Non-matrix varieties and nil-generated algebras whose units satisfy a group identity

Yuly Billig  David Riley  Vladimir Tasić

Abstract

Let R^{×} denote the group of units of an associative algebra R over an infinite field F. We prove that if R is unitarily generated by its nilpotent elements, then R^{×} satisfies a group identity precisely when R satisfies a non-matrix polynomial identity. As an application, we examine the group algebra F G of a torsion group G and the restricted enveloping algebra u(L) of a p-nil restricted Lie algebra L. Giambruno, Sehgal and Valenti recently proved that if the group of units (F G)^{×} satisfies a group identity then F G satisfies a polynomial identity, thus confirming a conjecture of Brian Hartley. We show that, in fact, (F G)^{×} satisfies a group identity if and only if F G satisfies a non-matrix polynomial identity. In the case of restricted enveloping algebras, we prove that u(L)^{×} satisfies a group identity if and only if u(L) satisfies the Engel condition.

1 Introduction

Let R be an associative unitary algebra over a field F of characteristic p ≥ 0. Recall that R is said to satisfy a polynomial identity whenever there exists a nontrivial element f(x_{1}, …, x_{m}) of the free F -algebra generated by {x_{1}, x_{2}, …} such that f(r_{1}, …, r_{m}) = 0 for all r_{i} ∈ R; whereas, the group of units R^{×} of R is said to satisfy a group identity if there exists a nontrivial word w(y_{1}, …, y_{m}) in the free group generated by {y_{1}, y_{2}, …} such that w(u_{1}, …, u_{m}) = 1 for all u_{i} ∈ R^{×}. There is ample evidence in the literature to suggest that there may be some general underlying relationship between group identities and polynomial identities. For example, Gupta and Levin proved that R^{×} is nilpotent whenever R is Lie nilpotent; Smirnov and Zalesskii established the fact that R^{×} is soluble whenever R is Lie soluble (if p ≠ 2); and Shalev proved that R^{×} satisfies the Engel condition whenever R does ([GL],[ZSm],[Sm],[Sh1]). In the other direction, it follows from a result of Valitskas ([V]) that a radical algebra over an infinite field is a PI-algebra whenever its adjoint group satisfies a group identity. In radical algebras the adjoint group R^{∘} of R with the group operation given by x ∘ y = x + y + xy plays the role of the unit group in a unitary algebra. Also along this vein, Giambruno, Sehgal and Valenti ([GSV]) recently confirmed a conjecture of Brian Hartley by proving that the group algebra F G of a torsion group G over an infinite field F satisfies a polynomial identity whenever (F G)^{×} satisfies a group identity. Subsequently, Passman ([Pa2]) gave necessary and sufficient conditions for (F G)^{×} to satisfy an identity (cf. Theorem 4.1).

Two questions thus seem immediately relevant:

1. Does R^{×} satisfy a group identity whenever R satisfies a PI?

2. Does R satisfy a PI whenever R^{×} satisfies a group identity?

However, it quickly becomes clear that these questions are posed too generally. Indeed, free algebras have only the trivial units corresponding to F^{×}; and M_{2}(F), the algebra of 2 × 2 matrices over F, satisfies the standard polynomial identity of degree 4 even though GL_{2}(F) contains a non-abelian free group whenever F contains a transcendental element. Thus it is natural to concentrate on algebras containing many units, and on polynomial identities not satisfied by matrices. A polynomial identity not satisfied by M_{2}(F) is called a non-matrix identity. It follows from well-known results that whenever R satisfies a non-matrix identity, then R^{×} satisfies a group identity. Namely, we have the following (we use brackets to denote Lie commutators and parentheses to denote group commutators):

Proposition 1.1 Let R be a unitary algebra over a field of characteristic p ≥ 0. Suppose that R satisfies a non-matrix identity. Then

1. R^{×} is soluble if p = 0; and,

2. R^{×} satisfies an identity of the form (y_{1}, y_{2})^{p^{t}} = 1 if p > 0.

In any case, R^{×} satisfies a group identity.

