Algebras with Laurent polynomial identity

TL;DR

This paper investigates algebraic structures where the group of units satisfies specific Laurent polynomial identities, providing new insights into their properties and classifications.

Contribution

It introduces new results on algebras with units obeying Laurent polynomial identities, expanding understanding of their algebraic behavior.

Findings

Characterization of algebras with Laurent polynomial identities

Conditions under which units satisfy specific polynomial identities

Examples illustrating the algebraic structures studied

Abstract

In this article we shows some results about algebra with the group of units having special polynomial identity.

Equations

$$w_i=x_1^{r_1}{x}_2^{s_2}\cdots x_1^{r_k}{x}_2^{s_k},k\ge 1,r_i,s_i$$

$$\sum exp{w^{\prime }}_i=\sum ex{p}_{x_1}{w^{\prime }}_i+\sum ex{p}_{x_2}{w^{\prime }}_i=k\sum ex{p}_{x_1}{w}_i+\sum ex{p}_{x_2}{w}_i$$

$$\sum exp{w^{\prime }}_i=\sum ex{p}_{x_1}{w}_i+k\sum ex{p}_{x_2}{w}_i.$$

$$P(\alpha ,\alpha )=a_1+a_2{\alpha }^{\sum exp{w}_2}+...+a_k{\alpha }^{\sum exp{w}_k}=0,1<k\le r,$$

$${\alpha }^n=1+aua+bauab+\sum _{t\in Q_n}t+(auabauab{)}^n$$

$$f_0(\alpha )$$

$$f_1(x_1,...,x_{2n})=S_{2n}\cdot (x_1\cdots x_{2n}{)}^{-1}$$

$$f_2(x_1,...,x_{2n})=f_1(x_1,...,x_{2n})+S_{2n}$$

$$f(\overline{u_1},...,\overline{u_{2n}})\ne 0.$$

$$g(xy)=f(u_1,...,u_{2n})\ne \sum _i{f}_i{u}^2{g}_i,$$

$$S_3(X,Y,XY)=(YX{)}^2-X^2{Y}^2-Y{X}^{2}Y+X{Y}^{2}X.$$

Full text

Keywords: Laurent polynomial identity, unit group, group identity, nil ideal, Brian Hartley Conjecture.

1. Introduction

A Laurent polynomial $f=f(x_{1},...,x_{l})$ in the noncommutative variables $x_{i},\,i=1,\ldots,l$ is an element non-zero in the group algebra $RF_{l}$ over a ring $R$ with free group $F_{l}=<x_{1},...,x_{l}>$. One says $f$ in $RF_{l}$ is a Laurent Polynomial Identity ($LPI$) for an $R$ algebra $A$ (respc. for $U(A)$ the group of units in A) if $f(a_{1},...,a_{l})=0$ for all sequence $a_{1},...,a_{l}$ in $A$ (respc. in U(A)). A word $w$ in $RF_{l}$ is group identity for all $U(A)$ if $w(a_{1},...,a_{l})=1$, for $a_{1},...,a_{l}$ in $U(A)$ .

Brian Hartley´s Conjecture: Let G be a torsion group and R a field. If the unit group $U(RG)$ of $RG$ satisfies a group identity $w=1$, then RG satisfies a polynomial identity.

In [2] A.Giambruno, E.Jespers and A.Valenti shows that the Brian Hartley´s Conjecture is true for a group algebra $RG$ over an infinite commutative domnain $R$ and a torsion group $G$ and $G$ has no divisible order elements by $p$ with characteristic of $R$ is $p$. For this purpose they proved the following crucial result.

Proposition 1.1**.**

Let $A$ be an algebra over an infinite commutative domain $R$ and suppose that $U(A)$ satisfies a group identity. There exists a positive m such that if $\,a,b,c,u$ in $A$ and $a^{2}=bc=0$, then $bacA$ is nil right ideal of bounded exponent less or equal than m.

