Nil-generated algebras and group algebras whose units satify a Laurent polynomial identity
aa r X i v : . [ m a t h . R A ] F e b NIL-GENERATED ALGEBRAS AND GROUP ALGEBRASWHOSE UNITS SATISFY A LAURENT POLYNOMIALIDENTITY
CLAUDENIR FREIRE RODRIGUES
Abstract.
Let A be an algebra whose group of units U ( A ) satisfies a Lau-rent polynomial identity (LPI). We established conditions at these polynomi-als in such a way that nil-generated algebras and group algebras with torsiongroups over infinite fields in characteristic p > Introduction
Let N A the set of nilpotents of a R algebra A . If A is generated by { } ∪ N A we say that A is nil-generated.A Laurent polynomial in the noncommutative variables x i , i = 1 , . . . , l is anelement non-zero in the group algebra RF l over a ring R with free group F l =
LP I instead of GI to guaranteethe existence of polynomial identities. Besides that, results were obtained for nil-generated unitary algebras in [12] confirming that existence of GI for U ( A ) impliesequivalently at the existence of nonmatrix identity for A . In this paper, firstsection, we aboard these questions with LP I for U ( A ), obeying a condition thatguarantees the existence of nonmatrix identity for nil-generated unitary algebras(Theorem 2.7), which is equivalent to the existence of group identity for the groupof units of these algebras. This condition will be justified now. For a LP I , P = a + a w + · · · + a n w n , the words w i ≥ are the nonconstant words. Accordingto Amitsur − Levitzki [5, Theorema V. 1.9], U ( M n ( K ) n >
1, satisfies the
LP If ( x , ..., x n ) = S n · ( x · · · x n ) − where S n is the standard identity and f is an LP I with all non-constant words having sum of exponents zero at every variable.So if K is infinite by [9, Lemm3.3] U ( M n ( K ) can not satisfies a group identity.From this, we impose that sums of exponents at non-constant words in the LP I benon-zero in at least one of the variables. In the second section, this characteristicis necessary for the determination of polynomial identities for group algebras witha finite group over an infinite field (Theorem 3.2.(2)). From the proof contextof Theorem 3.2.(1), we get the confirmation of Hartley Conjecture with arbitraryLPI for the group of units U ( KG ) with G a torsion group and K a field (Theorem3.3). And finally, in the third section, for the existence of nonmatrix identityfor group algebras with torsion group over infinite field in characteristic p > The Nil-Generated Case
We need some results for adaptation of classical known ideas. The next tworesults are taken from [12, Theorems 1.3-1.4].
Theorem 2.1.
Let A be a nil-generated unitary algebra over a field of character-istic 0 and U(A) the group of units. Then the following statements are equivalent: U ( A ) satisfies a group identity; A satisfies a nonmatrix identity; A is Lie soluble; U ( A ) is a soluble group. Theorem 2.2.
Let A be a nil-generated unitary algebra over a field of charac-teristic p > and U(A) the group of units. Then the following statements areequivalent: U(A) satisfies a group identity; A satisfies a nonmatrix identity; A satisfies a polynomial identity ([ x , x ] x ) p t = 0 for some t ; U(A) satisfies the group identity ( y , y ) p t = 1 for some t . From [4, Theorem 2.1] we have
Theorem 2.3.
Let A be an algebra whose group of units U ( A ) admits a LP I P over a ring R whith unit and non-constants words w i in P has the sum of exponentsnon-zero in at least one of the variables. Then there exists a polynomial f ∈ R [ X ] with degree d determined by l = min { P exp w i } and r = max { P exp w i } suchthat for all a, b, c, u in A with a = bc = 0 , f ( bacu ) = 0 . Thus, bacA has apolynomial identity. From now one, unless otherwise started, LPI are as in the Theorem 2.3. Andaccording to [4, Corollary 3.3] we have
Lemma 2.4.
Let A be a semiprime algebra over an infinite commutative domain R . If U ( A ) satisfies a LP I , then every idempotent element of R − A is centraland for all b, c ∈ A such that bc = 0 we have bac = 0 for every a nilpotent in A. A known result in the literature is
IL-GENERATED ALGEBRAS AND GROUP ALGEBRAS 3
Lemma 2.5. If f : R → S is an epimorphism of rings with kernel a nil ideal.Then f induces an epimorphism U ( R ) → U ( S ) . Now we are ready to begin our adaptations.
Theorem 2.6.
