CHAPTER 2. AD-NILPOTENT ELEMENTS IN AN ASSOCIATIVE ALGEBRA

This chapter has been published in the journal Bulletin of the Malaysian Mathematical Sciences Society and can be found in [12].

Throughout all this chapter R is an associative algebra over Φ with $\frac{1}{2}$ ϵ Φ.

The main goal of this chapter is to deepen into the description of ad-nilpotent elements of R and K where R is a semiprime associative algebra with involution. In the spirit of Martindale and Miers’ result [56, Main Theorem], we will obtain different types of ad-nilpotent elements of K of index n depending on the equivalence class of n modulo 4. In this chapter we will also study ad-nilpotent elements in semiprime associative algebras, as T.K. Lee did in [54], but we introduce a new concept called pure ad-nilpotent, that it will allow us to weaken torsion conditions and to obtain a more detailed classification. We say that an ad-nilpotent element a of index n in R⁻ is pure if λa remains ad-nilpotent of the same index for every λ in the extended centroid such that λa ≠ 0. An ad-nilpotent element a of index n in K is pure if for every symmetric λ in the extended centroid such that λa ≠ 0, λa is ad-nilpotent of the same index n. This is just a technical condition, since every ad-nilpotent element of R⁻ can be expressed as an orthogonal sum of pure ad-nilpotent elements of the central closure R̂ of R with decreasing indices of ad-nilpotency.

As a first step we focus on ad-nilpotent elements of R⁻. In this case, under the hypothesis of pure ad-nilpotence, the condition on the torsion of the algebra can be weakened when compared with the result of T.K. Lee in [54, Theorem 1.3].

From Theorems 2.2.4 and 2.3.6 we easily recover Lee’s results [54, Theorem 1.3 and Theorem 1.5]. Furthermore, we also describe ad-nilpotent elements of Lie algebras of the form R/Z(R) and K/(K∩Z(R)), and of their derived Lie algebras [R, R]/([R, R]∩Z(R)) and [K,K]/([K,K] ∩ Z(R)).

Let us write down some useful results where the extended centroid C(R) plays a really important role. We will use the following results due to Beidar, Martindale and Mikhalev.

Theorem 2.0.1. ([57, Theorem 2(a)]) Let R be a prime associative algebra. Let a_i, b_i ϵ R for i = 1, 2,..., n with b₁ ≠ 0 be such that $\sum _{i=1}^n{a}_{i}x{b}_i=0$ for every x ϵ R. Then there exist λ_i ϵ C(R) for i = 2,..., n such that $a_1=\sum _{i=2}^n{\lambda }_i{a}_i\text{ in }\hat{R}$.

Theorem 2.0.2. ([7, Theorem 2.3.3]) Let R be a semiprime associative algebra and let a₁,a₂,..., a_n ϵ R. If a₁ ∉ $\sum _{i=2}^{n}C\left(R\right)a_i$ in R̂ then there exist r_j, s_j ϵ R for j = 1, 2,..., m such that $\sum _{j=1}^m{r}_j{a}_1{s}_j\ne 0$ and $\sum _{j=1}^m{r}_j{a}_k{s}_j=0$ for k = 2,..., n.

The next corollary can be found in [13]. For the sake of completeness we include its proof here.

Corollary 2.0.3. Let R be a semiprime associative algebra. Let a_i, b_i ϵ R for i = 1, 2,..., n be such that Id_R(a1) ⊂ Id_R(b1)(b1) and $\sum _{i=1}^n{a}_{i}x{b}_i=0$ for every x ϵ R. Then there exist λ_i ϵ C(R) for i = 2,..., n such that $a_1=\sum _{i=2}^n{\lambda }_i{a}_i$ in R̂.

Proof. By Theorem 2.0.2, if a₁ $\sum _{i=2}^{n}C\left(R\right)a_i$ there exist r_j, s_j ϵ R, j = 1,..., m, such that $\sum _{j=1}^m{r}_j{a}_1{s}_j\ne 0$ and $\sum _{j=1}^m{r}_j{a}_k{s}_j=0$ for k = 2,3,..., n. Replace x by s_jx and multiply $\sum _{i=1}^n{a}_{i}x{b}_i=0$ on the left by r_j. We have

$$0=\sum _{i=1}^n\sum _{j=1}^m{r}_j{a}_i{s}_{j}x{b}_i=\sum _{j=1}^m{r}_j{a}_1{s}_{j}x{b}_1$$

which implies that the ideal generated by $\sum _{j=1}^m{r}_j{a}_1{s}_j$ is orthogonal to the ideal generated by b₁ and therefore, since Id_R(a1) ⊂ Id_R(b1), the ideal generated by $\sum _{j=1}^m{r}_j{a}_1{s}_j$ has zero square, a contradiction because R is semiprime.

The following proposition is an easy generalization of [7, Theorem 2.3.9(i)].

Proposition 2.0.4. Let R be a centrally closed semiprime associative algebra. For any subset V ⊂ R there exists a unique idempotent ∊ ϵ C(R) such that ∊υ = υ for all υ ϵ V, the annihilator in C(R) of V is Ann_C(R)(V) = (1 – ∊)C(R), the annihilator in R of the ideal generated by V is Ann_R(Id_R(V)) = (1 – ∊)R, and the ideal generated by V is essential in ∊R. Moreover, when R has an involution * and V ⊂ H or V ⊂ K, then ∊ ϵ H(C(R), *).

Proof. The first part of the proof follows as in [7, Theorem 2.3.9(i)] with the obvious changes. Let V ⊂ H or V ⊂ K, and consider the unique idempotent ∊ ϵ C(R) such that ∊υ = υ for all υ ϵ V, the annihilator in C(R) of V is Ann_C(R)(V) = (1 – ∊)C(R) and the annihilator in R of the ideal generated by V is Ann_R(Id_R(V)) = (1 – ∊)R. When R has an involution we can decompose ∊ = ∊_k + ∊_h with ∊_k ∊ Skew(C(R), *) and ∊_h ∊ H(C(R), *). We have that ∊υ = υ implies ∊_h∊ = 0. Therefore, ∊_k ∊ Ann_C(R)(V) = (1 – ∊)C(R), i.e., ∊_k∊ = 0 and ${ϵ}_k^2={ϵ}_k{ϵ}_h$ and therefore ∊ = ∊² = (∊^k + ∊^h)² = ${∊}_h^2∊H(C(R),*).$

Lemma 2.0.5. Let R be a centrally closed semiprime associative algebra and let {v_i}_i∊I be a family of idempotent elements in C(R). Suppose there exists a family {λ_i}_i∊I of elements in C(R) such that for every i, j ∊ I, λ_iV_iV_j = λ_jV_iV_j. Then there exists λ ∊ C(R) such that λV_i = λ_iV_i for every i ∊ I. Moreover, if the ideal generated by the family {V_i}_i∊I is essential in R, such X is unique.

