1.1.1. Let R be an algebra over Φ. We say that R is a superalgebra if R is ℤ₂-graded, i.e., R = R₀ ⊕ R₁ such that R_i · R_j ⊆ R_i+j with i, j ϵ ℤ₂. R_o it is the even part and it is a subalgebra of R and R₁ is the odd part and it is a bimodule over R₀. Any element of R₀ ∪ R₁ is called a homogeneous element and we define the parity of a homogeneous element as |a| = 0 if a ϵ R₀ and |a| = 1 if a ϵ R₁.

Let f : R → R′ be a linear map where R and R′ are both superalgebras. We say that f is homogeneous of degree γ ϵ ℤ₂ if f (R_i) ⊂ R′_i+γ. In addition, we say that f is a superalgebra homomorphism if it is an algebra homomorphism and it is homogeneous of degree 0, i.e., f (R_i) ⊂ R′_i.

1.1.2. Let X = {x₁, x₂, …} be a countable infinite set of variables and let Φ 〈X〉 be the free unital associative algebra generated by X over Φ. If I is the two-sided ideal of Φ (X) generated by the set of elements {x_ix_j + x_jx_i | i, j ≥ 1}, we set G := Φ 〈X/I. We call G the (infinite dimensional) Grassmann algebra. We denote by ξ_i := x_i + I. With this notation G has the following presentation:

Notice that ξi2=0 since 12 ϵ Φ. The set B={1,ξi1,…ξik∣1<i1<…<ik,for all k∈ℕ} is a basis of G over Φ. In addition, G is a ℤ₂-graded module over Φ :

Thus, G is an associative superalgebra. Moreover, if R = R₀ ⊕ R₁ is a superalgebra over Φ, we can define the Grassmann envelope of R, G(R), as the even part of the tensor product G ⊗ R, i.e., G(R) = (G ⊗ R)₀ = G₀ ⊗ R₀ + G₁ ⊗ R₁. Notice that G(R) is an algebra.

The Grassmann envelope allows us to define varieties of superalgebras. Let R = R₀ + R₁ be a superalgebra. We say that R belongs to a certain variety of superalgebras (Lie, Jordan, associative,…) if G(R) belongs to the same variety of algebras.

Notice that if R = R₀ ⊕ R₁ is a superalgebra (i.e., ℤ₂-graded) such that it is associative as algebra then it is easy to check that G(R) is associative as well. Hence R is an associative superalgebra if and only if R is an associative ℤ₂-graded algebra. But in general a Lie or Jordan superalgebra is not a Lie or Jordan ℤ₂-graded algebra.

1.1.3. Let R = R₀ ⊕ R₁ be an associative superalgebra over Φ. In these conditions the map σ : R → R defined by σ(x₀ + x₁) = x₀ − x₁, for every x₀ ϵ R₀, x₁ ϵ R₁, is an algebra automorphism with σ² = id. Conversely, given an associative algebra R, every algebra automorphism σ : R → R with σ² = id defines a ℤ₂-graduation on R given by R₀ = {a ϵ R | σ(a) = a} and R₁ = {a ϵ R | σ(a) = −a}. Therefore, a ℤ₂-graduation on R is equivalent to an algebra automorphism σ with σ² = id.

Notice that a Φ-module S of R is graded if and only if σ(S) ⊂ S.

1.1.4. Let R be an associative algebra or superalgebra. We say that * is an involution if it is a linear map * : R → R such that, for every a, b ϵ R, (a*)* = a and (ab)* = b*a*, and we say that * is a superinvolution if it is a homogeneous, 0-degree, linear map such that for every homogeneous a, b ϵ R, (a*)* = a and (ab)* = (−1)^|a||b|b*a*. We denote the symmetric and skew-symmetric sets with respect an involution or a superinvolution * as H := Sym(R, *) = {a ϵ R | a* = a} and K := Skew(R, *) = {a ϵ R | a* = −a} respectively.

1.1.5. An associative algebra R is semiprime (resp. *-semiprime) if for every nonzero ideal (resp. *-ideal) I of R,I2:={∑ixiyi∣xi,yi∈I}≠0, and it is prime (resp. *-prime) if IJ:={∑ixiyi∣xi∈I,yi∈J}≠0 for every pair of nonzero ideals (resp. *-ideals) I, J of R.

We recall that a *-ideal is an ideal I such that I* ⊂ I.

It is easy to prove that R is semiprime if and only if is *-semiprime: It is clear that if R is semiprime then is *-semiprime. Conversely, let R be a *-semiprime algebra and let I be an ideal of R such that I² = 0. Notice that I ∩ I* is a *-ideal whose square is zero. Then I ∩ I* = 0, hence II* = I*I = 0. Thus (I + I*)² = 0, and since I + I* is a *-ideal, we have that I = 0. Therefore R is semiprime. However, an algebra can be *-prime but not prime: Let S be a prime associative algebra over Φ and let us consider R = S × S with involution (a, b)* = (b, a). Then R is a *-prime algebra but it is not prime. It is interesting to remark that the symmetric elements are of the form (a, a) and the skew-symmetric are of the form (a, −a).

We can prove that an associative algebra R is prime if and only if aRb≠0 for arbitrary nonzero elements a, b ϵ R, and it is semiprime if and only if it is nondegenerate, i.e., aRb≠0 for every nonzero element a ϵ R (see [53, §10]).

We are going to study these concepts in super setting. Let R = R₀ ⊕ R₁ be an associative superalgebra and let σ be the automorphism associated to the ℤ₂-graduation. We say that an ideal I is graded if I = I₀ ⊕ I₁ where I₀ = I ∩ R₀ and I₁ = I ∩ R₁ or, as we remarked in 1.1.3, if σ(I) ⊂ I.