Let A be an algebra over R whose group of units U ( A ) admits a LP I P . Then There are identities in R [ x ] determined by P for the group of units U ( A ) and the Jacobson radical J ( A ) . These are nonmatrix identities if R is aninfinite field. There is a polynomial g ( x ) ∈ R [ x ] determined by P such that for all a, b ∈ A with a = b = 0 , g ( ab ) = 0 . And, conversely the ideal bacA where a = bc = 0 in A , has a polynomial identity. If A is a nil generated algebra and R is an infinite domain. Then the set N A of nilpotents in A it is a locally nilpotent ideal and A has a polynomialidentity. This identity is nonmatrix if R is an infinite field. And everyfinite set of nilpotent elements generates a nilpotent subalgebra.Proof. Let P = a + a w + ... + a t w t the LP I with w i in the form w i = x r x s · · · x r k x s k with k ≥ , r i , s i integers.If P exp w i = P exp x w i + P exp x w i = 0 then we substitute x = x k or x = x k with k > LP I in the form P = a + a w ′ + ... + a r w ′ r , where X exp w ′ i = X exp x w ′ i + X exp x w ′ i = k X exp x w i + X exp x w i or X exp w ′ i = X exp x w i + k X exp x w i . In any case that is not zero. So we suppose P exp w i not zero for all i = 2 , ..., t .From this P ( α, α ) = a + a α l + ... + a i α P exp w i + ... + a t α r = 0with all powers of α not zero for all α ∈ U ( A ) and l and r are respectively theminimum and maximum of the sum of the exponents of the w ′ i s .After organizing the common powers we will have a polynomial expression ofthe form f ( α ) = a + b α l + ... + b j α r = 0where b ′ j s are parcials sums of a ′ i s, i = 2 , ..., t . In general, P (1 ,
1) = a + a + ... + a t = 0, so a + ... + a t = 0 , because a = 0. So we can not have all b ′ i s = 0because in this case a + ... + a r = b + ... + b j = 0. Follows that f is a non-zeropolynomial over R with integers exponents not all necessarily positives.If l < α − l f ( α ) = f ( α ) = 0, where f = 0 is a polynomialidentity for U ( A ) over R of degree at most r − l . Suppose l >
0, then f is apolynomial identity for U ( A ). Now, with a in J ( A ), 1 + a is a unit and f (1 + a ) =0. It follows that g ( x ) = f (1 + x ) is an identity for J ( A ). For all λ ∈ R the C. F. RODRIGUES minimal polynomial of matrix λI × where I × is the identity matrix is x − λ . Ifa polynomial in F [ x ] is a matrix identity then, λ is a root for it and R would befinite field.In the sequel, by Theorem 2.3, at b = c case, there is a nonzero polynomial f ( x ) ∈ R [ x ] with f ( babu ) = for all u ∈ A . Now take u = 1, af ( bab ) = g ( ab ) = 0.The conversely is known, see [3, lemma 3.1]. Let g ( x ) ∈ R [ x ] such that g ( ab ) = 0,if a = b = 0 R . For all u ∈ A , ( cub ) = 0 then, bg ( acub ) acu = g ( bacu ) = 0.Finally, according to [7] A/L , where L is the lower nil radical of A (whichcoincides with the intersection of all prime ideals of A ), is a semiprime ring and L isa locally nilpotent ideal in A . It follows that U ( A ) → U ( A/L ) is an epimorphis byLemma 2.5 and so U ( A/L ) satisfies P . From this using Lemma 2.4, for all a, x ∈ A/L with a = 0 and x nilpotent, axa = 0. And by induction ax . . . x n a = 0 withthe x ′ i s nilpotents in A . As L is a nil ideal A/L is nil generated so a ( A/L ) a = 0and ( A/L ) a ( A/L ) is a nilpotent ideal. By semipriminess of
A/L it follows that a = 0 given us there is no nilpotents in A/L . Then L is the set of nilpotents in A and A = R + L . As for all x, y ∈ A, [ x, y ] ∈ L ⊂ J ( A ), A satisfies the polynomialidentity f ([ x, y ]) where f is the identity determined by the LP I in the first partwe have proved. With a = λe , b = e in M ( R ), we have that f ([ a, b ]) = (cid:20) f ( λ ) 00 ∗ (cid:21) for all λ ∈ R . Follows that f ([ x, y ]) is a nonmatrix identity if R is an infinite field.With L locally nilpotent we finish the proof. (cid:3) Theorem 2.7.
