CHAPTER 3. AD-NILPOTENT ELEMENTS IN A PRIME ASSOCIATIVE SUPERALGEBRA

This chapter is part of an article that has been published in the journal Linear and Multilinear Algebra [28].

In this chapter we are going to study nilpotent inner superderivations in prime associative superalgebras with and without involution.

The goal is to extend the results of the previous chapter to the prime super setting. In the first section we will give a detailed description of a homogeneous ad-nilpotent element a of index n in a prime associative superalgebra R free of $(\binom{n}{s})$ and s-torsion, where $s = [\frac{n + 1}{2}]$ , depending on the degree of the element and the equivalence class of n modulo 4. If a belongs to R₀ we can adjust the techniques and use the results from the previous chapter because R₀ is an algebra. On the other hand, if a ϵ R₁ we will work with a² ϵ R₀ and we will show that the only possible indexes of ad-nilpotency of a are n ≡₄ 1, 2. These two cases correspond to a nilpotent element of index $\frac{n + 1}{2}$ , when n ≡₄ 1, or to an element a for which there exists λ ϵ C(R)₀ with ${(a^{2} - λ)}^{\frac{n + 2}{4}} = 0$ , when n ≡₄ 2.

In the second section we will study ad-nilpotent elements of the skew-symmetric elements K of a prime superalgebra with superinvolution and characteristic p > n, i.e., elements a ϵ K₀ ∪ K₁ such that ${ad}_{a}^{n} K = 0 and {ad}_{a}^{n - 1} K \neq 0$ . The key point is the fact proven in Proposition 3.2.3 that any ad-nilpotent element a of K of index n is either nilpotent or ad-nilpotent of the whole R with the same index n. When a ϵ K is an ad-nilpotent homogeneous even element, it will be classified depending on its index of ad-nilpotency modulo 4 (see Theorem 3.2.4), and when a ϵ K₁ is ad-nilpotent of index n, its description will depend on the congruence class of n modulo 8 (see Theorem 3.2.5): if n ≡₈ 1, 2, 5,6 then a behaves as an ad-nilpotent element of R and if n ≡₈ 0, 7 then a is nilpotent of index s + 1 for $s = [\frac{n + 1}{2}]$ , and a^sKa^s = 0, implying that a^sRa^s is commutative as a local superalgebra at a^s. We will also show that the indexes of ad-nilpotency n ≡₈ 3,4 are not possible.

3.0.1. Let R be an associative superalgebra. We recall that a homogeneous 0-degree linear map * : R → R is a superinvolution in R if (a*)* = a and (ab)* = (-1)^|a||b|b*a* for every homogeneous a,b ϵ R₀ ∪ R₁ In particular

(abc)* = (-1)^{|a||b|+|a||c|+|b||c|}c*b*a*

for homogeneous a, b, c ϵ R₀ ∪ R₁ and and

(abc)* = (-1)^{|a||b|+|a||a|+|b||c|}a*b*a* = (-1)^|a|a*b*a*.

the set of skew-symmetric elements K := {a ϵ R | a* = -a} and the set of symmetric elements H := {a ϵ R | a* = a} are graded submodules of R. Since ½ ϵ Φ, R = H ⊕ K. We will denote H_i = H ⋂ R_i and K_i = K ⋂ R_i, i = 0,1. Notice that

$a \in K_{0} ⟹ {\begin{cases} a^{s} \in H_{0}, & when s is even, \\ a^{s} \in K_{0}, & when s is odd, \end{cases}$

$a \in K_{1} ⟹ {\begin{cases} a^{s} \in H_{0}, & when s \equiv_{4} 0, \\ a^{s} \in K_{1}, & when s \equiv_{4} 1, \\ a^{s} \in K_{0}, & when s \equiv_{4} 2, \\ a^{s} \in H_{1}, & when s \equiv_{4} 3 . \end{cases}$

Moreover, if R is a prime superalgebra and Skew(C(R), *) ≠ 0, then R = K + μK for any nonzero homogeneous μ ϵ Skew(C(R), *) (indeed, μ² ϵ C(R)₀ is invertible because C(R)0 is field, and therefore R ⊆ K + μ²H ⊆ K + μK ⊆ R).

3.0.2. Let a ϵ R₁. Taking into account that ${ad}_{a}^{2} = {ad}_{a^{2}}$ , it is convenient to compute the adjoint map depending on n modulo 4 and focus in the central terms because if a is nilpotent these will remain:

n ≡₄ 0

${ad}_{a}^{2 s} x = {ad}_{a^{2}}^{s} x = \sum_{i = 0}^{s} (\binom{s}{i}) (- 1)^{s - i} a^{2 i} x a^{2 s - 2 i} =$

$= \dots + (\binom{s}{\frac{s}{2} - 1}) (- 1)^{\frac{s}{2} - 1} a^{s - 2} x a^{s + 2} + (\binom{s}{\frac{s}{2}}) (- 1)^{\frac{s}{2}} a^{s} x a^{s} + (\binom{s}{\frac{s}{2} + 1}) (- 1)^{\frac{s}{2} + 1} a^{s + 2} x a^{s - 2} + \dots$

n ≡₄ 1

${ad}_{a}^{2 s - 1} x = {ad}_{a} {ad}_{a}^{2 s - 2} x = {ad}_{a} ({ad}_{a^{2}}^{s - 1} x) = {ad}_{a} (\sum_{i = 0}^{s - 1} (\binom{s - 1}{i}) (- 1)^{s - i} a^{2 i} x a^{2 s - 2 i - 2}) =$

$= {ad}_{a} (\dots + (\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} a^{s - 1} x a^{s - 1} + \dots) =$