Proof. Suppose first that p = 0 and R satisfies a non-matrix identity. Then by a theorem of Kemer, [Ke], the ideal in R generated by the Lie commutators of the form [[a, b], [c, d], e] is nilpotent. So, in particular, R is Lie soluble, and hence by Smirnov and Zalesskii’s theorem mentioned above we find that R^{×} is soluble.

Now suppose that p > 0 and R satisfies a non-matrix identity f, say. Let A be the relatively-free algebra of rank 3 in the variety satisfying f. Then A is a finitely generated PI-algebra, so that by a theorem of Razmyslov and Braun, [Br], the Jacobson radical J(A) of A is nilpotent. Also, A/J(A) is a semiprimitive PI-algebra satisfying a polynomial identity not satisfied by M_{2}(F); consequently, A/J(A) is commutative. Therefore, A satisfies a polynomial identity of the form ([x_{1}, x_{2}]x_{3})^{p^{t}} = 0 for a suitable t. It follows that R also satisfies ([x_{1}, x_{2}]x_{3})^{p^{t}} = 0, and so R^{×} satisfies

(y_{1}, y_{2})^{p^{t}} − 1 = ((y_{1}, y_{2}) − 1)^{p^{t}} = (y_{1}y_{2}y_{1}^{−1}y_{2}^{−1} − 1)^{p^{t}} = ([y_{1}, y_{2}]y_{1}^{−1}y_{2}^{−1})^{p^{t}} = 0.

□

The primary goal of this paper is to demonstrate that the converse to Proposition 1.1 holds for the class of nil-generated algebras over an infinite field. Let N (R) denote the set of nilpotent elements in R. We say that a unitary algebra R is nil-generated if it is generated by {1} ∪ N (R). For example, M_{n}(F) is nil-generated, as is the group algebra of a group generated by p-elements.

Theorem 1.2 Let R be a nil-generated unitary algebra over an infinite field of characteristic p ≥ 0. If R^{×} satisfies a group identity, then N (R) forms a locally nilpotent ideal and R satisfies a non-matrix identity.

As a consequence of Proposition 1.1 and Theorem 1.2, we obtain the following characterisations:

Theorem 1.3 Let R be a nil-generated unitary algebra over a field of characteristic 0. Then the following statements are equivalent:

1. R^{×} satisfies a group identity;

2. R satisfies a non-matrix identity; 3. R is Lie soluble; and,

4. R^{×} is a soluble group.

Theorem 1.4 Let R be a nil-generated unitary algebra over an infinite field of characteristic p > 0. Then the following statements are equivalent:

1. R^{×} satisfies a group identity;

2. R satisfies a non-matrix identity;

3. R satisfies the polynomial identity ([x_{1}, x_{2}]x_{3})^{p^{t}} = 0 for some t; and,

4. R^{×} satisfies the group identity (y_{1}, y_{2})^{p^{t}} = 1 for some t.

The following corollary is an analogue of the classical theorem of Kaplansky ([Ka]), which states that every nil-algebra satisfying a PI is locally nilpotent.

Corollary 1.5 Let R be a nil-algebra over an infinite field. The adjoint group R^{∘} of R satisfies a group identity if and only if R satisfies a nonmatrix identity. In this case R is locally nilpotent.

Proof. It is well-known that R can be embedded into a unitary F -algebra R_{1} in such a way that {1} ∪ R generates R_{1} and R_{1}^{×} ≅ F^{×} × R^{∘}. Theorems 1.2-1.4 now imply the result. □

As a further application of Theorem 1.4, we are able to study group algebras F G where G is any torsion group, and restricted enveloping algebras u(L) where L is any p-nil restricted Lie algebra. In particular, we deduce that (F G)^{×} satisfies a group identity if and only if F G satisfies a non-matrix identity; whereas u(L)^{×} satisfies a group identity if and only if u(L) satisfies the Engel condition. These results are detailed in Sections 4 and 5.

2 Existence of a polynomial identity

The following result, Corollary 2.2 of [GSV] (cf. Proposition 1 of [GJV]), plays a crucial role in the proof of our Theorem 1.2.