An associative superalgebra R is semiprime if for every nonzero graded ideal I of R,I2≠0. And it is prime (as a superalgebra) if it does not have nonzero orthogonal graded ideals.

Notice that a *-ideal or a graded ideal satisfies I* ⊂ I or σ(I) ⊂ I, respectively. Then, arguing as before, the concepts of semiprime associative superalgebra and semiprime associative algebra coincide. An associative superalgebra can be prime but not prime as an algebra: for instance, let S be a prime associative algebra over Φ. Then R = S × S with R₀ = {(a, a) | a ϵ S} and R₁ = {(a, −a) | a ϵ S} is a prime associative superalgebra, which is not prime as an algebra (see [25]). We can say more: If R is prime as a superalgebra but not as an algebra we can consider a nonzero ideal P of R with P ∩ σ(P) = 0. Then P ⊕ σ(P) is an essential graded ideal of R, where (P ⊕ σ(P))₀ = {x + σ(x) | x ϵ P} ≅ P as an algebra and (P ⊕ σ(P))₁ = {x − σ(x) | x ϵ P}. Hence

Primeness in associative superalgebras can be also characterized by elements: for any two elements a, b of a prime associative superalgebra R where a and b are homogeneous, the condition aRb = 0 implies that either a or b is zero (see [25, pag. 693]). As we said before, semiprime associative superalgebras are semiprime as algebras hence the property aRa ≠ 0 for every nonzero homogeneous element a ϵ R holds in semiprime superalgebras.

Moreover, when dealing with superalgebras we can always consider the algebra R₀. In the next two lemmas F. Montaner states what happens on the even part when the whole superalgebra is semiprime or prime:

Lemma 1.1.6. [60, Lemma 1.2] If R = R₀ ⊕ R₁ is a semiprime associative superalgebra, then R and R₀ are semiprime algebras.

Lemma 1.1.7. [60, Lemma 1.3] If R = R₀ ⊕ R₁ is a prime associative superalgebra, then either R or R₀ are prime as algebras.

1.1.8. An ideal I of an associative algebra R (resp., an associative algebra with involution *) is prime (resp., *-prime) if R/I is a prime (resp. *-prime) associative algebra. If R is a semiprime associative algebra then there exists a family of prime ideals {I_α}_αϵ∆ such that ∩_αϵ∆ I_α = {0} and therefore R can be seen as a subdirect product of prime associative algebras (see [53, §12]). Similarly, if R is a semiprime associative algebra with involution * there exists a family of *-prime ideals {I_α}_αϵ∆ such that ∩α_ϵ∆ I_α = {0} and therefore R can be seen as a subdirect product of *-prime associative algebras. This is also true for superalgebras.

Moreover, if R is semiprime and free of n-torsion then the intersection of all prime ideals I_α such that R/I_α is free of n-torsion is zero (notice that the intersection of all prime ideals I_α such that R/I_α has n-torsion contains the essential ideal nR) and therefore R is a subdirect product of prime associative algebras, all of them free of n-torsion.

1.2.1. Given an ideal I of R, we can define the ideal Ann_R(I) := {z ϵ R | zI = Iz = 0}, which is called the annihilator of I in R. Moreover, when R is semiprime, Ann_R(I) = {z ϵ R | zIz = 0}. An ideal I of R is essential (for every nonzero ideal J of R, I ∩ J ≠ 0) if and only if Ann_R(I) = 0 (see [23, Proposition 1.6(1)]).

1.2.2. Given an associative algebra R, we define a permissible map of R as a pair (I, f) where I is an essential ideal of R and f : I → R is a homomorphism of right R-modules. For permissible maps (I, f) and (J, g) of R, define a relation ≡ by (I, f) ≡ (J, g) if there exists an essential ideal K of R, contained in I ∩ J, such that f(x) = g(x) for all x ϵ K. It is easy to see that this is an equivalence relation. If R is a semiprime associative algebra then Qmr(R) has an associative algebra structure coming from the addition of homomorphisms and from the composition of restrictions of homomorphisms, see [7, Chapter 2]:

The quotient set Qmr(R) with the operations defined above is called the Martindale algebra of quotients of R. Note that if R is a semiprime associative algebra then the map f:R→Qmr(R) defined by f(r) := [R, _λr], where _λr : R → R is defined by _λr(x) := rx, is a monomorphism of associative algebras, i.e., R can be considered as a subalgebra of its right Martindale algebra of quotients. The right Martindale algebra of quotients of R satisfies that for all q∈Qmr(R) there exists an essential ideal I of R such that qI ⊆ R. This facts allow us to prove that every subalgebra S of Qmr(R) which contains R is semiprime. Otherwise, if I is a nonzero nilpotent ideal of S and pick 0 ≠ q ϵ I. There exists an essential ideal J of R such that qJ ⊆ R, i.e., qJ = qJ ∩ I ⊆ R ∩ I is a nonzero nilpotent ideal of R which is a contradiction with the semiprimeness of R.

The symmetric Martindale algebra of quotients of R is defined as

Qms(R):={q∈Qmr(R)∣∃ an essential ideal I of R such that ql + Iq ⊂ R}

(if R has an involution one can replace the filter of essential ideals by the filter of essential *-ideals in the definition of the symmetric Martindale algebra of quotients, see [3, p. 858–859]). If R is semiprime then Qms(R), which is a subalgebra of Qmr(R) containing R, is also a semiprime algebra.

When R has an involution *, this involution can be extended to Qms(R) as follows: let us consider q ϵ q ϵ Qms(R) and I an essential *-ideal such that qI + Iq ⊆ R. We define f : I → R by the rule f(x) = (x*q)*. We set q* := [I, f] and note that q*x* = (xq)* for all x ϵ I (see [7, 2.5.4]).