$= \dots + (\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} a^{s} x a^{s - 1} - (\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2} + | x |} a^{s - 1} x a^{s} + \dots$

n ≡₄ 2

$\begin{aligned} {ad}_{a}^{2 s} x & = {ad}_{a^{2}}^{s} x = \sum_{i = 0}^{s} (\binom{s}{i}) (- 1)^{s - i} a^{2 i} x a^{2 s - 2 i} = \\ = \dots + (\binom{s}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} a^{s - 1} x a^{s + 1} + (\binom{s}{\frac{s + 1}{2}}) (- 1)^{\frac{s + 1}{2}} a^{s + 1} x a^{s - 1} + \dots \end{aligned}$

n ≡₄ 3

$\begin{aligned} {ad}_{a}^{2 s - 1} x & = {ad}_{a} {ad}_{a}^{2 s - 2} x = {ad}_{a} ({ad}_{a^{2}}^{s - 1} x) = {ad}_{a} (\sum_{i = 0}^{s - 1} (\binom{s - 1}{i}) (- 1)^{s - i} a^{2 i} x a^{2 s - 2 i - 2}) = \\ = {ad}_{a} (\dots + (\binom{s - 1}{\frac{s - 2}{2}}) (- 1)^{\frac{s - 2}{2}} a^{s - 2} x a^{s} + (\binom{s - 1}{\frac{s}{2}}) (- 1)^{\frac{s}{2}} a^{s} x a^{s - 2} + \dots) = \\ = \dots + (\binom{s - 1}{\frac{s}{2} - 1}) (- 1)^{\frac{s}{2} - 1} a^{s - 1} x a^{s} + (\binom{s - 1}{\frac{s}{2}}) (- 1)^{\frac{s}{2}} a^{s + 1} x a^{s - 2} - \\ - (\binom{s - 1}{\frac{s}{2} - 1}) (- 1)^{\frac{s}{2} - 1 + | x |} a^{s - 2} x a^{s + 1} - (\binom{s - 1}{\frac{s}{2}}) (- 1)^{\frac{s}{2} + | x |} a^{s} x a^{s - 1} + \dots \end{aligned}$

Throughout all this chapter we will use these calculations without mentioning them.

3.1. Ad-nilpotent elements of R⁻

In the following result we will relate the index of nilpotence of a homogeneous element of R with its index of ad-nilpotence in R. It will be useful in our study of ad-nilpotent elements of K.

Proposition 3.1.1. Let R = R₀ ⨁ R₁ be a semiprime associative superalgebra. If a ϵ R is a homogeneous nilpotent element of index s and

(1) a ϵ R₀ and R is free of $(\binom{2 s - 2}{s - 1})$ -torsion, then a is ad-nilpotent of R (and of R₀) of index n = 2s − 1,

(2a) a ϵ R₁, s is even and R is free of $(\binom{s - 2}{\frac{s - 2}{2}})$ -torsion, then a is ad-nilpotent of R of index n = 2s − 2 (n ≡₄ 2),

(2b) a ϵ R₁, s is odd and R is free of $(\binom{s - 1}{\frac{s - 1}{2}})$ -torsion, then a is ad-nilpotent of R of index n = 2s − 1 (n ≡₄ 1).

Proof. (1) Since a ϵ R₀, the operator ad_a behaves as the adjoint map in the non-super setting. From a^s = 0 we get that ${ad}_{a}^{2 s - 1} (R) = 0$ . On the other hand, a^s−1 ≠ 0, so by semiprimeness of R (and of R₀) (see Lemma 1.1.6) there exists x ϵ R (respectively, x ϵ R₀) such that a^s−1xa^s−1 ≠ 0 and, since R has no $(\binom{2 s - 2}{s - 1})$ -torsion, $(\binom{2 s - 2}{s - 1}) a^{s - 1} x a^{s - 1} \neq 0$ . Thus

${ad}_{a}^{2 s - 2} (x) = (\binom{2 s - 2}{s - 1}) (- 1)^{s - 1} a^{s - 1} x a^{s - 1} \neq 0 .$

We have shown that a is ad-nilpotent of R (and of R₀) of index n = 2s − 1.

(2a) Suppose that a ϵ R₁ is a nilpotent element of even index s. Since ${ad}_{a}^{2} = {ad}_{a^{2}}$ is nilpotent of index $\frac{s}{2}$ , we have by (1) that a² is ad-nilpotent of R of index $2 (\frac{s}{2}) - 1 = s - 1$ . Hence the index of ad-nilpotence of a is less or equal to 2s − 2.

Let x be any element in R₀ ∪ R₁:

$\begin{aligned} {ad}_{a}^{2 s - 3} (x) & = {ad}_{a}^{2 s - 4} {ad}_{a} (x) = {ad}_{a^{2}}^{s - 2} {ad}_{a} (x) = \\ = (\binom{s - 2}{\frac{s - 2}{2}}) (- 1)^{\frac{s - 2}{2}} a^{s - 2} (a x - (- 1)^{| x |} x a) a^{s - 2} = \\ = (\binom{s - 2}{\frac{s - 2}{2}}) (- 1)^{\frac{s - 2}{2}} a^{s - 1} x a^{s - 2} - (\binom{s - 2}{\frac{s - 2}{2}}) (- 1)^{\frac{s - 2}{2} + | x |} a^{s - 2} x a^{s - 1}, hence \\ {ad}_{a}^{2 s - 3} (x) a & = (\binom{s - 2}{\frac{s - 2}{2}}) (- 1)^{\frac{s - 2}{2}} a^{s - 1} x a^{s - 1} . \end{aligned}$

Therefore ${ad}_{a}^{2 s - 3} (R)$ cannot be zero, since otherwise a^s−1 = 0 because R is free of $(\binom{s - 2}{\frac{s - 2}{2}})$ -torsion and semiprime, a contradiction. We have shown that a is ad-nilpotent of index n = 2s − 2.