The case identity group $w=1$ in the conjecture says the unity group satisfy equivalently the Laurent polynomial $F=1-w$. We use this idea and study the case $F=\,a_{1}+a_{2}w_{2}+...+a_{n}w_{n}$. Clearly the case $F=1-w$ is a special case of our generalization.

In this paper we prove the following results.

Proposition 1.2**.**

Let $A$ be an algebra with the group of units $U(A)$ admits a $LPI$ over a ring R whith unit whose non-constants words has the sum of exponents non-zero at least one of the variables. Then there exists a polynomial $f\,\in\,R[X]$ with the limited degree $d\leq 4(-l+r)+3$, where $l=min\,\{\sum expw_{i}\}$ and $r=max\{\sum exp\,w_{i}\}$, determinated by the $LPI$ such that for all $\,a,b,c,u$ in $A$ with $a^{2}=bc=0$, $f(bacu)=0$. In particular, $bacA$ is an algebraic ideal.

As a consequence of Proposition 1.2 we prove the following:

Corollary 1.3**.**

Let $A$ be an algebra over a commutative domain R with $\mid R\mid>d$ whose unit group $U(A)$ has a $LPI$. If $a,\,b,\,c\in A$ with $a^{2}=bc=0$, then $bacA$ is a nil ideal with limited exponent.

Notice that the restriction on $LPI$ is necessary because we will show that without this assumption Proposition 1.2 i not valid. In this article we always adopt $LPI$ with this restriction. Also, by changin $x_{i}$ by $x^{-i}yx^{i}$ we may assume that $LPI$ in two variables.

**Proof of Proposition 1.2: **

Let $P\,=\,a_{1}+a_{2}w_{2}+...+a_{r}w_{t}$ the $LPI$ with $w_{i}$ in the form

[TABLE]

integers. If $\sum exp\,w_{i}=\sum exp_{x_{1}}\,w_{i}+\sum exp_{x_{2}}\,w_{i}=0$ then we substitute $x_{1}=x_{1}^{k}$ or $x_{2}=x_{2}^{k}$ with $k>1$ big enough, We get a new $LPI$ in the form $P\,=\,a_{1}+a_{2}{w^{\prime}}_{2}+...+a_{r}{w^{\prime}}_{r},$ where

[TABLE]

or

[TABLE]

In any case that is not zero.

In general, $P(1,1)\,=\,a_{1}+a_{2}+...+a_{r}=0$, so $a_{2}+...+a_{r}\neq 0$ , because $a_{1}\neq 0$. Hence any collection of sums with distinct parcels involving all coefficients $a_{2},...,a_{r}$ has a non-zero term.

From this

[TABLE]

is a non-zero polynomial with integers exponents over $R$, for all $\alpha\in U(A)$. In the following we see that all exponents are not necessarily positive.

Let $\alpha^{-l}$ be where $l=min\,\{\sum expw_{i}\}$, so $\alpha^{-l}P(\alpha,\alpha)=f_{0}(\alpha)=0$, where $f_{0}\neq 0$ is a polynomial with degree $\leq-l+r$, over $R$. Therefore $U(A)$ is algebraic over $R$. Now, we are going to use this polynomial with a particular unit in $U(A)$. First let $b=c$ be, so $a^{2}=b^{2}=0$. Then we have $(1+aua),\,(1+bauab)\in U(A)$, for all $u\in A$. Let $Q_{n}$ be the set of all products in $aua$ and $bauab$ with at most $n-1$ factors, and let $W$ be the set of all products in coeficients $a_{i}`s$, $aua$ and $bauab$.

Thus $\alpha=(1+aua).(1+bauab)\,\in\,U(A)$, and for this unit,

[TABLE]

for all $n\geq 1$.

