On the Lie Structure of the Skew Elements of a Prime

Superalgebra with Superinvolution

Carlos Gómez-Ambrosi

Department of Mathematics, University of Zaragoza, 50009 Zaragoza, Spain

Jesús Laliena Department of Mathematics and Computation, University of La Rioja, 26004 Logroño, Spain

Ivan P. Shestakov

Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia

Abstract: We investigate the Lie structure of the Lie superalgebra K of skew elements of a prime associative superalgebra A with superinvolution. It is proved that if A is not a central order in a Clifford superalgebra of dimension at most 16 over the center then any Lie ideal of K or [K, K] contains [J ∩ K, K] for some nonzero ideal J of A or is contained in the even part of the center of A.

The first and second authors have been supported by the Spanish DGES (PB 97-1291-C03-03 and 02), and the second also by the University of La Rioja (API-98/B15). The third author has been partly supported by the Russian Ministry of General and Professional Education (Grant for Fundamental Researches in Mathematics).

1991 Mathematics Subject Classification: 17A70, 17B70.

1. Introduction.

The Lie structure of the skew elements of a simple associative superalgebra with superinvolution has been investigated by Gómez-Ambrosi and Shestakov in [2]: if A is a simple associative nontrivial superalgebra with superinvolution over a field of characteristic not 2 and U is a Lie ideal of K, then either U ⊆ Z, the even part of the center of A, or U ⊇ [K, K], unless A is a Clifford superalgebra of dimension 4 or 16 over Z (Theorem 5.7 of [2]). Moreover, if U is a proper Lie ideal of [K, K], then U ⊆ Z, unless A is a Clifford superalgebra of dimension 4, 8 or 16 over Z (Theorem 6.4 of [2]).

In this paper, we extend these results to prime associative superalgebras A with superinvolution. In this case, K and [K, K] have obvious Lie ideals of the kind [J ∩ K, K] for a nonzero ideal J of A. We show that these ideals are always nonzero if A is nontrivial, that is, with a nonzero odd part. Moreover, up to certain exceptional cases, all noncentral Lie ideals of K and of [K, K] are of this kind. Specifically, we prove

Theorem A. Let A be a prime associative nontrivial superalgebra with superinvolution. If U is a Lie ideal of K, then either U ⊆ Z or U ⊇ [J ∩ K, K] for some nonzero ideal J of A, except if A is a central order in a Clifford superalgebra of dimension 4 or 16.

Theorem B. Let A be a prime associative nontrivial superalgebra with superinvolution. If U is a Lie ideal of [K, K], then either U ⊆ Z or U ⊇ [J ∩ K, K] for some nonzero ideal J of A, except if A is a central order in a Clifford superalgebra of dimension 4, 8 or 16.

Classically, the situation has been studied in the context of prime rings with involution by Erickson [1], who followed the path initiated by Herstein in the early 1950s (see [4]). However, our approach differs from that of Erickson, which makes use of Martindale’s results for rings satisfying generalized polynomial identities, and simply modifies the arguments in [2] to the more general case which is considered here.

Observe that our results do not generalize but just complement Erickson’s results since we consider only nontrivial superalgebras which do not include ordinary (even) algebras.

Throughout the paper, if otherwise is not stating, A = A_{0} + A_{1} will denote a nontrivial prime associative superalgebra with superinvolution ∗ over a commutative unital ring Φ of scalars with \frac{1}{2} ∈ Φ. The reader is referred to [2] and [3] for a complete introduction to the basic definitions and examples in this context.

Z will denote the even part of the center of A, H the Jordan superalgebra of symmetric elements of A, and K the Lie superalgebra of skew elements of A. If P is some subset of A, we will write P_{H} = P ∩ H and P_{K} = P ∩ K.

Remind that the superinvolution ∗ is said to be of the first kind if Z_{H} = Z, and it is said to be of the second kind if Z_{H} ≠ Z.

In case Z ≠ 0, we can consider the localization Z^{−1}A = {z^{−1}a | 0 ≠ z ∈ Z, a ∈ A}. It is a central prime associative superalgebra over the field Z^{−1}Z, which we call the central closure of A (and we say that A is a central order in Z^{−1}A). Notice that Z(Z^{−1}A)_{0} = Z^{−1}Z. In particular, we say that A is a central order in C(n) if Z ≠ 0 and Z^{−1}A is isomorphic to the Clifford superalgebra of a nondegenerate quadratic space over Z^{−1}Z (see Example 1.5 of [2]).

It will be more convenient for us to represent the central closure Z^{−1}A in the form V^{−1}A for the subset V = Z_{H} −{0} of regular symmetric elements. (Notice that Z_{H} ≠ 0 always when Z ≠ 0.) Namely, for any 0 ≠ z ∈ Z, a ∈ A we have z^{−1}a = (zz^{∗})^{−1}(z^{∗}a), hence Z^{−1}A = V^{−1}A. We may consider V^{−1}A as a superalgebra over the field V^{−1}Z_{H}; then the superinvolution on A extends to one of the same kind on V^{−1}A over V^{−1}Z_{H} via (v^{−1}a)^{∗} = v^{−1}a^{∗}, and it is easy to check that H(V^{−1}A, ∗) = V^{−1}H and K(V^{−1}A, ∗) = V^{−1}K. Moreover, Z(V^{−1}A)_{0} = V^{−1}Z and V^{−1}Z ∩ V^{−1}H = V^{−1}Z_{H}. We will call the superalgebra V^{−1}A (considered over the field V^{−1}Z_{H}) a ∗-central closure of A.

2. Density of K in A.

Let M be a subsupermodule of A. Denote by \overline{M} the subalgebra of A generated by M. We will say that M is dense in A if \overline{M} contains a nonzero ideal of A. If we want to specify that \overline{M} contains a nonzero ideal J of A, we will say that M is J-dense in A. We prove in this section that, with some exceptions, K is dense in A.

The following results will be used several times throughout the paper, so we give them for convenience of the reader.

Lemma 2.1 [5, Lemma 1.1.9]. If A is a semiprime algebra and [a, [a, A]] = 0 then a ∈ Z(A).