Lemma 2.1 Let R be a semiprime algebra over an infinite commutative domain, such that its group of units R^{×} satisfies a group identity. Then for every nilpotent element a ∈ R, bc = 0 ⇒ bac = 0.

Let L(R) denote the Levitzki radical of the algebra R; that is, the unique maximal locally nilpotent ideal in R. Then R/L(R) is semiprime (see Section 10 of [L], for example) and L(R) is contained in N (R), the set of all nilpotent elements in R.

Lemma 2.2 Let R be a nil-generated unitary algebra over an infinite field. If R^{×} satisfies a group identity, then L(R) = N (R). Consequently, R = F · 1 + N (R) and every finite subset of N (R) generates nilpotent subalgebra in R.

Proof. Let R̄ = R/L(R). Because L(R) is contained in the Jacobson radical J(R) of R, R̄^{×} is a homomorphic image of R^{×} and, hence, satisfies the same group identity. We claim that N (R̄) = 0.

Let b̄, c̄ ∈ R̄ be such that b̄c̄ = 0. By induction, it follows that whenever r̄_{1}, …, r̄_{k} ∈ R̄ then b̄r̄_{1} … r̄_{k}c̄ = 0. Indeed, we know that (b̄r̄_{1} … r̄_{k−1})c̄ = 0 and hence Lemma 2.1 yields the result. Since R̄ is nil-generated, it follows that b̄R̄c̄ = 0.

To show R̄ has no nontrivial nilpotent elements, it suffices to show that the only square-zero element in R̄ is 0. Suppose then there exists x̄ ∈ R̄ with x̄^{2} = 0. The above argument yields x̄R̄x̄ = 0, so that R̄x̄R̄ is a nilpotent ideal. Hence x̄ = 0 and it follows that N (R̄) = 0, as claimed. This implies L(R) = N (R). □

Proposition 2.3 Let R be a nil-generated algebra over an infinite field. If R^{×} satisfies a group identity, then R satisfies polynomial identity.

Proof. Let w(y_{1}, …, y_{n}) = 1 be a group identity for R^{×}. Put k = F [t] and let A denote the completion of the free associative algebra k{x_{1}, …, x_{n}}. The free group F_{n} = ⟨y_{1}, …, y_{n}⟩ embeds into A by the well-known Magnus argument via the map φ induced by φ(y_{i}) = 1 + tx_{i} (cf. [MKS], Section 5.5). It follows easily that

1 ≠ φ(w(y_{1}, …, y_{n})) = 1 + ∑_{m=1}^{∞} t^{m}p_{m}(x_{1}, …, x_{n}),

where each p_{m}(x_{1}, …, x_{n}) is a homogeneous element of degree m in the free algebra F {x_{1}, …, x_{n}}. Not all these elements can be trivial, so assume p_{m_{0}}(x_{1}, …, x_{n}) ≠ 0 in F {x_{1}, …, x_{n}}.

Now consider S = ⟨r_{1}, r_{2}, …, r_{n}⟩, the subalgebra in R generated by arbitrary elements r_{1}, …, r_{n} in N (R). Then S is nilpotent by Lemma 2.2, and so for each λ ∈ F the map t → λ, x_{i} → r_{i} induces a well-defined epimorphism ψ from the augmentation ideal of A to S. Since w(y_{1}, …, y_{n}) = 1 is a group identity for R^{×}, upon application of ψ we obtain

1 = w(1 + λr_{1}, 1 + λr_{2}, …, 1 + λr_{n}) = 1 + ∑_{m=1}^{∞} λ^{m}p_{m}(r_{1}, …, r_{n})

for each λ ∈ F. (The sum is finite.) Now, using the fact that F is infinite, a routine Vandermonde matrix argument implies that each p_{m}(r_{1}, …, r_{n}) = 0. Thus p_{m_{0}}(x_{1}, …, x_{n}) = 0 is a polynomial identity for S, and hence for all of N (R). But R = F · 1 + N (R), so that p_{m_{0}}([x_{1}, x_{2}], …, [x_{2n−1}, x_{2n}]) is a nontrivial polynomial identity for R. □

3 Existence of a non-matrix identity

To complete the proof of Theorem 1.2, it remains to prove the following:

Proposition 3.1 Let R be a nil-generated algebra over an infinite field F. If R^{×} satisfies a group identity ω(y_{1}, …, y_{n}) = 1, then R satisfies a non-matrix identity.