It follows that there exists a polynomial $f_{0}\neq 0$ over $R$ such that

[TABLE]

As $a_{2}+...+a_{r}\neq 0$, after to organize the powers of $auabauab$ which appears in the last sum, has a non-zero coefficients. This term has the form $(a_{j_{1}}+...+a_{j_{k}})(auabauab)^{s}$ where $s\,\leq\,(-l+r)$. So, $abf_{0}(\alpha)au=0$ is a non-zero polynomial $g$ over R such that $g(abau)=0$ with degree $\leq\,2(-l+r)+1$. This finishs the case $b=c$.

In general, if $a,\,\alpha,\,\beta\in A$ and $a^{2}=\alpha\beta=0$, then we have $(\beta u\alpha)^{2}=0\,,$ for all $u\in A$. So $\beta u\alpha a\beta u\alpha A$ is algebraic and there exists $g$ over $R$ which is not zero and

$g((\beta u\alpha a\beta u\alpha)a)=0$, from this there exists a non-zero polynomial $f=(\alpha a)f_{0}(\beta u)$ with $f(\alpha a\beta u)=0$ such that $d=degree\,f\leq\,\,4(-l+r)+3$. Therefore $\alpha a\beta A$ is algebraic over $R$.

Proof of Corollary 1.3 :

By Proposition 1.2, there exists $f\in R[x]\setminus\{0\}$ with degree $d\leq m=4(-l+r)+3$, such that $f(bac\lambda u)=0$ for all $\lambda\in R,\,u\,\in\,A$. That is, $\lambda p_{1}+\lambda^{2}p_{2}+...+\lambda^{d}p_{d}=0$ where $p_{i}$ is polynomial on $bacu$ with $p_{d}=b_{d}(bacu)^{d}$ . Now, by the Vandermond argument, $p_{1}=...=p_{d}=0$, so $(bacu)^{d}=0$ .

The next results are LPI versions of Lema 3.2, Lema 3.3 of [6] and Corolary 3 of [2]. Their proofs are analogous to the group identity case changing GI by LPI.

Corollary 1.4**.**

Let $A$ be a semiprime algebra over an commutative domain $R$ with $|R|>d$. If $U(A)$ satisfies a $LPI$, then every idempotent element of $R^{-1}A$ is central.

Corollary 1.5**.**

Let $A$ be an algebra over a field $K$ with $U(A)$ satisfying a $LPI$, and $a,\,b\,\in A$ such that $a^{2}=b^{2}=0$.

1. (1)

If $\mid K\mid>2d$, then $(ab)^{2d}=0.$ 2. (2)

If $ab$ is nilpotent, then $(ab)^{2d}=0.$

Corollary 1.6**.**

Let $K$ be any field and $n\geq 2$. Then the following are equivalent

1. (1)

$U(M_{n}(K))$* satisfies the $LPI$.* 2. (2)

$K$* is a finite field.* 3. (3)

$U(M_{n}(K))$* satisfies a group identity.*

In addition, if one of these conditions is hold, then

$\mid K\mid\leq 2d$* and $n\leq 2log_{\mid K\mid}2d+2\leq 2log_{2}2d+2.$*

In general, Proposition 1.2 is not hold without the restrictions about exponents. If it does, then $bacu$ is algebraic for all $u\in A$, in particular, for $c=b$ and $u=a$, $ba$ is algebraic. It follows that by Vandermonde augument with $R$ a infinite commutative domain $ba$ is nilpotent. In general, $ba$ is not a nilpotent element. For example, let $A=M_{n}(R),\,n>1$. By $Amitsur-Levitzki$ theorem [3] $A$ satisfies a $LPI$

[TABLE]

where $s_{2n}$ is the polynomial standard. However $A$ with $a=e_{21}$ and $b=e_{12}$, $ba=e_{11}$ is not nilpotent. Note that this algebra also satisfies

[TABLE]

(an $LPI$ with some words having sum of exponents zero at every variable).