The extended centroid C(R) of a semiprime algebra R is defined as the center of Qms(R). The extended centroid of a prime algebra is a field (see [7, p. 70]), the set of symmetric elements of the extended centroid of a *-prime algebra is again a field (see [3, Theorem 4(a)]), and the extended centroid of a semiprime algebra is a commutative and unital von Neumann regular algebra (see [7, Theorem 2.3.9(iii)]). In particular, if R is semiprime, C(R) is a semiprime algebra without nilpotent elements.

The central closure of R, denoted by R̂, is defined as the unital subalgebra of Qms(R) generated by R and C(R), i.e., R̂ := C(R)R + C(R), and can be seen as a C(R)-algebra. Therefore we can consider R contained in R̂. Moreover, since R̂ contains R and it is contained in Qms(R), if R is semiprime then R̂ is semiprime. The algebra R̂ is centrally closed, i.e., it coincides with its central closure. In particular its center equals its extended centroid, Z(R̂) = C(R̂).

1.2.3. The notion of extended centroid for semiprime associative superalgebras was studied by M. Fošner, see [25]. Let R be a semiprime associative superalgebra. Since R is semiprime as algebra we can consider the symmetric Martindale algebra of quotients Qms(R). Let σ : R → R be the automorphism associated to the ℤ₂-grading of R (σ² = id). This automorphism, by [7, Proposition 2.5.3], can be extended to Qms(R) and we denote this extension by σ^. Therefore Qms(R) is an associative superalgebra such that Ri⊂Qms(R)i with i = 0, 1. Moreover, if R is endowed with a superinvolution *, this can be also extended to Qms(R) as follows: let us consider q∈Qms(R)i, with i = 0, 1, and I an essential graded *-ideal such that qI + Iq ⊆ R. We define f : I → R by f(x) = (−1)^|q||x|(x*q)*. We set q* := [I, f] and note that q*x* = (−1)^|x||q|(xq)* for all x ϵ I homogeneous. Indeed, * is a superinvolution on Qms(R): Let us consider qi∈Qms(R)i and qj∈Qms(R)j with i, j = 0, 1. Choose an essential graded *-ideal J of R such that Jq_i, q_iJ, Jq_j, q_jJ, Jq_iq_j, q_iq_jJ are all contained in R and let I = J². Then Iq_i, q_iI, Iq_j, Iq_i ⊆ J. For every homogeneous x ϵ I we have

Hence (q_iq_j)* = (−1)^ijq*_jq*_i for all qi∈Qms(R)i and qj∈Qms(R)j with i, j = 0, 1.

On the other hand, since R is semiprime as an algebra, we can consider the extended centroid C(R) of R, which it is also ℤ₂-graded because C(R) = Z(Q^r_m(R)). Let R̂ = C(R)R + C(R) be the central closure of R. We will say that R is centrally closed if R = R̂.

1.2.4. Let R be a prime associative superalgebra such that R is not prime as an algebra. Let σ denote the automorphism associated to the ℤ₂-grading of R and consider a nonzero ideal P of R with P ∩ σ(P) = 0. Then P ⊕ σ(P) is a graded essential ideal of R, where (P ⊕ σ(P))₀ = {x + σ(x) | x ϵ P} ≌ P as an algebra and (P ⊕ σ(P))₁ = {x − σ(x) | x ϵ P}. Since P ⊕ σ(P) is essential in R,

where the isomorphism is given by the restriction of permissible maps (for any λ = [I, f] ϵ C(R) we define λ^=(I∩(P⊕σ(P)))2, g] where g : (I∩(P⊕σ(P))² → P⊕σ(P) is the restriction of f to the essential ideal (I ∩ (P ⊕ σ(P)))² of P ⊕ σ(P)). Notice that the ℤ₂-grading of C(P)⊕σ^(C(P)) comes from the ℤ₂-grading of P⊕σ(P) (C(P)⊕σ^(C(P)))0={λ+σ^(λ)∣λ∈C(P)} and (C(P)⊕σ^(C(P)))1={λ-σ^(λ)∣λ∈ C(P)}. In particular,

On the other hand, by Lemma 1.1.7, R₀ is prime as an algebra, and therefore its nonzero ideals are essential. By restricting permissible maps from R₀ to (P ⊕ σ(P))₀ we get CR0≅C(P⊕σ(P))0≅C(P).

We have obtained that C(R)₀ ≌ C(R₀).

Lemma 1.2.5. Let R = R₀ ⊕ R₁ be a prime associative superalgebra, and let a ϵ R₀. If there exists λ ϵ C(R) such that a − λ is nilpotent of index n and R has no n-torsion then λ ϵ C(R)₀.

Proof. Let us consider a ϵ R₀ and suppose that there exists λ = λ₀ + λ₁ ϵ C(R) such that a − λ is nilpotent of index n. If λ₁ ≠ 0, it is invertible by Lemma 1.2.6 and there exists μ1 ϵ C(R)₁ such that λ₁μ₁ = 1. From the nilpotency of a − λ₀ − λ₁ we get that μ₁a − λ₀ − 1 is again nilpotent of index n, i.e., the element b = μ₁a − μ₁λ₀ ϵ R₁ satisfies a polynomial of the form p(X) = (X − 1)ⁿ ϵ C(R)₀[X]. Since C(R)₀ is a field, p(X) ϵ C(R)₀[X] is the minimal polynomial of b over C(R)₀. In particular

and by homogeneity

i.e., b satisfies the polynomial q(X)=∑i=1n+12n2i-1Xn-2i+1. But n − 1 = degq(X) < deg p(X) = n, a contradiction with the minimality of p(X). Therefore λ₁ = 0 and λ ϵ C(R)₀.