Proof. Let us consider the ideal $S=\sum R{\nu }_i$ generated by the family of idempotents {v_i}_i∊I and the essential ideal T = S ⊕ Ann_R(S). Define λ : T ↔ R by

$$\lambda \left(\sum x_i{\nu }_i+z\right):=\sum {\lambda }_i{x}_i{\nu }_i.$$

Let us prove that λ is well defined and an element in C(R). If $\sum x_i{\nu }_i+z=0$ then $\sum x_i{\nu }_i=0=z$ and for every v_k we have

$$\left(\sum {\lambda }_i{x}_i{\nu }_i\right){\nu }_k=\sum {\lambda }_k{x}_i{\nu }_i{\nu }_k={\lambda }_k\left(\sum x_i{\nu }_i\right){\nu }_k=0$$

Therefore $\sum {\lambda }_i{x}_i{\nu }_i\in S\cap$ Ann_R(S) = 0 which proves that λ is well defined. By construction [T, λ] ∊ C(R). Moreover, if the ideal S generated by the family {v_i}_i∊I is essential, Ann_R(S) = 0 and [S, λ] ∊ C(R) is uniquely defined.

2.1 Pure ad-nilpotent elements

Recall that an element a in a Lie algebra L is ad-nilpotent of index ${ad}_a^{n}L=0$ and ${ad}_a^{n-1}L\ne 0.$

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:

(i) a is a pure ad-nilpotent element of R⁻.

(ii) ${\mathrm{Id}}_R({\mathrm{ad}}_a^{n-1}(R))$ is an essential ideal of Id_R(a).

(iii) $${\mathrm{Ann}}_R\left({\mathrm{Id}}_R({\mathrm{ad}}_a^{n-1}(R))\right)={\mathrm{Ann}}_R({\mathrm{Id}}_R(a)).$$

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

(i) ⇒ (ii). Let us consider $V=\left\{{ad}_a^{n-1}x\mid x\in R\right\}$. 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 $0\ne {ad}_{(1-e)a}^{n-1}\left(R\right)=(1-e){ad}_a^{n-1}\left(R\right)=0$, a contradiction. So ea = a and ${Ann}_{I{d}_R\left(ea\right)}\left({Id}_R\left({ad}_a^{n-1}\left(R\right)\right)\right)\subset {Ann}_R\left({Id}_R\left({ad}_a^{n-1}\left(R\right)\right)\right)=(1-e)$R must be zero, i.e., ${Id}_R\left({ad}_a^{n-1}\left(R\right)\right)$ 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 ${ad}_{\lambda a}^n\left(R\right)=0$. Suppose that ${ad}_{\lambda a}^{n-1}\left(R\right)=0$: then ${\lambda }^{n-1}{ad}_{a\mid }^{n-1}\left(R\right)=0$, so λ^n-1 ∊ ${Ann}_R\left({Id}_R\left({ad}_a^{n-11}\left(R\right)\right)\right)={Ann}_R\left({Id}_R\left(a\right)\right)$, 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 ${ad}_{\mu \lambda a}^{n-1}R=0$, and let us prove that μλa = 0:

We have that ${\mu }^{n-1}{\mathrm{ad}}_{\lambda a\mid }^{n-1}R={ad}_{\mu \lambda a}^{n-1}R=0\text{, so }\mu {\mathrm{ad}}_{\lambda a}^{n-1}R=0$ because C(R) is regular von Neumann. In particular, $\mu {ad}_{\lambda a}^{n-1}H=\mu {ad}_{\lambda a}^{n-1}K=0$. Since μ = μ_h + μ^k, we have that ${\mu }_h{\mathrm{ad}}_{\lambda a}^{n-1}R={\mu }_k{\mathrm{ad}}_{\lambda a}^{n-1}R=0.$

From $0={\mu }_h^{n-1}{ad}_{\lambda a}^{n-1}R={ad}_{{\mu }_h\lambda a}^{n-1}R$ 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 $0={\left({\mu }_k^2\right)}^{n-1}{ad}_{\lambda a}^{n-1}R={ad}_{{\mu }_k^2\backslash a}^{n-1}R$ we get that ${\mu }_k^2\lambda a$ has index of ad-nilpotency in K lower than n, so again ${\mu }_k^2\lambda a=0$ (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 ${\left\{{ϵ}_i\right\}}_{i=1}^k\subset C\left(R\right)$ such that $a=\sum _{i=1}^k{ϵ}_{i}a$ 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 ${\left\{{ϵ}_i\right\}}_{i=1}^k\subset H\left(C\right(R),*)$ such that $a=\sum _{i=1}^k{ϵ}_i{a}_i$ 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 $V=\left\{{ad}_a^{n-1}x\mid x\in K\right\}$. 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 $x\in K,{ad}_{(1-ϵ)a}^{n-1}x=(1-ϵ){ad}_a^{n-1}x={ad}_a^{n-1}x-ϵ{ad}_a^{n-1}x=0.$

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 ${ad}_{\lambda ϵa}^n$(Skew(∊R, *)) = 0 and if ${\mathrm{ad}}_{\lambda ϵa}^{n-1}$(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, *).

2.2 Ad-nilpotent elements of 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

$$\sum _{i=0}^n{\lambda }_i{a}^i[x,y]a^{n-i}=0$$

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 $\sum _{i=0}^n{\lambda }_i{a}^i[x,y]a^{n-i}$ by a^s−1−k on the left and by a^{s−1−(n−k)} on the right, so that

$$0=a^{s-1-k}\left(\sum _{i=0}^n{\lambda }_i{a}^i[x,y]a^{n-i}\right)a^{s-1-(n-k)}={\lambda }_k{a}^{s-1}[x,y]a^{s-1}$$

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 $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ -torsion for $s:=\left[\frac{n+1}{2}\right]$, then t = s and n = 2t − 1.

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

$${ad}_a^{2t-1}x=\sum _{i=0}^{2t-1}\left(\genfrac{}{}{0ex}{}{2t-1}{i}\right)(-1{)}^{2t-1-i}{a}^{i}x{a}^{2t-1-i}=0$$

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

$${ad}_a^{t-1}x=\sum _{i=0}^{t-1}\left(\genfrac{}{}{0ex}{}{t-1}{i}\right)(-1{)}^{t-1-i}{a}^{i}x{a}^{t-1-i}=0$$

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 $0={ad}_a^n\left(x\right)=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{a}^{n-i}$, again by Lemma 2.2.1 $\left(\genfrac{}{}{0ex}{}{n}{s}\right)a^s=0$ for $s:=\left[\frac{n+1}{2}\right]$. If R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-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 $t\in \left\{\frac{p+1}{2},\dots ,p\right\}$ 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 $\stackrel{-}{\mathbb{F}}$ denote the algebraic closure of the field 𝔽 := C(R) and R̄ := R̂ ⊗ $\stackrel{-}{\mathbb{F}}$. Then:

1. There exists μ ϵ $\stackrel{-}{\mathbb{F}}$ such that a − μ is a nilpotent element of R̅.

2. If R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ -torsion for $s:=\left[\frac{n+1}{2}\right]$ then n is odd and the index of nilpotence of $a-\mu \text{ is }\frac{n+1}{2}$. If in addition R is free of s-torsion then μ ϵ C(R).

