On some products of commutators in an associative ring

TL;DR

This paper proves a simplified result showing that the product of certain commutators in an associative ring, multiplied by 3, always lies in a specific ideal when at least one of the commutator lengths is odd, refining previous results.

Contribution

The paper provides a simpler proof that the product of commutators, scaled by 3, belongs to a certain ideal when one of the lengths is odd, improving understanding of their algebraic structure.

Findings

Product scaled by 3 lies in the ideal for odd-length commutators

Simplifies previous proofs of commutator product properties

Defines precise conditions for commutator products in rings

Abstract

Let $A$ be a unital associative ring and let $T^{(k)}$ be the two-sided ideal of $A$ generated by all commutators $[a_1, a_2, \dots , a_k]$ $(a_i \in A)$ where $[a_1, a_2] = a_1 a_2 - a_2 a_1$, $[a_1, \dots , a_{k-1}, a_k] = \bigl[ [a_1, \dots , a_{k-1}], a_k \bigr]$ $(k >2)$. It has been known that, if either $m$ or $n$ is odd then \[ 6 \, [a_1, a_2, \dots , a_m] [b_1, b_2, \dots , b_n] \in T^{(m+n-1)} \] for all $a_i, b_j \in A$. This was proved by Sharma and Srivastava in 1990 and independently rediscovered later (with different proofs) by various authors. The aim of our note is to give a simple proof of the following result: if at least one of the integers $m,n$ is odd then, for all $a_i, b_j \in A$, \[ 3 \, [a_1, a_2, \dots , a_m] [b_1, b_2, \dots , b_n] \in T^{(m+n-1)}. \] Since it has been known that, in general, \[ [a_1, a_2, a_3] [b_1, b_2] \notin T^{(4)}, \] our result cannot be improved further for all $m, n$ such that at least one of them is odd.

Equations

$$6[a_1,a_2,\dots ,a_m][b_1,b_2,\dots ,b_n]\in T^{(m+n-1)}$$

$$3[a_1,a_2,\dots ,a_m][b_1,b_2,\dots ,b_n]\in T^{(m+n-1)}.$$

$$[a_1,a_2,a_3][b_1,b_2]\notin T^{(4)},$$

$$[a_1,\dots ,a_m][b_1,\dots ,b_n]\in T^{(m+n-2)}.$$

$$3[a_1,a_2,a_3][b_1,b_2,\dots ,b_n]\in T^{(n+2)}.$$

$$6[a_1,\dots ,a_m][b_1,\dots ,b_n]\in T^{(m+n-1)}.$$

$$[a_1,\dots ,a_{2m^{\prime }}][b_1,\dots ,b_{2n^{\prime }}]\notin T^{(2m^{\prime }+2n^{\prime }-1)}$$

$$\ell [a_1,\dots ,a_{2m^{\prime }}][b_1,\dots ,b_{2n^{\prime }}]\notin T^{(2m^{\prime }+2n^{\prime }-1)}$$

$$3[a_1,\dots ,a_m][b_1,\dots ,b_n]\in T^{(m+n-1)}.$$

$$[a_1,a_2,a_3][b_1,b_2]\notin T^{(4)}$$

$$[a_1,a_2,a_3][b_1,b_2,b_3]\in T^{(5)}$$

$$[a_1,\dots ,a_m][b_1,\dots ,b_n]\notin T^{(m+n-1)}$$

$$q=\left\{\begin{array}[]{ll}\ell&\mbox{ if $\ell<k$, that is, if at least one $n_{i}$ is even;}\\ \ell-1&\mbox{ if $\ell=k$, that is, if all $n_{i}$ are odd }\end{array}\right.$$

$$3^q[a_{11},\dots ,a_{1n_1}]\dots [a_{k1},\dots ,a_{k{n}_k}]\in T^{(N_{k,q})}.$$

$$r[a_{11},\dots ,a_{1n_1}]\dots [a_{k1},\dots ,a_{k{n}_k}]\notin T^{(N_{k,q}+1)}$$