(2b) Suppose that a ϵ R₁ is a nilpotent element of odd index s. For any homogeneous x ϵ R₀ ∪ R₁:

$\begin{aligned} {ad}_{a}^{2 s - 1} (x) & = {ad}_{a} {ad}_{a}^{2 s - 2} (x) = {ad}_{a} {ad}_{a^{2}}^{s - 1} (x) = {ad}_{a} ((\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} a^{s - 1} x a^{s - 1}) = \\ = (\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} (a^{s} x a^{s - 1} - (- 1)^{| x |} a^{s - 1} x a^{s}) = 0 \end{aligned}$

so ${ad}_{a}^{2 s - 1} (R) = 0$ . Let us see that ${ad}_{a}^{2 s - 2} (R) \neq 0 : a^{s - 1} \neq 0$ , so there exists x ϵ R such that

${ad}_{a}^{2 s - 2} (x) = {ad}_{a^{2}}^{s - 1} (x) = (\binom{s - 1}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} a^{s - 1} x a^{s - 1} \neq 0$

because R is semiprime and free of $(\binom{s - 1}{\frac{s - 1}{2}})$ -torsion. We have shown that a is ad-nilpotent of index n = 2s − 1.

In the following theorem we describe the homogeneous ad-nilpotent elements of R, depending on the equivalence class of their indexes of ad-nilpotence modulo 4.

Theorem 3.1.2. Let us consider a prime associative superalgebra R = R₀ ⨁ R₁, let R̂ denote the central closure of R, and let a ϵ R₀ ∪ R₁ be a homogeneous ad-nilpotent element of index n. If R is free of $(\binom{n}{s})$ -torsion and free of s-torsion, for $s = [\frac{n + 1}{2}]$ , then:

1. If a ϵ R₀, n is odd and exists λ ϵ C(R)₀ such that a − λ ϵ R̂ is nilpotent of index $\frac{n + 1}{2} .$

2. If a ϵ R₁, then

(a) if n ≡₄ 1 and R is free of $(\binom{\frac{n - 1}{2}}{\frac{s - 1}{2}})$ -torsion, then a is nilpotent of index $\frac{n + 1}{2} .$

(b) if n ≡₄ 2 then there is λ ϵ C(R)₀ such that (a² − λ) ϵ R̂ is nilpotent of index $\frac{n + 2}{4} .$

(c) the cases n ≡₄ 0 and n ≡₄ 3 do not occur.

Proof. We will suppose without loss of generality that R is centrally closed.

(1) Let a ϵ R₀ be an ad-nilpotent element of index n. By Lemma 1.1.6, R is semiprime as an algebra. Moreover, the element a is a pure ad-nilpotent element of R because every graded ideal of R is essential (see 2.1.2). Therefore, we can use Theorem 2.2.4 to obtain that n is odd and there exists λ ϵ C(R) such that a − λ is nilpotent of index $\frac{n + 1}{2}$ . Moreover, a ϵ R₀, R is prime and has no $\frac{n + 1}{2}$ -torsion, so λ ϵ C(R)₀ by Lemma 1.2.5.

(2) Let a ϵ R₁ be an ad-nilpotent element of index n. Let us split our argument in two cases:

(2a) If n is odd, n = 2s − 1 for some s. Then $0 = {ad}_{a}^{n + 1} (R) = {ad}_{a}^{2 s} (R) = {ad}_{a^{2}}^{S} (R)$ , and a² ϵ R₀ is ad-nilpotent of index s (notice that ${ad}_{a^{2}}^{s - 1} (R) = {ad}_{a}^{2 s - 2} (R) = {ad}_{a}^{n - 1} (R) \neq 0)$ . Therefore, by (1), s is odd (equivalently, n ≡₄ 1) and there exists λ ϵ C(R)₀ such that a² − λ is nilpotent of index $\frac{s + 1}{2}$ . Let us see prove that λ = 0: Let us denote $b = {(a^{2} - λ)}^{\frac{s - 1}{2}}$ . Then, for every x ϵ R₀ ∪ R₁,

$\begin{aligned} 0 & = {ad}_{a}^{n} (x) = {ad}_{a} ({ad}_{a^{2}}^{\frac{n - 1}{2}} (x)) = {ad}_{a} ({ad}_{a^{2} - λ}^{\frac{n - 1}{2}} (x)) = \\ = [a, \sum_{i = 0}^{\frac{n - 1}{2}} (\binom{\frac{n - 1}{2}}{i}) (- 1)^{\frac{n - 1}{2} - i} {(a^{2} - λ)}^{i} x {(a^{2} - λ)}^{\frac{n - 1}{2} - i}] = \\ = [a, (\binom{\frac{n - 1}{2}}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} {(a^{2} - ∣ λ)}^{\frac{s - 1}{2}} x {(a^{2} - λ)}^{\frac{s - 1}{2}}] = \\ = [a, (\binom{\frac{n - 1}{2}}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} b x b] = (\binom{\frac{n - 1}{2}}{\frac{s - 1}{2}}) (- 1)^{\frac{s - 1}{2}} (a b x b - (- 1)^{| x |} b x b a) . \end{aligned}$

Since R is free of $(\binom{\frac{n - 1}{2}}{\frac{s - 1}{2}})$ -torsion, we get that

$a b x b = (- 1)^{| x |} b x b a ∣, for every x \in R_{0} \cup R_{1} .$

Take any x ϵ R₀. Multiplying this last equality by a on the left and taking into account that ab = ba we have a²bxb = a(abxb) = a(bxba) = abxab; but a²bxb = ab(ax)b = −b(ax)ba = −abxab because ax ϵ R₁. Then a²bR₀b = abR₀ab = 0. Similarly, for any x ϵ R₁ we have that a²bxb = a(abxb) = −a(bxab), and we also have that a²bxb = ab(ax)b = b(ax)ba = abxab because ax ϵ R₀. Then a²bR₁b = abR₁ab = 0. We have obtained

a²bRb = abRab = 0.

From the definition of b we have that (a² − λ)b = 0, i.e., a²b = λb, so 0 = a²bRb = λbRb. If λ ≠ 0, we would have that bRb = 0 (notice that λ ϵ C(R)₀ and C(R)₀ is a field (Lemma 1.2.6)), leading to a contradiction with the semiprimeness of R and b ≠ 0.

Thus λ = 0, so 0 ≠ b = a^s−1, ab = a^s and 0 = abRab = a^sRa^s implies a^s = 0 by semiprimeness of R.