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

$$0={ad}_a^{n}x=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{a}^{n-i}$$

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 $\stackrel{-}{\mathbb{F}}$ be the algebraic closure of 𝔽 and let μ₁, …, μ_r ϵ $\stackrel{-}{\mathbb{F}}$ be the roots of p(X) in $\stackrel{-}{\mathbb{F}}$, i.e., p(X) = (X − μ₁)^k1 … (X - μr)^kr ∊ 𝔽[X].

Let us prove that p(X) has only one root in $\stackrel{-}{\mathbb{F}}$ 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 ${\mathrm{ad}}_a^2$ 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

$$0={ad}_a^{n}x={ad}_b^{n}x=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{b}^{i}x{b}^{n-i}$$

implies by Lemma 2.2.1 that $\left(\genfrac{}{}{0ex}{}{n}{m}\right)b^m=0$ and, since R̅ is free of $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$-torsion, that b^m = 0. Substituting in ${ad}_b^{n-1}x=\sum _{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{i}\right)(-1{)}^{n-1-i}{b}^{i}x{b}^{n-1-i}$ we get that ${ad}_b^{n-1}x=0$ for every x ϵ R, a contradiction.

Therefore n is odd and a − μ is nilpotent of R̅ of index $s:=\frac{n+1}{2}$ 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 $s:=\frac{n+1}{2}.$

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 $[a,[{\mathrm{ad}}_a^{n-1}x,[{\mathrm{ad}}_a^{n-1}x,y]]]=0$ for every x, y ϵ R. If 2, 3, …, r were invertible in R for $r\ge n+\left[\frac{n}{2}\right]+1$, 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 $s:=\left[\frac{n+1}{2}\right]$, and suppose that R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-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 $\frac{n+1}{2}$.

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 $s:=\frac{n+1}{2}$, hence for every x, y ϵ R we have

$$\begin{array}{rl}& \left({\mathrm{ad}}_a^{n-1}x\right)\left({\mathrm{ad}}_a^{n-1}x\right)=\left({\mathrm{ad}}_b^{n-1}x\right)\left({\mathrm{ad}}_b^{n-1}x\right)=0\text{, and}\\ & \begin{array}{rl}& [a,[{\mathrm{ad}}_a^{n-1}x,[{\mathrm{ad}}_a^{n-1}x,y]]]=[b,[{\mathrm{ad}}_b^{n-1}x,[{\mathrm{ad}}_b^{n-1}x,y]]]\\ & =-2\left(\genfrac{}{}{0ex}{}{n-1}{s-1}\right)\left(\genfrac{}{}{0ex}{}{n-1}{s-1}\right)[b,b^{s-1}x{b}^{s-1}y{b}^{s-1}x{b}^{s-1}]=0\end{array}\end{array}.$$

If R is semiprime, R is a subdirect product of prime associative algebras (without $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ and s-torsion) and in any of these prime quotients

$$\left({\mathrm{ad}}_a^{n-1}x\right)\left({ad}_a^{n-1}x\right)=\overline{0}\text{ and }\overline{\left[a,\left[{ad}_a^{n-1}x,\left[{ad}_a^{n-1}x,y\right]\right]\right]}=\overline{0}$$

which imply that

$$\left({ad}_a^{n-1}x\right)\left({ad}_a^{n-1}x\right)=0\text{, and }\left[a,\left[{ad}_a^{n-1}x,\left[{ad}_a^{n-1}x,y\right]\right]\right]=0$$

for every x,y ϵ R. For every x ϵ R, let $z_x:={ad}_a^{n-1}x$. By the identity above,

$$0=\frac{1}{2}\left[a,\left[z_x,\left[z_x,y\right]\right]\right]=-a{z}_{x}y{z}_x+z_{x}y{z}_{x}a.$$

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

$$\begin{array}{rl}0& =z_x{\mathrm{ad}}_a^{n}y=z_x\left(\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\right(-1{)}^{n-i}{a}^{i}y{a}^{n-i})=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{z}_x{a}^{i}y{a}^{n-i}\\ & =\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{z}_x{\lambda }_x^{i}y{a}^{n-i}=z_{x}y\left(\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\right(-1{)}^{n-i}{\lambda }_x^i{a}^{n-i})=z_{x}y{(a-{\lambda }_x)}^n\end{array}$$

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 $0={ϵ}_x{z}_x{\mathrm{ad}}_a^{n}y=z_{x}y(a-\lambda {)}^n$ for every y ∈ R thus (a – λ)ⁿ ∈ Ann_R(Id_R(z_x)) (see 1.2.1). Moreover $\underset{x\in R}{\cap }{\mathrm{Ann}}_R\left({\mathrm{Id}}_R\left(z_x\right)\right)={\mathrm{Ann}}_R\left({\mathrm{Id}}_R\left({\mathrm{ad}}_a^{n-1}\right(R\left)\right)\right)$ by definition of z_x, and ${\mathrm{Ann}}_R\left({\mathrm{Id}}_R\left({\mathrm{ad}}_a^{n-1}\right(R\left)\right)\right)={\mathrm{Ann}}_R\left({\mathrm{Id}}_R\right(a\left)\right)$ 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 $s:=\frac{n+1}{2}$ and take any prime quotient without s and $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion in which $\overline{a-ϵ\lambda }$ is still ad-nilpotent of index n. By Proposition 2.2.3(2) we get that n must be odd and $\overline{a-ϵ\lambda }$ is nilpotent of index s. Since in any prime quotient $(\overline{a-ϵ\lambda }{)}^s=\bar{0}$ 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 $\frac{n+1}{2}$.

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 ${\left\{{ϵ}_i\right\}}_{i=1}^k\subset$ C(R) such that $a=\sum _{i=1}^k{ϵ}_{i}a$ 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 $s_i:=\left[\frac{n_i+1}{2}\right]$ and for all i = 1, …, k. Hence $\lambda =\sum _{i=1}^n{ϵ}_i{\lambda }_i$ 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 ${\mathrm{ad}}_a^n(R)\subset Z(R)$. Then ${\mathrm{ad}}_a^n\left(R\right)=0.$

Proof. For every x ∈ R we have

$$0=\left[{\mathrm{ad}}_a^n\right(xa),x]=[\left({\mathrm{ad}}_a^{n}x\right)a,x]=\left({\mathrm{ad}}_a^{n}x\right)[a,x]$$

Therefore $0={\mathrm{ad}}_a^{n-1}\left(\left({\mathrm{ad}}_a^{n}x\right)\right[a,x\left]\right)={\left({\mathrm{ad}}_a^{n}x\right)}^2$ which implies, since R is semiprime and ${\mathrm{ad}}_a^{n}x\in Z\left(R\right)$, that ${\mathrm{ad}}_a^{n}x=0.$

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 $x\in R,{\mathrm{ad}}_a^{n+1}x={\mathrm{ad}}_a^n\left(\right[a,x\left]\right)\in {\mathrm{ad}}_a^n\left(\right[R,R\left]\right)\subset Z\left(R\right)$ so, by Lemma 2.2.6, ${\mathrm{ad}}_a^{n+1}x=0$ 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

$$0={\mathrm{ad}}_a^n\left(\right[x,y\left]\right)={\mathrm{ad}}_b^n\left(\right[x,y\left]\right)=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{b}^i[x,y]b^{n-i}$$

for every x, y ∈ R. By Lemma 2.2.1, for every k ∈ $\{0,1,\dots ,\left[\frac{n+1}{2}\right]\}$ we have $\left(\genfrac{}{}{0ex}{}{n}{k}\right)b^{max(k,n-k)}=0$, so

$${\mathrm{ad}}_a^{n}x={\mathrm{ad}}_b^{n}x=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{b}^{i}x{b}^{n-i}=0$$

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 $\frac{n+1}{2}$.