$$[a_1{a}_2,b]=a_1[a_2,b]+[a_1,b]a_2,$$

$$[a_1{a}_2,b_1,b_2]=a_1[a_2,b_1,b_2]+[a_1,b_1][a_2,b_2]+[a_1,b_2][a_2,b_1]+[a_1,b_1,b_2]a_2$$

$$\begin{gathered} [a_1{a}_2,b_1,b_2,b_3]=a_1[a_2,b_1,b_2,b_3]+[a_1,b_1][a_2,b_2,b_3]+[a_1,b_2][a_2,b_1,b_3] \\ +[a_1,b_3][a_2,b_1,b_2]+[a_1,b_1,b_2][a_2,b_3]+[a_1,b_1,b_3][a_2,b_2] \\ +[a_1,b_2,b_3][a_2,b_1]+[a_1,b_1,b_2,b_3]a_2 \end{gathered}$$

$$[g_1,\dots ,g_{k-1},f_1][f_2,f_3,f_4]+[g_1,\dots ,g_{k-1},f_2][f_1,f_3,f_4]\in T^{(k+2)},$$

$$[g_1,\dots ,g_{k-1},f_1][f_2,f_3,f_4]+[g_1,\dots ,g_{k-1},f_4][f_2,f_3,f_1]\in T^{(k+2)}.$$

$$[h_1{h}_2,c_1,h_3]=h_1[h_2,c_1,h_3]+[h_1,c_1][h_2,h_3]+[h_1,h_3][h_2,c_1]+[h_1,c_1,h_3]h_2\in T^{(k+2)}.$$

$$[h_1,c_1][h_2,h_3]+[h_1,h_3][h_2,c_1]\in T^{(k+2)}.$$

$$[c_{1},h_{1}][h_{2},h_{3}]+[c_{1},h_{2}][h_{1},h_{3}]-\bigl{[}[c_{1},h_{2}],[h_{1},h_{3}]\bigr{]}=-\bigl{(}[h_{1},c_{1}][h_{2},h_{3}]+[h_{1},h_{3}][h_{2},c_{1}]\bigr{)}\in T^{(k+2)}.$$

$$\bigl{[}[c_{1},h_{2}],[h_{1},h_{3}]\bigr{]}=[c_{1},h_{2},h_{1},h_{3}]-[c_{1},h_{2},h_{3},h_{1}]\in T^{(k+3)}\subseteq T^{(k+2)},$$

$$[c_1,h_1][h_2,h_3]+[c_1,h_2][h_1,h_3]\in T^{(k+2)}.$$

$$\begin{gathered} [f_1{f}_2,c_2,f_3,f_4]=f_1[f_2,c_2,f_3,f_4]+[f_1,c_2][f_2,f_3,f_4]+[f_1,f_3][f_2,c_2,f_4] \\ +[f_1,f_4][f_2,c_2,f_3]+[f_1,c_2,f_3][f_2,f_4]+[f_1,c_2,f_4][f_2,f_3] \\ +[f_1,f_3,f_4][f_2,c_2]+[f_1,c_2,f_3,f_4]f_4\in T^{(k+2)}. \end{gathered}$$

$$[f_{1},f_{3}][f_{2},c_{2},f_{4}]+[f_{1},f_{4}][f_{2},c_{2},f_{3}]=\bigl{(}[c_{2},f_{2},f_{4}][f_{3},f_{1}]+[c_{2},f_{2},f_{3}][f_{4},f_{1}]\bigr{)}\\ -\bigl{[}[c_{2},f_{2},f_{4}],[f_{3},f_{1}]\bigr{]}-\bigl{[}[c_{2},f_{2},f_{3}],[f_{4},f_{1}]\bigr{]}\in T^{(k+2)}$$

$$[f_1,c_2][f_2,f_3,f_4]+[f_1,f_3,f_4][f_2,c_2]\in T^{(k+2)}$$

$$[c_{2},f_{1}][f_{2},f_{3},f_{4}]+[c_{2},f_{2}][f_{1},f_{3},f_{4}]-\bigl{[}[c_{2},f_{2}],[f_{1},f_{3},f_{4}]\bigr{]}\\ =-\bigl{(}[f_{1},c_{2}][f_{2},f_{3},f_{4}]+[f_{1},f_{3},f_{4}][f_{2},c_{2}]\bigr{)}\in T^{(k+2)}.$$