Lemma 1.2.6. [25, Lemma 3.1] Let R be a semiprime associative superalgebra. Then the following assertions are equivalent:

1.3.1. We will work with Lie algebras and superalgebras arising from associative algebras and superalgebras. A Lie algebra L over a ring of scalarsΦ is a Φ-module with a bilinear product [ , ] satisfying, for every x, y, z ϵ L, the anticommutativity property and the Jacobi identity:

Let L = L₀ + L₁ be a superalgebra over Φ with bilinear product denoted by [ , ]_s. By using the Grassmann envelope, L is a Lie superalgebra if G(L) is a Lie algebra. Let us suppose that L = L₀ + L₁ is a Lie superalgebra, i.e., G(L) = L₀ ⊗ G₀ + L₁ ⊗ G₁ is a Lie algebra. We can deepen into which identities L satisfies: let us pick x ⊗ ξ_i, y ⊗ ξ_j ϵ (L₀ ⊗ G₀) ∪ (L₁ ⊗ G₁), then

so we can assure, by linearity, that [x, y]_s = − (−1)^|x||y| [y, x]_s for every x, y ϵ L₀ U L₁. Notice that the factor (−1) in the identity naturally arises from the property ξ_iξ_j + ξ_jξ_i = 0, i.e, ξ_iξ_j = − ξ_jξ_j of the generators of the Grassman algebra. Therefore, the identities (i) and (ii) can be translated to super setting as follows: Let L be a ℤ₂- graded module over Φ with a bilinear product [ , ]_s such that for every homogeneous x, y, z ϵ L:

Conversely, a superalgebra is a Lie superalgebra if both identities above are satisfied (see [?, Section 1]).

Recall that the adjoint map determined by any a ϵ L (resp. any homogeneous a ϵ L) is ad_a(x) := [a, x] (resp. ad_a(x) := [a, x]_s in super setting) for every x ϵ L. We say that an element a ϵ L is ad-nilpotent of index n ≥ 1 if adanL=0 L = 0 and adan–1L≠0. We say that an element a in L is a Jordan element if ada3L=0 (see [23, Chapter 4]). Since in superalgebras we will always consider homogeneous elements, we will define Jordan element in superalgebras for even elements as an even element which is ad-nilpotent of index less or equal to 3 of the whole Lie superalgebra. For odd elements we will work with ad-nilpotency of index less or equal to 4.

Typical examples of Lie algebras and superalgebras come from the associative setting: if R is an associative algebra (resp. superalgebra) over a ring of scalars Φ, then R with product, called bracket, [x, y] := xy − yx for every x, y ϵ R (resp. [x, y]_s = xy − (−1)^|x||y|yx, called super-bracket, for every homogeneous x, y ϵ R) is a Lie algebra (resp. a Lie superalgebra) denoted by R⁻. When dealing with R⁻ as a superalgebra, if a ϵ R₀ then ad_a behaves as the usual adjoint map in the non-super setting; when a ϵ R₁, ada2=ada2.

We will deal with Jordan algebras and superalgebras in Chapter 5. A linear Jordan algebra J over a ring of scalars Φ, with 12∈Φ, is a Φ-module with a bilinear product • satisfying, for every x, y ϵ J, the commutativity property and Jordan identity:

We already know that a superalgebra is a Jordan superalgebra if its Grassmann envelope is a Jordan algebra. But to translate the Jordan identity to super setting first we need to linearize it because the generatos in the Grassman algebra satisfy ξi2=0. We can prove that a ℤ₂-graded module J over Φ with a bilinear product •_s is a Jordan superalgebra if it satisfies

for every homogeneous x, y, z, t ϵ J. As above, if R is an associative algebra (resp. superalgebra) over a ring of scalars Φ, then R with product, called bullet, x • y = xy + yx for every x, y ϵ R (resp. x •_s y = xy + (−1)^|x||y|yx, called super-bullet, for every homogeneous x, y ϵ R) is a Jordan algebra (resp. Jordan superalgebra) denoted by R⁺.

The algebras R⁻ and R⁺ are well-known and it was I.N. Herstein the first one to study the relations between R and both of them in the non-super case (see for example [43]). Moreover, K is a Lie subalgebra (resp. subsuperalgebra) of R⁻ and H is a Jordan subalgebra (resp. subsuperalgebra) of R⁺. We refer the reader to [25], [35], [36], [37], [52], [60] and [62] for further information on associative superalgebras and on the Herstein theory on superalgebras. Although we have denoted super bracket as [ , ]_s, in Chapter 3, in order to simplify the notation, we will denote it as [ , ] (we will just work with the super bracket and there will not be any confusion).

1.3.2. If R is a centrally closed *-prime algebra and Skew(C(R), *) ≠ 0 then for any 0 ≠ λ ϵ Skew(C(R), *) we have R = H + K = λ²H + K ⊆ λK + K ⊆ R because 0 ≠ λ² is invertible, so R = λK + K for every 0 ≠ λ ϵ Skew(C(R), *). This occurs in particular when R is *-prime but not prime, because in this situation there exists a nonzero ideal I of R such that I ∩ I * = 0, and so we can define a nonzero skew element λ : I ⊕ I* → R in C (R) given by λ(x + y) := x − y.

If R is a centrally closed semiprime ring then R⁻ is a Lie algebra over the ring of scalars C(R); if in addition R has an involution *, then K is a Lie algebra over H(C(R), *).

Lemma 1.3.3. ([13, Lemma 2.11]) Let (R, *) be a semiprime associative algebra with involution and let a ϵ R. If there exist λ ϵ C(R) such that a − λ is nilpotent then λ is the unique element of C(R) such that a − λ is nilpotent. Moreover, if a ϵ K then λ ϵ Skew(C(R), *).