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.

2.3 Ad-nilpotent elements of K

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:

1. kKk = 0 implies k = 0.

2. hKh = 0 implies hRh ⊂ H(C(R), *)h. In particular, R satisfies

hxhyh = hyhxh for every x,y ∊ R,

and if Id_R(h) is essential then Skew(C(R), *) = 0.

3. hKh = 0 and hKk = 0 imply hRk = 0. In particular, if Id_R(h) is essential then k = 0, while if h ∊ Id_R(k) then h = 0 (resp. if k ∊ Id_R(h) then k = 0).

4. k[K, K]k = 0 and k² = 0 imply k = 0.

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

0 = —kxk(xky)k = –kxk(xky)*k = kxky*kx*k = kxkykxk,

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

0 = h(xhy – (xhy)*)h = hxhyh – hy*hx*h = hxhyh – (hy*h)x*h =

= hxhyh – hy(hx*h) = hxhyh – hyhxh = (hxh)yh – hy(hxh),

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 ${\mu }_x\in C\left(R\right)$ such that $hxh={\mu }_{x}h$ Hence 0 ≠ hRh ⊂ C(R)h. Moreover, since $h{x}^{\ast }h=hxh,2hxh=hxh+h{x}^{\ast }h=({\mu }_x+{\mu }_x^{\ast })h\in H\left(C\right(R),\ast )h$, 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.

(3) Suppose first that R is *-prime and, without loss of generality, that it is centrally closed. If R is not prime then there is λ ∈ Skew(C(R), *) such that R = K + λK (see 1.3.2), hence hKh = 0 implies hRh = 0 and h = 0 since R is semiprime, so trivially hRk = 0. Now assume R is prime. Since R = H + K we only need to show that hHk = 0. Let x ∈ H and y ∈ R. Then

0 = h(xky – (xky)*)h = hxkyh + hy*kxh = hxkyh + hykxh

since h(y* – y)k = 0 for every y ∈ R. By Corollary 2.0.3, since Id_R(hxk) ⊂ Id_R(h), for each x ∈ R there exists μ_x ∈ C(R) such that hxk = μ_xh. If μ_x = 0 then hxk = 0 and we are done. Otherwise, $0=hxkxk={\mu }_{x}hxk={\mu }_x^{2}h$, hence h = 0 and we are also done.

Suppose now that R is semiprime. Then there exists a family of *-prime ideals {I_α}_α∈Δ such that ∩_α∈Δ I_α = 0. In each *-prime quotient R/I_α we have ћR/I_αk̄ = 0̄, so hRk ⊂ I_α for all α, hence hRk = 0.

(4) Since k² = 0 and k[K, K]k = 0, for all x, y ∈ K we get

0 = k[[x, k], y]k = kxkyk + kykxk,                     (a)

thus kxkyk = —kykxk and 2kxkxk = 0 for all x ∈ K, hence kxkxk = 0 since R is free of 2-torsion. Now, by (a),

0 = (kxkxk)yk = kx(kxkyk) = —kxkykxk

for all x, y ∈ K. Thus (kxk)K(kxk) = 0 for all x ∈ K, kKk = 0 and k = 0 by item (1) applied twice.

Remark 2.3.2. Let R be a semiprime associative algebra with involution. If a ∈ K is an ad-nilpotent element of K of index n, then for every x = x_h + x_k ∈ R with x_h ∈ H and x_k ∈ K:

$$\begin{array}{rl}{\mathrm{ad}}_a^n(ax+xa)& ={\mathrm{ad}}_a^n(a{x}_k+x_{k}a)+{\mathrm{ad}}_a^n(a{x}_h+x_{h}a)\\ & =a{\mathrm{ad}}_a^n\left(x_k\right)+{\mathrm{ad}}_a^n\left(x_k\right)a+{\mathrm{ad}}_a^n(a{x}_h+x_{h}a)=0,\end{array}$$

since ax_h + x_ha ∈ K. On the other hand, expanding this expression,

$$0={\mathrm{ad}}_a^n(ax+xa)=(-1{)}^{n}x{a}^{n+1}+\sum _{i=1}^n(\left(\genfrac{}{}{0ex}{}{n}{i}\right)-\left(\genfrac{}{}{0ex}{}{n}{i-1}\right))(-1{)}^{n-i}{a}^{i}x{a}^{n+1-i}+a^{n+1}x.$$

Observe that a nilpotent element in K is ad-nilpotent of both K and R, but its index of ad-nilpotence in R may be higher than the one found in K.

In the following proposition we describe the ad-nilpotent elements of K of index n that are already nilpotent of certain index s. The description depends on the equivalence class of the index of ad-nilpotence modulo 4 and relates the index of nilpotence to the index of ad-nilpotence.

Proposition 2.3.3. Let R be a semiprime associative algebra with involution *, let R̂ be its central closure, and let a ∈ K be a nilpotent element of index of nilpotence t. Then a is ad-nilpotent in R. If the index of ad-nilpotence of a in K is n and R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ -torsion for $s:=\left[\frac{n+1}{2}\right]$, then:

1. If n ≡₄ 0 then t = s + 1 and a^sKa^s = 0.

2. If n ≡₄ 1 then t = s and the index of ad-nilpotence of a in R is also n.

3. The case n ≡₄ 2 is not possible.

4. If n ≡₄ 3 then there exists an idempotent ∊ ∈ C(R) such that ∊a^s = a^s. Moreover, when we write a = ∊a + (1 – ∊)a, we have:

(4.1) If 0 ≠ ∊a ∈E R̂ then ea is nilpotent of index s + 1, ∊a^s = a^s generates an essential ideal in ∊R̂. and (∊a)^s–1k(∊a)^s = (∊a)^sk(∊a)^s–1 for every k ∈ Skew(R̂, *).

(4.2) If 0 ≠ (1 – ∊)a ∈ R̂, then the index of ad-nilpotence of (1 – ∊)a in R̂ is not greater than n, and (1 – ∊)a^s = 0.

Furthermore, if a is a pure ad-nilpotent element of K then in (2) and in (4.2) we obtain pure ad-nilpotent elements of R (respectively of R̂) of index n.