Proof. Recall that R is a PI-algebra by Proposition 2.3. To establish the statement about the existence of a non-matrix identity is more involved and we shall require some reductions (cf. proof of Proposition 1 in [GJV]).

First let us point out that it is enough to show that N (R) satisfies a non-matrix identity as R is a commutative extension of N (R) by Lemma 2.2.

Next, using that the fact that the derived subgroup of a free group of rank 2 is free of countably-infinite rank, we may also assume that ω is a word in two variables only. Furthermore, the substitution y_{1} = y_{1}y_{2} and y_{2} = y_{2}y_{1} allows us to assume that R^{×} satisfies a group identity of the form

ω(y_{1}, y_{2}) = (y_{1}y_{2})^{α_{1}}(y_{2}y_{1})^{β_{1}} ⋯ (y_{1}y_{2})^{α_{j}}(y_{2}y_{1})^{β_{j}}(y_{1}y_{2})^{α_{j+1}} = 1,

where j ≥ 1 and the integers α_{i} and β_{i} are nonzero with the possible exception of α_{j+1}. The Magnus representation of ω now becomes

ω(1 + x_{1}, 1 + x_{2}) = 1 + ∑_{m=1}^{∞} p_{m}(x_{1}, x_{2}).

As is in the proof of Proposition 2.5, it follows that N (R) satisfies each of the polynomial identities p_{m}(x_{1}, x_{2}) = 0 (some of which may be trivial). In order to prove the proposition, it suffices for us to show that at least one of the p_{m} is not also satisfied by M_{2}(F). Let us suppose then to the contrary. It follows from the Magnus representation of ω that ω(1 + a, 1 + b) = 1 for each choice of nilpotent a, b in M_{2}(F). Notice as well that a^{2} = 0 implies that (1 + a)^{n} = 1 + na, for each integer n. It is easy to see that the reduced form of ω is of the type

ω(y_{1}, y_{2}) = y_{1}^{γ_{1}}y_{2}^{δ_{1}} ⋯ y_{1}^{γ_{k}}y_{2}^{δ_{k}}y_{1}^{γ_{k+1}},

where k ≥ 1 and the integers γ_{i} and δ_{i} are one of 1, −1, 2 or −2 with the possible exception of γ_{k+1}, which is one of 1, −1 or 0. Now fix two square-zero elements a, b ∈ M_{2}(F). Then for every λ ∈ F we have

ω(1 + λa, 1 + λb) = (1 + γ_{1}λa)(1 + δ_{1}λb) ⋯ (1 + γ_{k}λa)(1 + δ_{k}λb)(1 + γ_{k+1}λa).

But we also have

ω(1 + λa, 1 + λb) = 1 + ∑_{m=1}^{l} λ^{m}p_{m}(a, b),

where l = 2k+1 unless γ_{k+1} = 0, in which case l = 2k. Comparing coefficients of λ^{l} we find that

(γ_{1}γ_{2} ⋯ γ_{k})(δ_{1}δ_{2} ⋯ δ_{k})(ab)^{k+1} = 0.

In the case of characteristic p ≠ 2, this yields (ab)^{k+1} = 0, which leads to the desired contradiction; for example, set a = e_{12} and b = e_{21}.

It remains to consider the case of characteristic p = 2. Making the substitution y_{1} = y_{1}y_{2} and y_{2} = y_{1}y_{3} into the reduced form of ω above allows us to assume that R^{×} satisfies the following word:

ω_{2}(y_{1}, y_{2}, y_{3}) = (y_{1}y_{2})^{γ_{1}}(y_{1}y_{3})^{δ_{1}} ⋯ (y_{1}y_{2})^{γ_{k}}(y_{1}y_{3})^{δ_{k}}(y_{1}y_{2})^{γ_{k+1}}.