Lemma 2.2 [8, Lemmas 1.2, 1.3]. If A = A_{0} + A_{1} is a prime superalgebra, then A and A_{0} are semiprime and either A is prime or A_{0} is prime (as algebras).

Lemma 2.3 [8, Lemma 1.8]. Let A = A_{0} + A_{1} be a prime superalgebra. Then (i) If x_{1} ∈ A_{1} centralizes a nonzero ideal I of A_{0}, then x_{1} ∈ Z(A). (ii) If x_{1}^{2} belongs to the center of a nonzero ideal I of A_{0}, then x_{1}^{2} ∈ Z(A).

Let B be an associative superalgebra and M, S be subsupermodules of B. Define (M : S) = {a ∈ B | aS ⊆ M}. Notice that if M (resp. S) is a left ideal of B, then (M : S) is a left (resp. right) ideal of B. In particular, if M and S are ideals then (M : S) is also an ideal. Moreover, if B is prime then for any nonzero ideal I of B it holds (0 : I) = 0.

Lemma 2.4. Let M and S be subsupermodules of B such that [M, S] ⊆ S. Then (i) For any right ideal I of B hold the inclusions

(I : S)M ⊆ (I : S), M(S : I) ⊆ (S : I).

(ii) If B is prime and M is dense in B, then S ≠ (0) implies (0 : S) = 0. (iii) Under the same conditions as in (ii), if (\overline{S} : I) ≠ 0 for some ideal I ≠ 0 then S

is dense in B. (iv) If B is semiprime, M = B, and (\overline{S} : B) ≠ 0, then S is dense in B.

Proof. (i) Let s ∈ S, a ∈ (I : S), m ∈ M, then am · s = a[m, s] + (−1)^{\overline{m}\overline{s}}asm ∈ aS + Im ⊆ I. Thus, (I : S)M ⊆ (I : S). Furthermore, let b ∈ (S : I), i ∈ I, then mb · i = m · bi = (−1)^{\overline{m}(\overline{b}+\overline{i})}bi · m + [m, bi] ∈ bI + [m, S] ⊆ S, which proves the second inclusion in (i).

(ii) It is clear that (0 : S) is a left ideal of B. On the other hand, it follows from (i) that (0 : S) · \overline{M} ⊆ (0 : S). Since M is dense in B, there exists an ideal J ≠ 0 of B which lies in \overline{M}. In particular, (0 : S)J ⊆ (0 : S)\overline{M} ⊆ (0 : S). Let L = (0 : S)J, then L is an ideal of B and LS = 0, hence L = 0 and (0 : S) = 0, since J ≠ 0.

(iii) Notice that [\overline{S}, M] ⊆ \overline{S}, therefore, by (i) we have M(\overline{S} : I) ⊆ (\overline{S} : I). Let J ≠ 0 be the same as in (ii), then J(\overline{S} : I) ⊆ \overline{M}(\overline{S} : I) ⊆ (\overline{S} : I). It is clear that (\overline{S} : I) is a right ideal, hence L = J(\overline{S} : I)I is an ideal of B contained in \overline{S}. Since I, J, (\overline{S} : I) ≠ 0, it follows easily that L ≠ 0 and so \overline{S} is dense in B.

(iv) In this case, as in (iii), we have an ideal L = B(\overline{S} : B)B ⊆ \overline{S} which is clearly nonzero.

Corollary 1. Let B be prime and x ∈ B_{1} such that [x, B_{1}] = 0. Then x = 0.

Proof. Assume that x ≠ 0 and set S = Φx, M = B_{1}, then [S, M] = 0. It is easy to see that I = B_{1} + B_{1}B_{1} is an ideal of B which is nonzero since B is nontrivial. Thus M = B_{1} is dense in B. Furthermore, 2x^{2} = [x, x] = 0, hence xS = 0 and by Lemma 2.4.(ii) x = 0, a contradiction.

Corollary 2. Let B be semiprime and L be a Lie ideal of B. Then either [L, L] = 0 or L is dense in B.

Proof. Assume that [L, L] ≠ 0, and let u, v ∈ L such that [u, v] ≠ 0. For any a ∈ B, [u, va] is in L, that is [u, v]a + (−1)^{\overline{u} \overline{v}}v[u, a] is in L. Observe that v[u, a] ∈ LL ⊆ \overline{L}, therefore [u, v]B ⊆ \overline{L} and hence 0 ≠ [L, L] ⊆ (\overline{L} : B). By Lemma 2.4.(iv) applied to S = L, we have that L is dense in B.

Theorem 2.1. Let A be a prime nontrivial associative superalgebra. If L is a Lie ideal of A then either L ⊆ Z or L is dense in A, except if A is a central order in C(2).

Proof. If [L, L] ≠ 0 then L is dense in A by corollary 2 of Lemma 2.4. Suppose that [L, L] = 0. For any a ∈ L_{0} holds [a, [a, A]] ⊆ [L, [L, A]] ⊆ [L, L] = 0, hence a ∈ Z by Lemma 2.1 and L_{0} ⊆ Z. Furthermore, by the Jacobi superidentity, [L_{1}, [A_{1}, A_{1}]] ⊆ [[L_{1}, A_{1}], A_{1}] ⊆ [L_{0}, A_{1}] = 0. The subspace [A_{1}, A_{1}] is a Lie ideal of the semiprime algebra A_{0}, so again by corollary 2 of Lemma 2.4 (the proof of this lemma never used the condition A_{1} ≠ 0) either [[A_{1}, A_{1}], [A_{1}, A_{1}]] = 0 or [A_{1}, A_{1}] is dense in A_{0}.

Consider first the second case when there exists an ideal I ≠ 0 of A_{0} which lies in \overline{[A_{1}, A_{1}]}. In this case [L_{1}, I] = 0, hence by Lemma 2.3,(i) L_{1} ⊆ Z(A). For any x ∈ L_{1} we have 2x^{2} = [x, x] ∈ [L, L] = 0, hence x = 0 by the semiprimeness of A and L_{1} = 0. Thus, L ⊆ Z.

Notice that the same arguments prove that L ⊆ Z if A is commutative since in this case also L_{1} ⊆ Z(A) = A. So we may assume that A is not commutative.