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

Let a ∈ K be a nilpotent element of index of nilpotence t. Then a is ad-nilpotent of K of a certain index n. If we apply Lemma 2.2.1 to the second formula obtained in Remark 2.3.2 we get that all the monomials appearing in it are zero. We will now focus on certain monomials depending on the parity of n.

• If n is even, n = 2s. Let us see that t = s + 1: on the one hand, for any x ∈ R we know that

$$(\left(\genfrac{}{}{0ex}{}{n}{s}\right)-\left(\genfrac{}{}{0ex}{}{n}{s-1}\right))(-1{)}^s{a}^{s}x{a}^{s+1}=0$$

and, since $\left(\genfrac{}{}{0ex}{}{n}{s-1}\right)-\left(\genfrac{}{}{0ex}{}{n}{s-2}\right)$ is a divisor of $2\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ and R is free of $2\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion, we have that a^sxa^s+1 = 0 for all x. Therefore a^s+1 = 0 by semiprimeness, hence t < s + 1. On the other hand, if t = s then a^s = 0 and ${\mathrm{ad}}_a^{2s-1}\left(R\right)=0$, a contradiction.

Let us see that n ≡₄ 0: For any k ∈ K,

$$0={\mathrm{ad}}_a^{2s}\left(k\right)=\sum _{i=1}^{2s}\left(\genfrac{}{}{0ex}{}{2s}{i}\right)(-1{)}^{2s-i}{a}^{i}k{a}^{2s-i}=\left(\genfrac{}{}{0ex}{}{2s}{s}\right)(-1{)}^s{a}^{s}k{a}^s,$$

so a^s ka^s = 0 for every k ∈ K, which implies that s has to be even, since otherwise a^s ∈ K and a^s Ka^s = 0 imply a^s = 0 by Lemma 2.3.1(1), a contradiction. We have shown that, if n is even, n ≡₄ 2 is not possible.

• If n is odd, n = 2s – 1, and for any x ∈ R,

$$(\left(\genfrac{}{}{0ex}{}{n}{s-1}\right)-\left(\genfrac{}{}{0ex}{}{n}{s-2}\right))a^{s-1}x{a}^{s+1}=0.$$

Since $\left(\genfrac{}{}{0ex}{}{n}{s-1}\right)-\left(\genfrac{}{}{0ex}{}{n}{s-2}\right)$ is a divisor of $2\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ and R is free of $2\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion, we have that a^s–1xa^s+1 = 0 for all x. Therefore a^s+1 = 0 by semiprimeness, hence t ≤ s + 1. On the other hand t > s – 1 since otherwise ${\mathrm{ad}}_a^{2s-2}\left(R\right)=0$, a contradiction.

If a^s = 0 then a is already an ad-nilpotent element of R of index n. In this case n ≡₄ 1 or n ≡₄ 3 by Proposition 2.2.3(2). Furthermore, if a is pure in K then a is pure in R by Lemma 2.1.3.

Suppose from now on that a^s ≠ 0. Let us show that n ≡₄ 3. By Proposition 2.0.4 there exists an idempotent ∊ ∈ H(C(R), *) such that ∊a^s = a^s and Ann_R(Id_R(a^s)) = (1 – ∊)R (so a^s = ∊a^s generates an essential ideal in ∊R). Notice that ∊a ≠ 0 (otherwise 0 = (∊a)^s = ∊a^s = a^s, a contradiction). For every k ∈ K we have

$$0={\mathrm{ad}}_{ϵa}^{n}k=\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}ϵa^{i}k{a}^{n-i}=$$

$$=\left(\genfrac{}{}{0ex}{}{n}{s-1}\right)(-1{)}^sϵa^{s-1}k{a}^s+\left(\genfrac{}{}{0ex}{}{n}{s}\right)(-1{)}^{s-1}ϵa^{s}k{a}^{s-1}=$$

$$=\left(\genfrac{}{}{0ex}{}{n}{s}\right)(-1{)}^{s-1}(-ϵa^{s-1}k{a}^s+ϵa^{s}k{a}^{s-1}).$$

Since R has no $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion, ∊a^s–1ka^s = ∊a^ska^s–1 for every k ∈ K. Moreover, multiplying by a on the right we get ∊a^ska^s = a^ska^s = 0, so a^sKa^s = 0, which by Lemma 2.3.1(1) is only possible if a^s ≠ 0 is symmetric, hence s is even and n ≡₄ 3.

If (1 – ∊)a ≠ 0 then ${\mathrm{ad}}_{(1-ϵ)a}^{2s-1}\left(R\right)=0$ and (1 – ∊)a is an ad-nilpotent element of R of index not greater than 2s – 1.

If a is a pure ad-nilpotent element of index n in K then (1 – ∊)a is ad-nilpotent of K of index n and therefore (1 – ∊)a^s–1 ≠ 0. From this the index of ad-nilpotence of (1 – ∊)a in R must be n = 2s – 1. Then by Lemma 2.1.3 (1 – ∊)a is a pure ad-nilpotent element of R of index n.

Remark 2.3.4. Let a ∈ K be a nilpotent element of index t. If we denote its index of ad-nilpotence in K by n, we obtain from Proposition 2.3.3 that, under the right torsion hypothesis, $2t-3\le n\le 2t-1$ and $\frac{n+1}{2}\le t\le \frac{n+3}{2}.$

The next two results can be joint in one, but in order to clarify our proof we have decided split them in two. Firstly, in the next proposition we prove that a pure ad-nilpotent element of K can be descomposed into two parts, where one part is ad-nilpotent of R and the other part is nilpotent. After that, in Theorem 2.3.6, we apply Proposition 2.3.3 to obtain the classification of a pure ad-nilpotent element of K depending on its index of ad-nilpotence modulo 4.

Proposition 2.3.5. Let R be a semiprime associative algebra with involution *, let R̂ be its central closure, and let a ∈ K be a pure ad-nilpotent element of K of index n > 1. Then:

(1) There exists an idempotent ∊ ∈ H(C(R), *) such that (1 – ∊)a is an ad-nilpotent n element of R̂ of index ≤ n and ∊a is nilpotent with ${\mathrm{ad}}_{\mu ϵa}^n\left(\hat{R}\right)\ne 0$ for every μ ∈ C(R) such that μ∊a ≠ 0.

(2) Moreover, if a is pure ad-nilpotent in K and R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion and t-torsion for $s:=\left[\frac{n+1}{2}\right]$, when we write a = ∊a + (1 – ∊)a we have:

(2.1) If ∊a ≠ 0 then ∊a is nilpotent of index s + 1.

(2.2) If (1 – ∊)a ≠ 0 then (1 – ∊)a is pure ad-nilpotent in R̂ of index n. In this case n is odd and there exists λ ∈ Skew(C(R), *) such that ((1 – ∊)a – λ)^s = 0.

Proof. Notice that n ≥ 3 since ${\mathrm{ad}}_a^2\left(K\right)=0$ implies a ∈ Z(R) by [27, Corollary 4.8] and so ad_a(K) = 0, which is not possible because n > 1 by hypothesis.

