Jacobson's lemma via Groebner-Shirshov bases

TL;DR

This paper presents a constructive method using Groebner-Shirshov bases to derive the inverse of 1 - ab from the inverse of 1 - ba in rings, extending Jacobson's lemma to generalized inverses.

Contribution

It introduces a new algebraic approach to explicitly compute inverses related to Jacobson's lemma using Groebner-Shirshov basis theory.

Findings

Provides a constructive algebraic procedure for inverse calculation

Extends Jacobson's lemma to generalized inverses

Offers explicit formulas for inverses in ring theory

Abstract

Let $R$ be a ring with identity $1$. Jacobson's lemma states that for any $a,b\in R$, if $1-ab$ is invertible then so is $1-ba$. Jacobson's lemma has suitable analogues for several types of generalized inverses, e.g., Drazin inverse, generalized Drazin inverse, and inner inverse. In this note we give a constructive way via Groebner-Shirshov basis theory to obtain the inverse of $1-ab$ in terms of $(1-ba)^{-1}$, assuming the latter exists.

Equations

$$(1+b(1-ab{)}^{-1}a)(1-ba)=1-b(1-(1-ab{)}^{-1}+(1-ab{)}^{-1}ab)a=1-b0a=1$$

$$(1-ba)(1+b(1-ab{)}^{-1}a)=1+b((1-ab{)}^{-1}-1-ab(1-ab{)}^{-1}a)a=1+b0a=1$$

$$(1-ba{)}^{-1}$$

$$=1+b(1+ab+abab+\cdots )a$$

$$=1+b(1-ab{)}^{-1}a.$$

$$f{\to }_{g_1}{f}_1{\to }_{g_2}{f}_2{\to }_{g_3}\cdots {\to }_{g_t}{f}_t=h,$$

$$\mathrm{Irr}(G)=\{w\in X^{\ast }:w\ne a\overline{g}b\text{mbox}foranyg\in G,a,b\in X^{\ast }\}.$$

$$(1-ba)x=1$$

$$x(1-ba)=1.$$

$$fab-abg=(abc-c+1)ab-ab(cab-c+1)=-cab+abc{\to }_{G}0$$

$$gc-cf=(cab-c+1)c-c(abc-c+1)=0.$$

$$bax-x+1=\sum _{i=1}^m{\alpha }_{i}ba{u}_i-\sum _{i=1}^m{\alpha }_i{u}_i+1{\to }_{G}0.$$

$$ba({\alpha }_{1}bca)-{\alpha }_{1}bca+1{\to }_f{\alpha }_{1}b(c-1)a-{\alpha }_{1}bca+1=-{\alpha }_{1}ba+1,$$

$$bax-x+1$$

$${\to }_f{\alpha }_{1}b(c-1)a+\sum _{i=2}^m{\alpha }_{i}ba{u}_i-{\alpha }_{1}bca-\sum _{i=2}^m{\alpha }_i{u}_i+1$$

$$=\sum _{i=2}^m{\alpha }_{i}ba{u}_i-{\alpha }_{1}ba-\sum _{i=2}^m{\alpha }_i{u}_i+1.$$

$$\sum _{i=2}^m{\alpha }_{i}ba{u}_i-{\alpha }_{1}ba-\sum _{i=2}^m{\alpha }_i{u}_i+1$$

$$=$$

$${\to }_f$$

$$=$$

$$\overline{h}=\{\begin{array}{ll}ba{u}_3& \text{ if }{u}_3>1\\ ba& \text{ if }{u}_3=1\text{ and }{\alpha }_3\ne {\alpha }_1\\ b& \text{ if }{u}_3=1\text{ and }{\alpha }_3={\alpha }_1\end{array}.$$