Let now [[A_{1}, A_{1}], [A_{1}, A_{1}]] = 0. In this case for any a ∈ [A_{1}, A_{1}] we have [a, [a, A_{0}]] = 0 since [[A_{1}, A_{1}], A_{0}] ⊆ [A_{1}, A_{1}]. But A_{0} is semiprime, hence a ∈ Z(A_{0}) by Lemma 2.1 and [[A_{1}, A_{1}], A_{0}] = 0. For any x ∈ A_{1} it holds 2x^{2} = [x, x] ∈ Z(A_{0}), therefore, by Lemma 2.3,(ii), x^{2} ∈ Z. Since the product [x, y] is symmetric in A_{1}, it follows that [x, y] ∈ Z for any x, y ∈ A_{1}. If [A_{1}, A_{1}] = 0 then by corollary 1 of Lemma 2.4 A_{1} = 0, a contradiction. Therefore, Z ⊇ [A_{1}, A_{1}] ≠ 0, and we may consider the central closure Z^{−1}A, which is a central noncommutative prime superalgebra over the field F = Z^{−1}Z. We will keep for this superalgebra the same notation A.

The mapping (x, y) ↦ [x, y] is a symmetric bilinear form on A_{1}, which is nondegenerate by corollary 1 of Lemma 2.4. Choose in A_{1} an orthogonal base v_{1}, v_{2}, … with respect to this form, then [v_{i}, v_{j}] = 0 if i ≠ j and v_{i}^{2} = 1/2[v_{i}, v_{i}] = α_{i} ≠ 0 ∈ F. For any a ∈ A_{0} it holds v_{1}[a, v_{2}] = [v_{1}a, v_{2}] − [v_{1}, v_{2}]a = [v_{1}a, v_{2}] = β ∈ F, hence v_{1}[a, v_{2}] = βα_{1}^{−1}v_{1}^{2}, and since v_{1} is invertible, [a, v_{2}] ∈ F v_{1}. Now, for any i ≠ 2, F v_{1} ∋ [v_{i}v_{2}, v_{2}] = v_{i}[v_{2}, v_{2}] − [v_{i}, v_{2}]v_{2} = v_{i}[v_{2}, v_{2}] = 2α_{2}v_{i}, hence i = 1 and dim_{F} A_{1} ≤ 2.

If A_{1} = F x, then F x + F x^{2} = A_{1} + A_{1}^{2} is an ideal of A which contains an invertible element x, hence A = F x^{2} + F x and A is commutative, a contradiction.

If A_{1} = F v_{1} + F v_{2}, then again A_{0} = A_{1}^{2}, and since v_{i}^{2} ∈ F and [v_{1}, v_{2}] = 0, we have A_{0} = F + F (v_{1}v_{2}). If we had also v_{1}v_{2} = β ∈ F, then we would have α_{1}v_{2} = v_{1}v_{1}v_{2} = βv_{1}, a contradiction. Therefore, dim_{F} A_{0} = 2 and it could be easily seen that A is C(2).

Observe that the algebra C(2) may contain Lie ideals which are not dense and do not lie in the center; for instance, if α_{1} = −α_{2}, then L = F + F (v_{1} + v_{2}) is such a Lie ideal.

Theorem 2.2. Let A be a prime nontrivial associative superalgebra with a superinvolution ∗. Then K is dense in A, except if A is a central order in C(2).

Proof. We assume that A is not a central order in C(2). By [2, Lemma 4.1], K^{2} is a Lie ideal of A, hence, by Theorem 2.1, either K^{2} is dense in A or K^{2} ⊆ Z. In the first case it is obvious that K is dense in A too and we are finished.

So, suppose that K^{2} ⊆ Z. If K^{2} = 0, then for any k ∈ K, h ∈ H we have kKk = 0, khk = (k ∘ h)k − (−1)^{\overline{h}\overline{k}}hkk = (k ∘ h)k ∈ KK = 0. Thus kAk = 0, and since A is semiprime, k = 0. Therefore, A = H and [A, A] = [H, H] ⊆ K = 0, that is, A is supercommutative. Now for any a ∈ A_{1}, x ∈ A we have 2a^{2} = [a, a] = 0, axa = (−1)\overline{x}xaa = 0. Thus, aAa = 0 and a = 0, again by the semiprimeness of A. Therefore, A_{1} = 0, a contradiction.

Thus, 0 ≠ K^{2} ⊆ Z, and we may consider the ∗-central closure V^{−1}A, which is a prime superalgebra over the field F = V^{−1}Z_{H}. We keep for this superalgebra the same notation A. Observe that K_{1}K_{0} + K_{0}K_{1} ⊆ Z ∩ A_{1} = 0. Assume that K_{i}^{2} ≠ 0 for certain i ∈ {0, 1} and let k, s ∈ K_{i} such that 0 ≠ ks ∈ Z. Then k is invertible, and since kK_{1−i} = 0, we have K_{1−i} = 0. Moreover, for any r ∈ K_{i} it holds (ks)r = k(sr) ∈ Zk, that is, K = K_{i} ⊆ Zk. If k were odd, then we would have 2k^{2} = [k, k] ∈ K_{0} = 0, a contradiction (k is invertible). Therefore, K = K_{0} ⊆ Zk with 0 ≠ k^{2} ∈ Z ∩ H = F.

Assume first that the superinvolution ∗ is of the second kind, that is, Z∩K ≠ 0. Let 0 ≠ j ∈ K ∩ Z; then j^{2} ∈ F = Z ∩ H, and it could be easily seen that K = F j, H = Kj = F. In particular, in this case A_{1} = 0, a contradiction.