(1) Let us suppose first that R is a *-prime associative algebra and, without loss of generality, that it is centrally closed.

(1.a) Case 1: ${\mathrm{ad}}_a^n\left(R\right)=0$ and we get the claim for the idempotent ∊ = 0.

(1.b) Case 2: ${\mathrm{ad}}_a^n\left(R\right)\ne 0$ implies that there are no nonzero skew elements λ in C(R), since otherwise (by 1.3.2) R = K + λK would imply ${\mathrm{ad}}_a^n\left(R\right)=0$; in particular R is prime. Since ${\mathrm{ad}}_a^n\left(K\right)=0$, by the second formula of Remark 2.3.2 and Corollary 2.0.3, a is an algebraic element of R over the field 𝔽 := C(R). Let us consider the minimal polynomial p(X) ∈ 𝔽[X] of a. Let $\stackrel{-}{\mathbb{F}}$ be the algebraic closure of C(R) and let μ₁,…, μ_t ∈ $\stackrel{-}{\mathbb{F}}$ such that p(X) = (X – μ₁)^k1 … (X – μ_s)^ks. Let q₁(X) := p(X)/(X – μ₁), so q₁ (a)a = μ₁q₁(a). Now, for any x ∈ R ⊕ $\stackrel{-}{\mathbb{F}}$,

$$\begin{array}{rl}0& ={\mathrm{ad}}_a^n(ax+xa)q_1\left(a\right)\\ & =a\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{a}^{n-i}{q}_1(a)+\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{a}^{n-i}a{q}_1\left(a\right)\\ & =a\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{\mu }_1^{n-i}{q}_1(a)+\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^{i}x{\mu }_1^{n-i}{\mu }_1{q}_1\left(a\right)\\ & =a\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^i{\mu }_1^{n-i}x{q}_1(a)+\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(-1{)}^{n-i}{a}^i{\mu }_1^{n-i}{\mu }_{1}x{q}_1\left(a\right)\\ & =a{(a-{\mu }_1)}^{n}x{q}_1\left(a\right)+{(a-{\mu }_1)}^n{\mu }_{1}x{q}_1\left(a\right)={(a-{\mu }_1)}^n(a+{\mu }_1)x{q}_1\left(a\right)\end{array}$$

and therefore, since R ⊕ F is a centrally closed prime algebra (see [7, pp. 445–446]), (a – μ₁)ⁿ(a + μ₁) = 0. If μ₁ = 0 then a is nilpotent of index at most n + 1. If μ₁ ≠ 0, since the involution is the identity over C(R) because Skew(C(R), *) = 0, it extends to R ⊕ $\stackrel{-}{\mathbb{F}}$ via (r ⊕ λ)* := r* ⊕ λ, hence 0 = ((a – μ₁)ⁿ)*(a + μ₁)* = (a* – μ₁)ⁿ(a* + μ₁) = (–a – μ₁)ⁿ(–a + μ₁) implies (a + μ₁)ⁿ(a – μ₁) = 0. From the conditions (a – μ₁)ⁿ(a + μ₁) = 0 and (a + μ₁)ⁿ(a – μ₁) = 0 we obtain p(X) = (X – μ₁)(X + μ₁). Thus $a^2={\mu }_1^2$, but then ${\mathrm{ad}}_a^3\left(k\right)=4{\mu }_1^2[a,k]$ for every k ∈ K, a contradiction with n ≥ 3.

Let us study the semiprime case, and suppose without loss of generality that R is centrally closed: If a is already ad-nilpotent in R of index n, take ∊ = 0 and the claim holds. Suppose from now on that ${\mathrm{ad}}_a^n\left(R\right)\ne 0$. By Proposition 2.0.4 let ∊ ∈ H(C(R), *) be an idempotent such that ${ϵ\mathrm{ad}}_a^n\left(x\right)={\mathrm{ad}}_a^n\left(x\right)$ for every x ∈ R, ${\mathrm{Ann}}_R\left({\mathrm{Id}}_R\left({\mathrm{ad}}_a^n\right(R\left)\right)\right)=(1-ϵ)R\text{and}{\mathrm{Ann}}_{C\left(R\right)}\left({\mathrm{ad}}_a^n\right(R\left)\right)=(1-ϵ)C\left(R\right).$. Then ${\mathrm{ad}}_{(1-ϵ)a}^n\left(R\right)=(1-ϵ){\mathrm{ad}}_a^n\left(R\right)=0.$

Let us study the element ∊a: First notice that ${\mathrm{ad}}_{\mu \in a}^{n}R\ne 0$ for every μ such that μ∊a ≠ 0, since otherwise $\mu ϵ{\mathrm{ad}}_a^n\left(R\right)={\mathrm{ad}}_{\mu ϵa}^{n}R=0$ implies μ∊ ∈ ${\mathrm{Ann}}_{C\left(R\right)}\left({\mathrm{ad}}_a^n\right(R\left)\right)=$ = (1 – ∊)C(R) and hence μ∊ = 0, a contradiction. Let us see that ∊a is nilpotent. Since R is semiprime, the intersection of all *-prime ideals of R is zero. Consider the essential *-ideal $S:={\mathrm{Id}}_R\left({\mathrm{ad}}_a^n\right(R\left)\right)\oplus {\mathrm{Ann}}_R\left({\mathrm{Id}}_R\left({\mathrm{ad}}_a^n\right(R\left)\right)\right)={\mathrm{Id}}_R\left({\mathrm{ad}}_a^n\right(R\left)\right)\oplus (1-ϵ)R$. Let us consider the families

$${\mathrm{\Delta }}_1:=\{I{◃}^{\ast }R\mid R/I\text{is}\ast \text{-prime and}S\not\subset I\}$$

and

$${\mathrm{\Delta }}_2:=\{I{◃}^{\ast }R\mid R/I\text{is}\ast \text{-prime and}S\subset I\}.$$

Since S ⊂ ∩_IϵΔ2 I and S is essential, ∩_IϵΔ1 I = 0 and R is a subdirect product of R/I with I ϵ Δ₁. Let us see that in any *-prime quotient ∊a is nilpotent of index not greater than n + 1. Take I ϵ Δ₁ and consider R̄ := R/I. We may have two cases:

• If $\overline{∊}=\stackrel{-}{0}$ then $\stackrel{-}{∊a}=\stackrel{-}{0}.$

• If $\overline{∊}\ne \stackrel{-}{0}$ then ∊̄ = 1̄ ϵ R/I and 1̅ –̅ ∊̅ = 0, so (1 − ∊)R ⊂ I. Moreover, ${\mathrm{ad}}_{\overline{ϵa}}^n(R/I)\ne \bar{0}$ since otherwise ${\mathrm{ad}}_{\overline{ϵa}}^n(R/I)=\bar{0}$ would imply S ⊂ I, a contradiction. Let us see that R/I is prime: if R/I is *-prime and not prime there would exist a nonzero skew element λ in C(R/I), which implies that R/I = Skew(R/I, *) ⊗ λSkew(R/I, *) (see 1.3.2), so ${\mathrm{ad}}_{\overline{ϵa}}^n(R/I)={\mathrm{ad}}_{\overline{ϵa}}^n\left(\mathrm{Skew}\right(R/I,\ast )\oplus \lambda \mathrm{Skew}(R/I,\ast \left)\right)=0$ a contradiction. So R/I is a prime algebra with involution and ${\mathrm{ad}}_{\overline{ϵa}}^n(R/I))\ne \bar{0}$ which implies, by the case (1.b), that $\stackrel{-}{∊a}$ is nilpotent of index not greater than n + 1.