Let us represent ω_{2} by

ω_{2}(1 + x_{1}, 1 + x_{2}, 1 + x_{3}) = 1 + ∑_{m=1}^{∞} q_{m}(x_{1}, x_{2}, x_{3}).

As argued above, it suffices to show that M_{2}(F) does not satisfy some q_{m}. Therefore, let us suppose otherwise and fix two square-zero elements a, b ∈ M_{2}(F). Using the fact that the characteristic is 2, it is easy to check that a + ba + ab + bab also has square zero. Evaluating the Magnus representation tells us

ω_{2}(1 + λa, 1 + λb, 1 + λ(a + ba + ab + bab)) = 1.

Notice that in the reduced form of ω_{2}(y_{1}, y_{2}, y_{3}) the variables appear with exponents 1 or −1 only. Then because (1 + a)^{−1} = 1 − a = 1 + a, etc., it follows that ω_{2}(1 + λa, 1 + λb, 1 + λ(a + ba + ab + bab)) is merely an ordered product of the terms 1 + λa, 1 + λb, and 1 + λ(a + ba + ab + bab) in which no two consecutive terms are equal. The triviality of the coefficient of the highest power of λ appearing in the resulting expansion leads to the fact that (ab)^{t} = 0 for some suitable t > 0. This in turn gives the desired contradiction. □

4 Group algebras

Let us now consider group algebras F G of a torsion group over an infinite field F of prime characteristic. In [GSV] it was shown that whenever (F G)^{×} satisfies a group identity, then F G satisfies a polynomial identity. Passman ([Pa2]) subsequently characterised all torsion groups G such that (F G)^{×} satisfies a group identity. We are able to extend these results as follows:

Theorem 4.1 Let F G be a group algebra of a torsion group over an infinite field F of characteristic p > 0. Then the following are equivalent:

1. (F G)^{×} satisfies a group identity;

2. F G satisfies a non-matrix identity;

3. G contains a normal subgroup A such that G/A and (A, A) are finite, and (G, G) is a p-group of finite exponent;

4. [F G, F G]F G is nil of bounded index; and,

5. ((F G)^{×}, (F G)^{×}) is a p-group of finite exponent.

Proof. Assume that (1) holds; we shall deduce (2). Let X be the set of pelements in G and write P for the subgroup of G generated by X. Then the group algebra F P is nil-generated and (F P)^{×} satisfies a group identity. From Lemma 2.2 it follows that N (F P) is a locally nilpotent maximal ideal in F P. Therefore N (F P) coincides with the augmentation ideal of F P and P is a locally finite p-group. Using the normality of P in G, it follows that the ideal N (F P)F G in F G is also locally nilpotent. Now, according to Corollary 1.5, N (F P)F G satisfies a non-matrix identity as the adjoint group of N (F P)F G satisfies the identities of (F G)^{×}. Also, because the kernel of the canonical projection F G → F (G/P) is N (F P)F G, it follows that (F (G/P))^{×} satisfies a group identity. Since G/P is a p′-group, it follows that G/P is abelian as is shown in the semiprime case of [GSV]. Now F G satisfies some non-matrix identity by the fact that it is a commutative extension of an algebra satisfying a non-matrix identity.

The equivalence (1) ⇔ (2) now follows from Proposition 1.1. (1) ⇒ (3) is the statement of Lemma 2.4 in [Pa2]. The implication (3) ⇒ (4) follows as in the proof Lemma 3.3 in [Pa2]. The remaining implications, (4) ⇒ (5) and (5) ⇒ (1), are clear. □