Proof. If a−λ and a−μ are nilpotent elements of the central closure R̂ of R, a−λ−(a− μ) = μ − λ is a nilpotent element in the semiprime commutative ring C(R). Therefore λ = μ. Now, if a ϵ K and a − λ is nilpotent then (a − λ)* = − (a + λ*) is nilpotent and therefore a + λ* is nilpotent, which implies that λ = − λ* ϵ Skew(C(R), *).

We will need also this result in superalgebras. With the same argument as in the above lemma we have:

Lemma 1.3.4. Let R = R₀ ⊕ R₁ be a semiprime associative superalgebra with superinvolution *, and let a ϵ R₀ ∪ R₁. If there exists λ ϵ C(R) such that a − λ is nilpotent then λ is the unique element of C(R) such that a − λ is nilpotent. Moreover, if a ϵ K then λ ϵ Skew(C(R), *).

Recall that an element a in a Lie algebra L is ad-nilpotent of index adanL=0 and adan-1L≠0.

2.1.1. (i) Let us consider R⁻: we say that an element a is a pure ad-nilpotent element of R⁻ of index n if for every λ ∊ C(R) with λa ≠ 0, λa is ad-nilpotent in R̂⁻ of index n, where R̂ is the central closure of R.

(ii) Let us consider K: we say that an element a is a pure ad-nilpotent element of K of index n if for every λ ∊ H(C(R)), *) with λa ≠ 0, λa is ad-nilpotent in Skew(R̂, *) of index n, where R̂ is the central closure of R.

Lemma 2.1.2. If R is a semiprime associative algebra and a is an ad-nilpotent element of R of index n, the following conditions are equivalent:

Proof. Suppose that R is semiprime and centrally closed (otherwise, substitute R by its central closure R̂).

(i) ⇒ (ii). Let us consider V=adan-1x∣x∈R. By Proposition 2.0.4 there exists e ∊ C(R) such that ∊υ = υ for every υ ∊ V and Ann_R(Id_R(V)) = (1 – e)R. Suppose that (1 – e)a ≠ 0. By hypothesis (1 – e)a is ad-nilpotent of index n, hence 0≠ad(1-e)an-1(R)=(1-e)adan-1(R)=0, a contradiction. So ea = a and AnnIdR(ea)IdRadan-1(R)⊂AnnRIdRadan-1(R)=(1-e)R must be zero, i.e., IdRadan-1(R) is essential in Id_R(ea).

(ii) ⇒ (iii). This holds in general if I and J are ideals of R with I essential in J: 0 = Ann_J(I) = Ann_R(I) ∩ J implies Ann_R(I)J = 0, so Ann_R(I) ⊂ Ann_R(J).

(iii) ⇒ (i). Let λ ∊ C(R) be such that λa ≠ 0. Clearly a adλan(R)=0. Suppose that adλan-1(R)=0: then λn-1ada∣n-1(R)=0, so λ^n-1 ∊ AnnRIdRadan-11(R)=AnnRIdR(a), which is not possible because R is semiprime and λa ≠ 0.

Lemma 2.1.3. Let R be a centrally closed semiprime associative algebra with involution *, and let a ∊ K be a pure ad-nilpotent element of K of index n. If there exists λ ∊ H(C(R), *) such that λa is ad-nilpotent of R of index n, then λa is a pure ad-nilpotent element of R of index n.

Proof. Let us see that for every μ ∊ C(R) with μλa ≠ 0, the element μλa has index of ad-nilpotency in R equal to n. Suppose that there exists μ ∊ C(R) with adμλan-1R=0, and let us prove that μλa = 0:

We have that μn-1adλa∣n-1R=adμλan-1R=0, so μadλan-1R=0 because C(R) is regular von Neumann. In particular, μadλan-1H=μadλan-1K=0. Since μ = μ_h + μ^k, we have that μhadλan-1R=μkadλan-1R=0.

From 0=μhn-1adλan-1R=adμhλan-1R we get that μ_hλa index of ad-nilpotency in K lower than n, implying μ_hλa = 0 because a is a pure ad-nilpotent element of K.

From 0=μk2n-1adλan-1R=adμk2\an-1R we get that μk2λa has index of ad-nilpotency in K lower than n, so again μk2λa=0 (because a is a pure ad-nilpotent element of K), and by regularity of C(R), μ_kλa = 0.

This implies μλa = 0.

The next proposition shows that every ad-nilpotent of R⁻ or of K can be expressed as an orthogonal sum of pure ad-nilpotent elements of decreasing indices.

Proposition 2.1.4. Let R be a centrally closed semiprime associative algebra and let a ∊ R be an ad-nilpotent element of R⁻ of index n. There exists a family of orthogonal idempotents ϵii=1k⊂C(R) such that a=∑i=1kϵia a pure ad-nilpotent element of index n_i in ∊_iR for n = n1 > n > • • • > n_k.

Similarly, if R has an involution * and a is an ad-nilpotent element of K of index n, then there exists a family of orthogonal idempotents ϵii=1k⊂H(C(R),*) such that a=∑i=1kϵiai with ϵ_ia a pure ad-nilpotent element of index n_i in Skew (ϵ_iR, *) for n = n₁ > n₂ > … > n_k.

Proof. Let us prove the result for Lie algebras of skew-symmetric elements. We will proceed by induction on n. If n = 1 there is nothing to prove. Let us suppose that the result is true for every ad-nilpotent element of index less than n and let a ϵ K be an ad-nilpotent element of index n ≥ 2. Let us consider V=adan-1x∣x∈K. By Proposition 2.0.4 there exists ∊ ϵ H(C(R), *) such that ∊υ = υ for every υ ϵ V and AnnR(IdR(V)) = (1 − ∊)R. Then a = ∊a + (1 − ∊)a.