Let now ∗ be of the first kind, then Z = F and K = K_{0} = F k with 0 ≠ k^{2} ∈ F. We have [H_{0}, A_{1}] = [H_{0}, H_{1}] ⊆ K_{1} = 0, which implies [H_{0}, I] = 0 for the ideal I = A_{1} + A_{1}^{2}. Furthermore, [H_{0}, A]I ⊆ [H_{0}, AI] + A[H_{0}, I] = 0, which implies, by the primeness of A, that [H_{0}, A] = 0 and H_{0} ⊆ Z(A)_{0} = F. In particular, A_{0} = F + F k is commutative, and for any x ∈ A_{1} we have by Lemma 2.3,(ii) that x^{2} ∈ F. On the other hand, 2x^{2} = [x, x] ∈ [H_{1}, H_{1}] ⊆ K_{0} = F k. Therefore, x^{2} = 0 for any x ∈ A_{1}. Linearizing, we have [x, y] = 0 for any x, y ∈ A_{1}. Now, xyx = −yxx = 0, xkx = (x ∘ k)x − kxx = (x ∘ k)x. But x ∘ k ∈ H_{1} ∘ K_{0} ⊆ K_{1} = 0, hence xAx = 0 and x = 0 by the semiprimeness of A. Thus A_{1} = 0, a contradiction.

3. Lie structure of K.

Recall that A denotes a prime nontrivial associative superalgebra with a superinvolution ∗.

We first prove that the Lie ideals of K of the kind [I ∩K, K] are always nonzero for nonzero ideals I of A.

Proposition 3.1. For any nonzero ideal I of A the set [I ∩ K, K] is a nonzero Lie ideal of K, unless A is a central order in C(2).

Proof. Prove first that I_{K} = I ∩ K ≠ 0. Considering the ideal II^{∗} if necessary, we may assume without loss of generality that I is a ∗-ideal, that is, I^{∗} = I. Assume that I_{K} = 0, then for any i ∈ I we have i − i^{∗} ∈ I_{K} = 0, hence I ⊆ H. Therefore, [I, I] ⊆ I ∩ K = 0. Now [A, I]I ⊆ [AI, I] + A[I, I] = 0, hence [A, I] = 0 by the primeness of A. Furthermore, [A, A]I ⊆ [AI, A] + A[I, A] = 0 and [A, A] = 0. Now by Corollary 1 of Lemma 2.4 we have A_{1} = 0, a contradiction.

Assume now that [I_{K}, K] = 0. Then [I_{K}, \overline{K}] = 0 and, by Theorem 2.2, there exists a nonzero ideal J of A such that [I_{K}, J] = 0. Now again [A, I_{K}]J ⊆ [AJ, I_{K}]+ A[J, I_{K}] = 0, hence [A, I_{K}] = 0. By Corollary 1 of Lemma 2.4, (I_{K})_{1} = 0 and I_{K} ⊆ Z. Consider the ∗-central closure V^{−1}A of A, then evidently V^{−1}I = V^{−1}A and so V^{−1}I_{K} = V^{−1}K. This implies that [A, K] = 0 and [A, J] = 0, which is impossible by the above.

We will need later the following fact obtained during the proof of the Proposition.

Corollary. For any nonzero ideal I of A we have [I, I] ≠ 0.

Lemma 3.1. Let B be a subsupermodule of A which is J-dense in A. Then [J, A] ⊆ [B, A].

Proof. It is straightforward to check, by induction on n, that if a ∈ A and b_{1}, …, b_{n} ∈ B (n ≥ 1), then

[a, b_{1} … b_{n}] = − ∑_{i=1}^{n} ±[b_{i}, b_{i+1} … b_{n}ab_{1} … b_{i−1}].

We deduce from this that [B^{n}, A] ⊆ [B, A] for all n ≥ 1. Consequently, [J, A] ⊆ [\overline{B}, A] ⊆ [B, A].

Lemma 3.2. Let U be a subsupermodule of K which is J-dense in A. Then [J_{K}, K] ⊆ [U, K]. In particular, if U is a Lie ideal of K which is J-dense in A, then [J_{K}, K] ⊆ U.

Proof. By Lemma 3.1, [J_{K}, K] ⊆ [J, A] ⊆ [U, A], and then note that [U, K] is the skew part of [U, A] = [U, H] + [U, K].

Lemma 3.3. Suppose that K is dense in A, and let U be a Lie ideal of K. Then either U is dense in A or [u ∘ v, w] = 0 for all u, v, w ∈ U.

Proof. Let u, v ∈ U, h ∈ H, k ∈ K. By Lemma 5.1 of [2], [u ∘ v, h] ∈ U. On the other hand,

[u ∘ v, k] = u ∘ [v, k] + (−1)^{\overline{k} \overline{v}}[u, k] ∘ v ∈ \overline{U},

since [u, k], [v, k] ∈ [U, K] ⊆ U. Since A = H + K, this proves that [u ∘ v, A] ⊆ \overline{U}. Now let w ∈ U, a ∈ A. By the above, [u ∘ v, wa] ∈ \overline{U}. But

[u ∘ v, wa] = [u ∘ v, w]a + (−1)^{(\overline{u}+\overline{v})\overline{w}}w[u ∘ v, a],

and clearly w[u∘v, a] ∈ \overline{U} since [u∘v, a] ∈ \overline{U} and w ∈ U. Consequently, [u∘v, w]a ∈ \overline{U}; that is, [u ∘ v, w] ∈ (\overline{U} : A).

Now applying Lemma 2.4.(iii) to M = K, S = U, we get that either U is dense in A or [u ∘ v, w] = 0 for all u, v, w ∈ U.

This lemma, together with Lemma 3.2, enables us to establish the key fact contained in the next theorem.

Theorem 3.1. Suppose that A is not a central order in C(2) and let U be a Lie ideal of K. Then either U ⊇ [J_{K}, K] for some nonzero ideal J of A or [u ∘ v, w] = 0 for all u, v, w ∈ U.

Proof. Since A is not a central order in C(2), K is dense in A by Theorem 2.2. Assume that [u ∘ v, w] ≠ 0 for some u, v, w ∈ U; then U is dense in A by Lemma 3.3, and U ⊇ [J_{K}, K] by Lemma 3.2.

In order to obtain further consequences from the second case of Theorem 3.1, we will need the following theorem, analogous to Theorem 5.3 of [2].

Theorem 3.2. Let U be a Lie ideal of K satisfying the condition [U ∘ U, U] = 0. Then

(i) u ∘ v ∈ Z for all u, v ∈ U_{i}, i = 0, 1; (ii) u ∘ v = 0 for all u, v ∈ U_{1}; Moreover, assertion (i) implies (ii) for any Lie subsuperalgebra U of K such that [U ∘ U, U] = 0.

