We prove that a Lie nilpotent one-sided ideal of an associative ring $R$ is
contained in a Lie solvable two-sided ideal of $R$. An estimation of derived
length of such Lie solvable ideal is obtained depending on the class of Lie
nilpotency of the Lie nilpotent one-sided ideal of $R.$ One-sided Lie nilpotent
ideals contained in ideals generated by commutators of the form $[... [ [r_1,
r_{2}], ... ], r_{n-1}], r_{n}]$ are also studied.Comment: 5 page

Luchko, V. S.

Petravchuk, A. P.

Algebra and Discrete Mathematics

English

arXiv

We prove that a Lie nilpotent one-sided ideal of an associative ring R is contained in a Lie solvable two-sided ideal of R. An estimation of derived length of such Lie solvable ideal is obtained depending on the class of Lie nilpotency of the Lie nilpotent one-sided ideal of R. One-sided Lie nilpotent ideals contained in ideals generated by commutators of the form […[[r₁,r₂],…],rn₋₁],rn] are also studied.

Luchko, V.S.

Petravchuk, A.P.

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Journal Algebra Discrete Math.Algebra and Discrete Mathematics RESEARCH ARTICLENumber 4. (2007). pp. 102 – 107c© Journal “Algebra and Discrete Mathematics”On one-sided Lie nilpotent ideals of associativeringsVictoriya S. Luchko and Anatoliy P. PetravchukDedicated to Professor V. V. Kirichenkoon the occasion of his 65th birthdayAbstract. We prove that a Lie nilpotent one-sided idealof an associative ring R is contained in a Lie solvable two-sidedideal of R. An estimation of derived length of such Lie solv-able ideal is obtained depending on the class of Lie nilpotencyof the Lie nilpotent one-sided ideal of R. One-sided Lie nilpotentideals contained in ideals generated by commutators of the form[. . . [[r1, r2], . . .], rn−1], rn] are also studied.IntroductionIt is well-known that if I is an one-sided nilpotent ideal of an associativering R then I is contained in a two-sided nilpotent ideal of R. Hence thefollowing question is of interest: for which one-sided ideal I of the ring Rthere exists a two-sided ideal J such that J ⊇ I and J has properties likeproperties of I. In [5] it was noted that for an one-sided commutativeideal I of a ring R there exists a nilpotent-by-commutative two-sidedideal J of the ring R such that J ⊇ I.Note that Lie nilpotent and Lie solvable associative rings were in-vestigated by many authors (see, for example [4], [6], [7], [1]) and thestructure of such rings is studied well enough.In this paper we prove that a Lie nilpotent one-sided ideal I of anassociative ring R is contained in a Lie solvable two-sided ideal J of2000 Mathematics Subject Classification: 16D70.Key words and phrases: associative ring, one-sided ideal, Lie nilpotent ideal,derived length.Journal Algebra Discrete Math.V. S. Luchko, A. P. Petravchuk 103R. An estimation (rather rough) of Lie derived length of the ideal Jdepending on Lie nilpotency class of I is also obtained (Theorem 1).In case when the Lie nilpotent one-sided ideal I is contained inthe ideal Rn of the ring R generated by all commutators of the form[. . . [[r1, r2], . . .], rn−1], rn] and the Lie derived length of I is less then n itis proved that I is contained in a nilpotent two-sided ideal of R (Theorem2).The notations in the paper are standard. If S is a subset of an associa-tive ring R then by AnnlR(S) (AnnrR(S)) we denote the left (respectivelyright) annihilator of S in R. We also denote by R(−) the adjoint Lie ringof the associative ring R. Further, by R(−)n we denote the n-th member ofthe lower central series of the Lie ring R(−). Then Rn = R(−)n +R(−)n ·R == R(−)n + R ·R(−)n is a two-sided ideal of the (associative) ring R. In par-ticular, R2 is a two-sided ideal of the ring R generated by all commutatorsof the form [r1, r2] = r1r2 − r2r1, r1, r2 ∈ R. If R is a Lie solvable ring(i.e. such that R(−) is a solvable Lie ring) then we denote by s(R) its Liederived length. Analogously, by c(R) we denote Lie nilpotency class of aLie nilpotent ring R.1. Lie nilpotent one-sided idealsLemma 1. Let I be an one-sided ideal of an associative ring R andZ = Z(I) be the center of I. Then there exists an ideal J in R such thatJ2 = 0 and [Z, R] ⊆ J .Proof. Let, for example, I be a right ideal from R. Take arbitrary ele-ments z ∈ Z, i ∈ I, r ∈ R. Then it holds z(ir)− (ir)z = 0 (since ir ∈ I).This implies the equality i(zr − rz) = 0 since z ∈ Z(I). As elementsz, i, r