$$[c_2,f_1][f_2,f_3,f_4]+[c_2,f_2][f_1,f_3,f_4]\in T^{(k+2)}$$

Full text

[TABLE]

for all $a_{i},b_{j}\in A$. This was proved by Sharma and Srivastava in 1990 and independently rediscovered later (with different proofs) by various authors. The aim of our note is to give a simple proof of the following result: if at least one of the integers $m,n$ is odd then, for all $a_{i},b_{j}\in A$,

[TABLE]

Since it has been known that, in general,

[TABLE]

our result cannot be improved further for all $m,n$ such that at least one of them is odd.

2010 AMS MSC Classification: 16R10, 16R40

Keywords: polynomial identity, commutator, Lie nilpotent associative ring

1. Introduction

Let $A$ be a unital associative ring. Define a left-normed commutator $[a_{1},\dots,a_{k-1},a_{k}]$ ($k>1$, $a_{i}\in A$ for all $i$) recursively as follows: $[a_{1},a_{2}]=a_{1}a_{2}-a_{2}a_{1}$, [a_{1},\dots,a_{k-1},a_{k}]=\bigl{[}[a_{1},\dots,a_{k-1}],a_{k}\bigr{]} $(k>2)$. Let $T^{(k)}=T^{(k)}(A)$ be the two-sided ideal of $A$ generated by all commutators $[a_{1},a_{2},\dots,a_{k}]$ $(a_{i}\in A)$.

For each unital associative ring $A$, all $m,n\in\mathbb{Z}$, $m,n>1$ and all $a_{i},b_{j}\in A$, we have

[TABLE]

This result was proved by Latyshev [15, Lemma 1] in 1965 in an article published in Russian and independently rediscovered by Gupta and Levin [13, Theorem 3.2] in 1983.

If $m=3$ then, by (1), we have $[a_{1},a_{2},a_{3}][b_{1},b_{2},\dots,b_{n}]\in T^{(n+1)}$. Note that $T^{(n+1)}\supseteq T^{(n+2)}$. In 1985 Levin and Sehgal [16, Lemma 2(a)] proved that, for all $n>1$ and all $a_{i},b_{j}\in A$,

[TABLE]

Earlier, in 1978, a similar result was proved by Volichenko [19, Lemma 1] in a preprint written in Russian. More recently, some particular cases of this result were independently rediscovered, with different proofs, in [10, Theorem 3.4] and [11, Lemma 1].

In 1990 Sharma and Srivastava [18, Theorem 2.8] proved that if $m,n\in\mathbb{Z}$, $m,n>0$ and at least one of $m,n$ is odd then, for all $a_{i},b_{j}\in A$,

[TABLE]

Recently this result was independently rediscovered, with different proofs, in [2, Corollary 1.4] and [12, Theorem 1].

Note that if $m=2m^{\prime}$, $n=2n^{\prime}$ both are even then the result similar to (2) does not hold: there exit associative algebras $A=A(m,n)$ over a field $F$ of characteristic [math] such that, for some $a_{i},b_{j}\in A$,

[TABLE]

(see [8, Theorem 1.4] or [12, Lemma 6]) so

[TABLE]

for all $\ell\in\mathbb{Z}$, $\ell\neq 0$.

The aim of the present note is to give a simple proof of the following theorem that improves the result (2) by Sharma and Srivastava.

Theorem 1.1**.**

Let $A$ be a unital associative ring. Let $m,n\in\mathbb{Z}$, $m,n>0$. Suppose that at least one of the integers $m,n$ is odd. Then, for all $a_{i},b_{j}\in A$,

[TABLE]

Note that, for some $A$ and some $a_{i},b_{j}\in A$,

[TABLE]

(see [14, Theorem 1.1]) so for $m=3$, $n=2$ Theorem 1.1 cannot be improved. On the other hand, if $m=n=3$ then, for each associative ring $A$ and all $a_{i},b_{j}\in A$,

[TABLE]

(see [5, Lemma 2.1]). There is some evidence that suggests that the case $m=n=3$ is exceptional and in all other cases Theorem 1.1 cannot be improved further.