Proof. Assertion (i) is proved like in Theorem 5.3 of [2]. So, it only remains to show that if U is a Lie subsuperalgebra of K such that [U ∘U, U] = 0, then u∘v = 0 for any u, v ∈ U_{1}. Suppose, on the contrary, that 0 ≠ γ = u ∘ v ∈ Z and notice that

2u^{2} = [u, u] ∈ U, [u^{2}, v] = u ∘ (u ∘ v) = 2γu.

Then

4γu^{2} = [2γu, u] = [[u^{2}, v], u] = [u^{2}, [v, u]] = [u^{2} ∘ v, u] + [u^{2} ∘ u, v] ∈ [U ∘ U, U] = 0.

Hence u^{2} = 0 and so 2γu = [u^{2}, v] = 0 and u = 0, a contradiction.

At this point, our study of Lie ideals will be divided into two cases separately. We will consider first the case in which u^{2} = 0 for all u ∈ U_{0}, and second the opposite case. Motivated by future applications to the Lie structure of [K, K], we will sometimes make slightly more general assumptions on U.

Theorem 3.3. Let K′ be a Lie ideal of K such that K′ is dense in A and U a Lie ideal of K′ satisfying the conclusions (i)-(ii) of Theorem 3.2. If u^{2} = 0 for all u ∈ U_{0}, then U = 0.

Proof. First notice that [U, [U, K]] ⊆ [U, [K′, K]] ⊆ [U, K′] ⊆ U, hence U^{4} = 0 by Lemma 5.7 of [2] (the conclusion (iii) of Theorem 5.3 was never used there). If U were nonzero, then by Lemma 2.4.(ii) applied to the subsupermodules M = K′ and S = U, we would get, successively, U^{3} = 0, U^{2} = 0 and U = 0, a contradiction.

Theorem 3.3 completes the case in which u^{2} = 0 for all u ∈ U_{0}. We will henceforth assume that U satisfies the condition [U ∘ U, U] = 0, and that there exists an element u ∈ U_{0} such that 0 ≠ u^{2} = λ ∈ Z (notice that u^{2} = \frac{1}{2}(u ∘ u) ∈ Z by part (i) of Theorem 3.2).

Lemma 3.4. Suppose that K′ is a Lie ideal of K such that K′ is dense in A and U is a Lie ideal of K′ satisfying the conditions

(i) [u ∘ v, w] = 0 for all u, v, w ∈ U; (ii) There exists an element u ∈ U_{0} such that 0 ≠ u^{2} = λ ∈ Z. Assume also that A is not a central order in C(2). Then either U ⊆ Z or V^{−1}U_{0} has a basis over the field F = V^{−1}Z_{H} of the form {u, v, [u, v]}, where 0 ≠ u^{2} = λ ∈ F, 0 ≠ v^{2} = μ ∈ F, and u ∘ v = 0.

Proof. We will keep the same notation A for the ∗-central closure V^{−1}A. Notice that by Theorem 3.2 the Lie subsuperalgebra U satisfies conclusions (i), (ii) of that theorem. Arguing as in the proof of Lemma 5.8 of [2], we conclude that if U_{0} has no basis with desired properties, then U_{0} = F u ⊕ V_{0}, where V_{0} = {v ∈ U_{0}| v ∘ u = 0}. Moreover, V_{0}^{2} = 0, uV_{0} ⊆ V_{0}, V_{0} = [V_{0}, u] = [K′_{0}, u].

Now let u_{1} ∈ U_{1}; then 8u_{1}^{4} = [[u_{1}, u_{1}] ∘ u_{1}, u_{1}] ∈ [U ∘ U, U] = 0. Let us write u_{1}^{2} = zu + v with z ∈ F, v ∈ V_{0}. 0 = u_{1}^{4} = (zu + v)^{2} = z^{2}u^{2} + v^{2} + z(u ∘ v) = z^{2}λ since v^{2} ∈ V_{0}^{2} = 0 and u ∘ v = 0; but z^{2} = 0 and z ∈ F imply z = 0, hence u_{1}^{2} = v ∈ V_{0}. Let also u_{2} ∈ U_{1}. Then 0 = u_{1} ∘ u_{2} (Theorem 3.2); therefore (u_{1} + u_{2})^{2} = u_{1}^{2} + u_{2}^{2} + 2u_{1}u_{2} ∈ V_{0} with u_{1}^{2}, u_{2}^{2} ∈ V_{0} and hence u_{1}u_{2} ∈ V_{0}. This proves that U_{1}U_{1} ⊆ V_{0}.

We now claim that [U, U] = 0. Indeed, consider W = [U, U], which is obviously a Lie ideal of K′. Using the preceding paragraphs, we deduce that W_{0} = [U_{0}, U_{0}] + [U_{1}, U_{1}] ⊆ [Zu + V_{0}, Zu + V_{0}] + V_{0} ⊆ [u, V_{0}] + V_{0} ⊆ V_{0}. However, V_{0}^{2} = 0, hence w^{2} = 0 for all w ∈ W_{0}. Clearly W verifies the hypotheses of Theorem 3.3, hence W = 0.

We arrive then at [K′_{0}, u] = V_{0} = [V_{0}, u] ⊆ [U, U] = 0, hence U = Zu + U_{1} with U_{1}^{2} = 0, [u, U_{1}] = 0 and [u, K′_{0}] = 0. Moreover, [u, [u, K′_{1}]] ⊆ [u, U_{1}] = 0, and if k ∈ K′_{1} then u ∘ [u, k] = [u^{2}, k] = 0. Therefore, u[u, k] = 0, and since u is invertible, [u, k] = 0; i.e. [u, K′] = 0. Since K′ is dense in A, [u, I] = 0 for some nonzero ideal I of A. As in the proof of Proposition 3.1, this implies [u, A] = 0, hence u ∈ Z.

Therefore, Z ∩ K ≠ 0, and the involution ∗ is of the second kind. In this case Z = F + F u, H = Ku, and it is easy to see, as in the proof of Theorem 5.1 in [2], that ZU = U + Uu is a Lie ideal of A. By Theorem 2.1, either ZU is contained in Z or \overline{ZU} contains some nonzero ideal I of A, except if A is C(2). In the first case it is obvious that U ⊆ Z in the original prime superalgebra A.