In conclusion, for any I ϵ Δ₁ we have ∊aⁿ⁺¹ ϵ I and therefore ∊aⁿ⁺¹ = 0.

(2) Suppose now that a is a pure element of K of index n and R is free of $2\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion and free of s-torsion for $s:=\left[\frac{n+1}{2}\right]$. If a is already ad-nilpotent of R of index n then a is pure in R by Lemma 2.1.3 and we can use Theorem 2.2.4 to find that n is odd and there exists λ ϵ Skew(C(R), *) such that (a − λ)^s = 0. Otherwise write a = ∊a + (1 − ∊)a as before. Since ∊a is nilpotent and ad-nilpotent of K of index n (because we are assuming that a is pure in K), ∊a is nilpotent of index s + 1 (it has index s or s + 1 by Proposition 2.3.3, but ${\mathrm{ad}}_{ϵ\alpha }^n\left(R\right)\ne 0)$. Moreover, (1 − ∊)a is a pure ad-nilpotent element of R of index n (if it is nonzero, its index of ad-nilpotence cannot be lower than n since (1 − ∊)a is ad-nilpotent in K of index n), and we can apply Theorem 2.2.4 and Lemma 1.3.3 to get that n is odd and there exists λ ϵ Skew(C(R), *) such that ((1 − ∊)a − λ)^s = 0.

Theorem 2.3.6. Let R be a semiprime associative algebra with involution *, let R̂ be its central closure, and let a ϵ K be a pure ad-nilpotent element of K of index n > 1. If R is free of $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$-torsion and s-torsion for $s:=\left[\frac{n+1}{2}\right]$ then:

1. If n ≡⁴ 0 then a^s+1 = 0, a^s = 0 and a^sKa^s = 0. Moreover, there exists an idempotent ∊ ϵ H(C(R), *) such that ∊a = a and the ideal generated by a^s is essential in ∊R̂. In addition ∊R̂ satisfies the GPI a^sxa^sya^s = a^sya^sxa^s for every x,y ∊R̂.

2. If n ≡₄ 1 then there exists λ ϵ Skew(C(R), *) such that (a − λ)^s = 0 (a is an ad-nilpotent element of R of index n).

3. It is not possible that n ≡₄ 2.

4. If n ≡₄ 3 then there exists an idempotent ∊ ϵ H(C(R), *) making a = ∊a + (1 − ∊)a ϵ R̂ such that:

(4.1) If ∊a ≠ 0 then ∊a^s+1 = 0, ∊a^s ≠ 0 and ∊a^skea^s−1 = ∊a^s−1kea^s for every k ϵ Skew(R̂, *). The ideal generated by ∊a^s is essential in ∊R̂ and ∊R̂ satisfies the GPI a^sxa^sya^s = a^sya^sxa^s for every x,y ϵ ∊R̂.

(4.2) If (1 − ∊)a ≠ 0 then there exists λ ϵ Skew(C(R), *) such that ((1 − ∊)a − λ)^s = 0 ((1 − ∊)a is a pure ad-nilpotent element of R̂ of index n).

In particular, for all n > 1 there exists λ ϵ Skew(C(R), *) such that (a − λ)^s+1 = 0, (a − λ)^s−1 ≠ 0.

Proof. We can suppose without loss of generality that R = R̂, i.e., R is centrally closed. By Proposition 2.3.5 there exists an idempotent ∊ ϵ H(C(R), *) such that $ϵ{\mathrm{ad}}_a^{n}x={\mathrm{ad}}_a^{n}x$ for every x ϵ R and ${\mathrm{Ann}}_R\left({\mathrm{Id}}_R({\mathrm{ad}}_a^n(R))\right)=(1-∊)R$, and moreover:

•If ∊a ≠ 0, it is nilpotent of index s + 1 and ad-nilpotent of K of index n. By Proposition 2.3.3 this may happen if either n ≡₄ 0, in which case a^s+1 = 0, as ≠ 0, asKa^s = 0 and (1 − ∊)a = 0 (because (1 − ∊)a is ad-nilpotent of R and

•its index cannot be even), or n = ₄ 3. The case n =₄ 1 is not possible because ∊a^s = 0.

•If (1 − ∊)a ≠ 0 then (1 − ∊)a is a pure ad-nilpotent element of R, n is odd and there exists λ ∊ Skew(R, *) with ((1 − ∊)a − λ)^s = 0. By Proposition 2.3.3 this may happen if either n =₄ 1 (in this case ∊a = 0) or n =₄ 3. The decomposition (1 − ∊)a − λ = a₁ + a₂ given by Proposition 2.3.3(4) occurs with a₁ = 0 since otherwise the index s + 1 of a₁ would contradict ((1 − ∊)a − λ)^s = 0.

In the particular case of n =₄ 3 with ∊a ≠ 0, the idempotent e1 produced in Proposition 2.3.3(4) for the nilpotent element ∊a satisfies ∊₁∊a^s = ∊a^s, so (1 − ∊¹)∊ ∊ ${\mathrm{Ann}}_R\left({\mathrm{Id}}_R({\mathrm{ad}}_a^n(R))\right)=(1-∊)R$, thus ∊₁∊ = ∊ and ∊a^s = ∊₁∊a^s generates an essential ideal in ∊R. On the other hand, we know from Proposition 2.3.5 that (∊a)^s-1 k(∊a)^s = (∊a)^sk(∊a)^s−1 for every k ∊ K; in particular (∊a)^sK(∊a)^s = 0. Therefore, by Lemma 2.3.1(2) the identity

a^sxa^sya^s = a^sya^sxa^s

holds in ∊R̂.

In the particular case of n ≡₄ 0 the idempotent ∊ produced in Proposition 2.3.5 satisfies ∊a^sxa^s = ∊a^s for every x ∊ R and Ann_RId_R(a^sRa^s) = (1 − ∊)R. On the other hand, (1 − ∊)^a must be zero because ${\mathrm{ad}}_{(1-∊)a}^n(R)=0$ and a is a pure ad-nilpotent element (so a = ∊a). Therefore, the ideal generated by a^s in ∊R̂ is essential in ∊R̂ and the identity a^sxa^sya^s = a^sya^sxa^s holds in ∊R̂ by Lemma 2.3.1(2).

Remark 2.3.7. It is worth noting that in the semiprime case, when n ≡₄ 3 there can exist elements a with two nonzero parts ∊a and (1 − ∊)a behaving as in Theorem 2.3.6(4.1) and Theorem 2.3.6(4.2). This is no longer true in the prime case, see [56, Main Theorem].