(2b) If n is even, then n = 2s for some s, so a² ϵ R₀ is ad-nilpotent of index s $({ad}_{a^{2}}^{s} (R) = {ad}_{a}^{n} (R) = 0 and {ad}_{a^{2}}^{s - 1} (R) = {ad}_{a}^{2 s - 2} (R) = {ad}_{a}^{n - 2} (R) \neq 0)$ . Then by (1) we obtain that s is odd (equivalently, n ≡₄ 2) and there exists λ ϵ C(R)₀ such that ${(a^{2} - λ)}^{\frac{s + 1}{2}} = 0 .$

Notice that the cases n ≡₄ 0 and n ≡₄ 3 do not occur.

3.2. Ad-nilpotent elements of K

As in the non-super setting, the associative local superalgebra at the ad-nilpotent element give us extra information about the structure. In non-super setting for example we get that the GPI a^txa^tya^t = a^tya^txa^t holds for an ad-nilpotent element a of index n ≡₄ 0 of K for every x, y ϵ K.

3.2.1. Let R be an associative superalgebra over Φ and take an element a ϵ R₀ ∪ R₁. Then R_a := aRa with (aRa)_i := aR_i+|a|a, i ϵ {0, 1}, is a ℤ₂-graded Φ-module. Moreover, the product (axa)(aya) := axaya for any x,y ϵ R induces an associative superalgebra structure in R_a, which is called the local superalgebra of R at a. When R is an associative superalgebra with superinvolution *, the superinvolution induces a superinvolution ⋆ in R_a given by (axa)^⋆ := (−1)^|a|ax^*a, for every x ϵ R.

We start with a technical lemma, which is also interesting by itself. For example, it claims that every semiprime superalgebra with superinvolution and no nonzero skew even elements is a trivial superalgebra, i.e., the odd part is zero.

Lemma 3.2.2. Let R = R₀ ⨁ R₁ be a semiprime associative superalgebra with superinvolution *.

(i) If K₀ = 0 then R₁ = 0 and R = R₀ = H₀ is commutative.

(ii) Let us consider h₀ ϵ H₀. If h₀K₀h₀ = 0 then h₀R₁h₀ = 0 and h₀Rh₀ = h₀R₀h₀ = h₀H₀h₀ is commutative as the (trivial) local superalgebra of R at h₀.

Proof. (i) Take any k₁, k′₁ ϵ K₁ and h₁, h′₁ϵ H₁. Then, since R₀ = H₀, we have that

k₁h₁ = (k₁h₁)* = h₁k₁, k₁k′₁ = (k₁k′₁)* = −k′₁k₁, h₁h′₁ = (h₁h′₁)* = −h′₁h₁.

In particular, $k_{1}^{2} = h_{1}^{2} = 0 .$

We claim that K₁ = 0. Take any k₁ ϵ K₁. Then for every h₀ ϵ H₀, k₁h₀k₁ = (k₁h₀k₁)* = −k₁h₀k₁ implies k₁h₀k₁ = 0, so k₁H₀k₁ = 0; similarly, for every h₁ ϵ H₁, (k₁h₁)k₁ = h₁k²₁ = 0, so k₁H₁k₁ = 0, and, for every k′₁ ϵ K₁, (k₁k′₁)k₁ = −k′₁k²₁ = 0, so k₁K₁k₁ = 0. We have shown that k₁Rk₁ = 0, so by semiprimeness of R, k₁ = 0.

Let us show that H₁ = 0. Take any h₁ ϵ H₁. For every h₀ ϵ H₀, since h₁h₀h₁ = (h₁h₀h₁)* = −h₁h₀h₁, we have that h₁h₀h₁ = 0, so h₁H₀h₁ = 0. Similarly, for every h′₁ ϵ H₁, h₁h′₁h₁ = −h′₁h²₁ = 0, so h₁H₁h₁ = 0, and, finally, for every k₁ ϵ K₁, h₁k₁h₁ = k₁h²₁ = 0, so h₁K₁h₁ = 0. We have shown that h₁Rh₁ = 0, so by semiprimeness of R, h₁ = 0.

Therefore, R₁ = H₁ + K₁ = 0.

Finally, H₀ is commutative because for every h₀, h′₀ ϵ H₀,

$h_{0} h_{0}^{'} = {(h_{0} h_{0}^{'})}^{*} = h_{0}^{'} h_{0}$

(ii) Take h₀ ϵ H₀ and let us consider the local algebra R_h0 = h₀Rh₀ as defined in 3.2.1, which is an associative superalgebra with induced superinvolution (h₀xh₀)^⋆ := h₀x*h₀, for every x ϵ R. Clearly Skew(h₀Rh₀, ⋆) = h₀Kh₀ and Sym(h₀Rh₀, ⋆) = h₀Hh₀. If we suppose that h₀K₀h₀ = 0 then Skew(h₀Rh₀, ⋆)₀ = 0 and by (i) we have

(R_h0)₁ = h₀R₁h₀ = 0 and R_h0 = h₀Rh₀ = (R_h0)₀ = h₀R₀h₀ = h₀H₀h₀.

Proposition 3.2.3. Let R be a prime associative superalgebra with superinvolution * and let a ϵ K be a homogeneous ad-nilpotent element of K of index n > 2. Suppose that R is free of $(\binom{n}{s})$ -torsion and free of s-torsion, for $s = [\frac{n + 1}{2}]$ . If Skew(C(R), *) ≠ 0 then a is ad-nilpotent of R of index n. Otherwise, a is nilpotent.

Proof. If there exists a homogeneous 0 ≠ λ ϵ Skew(C(R), *) then λ² is invertible in the field C(R)₀, and R = K + λ²H ⊆ K + λK so ${ad}_{a}^{n} (R) = 0$ . Suppose from now on that Skew(C(R), *) = 0. We split our proof in two cases, depending on the parity of a:

(I) Suppose that a ϵ K₀. Let us see that a is nilpotent. Every x ϵ R can be expressed as x = x_h + x_k, so for every x ϵ R

$\begin{aligned} {ad}_{a}^{n} (a x + x a) & = {ad}_{a}^{n} (a x_{k} + x_{k} a) + {ad}_{a}^{n} (a x_{h} + x_{h} a) = a {ad}_{a}^{n} (x_{k}) + {ad}_{a}^{n} (x_{k}) a \\ + {ad}_{a}^{n} (a x_{h} + x_{h} a) = 0 \end{aligned}$

because ax_h + x_ha ϵ K and $a {ad}_{a}^{i} (x) = {ad}_{a}^{i} (x)$ for every x ϵ R and any i ϵ ℕ.