Let us now consider the second case, namely \overline{ZU} ⊇ I for some nonzero ideal I of A. It is easy to see that [U +Uu, U +Uu] = 0; hence [\overline{ZU}, \overline{ZU}] = 0 and [I, I] = 0. By the Corollary of Proposition 3.1, then I = 0, a contradiction.

We can now prove the following complement to Theorem 3.3, in correspondence with Theorem 5.5 of [2].

Theorem 3.4. Let K′ be a Lie ideal of K such that K′ is dense in A and U be a Lie ideal of K′ satisfying the conditions (i),(ii) of Lemma 3.4. If A is not a central order in C(2), then either U ⊆ Z or the involution ∗ is of the first kind and A is a central order in C(4), where, for the field F = Z^{−1}Z, dim_{F} Z^{−1}K = 8, dim_{F} Z^{−1}[K, K] = 7, dim_{F} Z^{−1}U = 6, and Z^{−1}K′ is Z^{−1}K or Z^{−1}[K, K].

Proof. Assume that U ⊈ Z, then it suffices to prove that the involution ∗ is of the first kind and the ∗-central closure V^{−1}A is C(4) with the desired properties. We will keep the notation A for V^{−1}A, which is a central prime superalgebra over Z. Observe that F = Z_{H} is a field and dim_{F} Z ≤ 2.

Consider the Z-subspace Q = Z +ZU_{0} of A. It follows easily from the previous lemma (see also the proof of Corollary 5.2 of [2]), that Q is a subalgebra of A such that Q = Q(λ, μ), a generalized quaternion algebra over Z.

Let P be the centralizer of Q in A. Since Q is a central simple unital fourdimensional subalgebra of A, by [6, Theorem V.11.2], A = QP is isomorphic to Q⊗_{Z}P. Notice that P is a subsuperalgebra of A, and since Q^{∗} ⊆ Q, we have also P^{∗} ⊆ P.

If W is some subset of any superalgebra with superinvolution, we write W_{H} for the symmetric part of W, and W_{K} for the skew part of W. Then it is obvious that K ⊇ P_{K} + U_{0}P_{H} and H ⊇ P_{H} + U_{0}P_{K}. Now we will prove some facts about P.

We first claim that (P_{H})_{0} = F = Z_{H}. By Lemma 3.2, since K′ is dense in A, K′ ⊇ [J_{K}, K] for some nonzero ideal J of A. Thus J = QI for some nonzero ideal I of P, and hence J_{K} ⊇ I_{K} + U_{0}I_{H}. Therefore, since U_{0} = [U_{0}, U_{0}], we have U_{0}I_{H} = [U_{0}, U_{0}]I_{H} = [U_{0}, U_{0}I_{H}] ⊆ [K, J_{K}] ⊆ K′. Let y ∈ (I_{H})_{0}; then y[u, v] = [yu, v] ∈ [K′_{0}, U_{0}] ⊆ U_{0}. Thus y[u, v] = αu + βv + γ[u, v], where α, β, γ ∈ F. Now it is clear that y = γ ∈ F, and so (I_{H})_{0} = F, I = P, and (P_{H})_{0} = (I_{H})_{0} = F.

We also claim that (P_{H})_{1}^{2} = 0, which can be proved like in Theorem 5.5 of [2].

We know that P is a unital associative superalgebra with superinvolution ∗|_{P}. Since A is nontrivial and prime, necessarily so is P. Moreover, it is easy to see that Z(P)_{0} = Z. Furthermore, notice that ∗|_{P} is of the same kind as ∗. However, P_{H} = F + (P_{H})_{1} with (P_{H})_{1}^{2} = 0; hence P_{H} is not prime and so, by the second corollary to the Second Herstein Construction for Superalgebras given in [3], we may suppose that ^{∗} is of the first kind, and (P, ∗|_{P}) is isomorphic to (M(1), trp) as a Z-superalgebra with superinvolution (following the notation of [2] or [3]).

The remainder of the proof is accomplished as in Theorem 5.5 of [2]: A is shown to be C(4) with the desired properties for K and K′.

We finally arrive at the main theorem on the Lie structure of K.

Theorem A. Let A be a prime associative superalgebra with superinvolution. If U is a Lie ideal of K, then either U ⊆ Z or U ⊇ [J ∩ K, K] for some nonzero ideal J of A, except if A is a central order in C(2) or in C(4).

Proof. By Theorem 3.1, either U ⊇ [J_{K}, K] for some nonzero ideal J of A or [u ∘ v, w] = 0 for all u, v, w ∈ U. In the last case, U satisfies the hypotheses (and so the conclusion) of Theorem 3.2. Furthermore, K is dense in A by Theorem 2.2, hence by Theorems 3.3 and 3.4, with K′ = K, we get that either U ⊆ Z or A is a central order in C(4).

4. Lie structure of [K,K].

A result analogous to Theorem 2.2 holds for [K, K] too. To arrive at this result we first need the following lemma.

Lemma 4.1. If A is not a central order in C(2) then [K, K] ≠ 0.

Proof. Since A is not a central order in C(2), K is dense in A by Theorem 2.2. Suppose, contrary to our claim, that [K, K] = 0. Let 0 ≠ k ∈ K_{1}; then the subsupermodules S = Φk and M = K satisfy the hypothesis of Lemma 2.4. We have 2k^{2} = [k, k] = 0; hence by Lemma 2.4.(ii) k = 0, a contradiction. Thus K_{1} = 0 and K = K_{0}. Since K is dense in A, \overline{K} = \overline{K_{0}} contains a nonzero ideal J of A. But then J_{1} = 0 and A_{1}J = 0, a contradiction since A is prime and nontrivial.

Theorem 4.1. Let A be a prime associative superalgebra with superinvolution. Then [K, K] is dense in A, except if [K, K] ⊆ Z or A is a central order in C(2).

Proof. Suppose that A is not a central order in C(2); then K is dense in A by Theorem 2.2. Moreover, the Lie ideal [K, K] of K is nontrivial by Lemma 4.1.