In the next corollary we recover T.K. Lee’s main result by taking into account that every ad-nilpotent element can be expressed as a sum of pure ad-nilpotent elements of decreasing indices.

Corollary 2.3.8. ([54, Theorem 1.5]) Let R be a semiprime associative algebra with involution * and free of n!-torsion, let R̂ be its central closure, and let a ∊ K be an ad-nilpotent element of K of index n. Then there exist λ ∊ Skew(C(R), *) and an idempotent ∊ ∊ H(C(R), *) such that (∊a – λ)^s+1 = 0 and (∊a – λ)^s−1 ≠ 0 for $s:=\left[\frac{n+1}{2}\right]$, and (1 – e)R̂ is a PI-algebra satisfying the standard identity S4.

Proof. We can suppose without loss of generality that R = R̂, i.e., R is centrally closed. By Proposition 2.1.4 there exists a family of orthogonal symmetric idempotents ${\left\{{ϵ}_i\right\}}_{i=1}^k$ of the extended centroid such that $a=\sum _{i=1}^k{ϵ}_{i}a$ with ∊_ia a pure ad-nilpotent element of index n_i (n = n₁ > n₂ > ...) of Skew(∊_iR, *). If n_k = 1 then ∊_ka can be decomposed as ∊_ka = ∊_k1a + (1 – ∊_k1)a, where ∊_k1^a ∊ Z(R) and (1 – ∊_k1)R is a PI-algebra satisfying the standard identity S₄ by [13, Theorem 4.2(i),(ii) and (*)]. The claim follows now from Theorem 2.3.6.

Let us extend this last result to Lie algebras of the form K/(K ∩ Z(R)) and [K,K]/([K,K] ∩ Z(R)).

Corollary 2.3.9. Let R be a semiprime associative algebra with involution free of n!- torsion, let R̂ be its central closure, and consider the Lie algebra L := K/(K ∩ Z(R)). If ā is an ad-nilpotent element of L of index n then there exist λ ∊ Skew(C(R), *) and an idempotent ∊ ∊ H(C(R), *) such that (∊a – λ)^s+1 = 0 and (∊a – λ)^s−1 ≠ 0 for $s:=\left[\frac{n+1}{2}\right]$, and (1 – ∊)R̂ is a PI-algebra that satisfying the standard identity S₄.

Proof. Let us prove that ${\mathrm{ad}}_a^n(K)\subset Z(R)$ implies ${\mathrm{ad}}_a^n(K)=0$: Suppose first that R is *-prime and, without loss of generality, centrally closed. If ${\mathrm{ad}}_a^n\left(K\right)\ne 0$, there would exist $0\ne \lambda \in {\mathrm{ad}}_a^n\left(K\right)\cap Z\left(R\right)$, so R = K + λK by 1.3.2 and hence ${\mathrm{ad}}_a^n\left(R\right)\subset Z\left(R\right)$, which implies by Lemma 2.2.6 that ${\mathrm{ad}}_a^n\left(R\right)=0$, a contradiction. The same result follows for semiprime algebras because they can be expressed as subdirects product of *-prime quotients.

The claim follows now from Corollary 2.3.8.

Now we turn to Lie algebras of the form [K, K]/([K, K] ∩ Z(R)). We first need a technical lemma.

Lemma 2.3.10. Let R be a semiprime associative algebra with involution * and a ∊ K be such that ${\mathrm{ad}}_a^n\left(\right[K,K\left]\right)\subset Z\left(R\right)$, n > 1. If R is free of (n + 1)!-torsion then ${\mathrm{ad}}_a^n(K)=0.$

Proof. Let us first suppose that R is a *-prime associative algebra and, without loss of generality, that it is centrally closed. If Skew(C(R), *) ≠ 0 then R = K + λK for any 0 ≠ λ ∊ Skew(C(R), *) (see 1.3.2); thus ${\mathrm{ad}}_a^n\left(\right[R,R\left]\right)\subset Z\left(R\right)$, and by Lemma 2.2.7 a is an ad-nilpotent element of R of index n. Otherwise Skew(C(R), *) = 0, in which case R must be prime and K ∩ Z(R) = 0, so ${\mathrm{ad}}_a^n([K,K])=0$. From ${\mathrm{ad}}_a^{n+1}K\subset {\mathrm{ad}}_a^n\left(\right[K,K\left]\right)=0$ and Skew(C(R), *) = 0 we get from Proposition 2.3.5 that a is a nilpotent element of R. Let t be its index of nilpotence. If ${\mathrm{ad}}_a^{n}K=0$ we are done; suppose it is not and let us compare the index of ad-nilpotence of a in K with its index of nilpotence t (see Proposition 2.3.3) to get a contradiction:

(a) If n + 1 ≡₄ 0 then $t=\frac{n+3}{2}$ and a^t–1Ka^t–1 = 0. From $\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)=\left(\genfrac{}{}{0ex}{}{n}{t-1}\right)$ we get, for every x ∊ R, that ${\mathrm{ad}}_a^{n}x=(-1{)}^{t-1}\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)(a^{t-2}x{a}^{t-1}-a^{t-1}x{a}^{t-2})$. Then, for every k, k^’ ∊ K,

$$\begin{array}{rl}& 2\left({\mathrm{ad}}_a^{n}k\right)k^{\prime }\left({\mathrm{ad}}_a^{n}k\right)=\\ & =2\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)(a^{t-2}k{a}^{t-1}{k}^{\prime }{a}^{t-2}k{a}^{t-1}+a^{t-1}k{a}^{t-2}{k}^{\prime }{a}^{t-1}k{a}^{t-2})\\ & =2\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)a^{t-2}k(a^{t-1}{k}^{\prime }{a}^{t-2}-a^{t-2}{k}^{\prime }{a}^{t-1})k{a}^{t-1}+\\ & +2\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)a^{t-1}k(a^{t-2}{k}^{\prime }{a}^{t-1}-a^{t-1}{k}^{\prime }{a}^{t-2})k{a}^{t-2}=\\ & =2(-1{)}^{t-2}\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)(a^{t-2}k\left({\mathrm{ad}}_a^n{k}^{\prime }\right)k{a}^{t-1}-a^{t-1}k\left({\mathrm{ad}}_a^n{k}^{\prime }\right)k{a}^{t-2})=\\ & =(-1{)}^{t-2}\left(\genfrac{}{}{0ex}{}{n}{t-2}\right)(a^{t-2}{\mathrm{ad}}_k^2\left({\mathrm{ad}}_a^n{k}^{\prime }\right)a^{t-1}-a^{t-1}{\mathrm{ad}}_k^2\left({\mathrm{ad}}_a^n{k}^{\prime }\right)a^{t-2})=\\ & ={\mathrm{ad}}_a^n\left({\mathrm{ad}}_k^2\left({\mathrm{ad}}_a^n{k}^{\prime }\right)\right)\in {\mathrm{ad}}_a^n\left(\right[K,K\left]\right)=0\end{array}$$