5 Restricted enveloping algebras

Let u(L) be the restricted enveloping algebra of a restricted Lie algebra L over a field F of characteristic p > 0. Restricted enveloping algebras satisfying a polynomial identity were characterised independently by Passman and Petrogradski in [Pa1] and [Pe]: see Lemma 5.3 below. A restricted Lie algebra L is said to be p-nil if for every x ∈ L there exists a natural number n such that x^{p^{n}} = 0. We are interested here in characterising p-nil restricted Lie algebras L for which u(L)^{×} satisfies a group identity. It follows from Jacobson’s restricted analogue of the Poincaré-Birkhoff-Witt Theorem (see [J]) that for such an L, u(L) is nil-generated. Therefore, according to Theorem 1.4, u(L) satisfies a non-matrix identity precisely when u(L)^{×} satisfies a group identity. More specifically, we have:

Theorem 5.1 If L is a p-nil restricted Lie algebra over an infinite field of characteristic p > 0, then the following statements are equivalent:

1. u(L)^{×} satisfies a group identity;

2. u(L) satisfies a non-matrix identity;

3. [L, L] is bounded p-nil and L contains a restricted ideal A of such that L/A and [A, A] are finite-dimensional;

4. u(L) satisfies the Engel condition; and,

5. u(L)^{×} satisfies an identity of the form (y_{1}^{p^{t}}, y_{2}) = 1 for some t.

In fact, in Theorem 5.1 we need only assume that L can be generated by p-nil elements.

Corollary 5.2 Let L be a virtually-(p-nil) restricted Lie algebra over an infinite field. If u(L)^{×} satisfies a group identity, then u(L) satisfies a polynomial identity.

Observe that some precondition on L is required in Corollary 5.2; indeed, the restricted enveloping algebra of a free restricted Lie algebra is a free associative algebra, and so has only the trivial unit group F^{×}.

To prove Theorem 5.1, we shall make use of the result of Passman and Petrogradski mentioned above:

Lemma 5.3 Let L be a restricted Lie algebra. Then its restricted enveloping algebra u(L) satisfies a polynomial identity if and only if L possesses a restricted ideal (or subalgebra) A such that

1. A has finite codimension in L, and

2. [A, A] is finite dimensional and p-nil.

We shall also require Theorem 1.2 of [RS].

Lemma 5.4 Let L be a restricted Lie algebra. Then its restricted enveloping algebra u(L) satisfies the Engel condition if and only if

1. L is nilpotent,

2. [L, L] is bounded p-nil, and

3. L possesses a restricted ideal A such that L/A and [A, A] are finite dimensional.

Only implications (2) ⇒ (3) and (3) ⇒ (4) in Theorem 5.1 do not follow directly from Theorem 1.4. To prove (3) ⇒ (4), assume that (3) holds. Observe that the centraliser C of [A, A] in A is of finite codimension in A, and in hence of finite codimension in L. Thus we may replace A by C, to assume that A is nilpotent of class 2. Now L is nilpotent-by-(finite-dimensional and p-nil). It follows from a result of Shalev ([Sh2], Proposition 5.1) that L is nilpotent. Now Lemma 5.4 yields the fact that u(L) satisfies the Engel condition. It remains then to prove the following lemma.

Lemma 5.5 If L is a p-nil restricted Lie algebra such that u(L) satisfies a non-matrix identity, then [L, L] is bounded p-nil and L contains a restricted ideal A such that L/A and [A, A] are finite dimensional.

Proof. The existence of A follows immediately from Lemma 5.3. From Theorem 1.4, there exists some t such that u(L) satisfies an identity of the form ([x_{1}, x_{2}]x_{3})^{p^{t}} = 0. As argued above, L must be nilpotent. It remains to prove that any linear combination of commutators in L is p-nil of bounded index.

Claim 5.6 For a sufficiently large integer k, L satisfies the identity

(x + y)^{p^{k}} = x^{p^{k}} + y^{p^{k}}.