Clearly, by construction (1 − ∊)a is ad-nilpotent of index less than n in K: for every x∈K,ad(1-ϵ)an-1x=(1-ϵ)adan-1x=adan-1x-ϵadan-1x=0.

Let us prove that ∊a is pure ad-nilpotent of index n in Skew(∊R, *). For any λ ϵ H(C(R), *) such that λ∊a ≠ 0, λ∊a is ad-nilpotent of index n: clearly adλϵan(Skew(∊R, *)) = 0 and if adλϵan-1(Skew(∊R, *)) = 0 then λⁿ⁻¹∊ ϵ Ann_R(Id_R(V)) = (1 − ∊)R, which leads to a nilpotent ideal generated by the nonzero element λ∊a, a contradiction with the semiprimeness of R.

Apply now the induction hypothesis to (1 − ∊)a and the Lie algebra of skew- symmetric elements Skew((1 − ∊)R, *).

In this section we are going to prove that every nilpotent inner derivation is induced by a nilpotent element, generalizing to semiprime algebras Herstein’s result [42, Theorem in p. 84] for simple algebras. This result was already proved by Grzeszczuk ([38, Corollary 8]). Our techniques are rather elementary and, by adding the hypothesis of pure ad-nilpotence, we can describe such elements with less restrictions on the torsion of the algebra.

Lemma 2.2.1. Let R be a semiprime associative algebra and let a ϵ R be a nilpotent element. Suppose that there exist some λ_i ϵ ℤ, i = 0, …, n, such that

for all x, y ϵ R. Then for every i = 0, …, n we have λ_ia^max(i,n−i) = 0. In particular, each term in the identity above is zero.

Proof. First, let us suppose that R is prime and suppose that a ≠ 0 has index of nilpotence s. If the lemma is not satisfied, there exists some k with λ_ka^max(k,n−k) ≠ 0. In particular, max(k,n − k) < s. Let us multiply the expression ∑i=0nλiai[x,y]an-i by a^s−1−k on the left and by a^{s−1−(n−k)} on the right, so that

for every x,y ϵ R. Hence λ_ka^s−1yxa^s−1 = λ_ka^s−1yxa^s−1 for every x, y ϵ R. Since a^s−1 ≠ 0 for every x ϵ R we have by Theorem 2.0.1 that there exists α_x ϵ C(R) such that λ_ka^s−1x = α_xλ_ka^s−1. Multiplying this last expression by a on the right we get λ_ka^s−1xa = 0 for every x ϵ R. By primeness of R we get that either a^s−1 = 0 or λ_ka = 0, leading to a contradiction.

If R is semiprime then R is a subdirect product of prime quotients R/I_a with ∩_α I_α = 0. For any α and any i, by the prime case λ_ia^max(i,n−i) ϵ I_a, so λ_ia^max(i,n−i) = 0.

Lemma 2.2.2. Every nilpotent element of an associative algebra R is ad-nilpotent.

If a has index of nilpotence t and index of ad-nilpotence n then n ≤ 2t − 1. If R is semiprime then n ≥ t, and if in addition R is free of ns -torsion for s:=n+12, then t = s and n = 2t − 1.

Proof. Since a^t = 0, for every x ϵ R we have

because if i < t then 2t − 1 − i ≥ s. Therefore n ≤ 2t − 1.

Suppose now that R is semiprime and let us see that n ≥ t: if on the contrary

for every x ϵ R, focusing on the first summand of this expression (( − 1) ^t−1xa^t−1) we get that a^t−1 = 0 by Lemma 2.2.1, a contradiction.

Moreover, since for every x ϵ R we have 0=adan(x)=∑i=0nni(-1)n-iaixan-i, again by Lemma 2.2.1 nsas=0 for s:=n+12. If R is free of ns-torsion then a^s = 0 so s ≥ t, i.e., n ≥ 2t − 1, and therefore n = 2t − 1 (equivalently, t = s).

The next example shows that all possible cases in the lemma above can be realized: Let p be an odd prime number and R a prime associative algebra with characteristic p. If a ϵ R is a nilpotent element of index t∈p+12,…,p then a is ad-nilpotent of index p. In particular there are no ad-nilpotent elements of index between p + 1 and 2p − 1, and a nilpotent element of index p is ad-nilpotent of the same index p.

Proposition 2.2.3. Let R be a prime associative algebra and let a ϵ R be an ad-nilpotent element of R⁻ of index n. Let F- denote the algebraic closure of the field 𝔽 := C(R) and R̄ := R̂ ⊗ F-. Then:

Proof. (1) Since R is prime, 𝔽 = C(R) is a field and R̅ is a centrally closed prime algebra (see [7, pp. 445–446]). From

for every x ϵ R we have, by Theorem 2.0.1, that a seen as an element of R̂ is an algebraic element over 𝔽 of degree not greater than n. Let us consider the minimal polynomial p(X) ∊ 𝔽[X] of a. Let F- be the algebraic closure of 𝔽 and let μ₁, …, μ_r ϵ F- be the roots of p(X) in F-, i.e., p(X) = (X − μ₁)^k1 … (X - μr)^kr ∊ 𝔽[X].

Let us prove that p(X) has only one root in F- and therefore p(X) = (X − μ)^k ϵ 𝔽[X], whence a − μ is nilpotent in R̅: Suppose on the contrary that p(X) has different roots μ₁, …, μ_r, r > 1, and define q_i(X) := p(X)/(X − μ_i) for every i. Since p(X) is the minimal polynomial of a, q_i(a) ≠ 0 in R̅. Note that (a − μ_i)q_i(a) = p(a) = 0 and therefore aq_i(a) = μ_iq_i(a). Now, since we are in the prime case, there exists y ϵ R such that q₁(a)yq₂(a) ≠ 0 and therefore ad_a(q₁(a)yq₂(a)) = aq₁(a)yq₂(a) − q₁(a)yq₂(a)a = (μ₁−μ₂)q₁(a)yq₂(a) ≠ 0. This means that q₁(a)yq₂(a) is an eigenvector of the linear map ad_a associated to the eigenvalue μ₁ − μ₂, hence it is an eigenvector of ada2 associated to (μ₁ − μ₂)², etc. This is a contradiction because both q₁(a)yq₂(a) and each power of (μ₁ − μ₂) are nonzero, while ad_a is nilpotent. Therefore r = 1, p(X) = (X − μ)^k ∊ 𝔽[X] and (a − μ)^k = 0.