By Lemma 3.3, either [K, K] is dense in A or [u ∘ v, w] = 0 for all u, v, w ∈ [K, K]. In the first case we are finished, so suppose the second. If we write U = [K, K], then U satisfies the conclusions (i),(ii) of Theorem 3.2. Since U ≠ 0, by Theorems 3.3 and 3.4 applied to U and K′ = K, we conclude that either [K, K] ⊆ Z or A is a central order in C(4) and dim_{Z^{−1}Z} Z^{−1}[K, K] = 7, dim_{Z^{−1}Z} Z^{−1}U = 6. But the last case is impossible since Z^{−1}U = Z^{−1}[K, K].

Lemma 4.2. Suppose that A is not a central order in C(n) for n = 2 or 4. If U is a Lie ideal of [K, K] such that U ⊈ Z, then either U is dense in A or [U, U] = 0.

Proof. Let u, v ∈ [U, U], w ∈ U. Similarly to the proof of Lemma 6.4 of [2], we get that [u ∘ v, w]A ⊆ \overline{U}, that is, [u ∘ v, w] ∈ (\overline{U} : A).

Since A is not a central order in C(2) and evidently [K, K] ⊈ Z, by Theorem 4.1 [K, K] is dense in A. Putting M = [K, K] and S = U in Lemma 2.4.(iii), we conclude that either U is dense in A or [u ∘ v, w] = 0 for all u, v ∈ [U, U], w ∈ U. Finally, this last case leads to [U, U] = 0, as in the proof of Lemma 6.4 of [2].

Lemma 4.3. Suppose that A is not a central order in C(n) for n = 2 or 4. If U is a Lie ideal of [K, K] not lying in Z, then either U ⊇ [J_{K}, K] for some nonzero ideal J of A or [[U, U], [U, U]] = 0.

Proof. Assume that [[U, U], [U, U]] ≠ 0. Then, by Lemma 4.2, [U, U] is J-dense in A for some nonzero ideal J of A, hence [J_{K}, K] ⊆ [[U, U], K] by Lemma 3.2. Moreover, following the proof of Lemma 6.4 of [2], we get that [[U, U], K] ⊆ U.

Lemma 4.4. Suppose that the involution ∗ is of the second kind and A is not a central order in C(n) for n = 2, 3. If W is a Lie ideal of [K, K] such that [W, W] ⊆ Z, then W ⊆ Z.

Proof. Since ∗ is of the second kind, the center Z is nonzero, so we can pass to the ∗-central closure V^{−1}A of A, which is a central prime superalgebra over the field V^{−1}Z. Clearly, it suffices to prove the lemma for the Lie ideal V^{−1}W of [V^{−1}K, V^{−1}K], so we will assume, without loss of generality, that F = Z_{H} is a field.

As in the proof of Lemma 3.4 (or of Theorem 6.1 in [2]), it is easily seen that the set ZW is a Lie ideal in [A, A]. Since A is not a central order in C(n) for n = 2 or 3, we can apply Theorem 3.3 of [8] to conclude that either ZW is contained in Z or ZW contains [I, A] for some nonzero ideal I of A.

In the first case W ⊆ Z; so assume the second case holds. Then we have

[[[I, I], [I, I]], I] ⊆ [[ZW, ZW], A] = Z[[W, W], A] = 0.

It is clear that I_{0} is a nonzero ideal of A_{0}. An ideal of a semiprime algebra is itself a semiprime algebra, hence I_{0} is a semiprime algebra and is isomorphic to a subdirect sum of prime algebras. But, by the Posner-Markov-Rowen’s Theorem ( see for example [9, Theorem 1.7.9], any prime algebra that satisfies the identity [[[x, y], [z, t]], u] is commutative (it is a P I-algebra and hence is a central order in a finite dimensional central simple algebra that satisfies the same identity; and the only possibility for such an algebra is to be a field). Hence I_{0} is commutative. Now for any i ∈ I_{1} the element 2i^{2} = [i, i] centralizes the ideal I_{0}, hence, by Lemma 2.3.(ii), i^{2} ∈ Z. If there exists i ∈ I_{1} such that i^{2} ≠ 0, then I ∩ Z ≠ 0, so I = A and A_{0} is commutative. By [8, Lemma 1.9], then the algebra A is either commutative or C(2). The second case contradicts our assumption. Now, if A is commutative then [A_{0}, A] = 0, and so W ⊆ [K, K] ⊆ [A, A] = [A_{1}, A_{1}] ⊆ A_{0} = Z.

Assume now that i^{2} = 0 for all i ∈ I_{1}; then we have [I_{1}, I_{1}] = 0. Furthermore, by the Jacobi superidentity, [[I_{0}, I_{1}], A_{1}] ⊆ [I_{1}, I_{1}] + [I_{0}, I_{0}] = 0, hence by Corollary 1 of Lemma 2.4 we get [I_{0}, I_{1}] = 0 and finally [I, I] = 0, which contradicts the Corollary of Proposition 3.1.

Lemma 4.5. Suppose that the involution ∗ is of the first kind and A is not a central order in C(2). If V is a Lie ideal of [K, K] such that [V, V] = 0, then V = 0.

Proof. We sketch the proof for this case in 12 steps. If a proof is not provided for some step, this means that its proof is the same as in Lemma 6.6 of [2].

1. v^{2} = 0 for all v ∈ V_{1}.

2. v_{1}v_{2}[K, K] ⊆ Av_{1} + Av_{2} for all v_{1}, v_{2} ∈ V.
3. v_{1}v_{2}v_{3}K ⊆ Av_{1} + Av_{2} + Av_{3} for all v_{1}, v_{2}, v_{3} ∈ V.
4. v_{1}v_{2}v_{3}v_{4}A ⊆ Av_{1} + Av_{2} + Av_{3} + Av_{4} for all v_{1}, v_{2}, v_{3}, v_{4} ∈ V.