Expanding this expression

$0 = {ad}_{a}^{n} (a x + x a) = (- 1)^{n} x a^{n + 1} + \sum_{i = 1}^{n} ((\binom{n}{i}) - (\binom{n}{i - 1})) (- 1)^{n - i} a^{i} x a^{n + 1 - i} + a^{n + 1} x .$

Since R is semiprime as an algebra, by Lemma 2.0.3, a is an algebraic element of R over C(R).

(I.a) Let us suppose that R is prime as an algebra. The calculations of (1.b) in the proof of Proposition 2.3.5 [12, Proposition 5.5] show that a is nilpotent.

(I.b) If R is prime as a superalgebra but not prime as an algebra, R₀ is prime by 1.1.7, C(R)₀ ≅ C(R₀) by 1.2.4, the superinvolution * restricted to R₀ is an involution and Skew(C(R₀), *) = 0 because we are assuming that Skew(C(R), *) = 0. The element a is a pure ad-nilpotent element of K₀ because C(R₀) is a field, so we can apply Proposition 2.3.5(2) to the prime associative algebra R₀ to obtain that a is nilpotent.

(II) If a ϵ K₁, consider a² ϵ K₀ and by (I), a² is nilpotent, i.e., a is nilpotent.

In the following two theorems we will describe the homogeneous ad-nilpotent elements of K. Our goal is to relate the index of ad-nilpotence of a homogeneous element of K with its index of ad-nilpotence in R (and in R₀ and in K₀ when the element is even). Moreover, when these indexes in K and in R do not coincide, we will show that the element is nilpotent of an explicit index.

We begin with the description of even ad-nilpotent elements of K.

Theorem 3.2.4. Let R be a prime associative superalgebra of characteristic p > n with superinvolution *, let R̂ be its central closure, let a ϵ K₀ := Skew(R, *)₀ be an ad-nilpotent element of K of index n > 1 and let $s = [\frac{n + 1}{2}] .$

Then

(1) If n ≡₄ 0 then a is nilpotent of index s + 1, ad-nilpotent of R and of R₀ of index n + 1 and satisfies a^sKa^s = 0. Moreover, the index of ad-nilpotence of a in K₀ can be n − 1 or n.

(2) If n ≡₄ 1 then there exists λ ϵ Skew(C(R), *)₀ such that a − λ ϵ R̂ is nilpotent of index s and a is ad-nilpotent of R, of R₀ and of K₀ of index n.

(3) The case n ≡₄ 2 is not possible.

(4) If n ≡₄ 3 then either:

(4.1) a is nilpotent of index s + 1, ad-nilpotent of K₀ of index n, ad-nilpotent of R and of R₀ of index n + 2 and satisfies a^ska^s−1 − a^s−1ka^s = 0 for every k ϵ K. In particular R satisfies a^sKa^s = 0, or

(4.2) there exists λ ϵ Skew(C(R), *)₀ such that a − λ ϵ R̂ is nilpotent of index s and a is ad-nilpotent of R, of R₀ and of K₀ of index n.

Proof. Suppose without loss of generality that R is centrally closed. Let a ϵ K₀ be an ad-nilpotent element of K of index n.

-If Skew(C(R), *) ≠ 0, by Proposition 3.2.3, a is ad-nilpotent of index n of R and by Theorem 3.1.2 n has to be odd (n ≡₄ 1 or n ≡₄ 3) and there exists λ ϵ C(R)₀ such that a − λ is nilpotent of index s, so a is ad-nilpotent of R and of R₀ of the same index n = 2s − 1, see Proposition 3.1.1(1). Moreover, λ ϵ Skew(C(R), *)₀ by Lemma 1.3.4 and since Skew(C(R), *)₀ ⊂ Skew(C(R₀), *), the index of ad-nilpotence of a − λ in K₀ is again n = 2s − 1 (notice that, by Lemma 1.3.4, λ is the unique element of C(R₀) such that a − λ is nilpotent). These are the cases (2) and (4.1).

-If Skew(C(R), *) = 0, by Proposition 3.2.3, a is nilpotent. We are going to approach this case considering the index of ad-nilpotence of a in K₀ and comparing it with its index of ad-nilpotence in K and in R. Let us suppose that a is ad-nilpotent of K₀ of index m ≤ n and let $r = \frac{m + 1}{2}$ . Since R₀ is a semiprime algebra and the superinvolution * restricted to R₀ is an involution, by Proposition 2.3.3 we have four possibilities:

-m ≡₄ 0 then a is nilpotent of index r + 1 and a^rK₀a^r = 0, which, by Lemma 3.2.2(ii), implies that a^rR₁a^r = 0, so a is also ad-nilpotent of index m of K, i.e., m = n and a is nilpotent of index s + 1 with $s = \frac{n}{2} = r$ . Now, since s + 1 is the index of nilpotence of a, by Proposition 3.1.1(1) a is ad-nilpotent of index n + 1 of R and of R₀. This is the case (1) (n ≡₄ 0) with the index of ad-nilpotence of a in K₀ equal to the index of ad-nilpotence of a in K.

•m ≡₄ 1 then a is nilpotent of index r. This implies, by Proposition 3.1.1(1), that a is ad-nilpotent of R and of R₀ of index m. So n has to be equal to m and therefore the index of nilpotence of a is $s = \frac{n + 1}{2} = r$ . This is the case (2), i.e., n ≡₄ 1.

•m ≡₄ 2 does not occur.