Let A be a nil-generated unitary algebra over an infinite field F with the group of units U ( A ) satisfying an LPI. Then A satisfies a nonmatrixidentity. And equivalently in characteristic : U(A) satisfies a group identity; A is Lie soluble; U(A) is a soluble group.And equivalently in characteristic p > : A satisfies a nonmatrix identity; U(A) satisfies a group identity; A satisfies the polynomial identity ([ x , x ] x ) p t = 0 for some t ; U ( A ) satisfies the group identity ( y , y ) p t = 1 for some t .Proof. Immediate consequence of Theorems 2.1, 2.2 and 2.6. (cid:3) Group Algebras and Identities Standard
According to [3] a R-algebra A has the property P with respect a nonzeropolynomial g [ x ] ∈ R [ x ] if g ( ab ) = 0 for every a, b ∈ R , with a = b = 0. And aLaurent polynomial P over R has the property P if every R -algebra, for wich L is a LP I of U ( A ), has the property P with respect to some nonzero polynomial. IL-GENERATED ALGEBRAS AND GROUP ALGEBRAS 5
Lemma 3.1.
Let D be a noncommmutative division ring finite dimensional overits center. Then U(D) contains a free subgroup of rank two.Proof.
See [8]. (cid:3)
Theorem 3.2.
Let G be a finite group and F a field of characteristic p ≥ withU(FG) satisfying an LPI P = a + a w + · · · + a n w n .
1. If P is an arbitrary LP I , then there exist positive integers m, t such thatthe group algebra
F G satisfies the polynomial identity ( S m ) t = 0. And if F is of charactheristc 0 or characteristic p where p G | , then F G satisfiesthe identity standard S m .2. If F is infinite, then F G satisfies the nonmatrix identity ( S ) t = [ x, y ] t = 0.In charF = 0, G is an abelian group.If charF = p > U ( F G ) satisfies the group identity ( x, y ) p k = 1 for somek and equivalently, at this case, G is a p-abelian group. In particular, if G is a p’-group, then G is abelian. Proof. As G is finite, we have that F G and
F G/J are Artinian and the Jacobsonradical J = J ( F G ) is nilpotent. By Lemma 2.5 the epimorphism
F G → F G/J induces the epimorphis π : U ( F G ) → U ( F G/J ) and so U ( F G/J ) satisfies P .As F G/J is semisimple, by Wedderburn-Artin Theorem [11, pg 35]
F G/J ∼ = ⊕ ri =1 M n i ( D i ) with D i division rings and the center Z i of D i contains F , such that[ D i : Z i ] ≤ [ D i : F ] < ∞ because G is finite. Assume that D i is noncommutativethen, according to Lemma 3.1, U ( D i ) contains a non-abelian free group H of rank2, H = < g , g > where { g , g } is a free generator set of H . With the units g and g , as U ( D i ) satisfies P then we have P ( g , g ) = a + a w ( g , g ) + · · · + a n w n ( g , g ) = 0For every reduced word w i , w i ( g , g ) is unique and not trivial in H . The secondiguality above tells us that every a i is zero because the w i ( g , g ) are linearlyindependent in the subalgebra F H . With this contradiction D i is a field for each i . And with m = max { n i } , F G/J satisfies S m . Follows that for all α , . . . , α m in F G , S m ( α , . . . , α m ) is in J and so there exist t > S m ( α , . . . , α m )) t =0. And if the characteristic is 0 or p > p G | , then by Maschke ´ s Theorem, F G is semisimple satisfying S m .For the second part suppose that every nonconstant word w i has the sume ofexponents non zero in at least one of the variables. As F G/J is semiprime, byLemma 2.4 the idempotents are central and thus n i = 1 for every i . Follows that F G/J ∼ = ⊕ ri =1 D i and is commutative. Thus for all x, y ∈ F G , [ x, y ] ∈ J and[ x, y ] t = 0. As we saw in proof of Theorem 2.6(3), over an infinite field this is anonmatrix identity.Moreover for all x, y ∈ U ( F G ) π ( x, y ) = ( π ( x ) , π ( y )) = 1 = π (1) , and consequently ( x, y ) = 1 modJ. C. F. RODRIGUES
Thus G ′ ⊆ J, follows that △ ( G ′ ) ∈ J where △ ( G ′ ) is the augmentation ideal of F G ′ . And so △ ( G ′ ) is nilpotent. Fromthis G ′ is a finite p-group in charF = p > G is abelian if G is a p’-group. If charF = 0, △ ( G ′ ) is not nilpotent and G ′ = { } , that is, G isabelian.To finish the second part, with charF = p >
0. As we see before ( x, y ) =1 modJ . So there is k such that (1 − ( x, y )) p k = 0 and U ( F G ) satisfies ( x, y ) p k = 1 . The conclusion follows by [2, Lemma 2.3]. (cid:3)
At the next results, there is no restriction about the LPI.
Theorem 3.3.