$$bax-x+1{\to }_f{\alpha }_{2}ba{u}_2-{\alpha }_{1}ba-{\alpha }_2{u}_2+1.$$

$$bax-x+1$$

$${\to }_f{\alpha }_{2}b(c-1)-{\alpha }_{1}ba-{\alpha }_{2}bc+1$$

$$=-{\alpha }_{2}b-{\alpha }_{1}ba+1,$$

$$bax-x+1$$

$$=babca+ba-bca$$

$${\to }_{f}b(c-1)a+ba-bca$$

$$=0$$

$$xba-x+1$$

$$=bcaba+ba-bca$$

$${\to }_{g}b(c-1)a+ba-bca$$

$$=0$$

Full text

Jacobson’s lemma via Gröbner-Shirshov bases

Xiangui Zhao

Department of Mathematics, Huizhou University

Huizhou, Guangdong Province 516007, China

zhaoxg@hzu.edu.cn

Abstract. Let $R$ be a ring with identity $1$. Jacobson’s lemma states that for any $a,b\in R$, if $1-ab$ is invertible then so is $1-ba$. Jacobson’s lemma has suitable analogues for several types of generalized inverses, e.g., Drazin inverse, generalized Drazin inverse, and inner inverse. In this note we give a constructive way via Gröbner-Shirshov basis theory to obtain the inverse of $1-ab$ in terms of $(1-ba)^{-1}$, assuming the latter exists.

2010 Mathematics Subject Classification: 15A09, 13P10,

Keyword: inverse, Gröbner-Shirshov basis, Jacobson’s lemma

1 Introduction

Let $R$ be a ring with identity $1$. Recall that an element $x\in R$ is invertible (in $R$) if there exists $y\in R$ such that $xy=yx=1$. Jacobson’s lemma [9, 4] states that for any $a,b\in R$, if $1-ab$ is invertible then so is $1-ba$. Jacobson’s lemma has suitable analogues for several types of generalized inverses, e.g., Drazin inverse [2, 6], generalized Drazin inverse [11], and inner inverse [3].

Suppose $a,b\in R$ and $1-ab$ is invertible. It is easy to verify that

[TABLE]

and then Jacobson’s lemma follows. But, how is this formula constructed? As Halmos [7] observed, $(1-ba)^{-1}$ can be “constructed” by using the following “geometric series trick”, which is “usually ascribed to Jacobson” [9],

[TABLE]

Once the inverse is “constructed”, it is easy to verify that Formula (1) works as above, despite its unlawful derivation.

Now it is natural to ask: Does there exist a legal way to construct the inverse of $1-ba$? We shall answer this question in this note.

In this note we give a constructive way via Gröbner-Shirshov basis theory to obtain the inverse of $1-ab$ in terms of $(1-ba)^{-1}$, assuming the latter exists. A similar method is used in [8] to obtain Hua’s identity.

2 Gröbner-Shirshov bases for algebras over a unitary ring

In this section, we briefly review several properties of Gröbner-Shirshov basis for algebras over a unitary commutative ring.

Let $X\neq\emptyset$ and $k$ be a unitary ring. Let $X^{*}$ be the free monoid generated by $X$ and $A=k\langle X\rangle=kX^{*}$ be the free algebra over $k$ with generator set $X$. Similar to the case of associative algebras over a field (see [1], c.f., [5, 10]), we have concepts of monomial ordering, leading term/coefficient, composition, and Gröbner-Shirshov basis. Particularly, a Gröbner-Shirshov basis consists of only monic polynomials (i.e., polynomials with leading coefficient $1$). For convenience, we define the leading term of a nonzero element from $k$ to be $1$ and the leading term of [math] to be [math]. Fixing a monomial ordering, we denote the leading term and leading coefficient of $f\in A$ by $\overline{f}$ and $\operatorname{lc}(f)$ respectively.