•m ≡₄ 3 then there exists an idempotent ϵ ϵ C(R₀) such that ϵa^r = a^r and a decomposes as a = ϵa + (1 − ϵ)a (although the elements ϵa and (1 − ϵ)a do not belong to R but in central closure of R₀, this decomposition will be useful for our purposes):

⬦ If ϵa = 0 then a = (1 − ϵ)a is nilpotent of index r. By Proposition 3.1.1(1), this implies that a is ad-nilpotent of R and of R₀ of index m, so n = m and the index of nilpotence of a is $s = \frac{n + 1}{2} = r$ . This is the case (4.2), i.e., n ≡₄ 3.

⬦ If ϵa ≠ 0 then a is nilpotent of index r+1 and a^rk₀a^r⁻¹−a^r⁻¹k₀a^r = (ϵa)^rk₀(ϵa)^r−1 − (ϵa)^r−1k₀(ϵa)^r = 0 for every k₀ ϵ K₀. Since a^r+1 = 0, a^rK₀a^r = 0 and, by Lemma 3.2.2(ii), a^rR₁a^r = 0, so a^rKa^r = 0 and therefore ${ad}_{a}^{m + 1} K = 0$ . There are two possibilities:

– Either a^rka^r−1 − a^r−1ka^r = 0 for every homogeneous k ϵ K and therefore a is ad-nilpotent of index m of K. Then $n = m, r = \frac{n + 1}{2} = s$ , so a^ska^s−1 − a^s−1ka^s = 0 and a is nilpotent of index s+1 which, by Proposition 3.1.1(1), implies that a is ad-nilpotent of R and of R₀ of index n + 2 and fits with the case (4.1), i.e., n ≡₄ 3,

– or there exists k ϵ K such that a^rka^r−1 − a^r−1ka^r ≠ 0, so a is ad-nilpotent of K of index m + 1. Hence $n = m + 1, r = \frac{n}{2} = s$ , and a is nilpotent of index s + 1. Therefore, by Proposition 3.1.1(1), a is ad-nilpotent of R and of R₀ of index n + 1. This is again case (1) with the index of ad-nilpotence of a in K₀ equal to n − 1 and n ≡₄ 0.

In the following theorem we describe the odd ad-nilpotent elements of K. We will first distinguish whether C(R) has skew-symmetric elements, in which case a is ad-nilpotent of R of the same index, or Skew(C(R), *) = 0, which implies by Proposition

3.2.3 that a is nilpotent. In this second case, we will consider a² ∊ K₀ and use Theorem 3.2.4 applied to a² to obtain the description of a.

Theorem 3.2.5. Let R be a prime associative superalgebra of characteristic p > n with superinvolution *, let R̂ be its central closure, let a ∊ K₁ := Skew(R, *)₁ be an ad-nilpotent element of K of index n > 1 and let $s = [\frac{n + 1}{2}] .$

(1) If n ≡₈ 0 then a is nilpotent of index s + 1, ad-nilpotent of R of index n + 1 and a^sKa^s = 0 (so a^sRa^s is a commutative trivial local superalgebra).

(2) If n ≡₈ 1 then a^s−1 ∊ H₀, and a is nilpotent of index s and ad-nilpotent of R of index n.

(3) If n ≡₈ 2 then there exists λ ∊ Skew(C(R), *)₀ such that a² — λ ∊ R̂ is nilpotent of index $\frac{s + 1}{2}$ and a is ad-nilpotent of R of index n.

(4) If n ≡₈ 5 then a^s−1 e K₀, and a is nilpotent of index s and ad-nilpotent of R of index n.

(5) If n ≡₈ 6 then there exists λ ∊ Skew(C(R), *)₀ such that a² – λ ∊ R̂ is nilpotent of index $\frac{s + 1}{2}$ and a is ad-nilpotent of R of index n.

(6) If n ≡₈ 7 then a is nilpotent of index s + 1, ad-nilpotent of R of index n + 2 and a^ska^s−1 + (–1)^ka^s−1ka^s = 0 for every homogeneous k ∊ K (so a^sRa^s is a commutative trivial local superalgebra).

(7) The cases n ≡₈ 3 and n ≡₈ 4 do not occur.

Proof. Suppose without loss of generality that R is centrally closed.

Let a ∊ K₁ be an ad-nilpotent element of K of index n. If Skew(C(R), *) ≠ 0, by Proposition 3.2.3, a is ad-nilpotent of R of index n. By Theorem 3.1.2 n can be:

•n ≡₄ 1 and therefore a is nilpotent of index s (cases (2) and (4)), or

•n ≡₄ 2 and therefore there exists λ ∊ Skew(C(R)₀, *) such that a² – λ is nilpotent of index $\frac{s + 1}{2}$ (cases (3) and (5)).

Let us suppose that Skew(C(R), *) = 0. By Proposition 3.2.3, a is nilpotent. Then, since a² ∊ K₀ and ${ad}_{a}^{2} (x) = {ad}_{a^{2}} (x)$ , a² is an ad-nilpotent element of K. Let us denote by m the index of ad-nilpotence of a² in K and let $r = [\frac{m + 1}{2}]$ . By Theorem 3.2.4 applied to the element a² we have:

•If m ≡₄ 0 and $r = \frac{m}{2}, {(a^{2})}^{r} \neq 0, {(a^{2})}^{r + 1} = 0$ . We are going to show that a^2r+1 = 0: let x be any homogeneous element in R, so ax + (–1)^|x|x*a ∊ K_1+|x|,

$\begin{aligned} 0 & = {ad}_{a^{2}}^{m} (a x + (- 1)^{| x |} x^{*} a) a = (\binom{m}{\frac{m}{2}}) (- 1)^{\frac{m}{2}} (a^{m} (a x + (- 1)^{| x |} x^{*} a) a^{m}) a = \\ = (\binom{m}{r}) (- 1)^{r} a^{2 r} (a x + (- 1)^{| x |} x^{*} a) a^{2 r + 1} = (\binom{m}{r}) (- 1)^{r} a^{2 r + 1} x a^{2 r + 1} \\ + (\binom{m}{r}) (- 1)^{r} (- 1)^{| x |} a^{2 r} x^{*} a^{2 r + 2} = (\binom{m}{r}) (- 1)^{r} a^{2 r + 1} x a^{2 r + 1} . \end{aligned}$

Since R is semiprime and free of $(\binom{m}{r})$ -torsion, a^2r+1 = 0. Moreover, since $a d_{a^{2}}^{m - 1} (K) \neq 0$ , we have two possibilities:

•If ${ad}_{a}^{2 m - 1} (K) \neq 0$ , then a is an ad-nilpotent element of K of index n = 2m. In this case n ≡₈ 0 and for $s = \frac{n}{2}$ we have that a^s+1 = 0, a^s ≠ 0 and a^sKa^s = 0. Moreover, by Proposition 3.1.1, a is ad-nilpotent of R of index n + 1, case (1).

