ◦ r ◦ b ◦ v ∈ Bi−1.Hence(a ◦ b)r ◦ v = (a ◦ b ◦ v)r + (a ◦ b)(r ◦ v) ∈ Bi−1.It follows thatr(a ◦ b) ◦ v = −(a ◦ b ◦ r) ◦ v + (a ◦ b)r ◦ v ∈ Bi−1.Finally,s(a ◦ b)r ◦ v = s((a ◦ b)r ◦ v) + (s ◦ v)(a ◦ b)r ∈ Bi−1.Hence for all i ∈ N, 1 < i ≤ 2t we have(Bi ∩M) ◦R ⊆ (Bi ◦R) ∩ (M ◦R) ⊆ Bi−1 ∩M.Therefore0 = B1 ∩M ⊆ · · · ⊆ B2t ∩M = Mis a finite central series of R between 0 and M . QEDTheorem 1. Let R be a ring, I an ideal of R such that its strong centreF (I) is an ideal of R, and let M be the largest ideal of R contained in I ◦ I.If I and R/M are strongly Lie nilpotent rings, then R is strongly Lie nilpo-tent.Space curves not contained in low degree surfaces in positive characteristic 41Proof. We proceed by induction on the strongly Lie nilpotent class c ofI. If c = 1, then I = F (I) and I ◦ I = 0. It follows that M = 0. Hence R isstrongly Lie nilpotent.If c = 2, then I ◦ I ⊆ F (I). Hence M ⊆ F (I). As R/M is strongly nilpotent,it follows that R/F (I) is strongly Lie nilpotent. Now, I/F (I) is an ideal ofR/F (I), and therefore there is a finite central series of R between F (I) and I.By 1, there is a finite central series of R between 0 and M . It follows that R isstrongly Lie nilpotent.If c > 3 and M is the largest ideal of R/F (I) contained in I/F (I) ◦ I/F (I),then F (I) ⊆ M and M = M/F (I). Since (R/F (I))/M ∼= R/M , we have that(R/F (I))/M is strongly Lie nilpotent. Now I/F (I) is strongly Lie nilpotent ofclass c− 1 and, inductively R/F (I) is strongly Lie nilpotent.Proceeding as in the case of c = 2, we complete our proof. QEDReferences[1] Yu. A. Bahturin: Identical Relations in Lie Algebra, VNU Science Press, Utrecht, 1987[2] A. A. Bovdi: The group of units of a group algebra of characteristic p, Publ. Math.Debrecen 52 (1998), 193–244.[3] H. Laue, On the associated Lie ring and the adjoint group of a radical ring, Canad. Math.Bull. 27 (1984), 215–222[4] S. A. Jennings, Central chains of ideals in an associative ring, Duke Math. J. 9 (1942),341–355.[5] R. L. Kruse, D. T. Price, Nilpotent Rings, Gordon and Breach, New York, 1969[6] D. J. S. Robinson, A Course in the Theory of Groups, Springer, New York, 1982[7] S. K. Sehgal, Topics in group ring, Marcel Dekker, New York, 1978

MICCOLI M. M.

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Note on strongly Lie nilpotent rings

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