Let KG be the group algebra of the torsion group G over the fieldK. If U ( KG ) satisfies a LP I then KG satisfies a PI.Proof. Let F = K [ α, β : α = β = 0] the free algebra in two noncommutativevariables α, β relative to the relations α = β = 0. According to [3] the groupof units U = U ( F ) contains a non-abelian free group H , say us with g , . . . , g n free generators of H . Arguing as in the proof of Theorem 3.2.(1), we concludethat U ( F ) can not satisfies P . Follows by [3, Theorem 1.1] that KG satisfies apolynomial identity. (cid:3) Remark . According to Theorem 2.6.(2) if P = a + a w + · · · + a n w n is a LP I with all non-constant words having sum of exponents non-zero in at leastone of the variables, then P satisfies the property P with respect to a polynomial g ( x ) ∈ R [ x ] . But for this conclusion, this restriction in general is not necessarybecause with the last proof we guarantee more information. As we saw, no Laurentpolynomial can be a LPI for the units U of the free algebra K [ α, β : α = β = 0]in noncommutative variables α, β relative to the relations α = β = 0. Withthis, we will get to remove the expression “if” in [3, Lemma 1.3]. Lemma 3.5.
Let f be a Laurent polynomial over a field K. Then f is not a LPIof U . And equivalently, there is a non-zero polynomial g ∈ K [ x ] such that every K -algebra B for which f is a LP I of U ( B ) has the property P with respect to g.Proof. We know the first part by Remark 3.4 and the equivalent statement is by[3, Lema 1.3]. (cid:3)
Theorem 3.6.
If K is a field of characterist p and G is a locally finite p ′ -groupwith U(KG) satisfying a LPI. Then KG satisfies a standard polynomial identity.Proof. This is by Lemma 3.5 and [3, Proposition 1.4]. (cid:3)
IL-GENERATED ALGEBRAS AND GROUP ALGEBRAS 7
Now we return with LPI having restrictions about the exponents.Nil rings give rise to multiplicative groups, called adjoint groups. If R is nilthen its adjoint group R consists of the elements of R as underlying set, withmultiplication given by r ◦ s = r + s + rs . In the case of a nil-algebra A over afield F , it is known that we can embed A into the unitary nil-generated algebra A = F + A and by relation a + r → ( a, a − r ) where a is in F \{ } and r is in A we have that U ( A ) ∼ = ( F \{ } ) × A . Lemma 3.7.
Let A be a nil-algebra over an infinite field F with the adjoint group A satisfying a LP I , then A satisfies a nonmatrix identity. And equivalently A satisfies a group identity.Proof. As we see U ( A ) ∼ = ( F \{ } ) × A . By Theorem 2.6, A satisfies a nonmatrixidentity. The result follows by [12, Corollary 1.5]. (cid:3) With this last adaptation, we are going to finish these notes with a version of[12, Theorem 4.1] in the context of LPI.
Theorem 3.8.
Let FG be a group algebra of a torsion group over an infinite fieldF of characteristic p > whose group of units ( F G ) satisfies a LP I P . Then
F G satisfies a nonmatrix identity. And equivalently U ( F G ) satisfies a group identity; G contains a normal subgroup A such that G/A and A’ are finite and G’is a p-group of finite exponent;
3. [
F G, F G ] F G is nil of bounded index; and
4. ( U ( F G )) ′ is a p-group of finite exponent.Proof. According to [12, Theorem 4.1] we guarantee that result holds if FG hasa nonmatrix identity. With the subroup P G generated by the set of p -elements in G , F P G is nil-generated, U ( F P G ) satisfies P and according to Theorem 2.6 theset of nilpotent elements N F P G forms a locally nilpotent maximal ideal in F P G .Thus F P coincides with the augmentation ideal of
F P G and P G is a locally finite p -group. As P G is normal in G , N ( F P G ) F G is a locally nilpotent ideal in
F G . Byrelation a → a we identified the adjoint group of N ( F P G ) F G with a subgroupof U ( F G ). Follows by Lemma 3.7 that N ( F P G ) F G has a nonmatrix identity.And as N F P G is the Kernel of the projection F G → F ( G/P G ) then, by Lemma2.5, F ( G/P G ) satisfies P . Now we are going to show that G/P G is abelian. Withthis, we will finish the proof because FG will be a commutative extension of analgebra satisfying a nonmatrix identity.As G = G/P G is a p ′ -group then, by [5, pg. 130-131], F P is semiprime andaccording to Lemma 2.4, for all g ∈ G the idempotent n − (1+ g + g + . . . + g n − ) iscentral in F G . From this h g i is a normal cyclic group in G . Thus G is an abelian orHamiltonian group. With G Hamiltonian it contains a copy of Q the quarterniongroup of order 8 and F Q satisfies P . By Theorem 3.2.(2), Q would be abelian.So, we have G abelian. And we finish. (cid:3) C. F. RODRIGUES
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Departmento de Matem´atica, Universidade Federal do Amazonas, Amazonas, Man-aus, Brasil
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