are arbitrarily chosen then we have I[Z, R] = 0. Consider the rightannihilator T = AnnrR(I). It is clear that T is a two-sided ideal of thering R (since I is a right ideal of R) what implies that [Z, R] ⊆ T .Further, for any element of the form zr − rz from [Z, R] and for anyt ∈ T it holds (zr − rz)t = z(rt) − r(zt). Since rt ∈ T then z(rt) = 0.Besides, z ∈ I and therefore zt = 0 what brings the equality (zr− rz)t =0. It means that [Z, R] · T = 0.Consider the left annihilator J = AnnlT(T ). It is easy to see thatJ is a two-sided ideal of the ring R. From relations [Z, R] ⊆ T and[Z, R] · T = 0 we have the inclusion [Z, R] ⊆ J . It is also clear thatJ2 = 0. Analogously one can consider the case when I is a left ideal.Theorem 1. Let R be an associative ring and I be an one-sided ideal ofR. If the subring I is Lie nilpotent then I is contained in a Lie solvableJournal Algebra Discrete Math.104 On one-sided Lie nilpotent ideals of associative ringstwo-sided ideal J of R such that s(J) ⊆ m(m+1)/2+m where m = c(I)is Lie nilpotency class of the subring I.Proof. Let for example I be a right ideal. We prove our proposition by theinduction on the class of Lie nilpotency n = c(I) of the subring I. If n = 1then I is a commutative right ideal and by Lemma 1 the ring R containssuch an ideal T with zero square that it holds (I + T )/T ⊆ Z(R/T ) inthe quotient ring R/T where Z(R/T ) is the center of R/T . It meansthat I + T is a two-sided ideal of the ring R and s(I + T ) 6 2. Clearly2 = n+n(n+1)/2 if n = 1 and the statement of Theorem is rue in casen = 1. Assume that the statement is true in case c(I) 6 n− 1 and proveit when c(I) = n. Denote by Z the center of the subring I. By Lemma 1there exists an ideal T of R with T 2 = 0 such that [Z, R] ⊆ T . Considerthe quotient ring R = R/T . Then Z = (Z + T )/T lies in the center ofR and therefore Z + Z · R = Z + R · Z is a two-sided ideal of the ringR. Since Z ⊆ I = (I + T )/T the ideal Z + Z · R is Lie nilpotent of andits class of Lie nilpotency 6 m. Further, the quotient ring R/(Z + Z ·R)contains the right Lie nilpotent ideal I + (Z + Z ·R)/(Z + Z ·R) whichis Lie nilpotent of class of Lie nilpotency 6 m − 1. By the inductionassumption the last right ideal is contained in some Lie solvable ideal ofthe ring R/(Z + Z · R) of derived length 6 (m−1)m2 + (m − 1). SinceZ + Z ·R is Lie solvable and its derived length 6 m (even 6 [log2m] + 1but we take a rough estimation) and we consider the quotient ring R/Twhere T is Lie solvable of derived length 1, one can easily see that I iscontained in some Lie solvable (two-sided) ideal of derived length whichdoes not exceed(m− 1)m2+ (m− 1) + (m + 1) =(m + 1)m2+ m.Analogously one can consider the case when I is right ideal.It seems to be unknown whether a sum of two Lie nilpotent associativerings is Lie solvable. So the next statement can be of interest (see alsoresults about sums of PI-rings in [3]).Corollary 1. Let R be an associative ring which can be decomposed intoa sum R = A+B of its Lie nilpotent subrings A and B. If at least one ofthese subrings is an one-sided ideal of R then the ring R is Lie solvable.Remark 1. The statements of Theorem 1 and its Corollary become falsewhen we replace Lie nilpotency of one-sided ideals by Lie solvability.Really, consider full matrix ring R = M2(K) over an arbitrary field K ofcharacteristic 6= 2. It is clear thatI ={(x y0 0)∣∣∣∣x, y ∈ K}Journal Algebra Discrete Math.V. S. Luchko, A. P. Petravchuk 105is a right Lie solvable ideal of the ring R but I is not contained in anyLie solvable ideal of R since R is a non-solvable Lie ring. It is also clearthatR = I + J where J ={(0 0z t)∣∣∣∣ z, t ∈ K},i.e. the simple associative ring R is a sum of two right Lie solvable ideals.2. On embedding of Lie nilpotent ideals in ringsLemma 2. Let R be an associative ring, A be a Lie nilpotent subring of Rof Lie nilpotency class < m. If Z0 is a subring of A such that Z0 ⊆ Z(R)and Z0R ⊆ A then Zm0 Rm = 0.Proof. Consider the two-sided ideal J = Z0 + Z0R = Z0 + RZ0 of thering R. As J ⊆ A then [J, ..., J ]︸ ︷︷ ︸m= 0 by the condition c(A) < m. Further,it is easily to show that[J, J ] = [Z0 + Z0R, Z0 + Z0R] = Z20 [R, R].By induction on k one can also show that [J, ..., J ]︸ ︷︷ ︸k= Zk0 [R, ..., R]︸ ︷︷ ︸k. Thenwe have from the condition on J that [J, ..., J ]︸ ︷︷ ︸m= Zm0 [R, ..., R]︸ ︷︷ ︸m= 0. Thisimplies the equalityZm0 Rm = Zm0 ([R, ..., R]︸ ︷︷ ︸m+ [R, ..., R]︸ ︷︷ ︸m·R) = [J, ..., J ]︸ ︷︷ ︸m+ [J, ..., J ]︸ ︷︷ ︸m·R = 0.Lemma 3. Let R be an associative ring, I be an ideal of R. Then1) if J is a nilpotent ideal of the subring I then J lies in a nilpotentideal JI of the ring R such that JI ⊆ I;2) if S = AnnlI(I) (or AnnrI(I)) then S is contained in a nilpotentideal of the ring R which is contained in I.The proof of this Lemma immediately follows from Lemma 1.1.5 from[2].Theorem 2. Let R be an associative ring and I be a Lie nilpotent one-sided ideal of R. If I ⊆ Rn and Lie nilpotency class of I is less than nthen I is contained in an (associative) nilpotent ideal of R.Proof. Let for example I be a right ideal of the ring R and I ⊆ Rn. Onecan assume that that n > 2 because the statement of Theorem is obviousJournal Algebra Discrete Math.106 On one-sided Lie nilpotent ideals of associative ringsin case n = 1. We fix n > 2 and prove the statement of Theorem byinduction on the class of Lie nilpotency c = c(I) of the subring I. Ifc = 0 then I is the zero ideal and the proof is complete. Assume that thestatement is true for rings R with c(I) 6 c−1 and prove it in case c(I) = c.Since I is Lie nilpotent then by Lemma 1 there exists a nilpotent ideal Tof the ring R such that in the quotient ring R = R/T it holds [Z0, R] = 0where Z0 is the center of the subring I and Z0 = (Z0 + T )/T . Then byLemma 2 it holds the relation Zn0 ·Rn = 0. If Zn0 = 0 then Z0 + Z0R is anilpotent ideal of the ring R and then the subring Z0 is contained in thenilpotent ideal J = Z0 + T of the ring R. Since in the quotient ring R/Jfor the right ideal (I + J)/J it holds the inequality c((I + J)/J) 6 c− 1then by the inductive assumption (I + J)/J is contained in a nilpotentideal S/J of the ring R/J . But then I ⊆ S where S is nilpotent ideal ofthe ring R.Let now Zn0 6= 0. Then Zn0 ⊆ AnnlRn(Rn) and since Z0 ⊆ Rn thenZn0 is contained in a nilpotent ideal M of the ring R by Lemma 3. It isobvious that Z0 + Z0R is a nilpotent ideal of the ring R. Repeating theabove considerations we see that I ⊆ S where S is a nilpotent ideal ofthe ring R.Corollary 2. Let R be an associative ring with condition R = [R, R].If I is a Lie nilpotent one-sided ideal of R then there exists a nilpotent(two-sided) ideal J of the ring R such that I ⊆ JCorollary 3. Let R be a semiprime ring. Then every Lie nilpotent one-sided ideal is contained in the center Z(R) of the ring R and has trivialintersection with the ideal R2.Proof. Really since all nilpotent ideals of the ring R are zero then byLemma 1 every Lie nilpotent one-sided ideal I is contained in Z(R).Since IR ⊆ Z then [IR, R] = I[R, R] = 0. Then from this equality wehave IR2 = I([R, R] + [R, R] ·R) = 0. Denote J = I ∩R2. It is easily toshow that J ⊆ AnnlR2(R2) and by Lemma 3 the intersection J lies in anilpotent ideal of the ring R. Because the ring R is semiprime we haveJ = 0.References[1] B.Amberg and Ya.P.Sysak, Associative rings with metabelian adjoint group, Jour-nal of Algebra, 277 (2004), 456-473.[2] V.A.Andrunakievich and Yu.M.Ryabukhin, Radicals of algebras and structuretheory, Nauka, Moscow, 1979. (in Russian).[3] B.Felzenszwalb, A.Giambruno and G.Leal, On rings which are sums of two PI-subrings: a combinatorial approach, Pacific Journal of Math., 209, no.1 (2003),17-30.Journal Algebra Discrete Math.V. S. Luchko, A. P. Petravchuk 107[4] S.A.Jennigs, On rings whose associated Lie rings are nilpotent, Bull. Amer. Math.Soc., 53 (1947), 593-597.[5] A.P.Petravchuk, On associative algebras which are sum of two almost commu-tative subalgebras, Publicationes Mathematicae (Debrecen). 53, no.1-2, (1998),191-206.[6] R.K.Sharma and I.B.Srivastava, Lie solvable rings, Proc. Amer. Math. Soc., 94,no.1 (1985), 1-8.[7] W.Streb, Ueber Ringe mit aufloesbaren assoziirten Lie-Ringen, Rendiconti delSeminario Matematico dell’Universita` di Padova, 50, (1973), 127-142.Contact informationVictoriya S. Luchko Kiev Taras Shevchenko University,Faculty of Mechanics and Mathematics,64, Volodymyrska street,01033 Kyiv, UkraineE-Mail: vsluchko@gmail.comAnatoliy P. Pe-travchukKiev Taras Shevchenko University,Faculty of Mechanics and Mathematics,64, Volodymyrska street,01033 Kyiv, UkraineE-Mail: aptr@univ.kiev.uaReceived by the editors: 04.02.2008and in final form 10.04.2008.

On one-sided Lie nilpotent ideals of associative rings

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)