Proof. Let c be the nilpotency class of L, and choose k large enough that p^{k} ≥ p^{t}c. Consider λ ∈ F and expand (x + λy)^{p^{k}} to get:

(x + λy)^{p^{k}} − x^{p^{k}} − λ^{p^{k}}y^{p^{k}} = ∑_{i≥1} λ^{i}h_{i}(x, y),

where each h_{i}(x, y) in L is homogeneous in x, y of total degree p^{k}. Each h_{i} is a sum of elements of the form

[r_{1}^{p^{α_{1}}}, …, r_{l}^{p^{α_{l}}}]^{p^{α}},

where p^{α}(∑_{j} p^{α_{j}}) = p^{k} and r_{j} ∈ {x, y}. If α ≥ t then this restricted Lie monomial is zero. On the other hand, if α < t, then

∑_{j} p^{α_{j}} = p^{k−α} ≥ p^{k−t+1} ≥ pc ≥ c + 1.

Therefore

[r_{1}^{p^{α_{1}}}, …, r_{l}^{p^{α_{l}}}] = [r_{1}^{p^{α_{1}}}, \underbrace{r_{2}, …, r_{2}}_{p^{α_{2}}}, …, \underbrace{r_{l}, …, r_{l}}_{p^{α_{l}}}]
= [−r_{2}, \underbrace{r_{1}, …, r_{1}}_{p^{α_{1}}}, \underbrace{r_{2}, …, r_{2}}_{p^{α_{2}}−1}, …, \underbrace{r_{l}, …, r_{l}}_{p^{α_{l}}}]
= 0,

being a commutator of length greater than c. □

To finish the proof of Lemma 5.5, let r_{i}, s_{i} be arbitrary elements in L. Taking p^{k} ≥ p^{t}c as in the claim we have

(∑_{i} β_{i}[r_{i}, s_{i}])^{p^{k}} = ∑_{i} β_{i}^{p^{k}}[r_{i}, s_{i}]^{p^{k}} = 0,

as required. □

Corollary 5.2 follows by combining Theorem 5.1 with Lemma 5.3. □

6 Concluding remark

Let us close by observing that not every nil PI-algebra satisfies a non-matrix identity. In light of our results, this is equivalent to the fact that the adjoint group of a nil PI-algebra need not satisfy a group identity.

Proposition 6.1 Let F be an infinite field of characteristic p ≥ 0. Then there exists a locally nilpotent associative algebra R over F such that R satisfies a polynomial identity, and yet its adjoint group R^{∘} does not satisfy any group identity.

Proof. For the case of p > 0, consider the restricted Lie algebra L generated by the set {x,y_{1},y_{2},…,z_{1},z_{2},…}, subject to the relations: [x, y_{i}] = z_{i} is central, [y_{i}, y_{j}] = 0, and x^{p} = y_{i}^{p} = z_{i}^{p^{i}} = 0, for all i, j ≥ 1. Then the ideal of L generated by {y_{1}, y_{2}, …} is abelian and of codimension 1 in L. Hence, R = L(u(L)) satisfies a polynomial identity by Lemma 5.3. Furthermore, L is locally-(finite-dimensional and p-nil), so that R is locally nilpotent (see Lemma 2.4 of [RS], for example). However, R^{∘} does not satisfy any group identity for otherwise, by Theorem 5.1, [L, L] would be bounded p-nil.

Now suppose that p = 0. Consider the exterior algebra E of an infinite dimensional F -space. Then E satisfies a PI: it is Lie nilpotent of class 2. Therefore, R = E ⨂_{F} E ⨂_{F} E also satisfies a PI by a theorem of Regev, [Re]. Moreover, R is locally nilpotent since E is locally nilpotent. It was shown in [R], however, that R satisfies no non-matrix identity and so, by Theorem 1.3, R^{∘} cannot satisfy a group identity. □

Notice that by Kaplansky’s theorem any nil example to this effect cannot be finitely generated.

Acknowledgment: The authors are grateful to the referee for pointing out some errors and obfuscations in the original manuscript. The research of the first and the third authors was supported by NSERC of Canada; the second author received support from NSF-EPSCoR in Alabama and the CNR in Italy.

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Address of the first and the third author:

Department of Mathematics and Statistics University of New Brunswick Frederiction, NB Canada E3B 5A3

Address of the second author:

Department of Mathematics The University of Alabama Tuscaloosa, AL 35487-0350 U.S.A.