•If ${ad}_{a}^{2 m - 1} (K) = 0$ , then a is an ad-nilpotent element of K of index n = 2m – 1. So in this case we have got n ≡₈ 7 and for $s = \frac{n + 1}{2}$ we have that a^s+1 = 0, a^s ≠ 0. Moreover, for every homogeneous k ∊ K,

$\begin{aligned} 0 & = {ad}_{a}^{2 m - 1} (k) = (\binom{m - 1}{\frac{m}{2}}) (- 1)^{\frac{m}{2}} (a^{m} k a^{m - 1} + (- 1)^{| k |} a^{m - 1} k a^{m}) = \\ = (\binom{m - 1}{\frac{s}{2}}) (- 1)^{\frac{s}{2}} (a^{s} k a^{s - 1} + (- 1)^{| k |} a^{s - 1} k a^{s}) \end{aligned}$

and since R is free of $(\binom{m - 1}{\frac{s}{2}})$ -torsion we have that a^ska^s−1 + (–1)^|k|a^s−1ka^s = 0. In addition, by Proposition 3.1.1, a is ad-nilpotent element of R of index n + 2, case (6).

•If m ≡₄ 1 and $r = \frac{m + 1}{2}$ we have that (a²)^r = 0, (a²)^r–1 ≠ 0 and ${ad}_{a^{2}}^{m} (R) = 0$ . Since $a d_{a^{2}}^{m - 1} (K) \neq 0$ , we have two possibilities:

•If ${ad}_{a}^{2 m - 1} (K) \neq 0$ , then a is an ad-nilpotent element of K of index n = 2m and there exists a homogeneous k in K such that:

$\begin{aligned} 0 & \neq {ad}_{a}^{2 m - 1} (k) = {ad}_{a^{2}}^{m - 1} {ad}_{a} (k) = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} (a^{m} k a^{m - 1} - (- 1)^{| k |} a^{m - 1} k a^{m}) = \\ = (\binom{m - 1}{r}) (- 1)^{r} (a^{2 r - 1} k a^{2 r - 2} - (- 1)^{| k |} a^{2 r - 2} k a^{2 r - 1}) . \end{aligned}$

Therefore, since R is free of $(\binom{m - 1}{r})$ -torsion, a^2r−1 ≠ 0. In this case n ≡₈ 2 and for $s = \frac{n}{2}$ we have that a^s+1 = 0, a^s ≠ 0. By Proposition 3.1.1, a is ad-nilpotent of index n, case (3).

•If ${ad}_{a}^{2 m - 1} (K) = 0$ , then a is ad-nilpotent of K of index n = 2m – 1. Let x be any homogeneous element in R and let us consider ax + (–1)^|x|x*a ∊ K₁₊|_x|:

$\begin{aligned} 0 & = {ad}_{a}^{2 m - 1} (a x + (- 1)^{| x |} x^{*} a) = {ad}_{a}^{2 m - 2} {ad}_{a} (a x + (- 1)^{| x |} x^{*} a) = \\ = {ad}_{a^{2}}^{m - 1} {ad}_{a} (a x + (- 1)^{| x |} x^{*} a) = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m - 1} (a^{2} x + (- 1)^{| x |} a x^{*} a - (- 1)^{1 + | x |} (a x a + (- 1)^{| x |} x^{*} a^{2})) a^{m - 1} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} a^{2 r - 2} (a^{2} x + (- 1)^{| x |} a x^{*} a - (- 1)^{1 + | x |} (a x a + (- 1)^{| x |} x^{*} a^{2})) a^{2 r - 2} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{\frac{m - 1}{2} + | x |} a^{2 r - 1} (x^{*} + x) a^{2 r - 1} \end{aligned}$

and

$\begin{aligned} 0 & = {ad}_{a}^{2 m - 1} (x - x^{*}) a = {ad}_{a}^{2 m - 2} {ad}_{a} (x - x^{*}) a = {ad}_{a^{2}}^{m - 1} {ad}_{a} (x - x^{*}) a = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m - 1} (a x - a x^{*} - (- 1)^{| x |} (x a - x^{*} a)) a^{m} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} a^{2 r - 2} (a x - a x^{*} - (- 1)^{| x |} (x a - x^{*} a)) a^{2 r - 1} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} a^{2 r - 1} (x - x^{*}) a^{2 r - 1} \end{aligned}$

Therefore, since R is free of $(\binom{m - 1}{r - 1})$ -torsion, a^2r−1Ra^2r−1 = 0, and by semiprimeness of R, a2r−1 = 0 and a is an ad-nilpotent element of R of index n = 2m – 1. So n ≡₈ 1 and for $s = \frac{n + 1}{2}$ we have that a^s = 0, as⁻¹ ≠ 0. By Proposition 3.1.1, a is ad-nilpotent of R of index n, case (2).

•m ≡₄ 2 is not possible.