Conjecture 1.2**.**

Let $m,n\in\mathbb{Z}$, $m,n>1$ and either $m$ or $n$ is odd. If $(m,n)\neq(3,3)$ then there is a unital associative ring $A$ and $a_{i},b_{j}\in A$ such that

[TABLE]

It is easy to check that to prove (or disprove) Conjecture 1.2 one can assume that $A=\mathbb{Z}\langle X,Y\rangle$ is the free unital associative ring on a free generating set $X\cup Y$ where $X=\{x_{1},x_{2},\dots\}$, $Y=\{y_{1},y_{2},\dots\}$ and $a_{i}=x_{i}$, $b_{j}=y_{j}$ for all $i,j$.

The following assertion follows immediately from Theorem 1.1 (see [9, Prop. 1.3 and 1.4] for more details).

Corollary 1.3**.**

Let $A$ be a unital associative ring and let $n_{1},\dots,n_{k}\in\mathbb{Z}$, $n_{i}>0$ for all $i$. Suppose that $\ell$ of the integers $n_{1},\dots,n_{k}$ are odd. Let

[TABLE]

and let $N_{k,q}=n_{1}+\dots+n_{k}-2(k-1)+q$. Then, for all $a_{ij}\in A$,

[TABLE]

Note that, in general,

[TABLE]

for some unital associative ring $A$, some $a_{ij}\in A$ and all $r\in\mathbb{Z}$, $r\neq 0$. This was proved by Dangovski [6, Prop. 2.2] if $\ell=k$ and by the authors of the present article [9, Theorem 1.7] if $\ell<k$.

Remarks. 1. Theorem 1.1 and Conjecture 1.2 are closely connected to the description of the additive group of the ring $\mathbb{Z}\langle X\rangle/T^{(k)}$ where $\mathbb{Z}\langle X\rangle$ is the free unital associative ring with a free generating set $X=\{x_{1},x_{2},\dots\}$.

It is clear that the additive group of the ring $\mathbb{Z}\langle X\rangle/T^{(2)}$ is free abelian. It was shown in [3] that the additive group of $\mathbb{Z}\langle X\rangle/T^{(3)}$ is also free abelian. On the other hand, the additive group of the ring $\mathbb{Z}\langle X\rangle/T^{(4)}$ is a direct sum $G\oplus H$ of a free abelian group $G$ and an elementary abelian $3$-group $H$ (see [7, 14]). Computational data by Cordwell, Fei and Zhou presented in [4, Appendix A] suggest that for $k\geq 6$ the additive group of the ring $\mathbb{Z}\langle X\rangle/T^{(k)}$ is also a direct sum of a free abelian group $G$ and a non-trivial elementary abelian $3$-group $H$ while for $k=5$ this group is free abelian. However, it is still an open problem whether the torsion subgroup $H$ of the additive group of $\mathbb{Z}\langle X\rangle/T^{(k)}$ is indeed a non-trivial (elementary) abelian $3$-group if $k>5$ and $H=\{0\}$ if $k=5$.

If Conjecture 1.2 holds then the elements $[a_{1},\dots,a_{m}][b_{1},\dots,b_{n}]+T^{(m+n-1)}$ are non-trivial elements of $H\subset\mathbb{Z}\langle X\rangle/T^{(m+n-1)}$ whose order, by Theorem 1.1, is equal to $3$ (if $(m,n)\neq(3,3)$). If $m=3$, $n=2$ then such products of commutators generate $H$ as a two-sided ideal in $\mathbb{Z}\langle X\rangle/T^{(4)}$ (see [7, 14]). One might expect a similar situation if $k>5$.

2. The proof of (2) given in [2] can be modified to prove Theorem 1.1 (see [1, Remark 3.9] for explanation). This modification uses computer calculations in a free associative ring. Our proof of Theorem 1.1 does not require computer calculations and is much simpler then that modification of the proof given in [2].