Given $f=\alpha_{1}u_{1}+\cdots\alpha_{n}u_{n}\in A$, where $n\in\mathbb{N}$, each $\alpha_{i}\in k$, $u_{i}\in X^{*}$, and $\alpha_{i}\neq 0$, the support of $f$ is $\operatorname{Supp}(f)=\{u_{1},\ldots,u_{n}\}$. Let $f,h,g\in A$, where $g$ is a monic polynomial. Then $f$ reduces to $h$ modulo $g$, denoted by $f\rightarrow_{g}h$, if $\overline{f}=\overline{agb}$ for some $a,b\in X^{*}$, and $h=f-\operatorname{lc}(f)agb$. For a set $G$ consisting of monic polynomials, we say that $f$ reduces to $h$ modulo $G$, denoted by $f\rightarrow_{G}{h}$, if there exists a finite chain of reductions

[TABLE]

where each $g_{i}\in G$ and $t\in\mathbb{N}$. Denote the set of words that are irreducible w.r.t. $G$ by $\operatorname{Irr}(G)$, i.e.,

[TABLE]

If ${\operatorname{Supp}(h)}\subseteq\operatorname{Irr}(G)$, then we say $h$ is irreducible w.r.t. $G$.

The following lemma follows from the composition-diamond lemma for associative algebras over a commutative ring ([1], Theorem 1 and Remark 2).

Lemma 2.1**.**

Suppose $G$ is a Gröbner-Shirshov basis for $I\unlhd A$, where $I$ is the ideal of $A$ generated by $G$. Then, for any $f\in A$, $f\in I$ if and only if $f\to_{G}0$.

3 A constructive method to Jacobson’s lemma

Let $R$ be a unitary ring. Given $a,b\in R$ with $1-ab$ invertible (say, $c=(1-ab)^{-1}$), we give a constructive method in this section to find the inverse of $1-ba$.

Let $S$ be the $\mathbb{Z}$-algebra generated by $\{a,b,c\}$ with defining relations $\{(1-ab)c=1,c(1-ab)=1\}$, i.e., $S=\mathbb{Z}\langle a,b,c:(1-ab)c=1,c(1-ab)=1\rangle.$ We want to solve the following system for $x$ in $S$:

[TABLE]

It is obvious that the subring of $R$ generated by $\{a,b,c\}$ is a ring homomorphic image of $S$. Thus the image of $x$ is the inverse of $1-ba$ in $R$.

Let $<$ be the degree-lexicographic ordering on $\{a,b,c\}^{*}$ with $a<b<c$. Denote $f=abc-c+1$, $g=cab-c+1$, and $G=\{abc-c+1,cab-c+1\}=\{f,g\}$.

The following two lemmas will be used in the proof of our main theorem.

Lemma 3.1**.**

The set $G$ is a Gröbner-Shirshov basis (in $\mathbb{Z}\langle a,b,c\rangle$) for $S$ w.r.t. $<$.

Proof.

There exist only two compositions (both are intersection compositions) in $G$, i.e.,

[TABLE]

and

[TABLE]

Hence $G$ is a Gröbner-Shirshov basis. ∎

Lemma 3.2**.**

Suppose $\sum_{i=1}^{n}\alpha_{i}uv_{i}\to_{G}0$, where $1\leq n\in\mathbb{N}$, $0\neq\alpha_{i}\in\mathbb{Z}$, $u,v_{i}\in\operatorname{Irr}(G)$ for all $1\leq i\leq n$ and $v_{1}>v_{2}>\cdots>v_{n}$. Then $uv_{1}\not\in\operatorname{Irr}(G)$.

Proof.

If $uv_{1}\in\operatorname{Irr}(G)$, then $uv_{1}\neq 0$ is the leading term of $\sum_{i=1}^{n}\alpha_{i}uv_{i}$, contradicting the assumption that $\sum_{i=1}^{n}\alpha_{i}uv_{i}\to_{G}0$. ∎

The following theorem gives a new proof for Jacobson’s lemma.

Theorem 3.3**.**

System (3) has a unique solution in $S$.

Proof.

(Uniqueness) It follows from the uniqueness of an inverse in a ring.