•If m ≡₄ 3 and $r = \frac{m + 1}{2}$ , let us first see that (a²)^r = 0. Suppose otherwise that (a²)^r ≠ 0. Then (a²)^r+1 = 0 and a^2rka^2r−2 – a^2r−2 ka^2r = 0 for every k ∊ K. Let x be any homogeneous element in R and let us consider ax + (–1)^|x|x^*a ∊ K₁₊|_x|:

$\begin{aligned} 0 & = {ad}_{a^{2}}^{m} (a x + (- 1)^{| x |} x^{*} a) a^{3} = (\binom{m}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m + 1} (a x + (- 1)^{| x |} x^{*} a) a^{m + 2} + \\ + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{m - 1} a x a^{m + 4} + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{m - 1} (- 1)^{| x |} x^{*} a a^{m + 4} = \\ = (\binom{m}{r - 1}) (- 1)^{r - 1} a^{2 r} (a x + (- 1)^{| x |} x^{*} a) a^{2 r + 1} + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{2 r - 2} a x a^{2 r + 3} \\ + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{2 r - 2} (- 1)^{| x |} x^{*} a a^{2 r + 3} = (\binom{m}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{2 r + 1} x a^{2 r + 1} \end{aligned}$

and therefore, since R is free of $(\binom{m}{r - 1})$ -torsion and semiprime, a^2r+1 = 0. Then for every homogeneous x ∊ R

$\begin{aligned} 0 & = a {ad}_{a^{2}}^{m} (a x + (- 1)^{| x |} x^{*} a) = (\binom{m}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m + 2} (a x + (- 1)^{| x |} x^{*} a) a^{m - 1} + \\ + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{m} a x a^{m + 1} + (\binom{m}{\frac{m + 1}{2}}) (- 1)^{\frac{m + 1}{2}} a^{m} (- 1)^{| x |} x^{*} a a^{m + 1} = \\ = (\binom{m}{r - 1}) (- 1)^{r - 1} a^{2 r + 1} (a x + (- 1)^{| x |} x^{*} a) a^{2 r - 2} + \\ + (\binom{m}{r}) (- 1)^{r} a^{2 r - 1} a x a^{2 r} + (\binom{m}{r}) (- 1)^{r} a^{2 r - 1} (- 1)^{| x |} x^{*} a a^{2 r} = (\binom{m}{r}) (- 1)^{r} a^{2 r} x a^{2 r} \end{aligned}$

and therefore, since R is free of $(\binom{m}{r})$ -torsion and semiprime, a^2r = 0, a contradiction. Thus (a²)^r−1 = 0, (a²)^r−1 ≠ 0 and $({ad}_{a^{2}}^{m} (R) = 0 .$

•If ${ad}_{a}^{2 m - 1} (K) \neq 0$ , then a is ad-nilpotentof K of index n = 2m and there exists k ∊ K homogeneous such that

$\begin{aligned} 0 & \neq {ad}_{a}^{2 m - 1} (k) = {ad}_{a}^{2 m - 2} {ad}_{a} (k) = {ad}_{a^{2}}^{m - 1} {ad}_{a} (k) = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} (a^{m} k a^{m - 1} - (- 1)^{| k |} a^{m - 1} k a^{m}) = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} (a^{2 r - 1} k a^{2 r - 2} - (- 1)^{| k |} a^{2 r - 2} k a^{2 r - 1}) \end{aligned}$

Therefore, since R is free of $(\binom{m - 1}{r - 1})$ -torsion, a^2r−1 ≠ 0 so a is nilpotent of index 2r. So n ≡₈ 6 and with $s = \frac{n}{2}, a^{s + 1} = 0, a^{s} \neq 0$ and by Proposition 3.1.1 a is ad-nilpotent of R of index n, case (5).

•If ${ad}_{a}^{2 m - 1} (K) = 0$ , then a is ad-nilpotent of K of index n = 2m – 1. Let x be any homogeneous element in R and let us consider ax + (–1)^|x|x*a ∊ K₁₊|_x|:

$\begin{aligned} 0 & = {ad}_{a}^{2 m - 1} (a x + (- 1)^{| x |} x^{*} a) = {ad}_{a}^{2 m - 2} {ad}_{a} (a x + (- 1)^{| x |} x^{*} a) = \\ = {ad}_{a^{2}}^{m - 1} {ad}_{a} (a x + (- 1)^{| x |} x^{*} a) = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m - 1} (a^{2} x + (- 1)^{| x |} a x^{*} a - (- 1)^{1 + | x |} (a x a + (- 1)^{| x |} x^{*} a^{2})) a^{m - 1} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} a^{2 r - 2} (a^{2} x + (- 1)^{| x |} a x^{*} a - (- 1)^{1 + | x |} (a x a + (- 1)^{| x |} x^{*} a^{2})) a^{2 r - 2} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1 + | x |} a^{2 r - 1} (x^{*} + x) a^{2 r - 1}, \end{aligned}$

and

$\begin{aligned} 0 & = {ad}_{a}^{2 m - 1} (x - x^{*}) a = {ad}_{a}^{2 m - 2} {ad}_{a} (x - x^{*}) a = {ad}_{a^{2}}^{m - 1} {ad}_{a} (x - x^{*}) a = \\ = (\binom{m - 1}{\frac{m - 1}{2}}) (- 1)^{\frac{m - 1}{2}} a^{m - 1} (a x - a x^{*} - (- 1)^{| x |} (x a - x^{*} a)) a^{m} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{r - 1} a^{2 r - 2} (a x - a x^{*} - (- 1)^{| x |} (x a - x^{*} a)) a^{2 r - 1} = \\ = (\binom{m - 1}{r - 1}) (- 1)^{\frac{m - 1}{2}} a^{2 r - 1} (x - x^{*}) a^{2 r - 1} . \end{aligned}$

Therefore, since R is free of $(\binom{m - 1}{r - 1})$ -torsion, a^2r−1 = 0, and by semiprimeness of R, a^2r−1 = 0. So in this case n ≡₈ 5. For $s = \frac{n + 1}{2}$ we have that a^s = 0, a^s−1 ≠ 0 and, by Proposition 3.1.1, a is an ad-nilpotent element of R of index n, case (4).

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CHAPTER 3. AD-NILPOTENT ELEMENTS IN A PRIME ASSOCIATIVE SUPERALGEBRA

3.1. Ad-nilpotent elements of R−

3.2. Ad-nilpotent elements of K

BIBLIOGRAPHY

3.1. Ad-nilpotent elements of R⁻