(2) Let us consider b := a − μ ϵ R̅, which is ad-nilpotent of index n. Let us see that n is odd: Suppose on the contrary that n = 2m. Then

implies by Lemma 2.2.1 that nmbm=0 and, since R̅ is free of nm-torsion, that b^m = 0. Substituting in adbn-1x=∑i=0n-1n-1i(-1)n-1-ibixbn-1-i we get that adbn-1x=0 for every x ϵ R, a contradiction.

Therefore n is odd and a − μ is nilpotent of R̅ of index s:=n+12 by Lemma 2.2.2. Moreover, since the coefficient of degree s − 1 of p(X) = (X − μ)^s ∊ 𝔽[X] is −sμ ϵ 𝔽, if R is free of s-torsion then μ ϵ 𝔽, i.e., there exists μ ϵ C(R) such that a − μ is nilpotent of index s:=n+12.

In the following theorem we get the description of the pure ad-nilpotent elements of R⁻. In its proof, Proposition 2.2.3 is primarily used to find that any ad-nilpotent element a ϵ R of index n forces [a,[adan-1x,[adan-1x,y]]]=0 for every x, y ϵ R. If 2, 3, …, r were invertible in R for r≥n+n2+1, this identity would directly follow from the proof of [29, Theorem 2.3].

Theorem 2.2.4. Let R be a semiprime associative algebra, let R̂ be its central closure, and let a ϵ R be a pure ad-nilpotent element of R⁻ of index n. Put s:=n+12, and suppose that R is free of ns-torsion and s-torsion. Then n is odd and there exists λ ϵ C(R) such that a − λ ϵ R̂ is nilpotent of index n+12.

Proof. Let us suppose that R is a prime associative algebra and, without loss of generality, that it is centrally closed. Consider μ ϵ C(R) as given by Proposition 2.2.3. Putting b := a − μ, we know that b^s = 0 for s:=n+12, hence for every x, y ϵ R we have

If R is semiprime, R is a subdirect product of prime associative algebras (without ns and s-torsion) and in any of these prime quotients

which imply that

for every x,y ϵ R. For every x ϵ R, let zx:=adan-1x. By the identity above,

Therefore, since Id_R(z_xa) ⊂ Id_R(z_x), by Corollary 2.0.3 there exists λ_x ϵ C(R) such that z_xa = λ_xz_x and by Proposition 2.0.4 there exists ∊_x ϵ C(R) such that ∊_xz_x = z_x and Ann_R(Id_R(z_x)) = (1 − ∊_x)R. Therefore

for every y ∈ R, whence (a – λ_x)ⁿ ∈ Ann_R(Id_R(z_x)). So ∊_x(α – λ_x)ⁿ = 0. Now, for every x, xʹ ∈ R there exist λ_x,λ_x’ ∈ C(R) and idempotents ∊_x, ∊_xʹ ∈ C(R) such that 0 = (∊_x∊_xʹa – ∊_x∊_xʹaλ_x)_n = (∊_x∊_xʹa – ∊_x∊_xʹaλ_x)_n, so ∊_x∊_xʹaλ_x = ∊_x∊_xʹaλ_xʹ by Lemma 1.3.3. By Lemma 2.0.5 there exists λ ∈ C(R) such that ∊_xλ = ∊_xλ_x for every x ∈ R. Then for every x ∈ R we have z_x(a – λ)ⁿ = ∊_xz_x(a – λ_x)ⁿ = 0, so 0=ϵxzxadany=zxy(a−λ)n for every y ∈ R thus (a – λ)ⁿ ∈ Ann_R(Id_R(z_x)) (see 1.2.1). Moreover ∩x∈RAnnR(IdR(zx))=AnnR(IdR(adan−1(R))) by definition of z_x, and AnnR(IdR(adan−1(R)))=AnnR(IdR(a)) because a is pure (Lemma 2.1.2(iii)). Finally, let ∊ ∈ C(R) be such that ∊a = a and Ann_R(Id_R(a)) = (1 – ∊)∈. Then ∊(a – λ)ⁿ = (a – ∊λ)ⁿ = 0 because it is contained in (1 – ∊)R.

Hence a – ∊λ is nilpotent in addition to being ad-nilpotent of index n. Put s:=n+12 and take any prime quotient without s and ns-torsion in which a−ϵλ― is still ad-nilpotent of index n. By Proposition 2.2.3(2) we get that n must be odd and a−ϵλ― is nilpotent of index s. Since in any prime quotient (a−ϵλ―)s=0― by Proposition 2.2.3(2), we have that s is the index of nilpotence of a – ∊λ.

Lee’s description of ad-nilpotent elements of R^– is recovered when the hypothesis of being pure is removed.

Corollary 2.2.5. ([54, Theorem 1.3]) Let R be a semiprime associative algebra, let Ř be its central closure, let a ∈ R be an ad-nilpotent element of R^– of index n, and suppose that R is free of n!-torsion. Then n is odd and there exists λ ∈ C(R) such that a – λ ∈ R̂ is nilpotent of index n+12.

Proof. Suppose without loss of generality that R is centrally closed, i.e., R = R̂.