3. Theorem 1.1 and its corollary remain valid for a non-unital associative ring $A$; one can easily deduce this from the corresponding results for unital rings. We state and prove our results for a unital associative ring $A$ in order to simplify notation in the proof.

2. Proof of Theorem 1.1

It is straightforward to check that

[TABLE]

so the map $D_{b}:A\rightarrow A$ such that $D_{b}(a)=[a,b]$ $(a\in A)$ is a derivation of the ring $A$. It follows that

[TABLE]

[TABLE]

for all $a_{i},b_{j}\in A$.

The following lemma is a modification of well-known results (see, for instance, [14, Lemma 2.1], [15, Lemma 2 (3)], [17, Lemma 8.2]). We prove it here in order to have the article self-contained.

Lemma 2.1**.**

Let $A$ be a unital associative ring. Then, for all $k>1$ and all $g_{i},f_{j}\in A$, we have

[TABLE]

[TABLE]

Proof.

Let $c_{1}=[g_{1},\dots,g_{k-1},g_{k}]$. It is clear that $[c_{1},(h_{1}h_{2}),h_{3}]\in T^{(k+2)}$ for all $g_{i},h_{j}\in A$. We have $[c_{1},(h_{1}h_{2}),h_{3}]=-[h_{1}h_{2},c_{1},h_{3}]$ and, by (4),

[TABLE]

It is clear that $h_{1}[h_{2},c_{1},h_{3}]=-h_{1}[c_{1},h_{2},h_{3}]\in T^{(k+2)}$; similarly, $[h_{1},c_{1},h_{3}]h_{2}\in T^{(k+2)}$ so

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It follows that

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Since

[TABLE]

we have

[TABLE]

Now we check that (7) holds. Let $c_{2}=[g_{1},\dots,g_{k-1}]$. It is clear that $[c_{2},(f_{1}f_{2}),f_{3},f_{4}]\in T^{(k+2)}$ for all $g_{i},f_{j}\in A$. We have $[c_{2},(f_{1}f_{2}),f_{3},f_{4}]=-[f_{1}f_{2},c_{2},f_{3},f_{4}]$ and, by (5),

[TABLE]

It is clear that $f_{1}[f_{2},c_{2},f_{3},f_{4}]=-f_{1}[c_{2},f_{2},f_{3},f_{4}]\in T^{(k+2)}$ and similarly $[f_{1},c_{2},f_{3},f_{4}]f_{4}\in T^{(k+2)}$. Further, by (8),

[TABLE]

and similarly $[f_{1},c_{2},f_{3}][f_{2},f_{4}]+[f_{1},c_{2},f_{4}][f_{2},f_{3}]=[c_{2},f_{1},f_{3}][f_{4},f_{2}]+[c_{2},f_{1},f_{4}][f_{3},f_{2}]\in T^{(k+2)}$. It follows that

[TABLE]

so

[TABLE]

Since \bigl{[}[c_{2},f_{2}],[f_{1},f_{3},f_{4}]\bigr{]}\in T^{(k+3)}\subseteq T^{(k+2)}, we have

[TABLE]

so (7) holds.

It remains to check that (6) holds. Recall that $c_{2}=[g_{1},\dots,g_{k-1}]$. By the Jacobi identity,

[TABLE]

so

[TABLE]

By (7), we have $[c_{2},f_{1}][f_{2},f_{3},f_{4}]+[c_{2},f_{4}][f_{2},f_{3},f_{1}]\in T^{(k+2)}$, that is, (6) holds.

This completes the proof of Lemma 2.1. ∎

Corollary 2.2**.**

Let $A$ be a unital associative ring. Then, for all $k>1$ and all $g_{i},f_{j}\in A$ and for each permutation $\sigma$ on the set $\{1,2,3,4\}$, we have

[TABLE]

Proof.