5. For every v ∈ V_{0}, if v is nilpotent then v^{3} = 0. Indeed, given v ∈ V_{0}, v^{4}A ⊆ Av by step 4. If v^{4} ≠ 0 then, since A is prime, v^{m} ≠ 0 implies 0 ≠ Av^{4}Av^{m} ⊆ Av^{m+1}; hence v^{m+1} ≠ 0 and v is not nilpotent. Now suppose that v^{4} = 0. Then we get v^{3} = 0 as in step 5 of the proof of Lemma 6.6 of [2].

6. If char Φ = n ≥ 3 and v ∈ V_{0} then v^{3} = 0. Indeed, let v ∈ V_{0} and k ∈ [K, K]. It is clear that 0 = [v, [v, k]]; hence v[v, k] = [v, k]v. Using this, it is easily checked by induction on m that [v^{m}, k] = mv^{m−1}[v, k] for all m ≥ 1. Since char Φ = n, [v^{n}, k] = nv^{n−1}[v, k] = 0 for all k ∈ [K, K], thus [v^{n}, a] = 0 for all a ∈ \overline{[K, K]}. But [K, K] is dense in A by Theorem 4.1, hence v^{n} centralizes a nonzero ideal J of A. Now [v^{n}, A]J ⊆ [v^{n}, AJ] + A[v^{n}, J] = 0, hence [v^{n}, A] = 0 and v^{n} ∈ Z. On the other hand, (v^{n})^{∗} = (v^{∗})^{n} = (−v)^{n} = (−1)^{n}v^{n} = −v^{n} for n is odd because \frac{1}{2} ∈ Φ. Consequently, v^{n} ∈ Z ∩ K = 0; since v is nilpotent, it must be v^{3} = 0 by step 5.

7. If char Φ = 0 and A is a prime algebra, then v^{3} = 0 for all v ∈ V_{0}. Indeed, let v ∈ V_{0} and consider the derivation d = [v, .] of A. It is evident, by the hypothesis on V, that [V, [V, [V, K]]] = 0; in particular, d^{3}(K) = 0. Applying now the fundamental theorem of [7], we conclude that v^{3} = 0.

8. If A_{0} is prime, then V_{0} = 0. Simply notice that the hypotheses of Lemma 7 of [1] hold for the Lie ideal V_{0} of [K_{0}, K_{0}].

9. V_{0}^{8} = 0. Indeed, let us first take into account that if v ∈ V_{0}, then v^{3} = 0 by steps 6, 7, and 8 (use Lemma 2.2). Let v_{1}, v_{2}, v_{3}, v_{4} ∈ V_{0}. By step 4, assuming that v_{1}v_{2}v_{3}v_{4} ≠ 0 and taking into account the primeness of A, 0 ≠ Av_{1}v_{2}v_{3}v_{4}A ⊆ Av_{1} + Av_{2}+Av_{3}+Av_{4}. We deduce from this, since the v_{i} commute and v_{i}^{3} = 0 for all i, that Av_{1}v_{2}v_{3}v_{4}Av_{1}^{2}v_{2}^{2}v_{3}^{2}v_{4}^{2} ⊆ (Av_{1} + Av_{2} + Av_{3} + Av_{4})v_{1}^{2}v_{2}^{2}v_{3}^{2}v_{4}^{2} = 0, hence v_{1}^{2}v_{2}^{2}v_{3}^{2}v_{4}^{2} = 0. It is obvious that the same occurs if v_{1}v_{2}v_{3}v_{4} = 0. Linearizing this identity, we finally obtain that v_{1}w_{1}v_{2}w_{2}v_{3}w_{3}v_{4}w_{4} = 0 for all v_{1}, v_{2}, v_{3}, v_{4}, w_{1}, w_{2}, w_{3}, w_{4} ∈ V_{0}; that is, V_{0}^{8} = 0.

10. V_{1}^{4} = 0. Let v_{1}, v_{2}, v_{3}, v_{4} ∈ V_{1}. By step 4, since the v_{i} anticommute and v_{i}^{2} = 0 for all i (by step 1), v_{1}v_{2}v_{3}v_{4}Av_{1}v_{2}v_{3}v_{4} ⊆ (Av_{1}+Av_{2}+Av_{3}+Av_{4})v_{1}v_{2}v_{3}v_{4} = 0, hence v_{1}v_{2}v_{3}v_{4} = 0 by the semiprimeness of A.

11. V^{11} = 0.

12. V = 0. Indeed, let v ∈ V and k ∈ [K, K]. Notice that vk = (−1)^{\overline{v} \overline{k}}kv + [v, k] ∈ A′V, where A′ is the superalgebra obtained by adjoining a unit to A; hence V \overline{[K, K]} ⊆ A′V. Since [K, K] is dense in A by Theorem 4.1, \overline{[K, K]} contains a nonzero ideal J of A, and V J ⊆ A′V by the above. It is easy to obtain from this by induction that (JV J)^{m} ⊆ A′V^{m} for all m ≥ 1. In particular, for m = 11, (JV J)^{11} ⊆ A′V^{11} = 0 by step 11. Since A is prime, we conclude from this that JV J = 0, and hence that V = 0.

We arrive finally at our second main theorem on the ideal structure of [K, K].

Theorem B. Let A be a prime associative superalgebra with a superinvolution. If U is a Lie ideal of [K, K], then U ⊆ Z or U ⊇ [J ∩ K, K] for some nonzero ideal J of A, except if A is a central order in C(n) for n = 2, 3 or 4.

Proof. Suppose that A is not a central order in C(n) for n = 2, 3, 4, and let U be a Lie ideal of [K, K]. Assume that U ⊈ Z. By Lemma 4.3, either U ⊇ [J_{K}, K] for some nonzero ideal J of A or [[U, U], [U, U]] = 0. In the first case we are finished.

So, suppose that [[U, U], [U, U]] = 0. Applying Lemmas 4.4, 4.5 to the ideal [U, U] of [K, K], we obtain that [U, U] ⊆ Z. Finally, applying the same Lemmas again, now to the ideal U, we obtain that U ⊆ Z, a contradiction.

Acknowledgements.

The main part of the research has been carried out during the third author’s stay as a visiting professor at the University of La Rioja (Logroño, Spain), in October 1997 — February 1998. The welcome atmosphere at the Mathematical Department of the University is greatly appreciated.

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