By Proposition 2.1.4 there exists a family of orthogonal idempotents {ϵi}i=1k⊂ C(R) such that a=∑i=1kϵia with ∊_ia a pure ad-nilpotent element of index n_i (n = n₁ > n₂ > •••) of R∊_i. Then by Theorem 2.2.4 there exists λ_i ⊂ C(R∊_i) ⊂ C(R) such that (∊_ia – λ_i)^Si = 0 for si:=[ni+12] and for all i = 1, …, k. Hence λ=∑i=1nϵiλi satisfies the claim.

Interesting Lie algebras associated to simple associative algebras R are the quotient algebras [R, R]/([R, R] ⋂ Z( R)), which are simple unless R has 2-torsion and is 4-dimensional over its center ([44, Theorem 1.13]). Let us study ad-nilpotent elements in these associative algebras.

Lemma 2.2.6. ([23, Lemma 4.6]) Let R be a semiprime associative algebra and let a ∈ R be such that adan(R)⊂Z(R). Then adan(R)=0.

Proof. For every x ∈ R we have

Therefore 0=adan−1((adanx)[a,x])=(adanx)2 which implies, since R is semiprime and adanx∈Z(R), that adanx=0.

Lemma 2.2.7. Let R be a semiprime associative algebra, let L := [R, R]/([R, R] ⋂ Z(R)) and let ā : = a + ([R, R] ⋂ Z(R)) ∈ L be an ad-nilpotent element of L of index n. Then a is an ad-nilpotent element of index n in R^–.

Proof. For every x∈R,adan+1x=adan([a,x])∈adan([R,R])⊂Z(R) so, by Lemma 2.2.6, adan+1x=0 for every x ∈ R, i.e., a is ad-nilpotent in R^– of index n or n + 1.

Let us suppose that R is prime. Then, by Proposition 2.2.3, there exists μ ∈ 𝔽, the algebraic closure of 𝔽 := C(R), such that a – μ is nilpotent in R © 𝔽 of some index s. Moreover, by Lemma 2.2.2, s ≤ n + 1. Put b := a – μ. Then

for every x, y ∈ R. By Lemma 2.2.1, for every k ∈ {0,1,…,[n+12]} we have (nk)bmax(k,n−k)=0, so

i.e., a is an ad-nilpotent element of R¯ of index n.

Finally, since ā is ad-nilpotent of index not greater than n in any prime quotient, a is an ad-nilpotent element of R¯ of index n when R is semiprime.

In particular, from these last two lemmas we get that if R is semiprime then [R, R]/([R, R]∩Z(R)) and R/Z(R) are nondegenerate Lie algebras (see [44, Sublemma in p. 5]).

Corollary 2.2.8. Let R be a semiprime associative algebra, let R be its central closure, and let L := [R,R]/([R,R] ∩ Z (R)) or L := R/Z (R). If ā ϵ L is an ad-nilpotent element of L of index n and R is free of n-torsion, then n is odd and there exists λ ϵ C(R) such that a – λ ϵ R is nilpotent of index n+12.

Proof. If L = [R, R]/([R, R] ∩ Z(R)) the result follows by Lemma 2.2.7 and Corollary 2.2.5. If L = R/Z(R) the result follows by Lemma 2.2.6 and Corollary 2.2.5.

In this section we focus on semiprime algebras R with involution * and their set of skew-symmetric elements K. As in the previous section, we will first describe the pure ad-nilpotent elements of K, and then remove the hypothesis of being pure by decomposing each ad-nilpotent element into a sum of pure ad-nilpotent elements of decreasing indices.

The following lemma collects some results about *-identities. Item (1) is [44, Remark on p. 43] (with a different proof), item (2) is a generalization of [56, Lemma 5], and item (3) is a generalization of [13, Lemma 5.2].

Lemma 2.3.1. Let R be a semiprime associative algebra with involution *. Let k ∊ K and h ∊ H. Then:

Proof. We can suppose without loss of generality that R = R, i.e., R is centrally closed.

(1) Take x ∊ R. Note that k(x – x*)k = 0, so that kxk = kx*k. Then

k(xkx)k = k(xkx)*k = –kx*kx*k = –(kx*k)x*k = –kxkx*k = –kx(kx*k) = –kxkxk

and so we have kxkxk = 0 since R is free of 2-torsion. Therefore kxkxkyk = 0 for every y ∊ R, hence

so (kxk)R(kxk) = 0 and kxk = 0 since R is semiprime. Now kRk = 0 implies, again by semiprimeness, that k = 0.

(2) If h = 0 then the claim is trivially fulfilled, so assume h ≠ 0. Take x, y ∈ R. Note that h(x – x*)h = 0 and therefore hxh = hx*h. Then

i.e., hxhyh = hyhxh. By Corollary 2.0.3, since h ≠ 0 and Id_R(hxh) ⊆ Id_R(h), for each x ∊ R there exists μx∈C(R) such that hxh=μxh Hence 0 ≠ hRh ⊂ C(R)h. Moreover, since hx∗h=hxh,2hxh=hxh+hx∗h=(μx+μx∗)h∈H(C(R),∗)h, so hRh ⊆ H(C(R), *)h.

Let us suppose that Id_R(h) is essential in R and let us show that Skew(C(R), *) = 0: Take λ ∊ Skew(C(R), *) and y ∊ R. Then (λh)y(λh) = λh(yλ)h = λμ_λy^h ∊ K for some μ_λy ∊ H(C(R), *). On the other hand (λh)y(λh) = λ²hyh = λ²μ_yh ∊ H for some μ_y ∊ H(C(R), *). Therefore (λh)y(λh) = 0 for every y ∊ R, and by semiprimeness of R, λh = 0, so λ = 0 because Id_R(h) is essential.