(Existence) Suppose a solution $x\in S$ of the system exists and let $x=\sum_{i=1}^{m}\alpha_{i}u_{i}$ where $0\neq\alpha_{i}\in\mathbb{Z}$, $m\in\mathbb{N}$ and all $u_{i}\in{\operatorname{Irr}(G)}$ with $u_{1}>u_{2}>\cdots>u_{m}$. Then it follows from $(1-ba)x=1$ and Lemma 2.1 that

[TABLE]

If $bau_{1}\in\operatorname{Irr}(G)$ then $bau_{1}\neq 0$ is the leading term of $bax-x+1$, contradicting the fact that $bax-x+1\to_{G}0$. Thus $bau_{1}\not\in\operatorname{Irr}(G)$. Hence $u_{1}=bcv$ for some $v\in\{a,b,c\}^{*}$, where $\{a,b,c\}^{*}$ denotes the free monoid generated by $\{a,b,c\}$. Similarly, the assumption $1-x(1-ba)=0$ in $S$ implies that $u_{1}=wca$ for some $w\in\{a,b,c\}^{*}$.

Now there are two cases for $u_{1}$: (i) $u_{1}=bca$ and (ii) $u_{1}=bczca$ for some $z\in\{a,b,c\}^{*}$. We begin with Case (i). Suppose $u_{1}=bca$. First we notice that

[TABLE]

where $-\alpha_{1}ba+1$ is irreducible modulo $G$. Thus $ba(\alpha_{1}bca)-\alpha_{1}bca+1\not\to_{G}0$ and $x\neq\alpha_{1}bca$. Hence $m\geq 2$. Now we may suppose $x=\alpha_{1}bca+\sum_{i=2}^{m}\alpha_{i}u_{i}$. Then

[TABLE]

We claim that $m=2$. Otherwise, suppose $m\geq 3$. Then $u_{2}>u_{3}\geq 1$, $bau_{2}>bau_{i}$, $bau_{2}>ba$, and $bau_{2}>u_{j}$ for all $i\geq 3$ and $j\geq 2$. Namely, $bau_{2}$ is the leading monomial of (3). Thus the fact $bax-x+1\to_{G}0$ implies $bau_{2}\not\in\operatorname{Irr}(G)$. So $u_{2}=bcu^{\prime}$ for some $u^{\prime}\in\{a,b,c\}^{*}$. But $u_{2}<u_{1}=bca$. Hence $u^{\prime}=1$ and $u_{2}=bc$. If $bau_{i}\not\in\operatorname{Irr}(G)$ for some $i\geq 3$, then by the same argument we have $u_{i}=bc$, which contradicts that $u_{2}>u_{i}$. Therefore $bau_{i}\in\operatorname{Irr}(G)$ for $i\geq 3$. Thus we may reduce polynomial (3) as follows

[TABLE]

The last polynomial, denoted by $h$, is irreducible and its leading term is (note that if $u_{3}=1$ then $m=3$)

[TABLE]

In each of these three cases, $\overline{h}\neq 0$ and thus $h\neq 0$, which contradicts the assumption that $bax-x+1\rightarrow_{G}0$. This proves our claim that $m=2$.

Now we may write $x=\alpha_{1}bca+\alpha_{2}u_{2}$ and from (3) we have

[TABLE]

If $bau_{2}\not\in\operatorname{Irr}(G)$, then $u_{2}=bc$ as we proved in the last paragraph. Then we have

[TABLE]

which is irreducible modulo $G$ and nonzero, contradicting that $bax-x+1\to_{G}0$. Therefore, $bau_{2}\in\operatorname{Irr}(G)$. Now the only possibility in which $\alpha_{2}bau_{2}-\alpha_{1}ba-\alpha_{2}u_{2}+1\to_{G}0$ is that $u_{2}=1$ and $\alpha_{1}=\alpha_{2}=1$. That is, $x=bca+1$.

Now we check by reduction whether $x=bca+1$ is a solution of (3).

[TABLE]

Similarly, we have

[TABLE]

Thus, $x=bca+1$ is a solution of (3) in $S$. This implies that $x$ is an inverse of $1-ba$ in $S$. Since an inverse is unique if it exists in $S$, we do not need to consider Case (ii) for $x$. ∎

It is clear that Theorem 3.3 gives a new proof for Jacobson’s lemma.

Acknowledgements. This work was motivated by a talk by R. Padmanabhan in the Rings and Modules Seminar at the University of Manitoba.