It is clear that (9) is true if $\sigma=(23)$ is the transposition that permutes $2$ and $3$. By Lemma 2.1, (9) holds if $\sigma=(12)$ and $\sigma=(14)$. Hence, (9) is true for all permutations $\sigma$ that are products of the transpositions $(12),(23)$ and $(14)$. However, it is easy to check that these $3$ transpositions generate the group $S_{4}$ of all permutations on the set $\{1,2,3,4\}$. The result follows. ∎

The following corollary has been proved by Levin and Sehgal [16, Lemma 2(a)]. Our proof is different from one given in [16]; it shows that the coefficient $3$ in (10) appears because of the Jacobi identity.

Corollary 2.3** (see [16]).**

Let $A$ be a unital associative ring. Then, for all $a_{i},b_{j}\in A$,

[TABLE]

Proof.

By the Jacobi identity,

[TABLE]

On the other hand, by Corollary 2.2,

[TABLE]

It follows that

[TABLE]

as required. ∎

The following lemma is a modification of [12, Lemma 2]. Note that the proof given in [12] allows to prove only the inclusion $6\,[T^{(k)},A,A]\subseteq T^{(k+2)}$.

Lemma 2.4** (cf. [12]).**

Let $A$ be a unital associative ring. Then, for each $k\geq 1$, we have

[TABLE]

Remark**.**

In general, for $\ell\in\mathbb{Z}$, $\ell\neq 0$,

[TABLE]

More precisely, if $k=2k^{\prime}$ is even then, in general,

[TABLE]

for any $\ell\in\mathbb{Z}$, $\ell\neq 0$. This can be easily deduced from (3).

On the other hand, if $k=2k^{\prime}+1$ is odd then

[TABLE]

however, to prove this one has to use Theorem 1.1. **

Proof of Lemma 2.4.

By definition, $T^{(k)}$ is the two-sided ideal of $A$ generated by all commutators $[a_{1},a_{2},\dots,a_{k}]$ $(a_{i}\in A)$. However, one can easily check that $T^{(k)}$ is generated by the commutators $[a_{1},a_{2},\dots,a_{k}]$ $(a_{i}\in A)$ as a right ideal in $A$ as well. Hence, to prove the lemma it suffices to prove that

[TABLE]

where $u,v,w\in A$, $c=[a_{1},a_{2},\dots,a_{k}]$ $(a_{i}\in A)$. We have

[TABLE]

By Corollary 2.3, we have

[TABLE]

It is clear that

[TABLE]

Further,

[TABLE]

where \bigl{[}[c,w],[u,v]\bigr{]}\in T^{(k+3)}\subseteq T^{(k+2)} and, by (4),

[TABLE]

Hence,

[TABLE]

It follows from (11)–(14) that $3\,[cu,v,w]\in T^{(k+2)}$, as required. The proof of Lemma 2.4 is completed. ∎

Now we are in a position to prove the theorem. We need to check that

[TABLE]

if either $m$ or $n$ is odd. Since \bigl{[}[a_{1},\dots,a_{m}],[b_{1},\dots,b_{n}]\bigr{]}\in T^{(m+n)}\subseteq T^{(m+n-1)}, we may assume without loss of generality that $m$ is odd.

The proof is by induction on $m$. If $m=1$ then the theorem clearly holds. Let $m=2k+1$, $k>0$. Suppose that

[TABLE]

for all $n>0$ and all $a_{i},b_{j}\in A$.

Let $c_{1}=[a_{1},a_{2},\dots,a_{m-2}]$, $c_{2}=[b_{1},b_{2},\dots,b_{n}]$ $(a_{i},b_{j}\in A)$. By (4), we have

[TABLE]

so

[TABLE]

for all $v,w\in A$.

By the induction hypothesis, $3\,c_{1}c_{2}\in T^{(m+n-3)}$ so, by Lemma 2.4,

[TABLE]

Again, by the induction hypothesis,

[TABLE]

Further,

[TABLE]

where \bigl{[}[c_{1},w],[c_{2},v]\bigr{]}\in T^{(m+n)}\subseteq T^{(m+n-1)} and, by (4),

[TABLE]

Hence,

[TABLE]

It follows from (15)–(18) that $3\,[c_{1},v,w]c_{2}\in T^{(m+n-1)}$ for all $v,w\in A$. Theorem 1.1 follows.

Acknowledgment

The second author was partially supported by CNPq grant 310331/2015-3.