By the same arguments we can see that if $A=M_{n}(R)$ be $n>1$ and $R$ a infinite commutative domain, there is not exist a non-zero polynomial $g\,\in\,R[X]$ not zero such that $g(ab)=0$ for all $a,\,b\in\,A$ with $a^{2}=b^{2}=0$. Note that the other side of Proposition 1.2 is true. In fact, suppose that $R$ is finite, so $A$ is finite. From this for all matrices $x\in A$ there exist integers $r>t>0$ such that $x^{r}=x^{t}$. So $x$ satisfies the polynomial $p_{x}(X)=X^{r}-X^{t}$. Then defining $g=\displaystyle\prod_{x\in A}p_{x}$ one has $g(ab)=0$ for all $a,\,b\in\,A$ with $a^{2}=b^{2}=0$, which is a contradiction.

Lemma 1.7**.**

Let $A=M_{n}(R)$ be $n>1$ and $R$ a commutative domain. Then $R$ is infinite if and only if there is not any non-zero polynomial $g\,\in\,R[X]$ such that $g(ab)=0$ for all $a,\,b\in\,A$ with $a^{2}=b^{2}=0$.

Proposition 1.8**.**

Let $F=\displaystyle\frac{R[x,y]}{(x^{2},\,y^{2})}$ where $R$ is a infinite commutative domain. Then $F$ satisfies any polynomial identity $f$ of $M_{n}(R)$ where $n\geq 2$. Also, $U(F)$ does not satisfy the standard polynomial $S_{2}$ and $S_{3}$. In particular, $F$ satisfies $S_{2n}$ if only if $n\geq 2$.

Proof.

Suppose by contradiction that $F$ does not satisfy an identity $f$ of $M_{n}(R)$ where $n\geq 2$. Then there exist $\overline{u_{1}}=\overline{g_{1}(x,y)},...,\overline{u_{2n}}=\overline{g_{2n}(x,y)}\,\in\,F$ with

[TABLE]

So (multiplying by $x$ or $y$ on left or right if necessary) we obtain

[TABLE]

$u=x,\,y$. Then, for all $a,b\,\in\,A=M_{n}(R)$ such that $a^{2}=b^{2}=0$, we have $g(ab)=0$ it is a contradiction by Lemma 1.7. By evaluation of $S_{2}$ and $S_{3}$ on units $\overline{1+x},\,\overline{1+y}$ and $\overline{(1+x)(1+y)}$, we see that $U(F)$ does not satisfy $S_{2}$ and $S_{3}.$

∎

Corollary 1.9**.**

Let $B$ be an algebra over a commutative domain $R$ whose $U(A)$ satisfies the standard polynomial $S_{3}$. Then there exists a polynomial $g\in R[X]$ such that $g(ab)=0$ for all $a,\,b\in B$ with $a^{2}=b^{2}=0.$

Proof.

By applying $S_{3}$ on $X,Y$ and $XY$ we obtain

[TABLE]

So $S_{3}(1+x,1+y,(1+x)(1+y))\neq 0$ in $F$, then multiplying by $x$ on the left and $y$ on the right we obtain a polynomial $g(xy)=xS_{3}y$ such that $g(ab)=0.$

∎

Corollary 1.10**.**

Let $F$ be the algebra in Proposition 1.8. The $F$ does not satisfy the group identity $w=1$, with $R$ a infinite field.

Proof.

If $F$ satisfies a group identity $w=1$. By Lemma 3.1 [6] there exists a polynomial $g(t)\in R[t]$ such that $g(\overline{x}\,\overline{y})=0$. Thus $\overline{g(xy)}=0$, that is, $g(xy)=\sum f_{i}u^{2}g_{i},\,u=x,\,y$. Then, $g(ab)=0$ for all $a,b\in M_{n}(R)$ such that $a^{2}=b^{2}=0$ and this contradicts Lemma 1.1.

∎

With some adaptations Lemma 3.1 [6] holds whenever $R$ is an infinite commutative domain. It follows that Corollary 1.5 holds if $R$ is an infinite commutative domain. Note that $F$ is an algebra with $LPI$ ( having more than one word which not constant) with $\sum\,expw_{i}=0$ without group identity ($LPI$ with only one non-constant word).

Index