Idempotents and Peirce Gradings of Jordan Algebras^{1}

José A. Anquela^{∗}, Teresa Cortés

    anque@orion.ciencias.uniovi.es, cortes@orion.ciencias.uniovi.es

Departamento de Matemáticas, Universidad de Oviedo, C/ Calvo Sotelo s/n, 33007 Oviedo, Spain

Dedicated to Professor Ivan P. Shestakov on the occasion of his 60th birthday

Abstract: In this paper we find a necessary and sufficient condition for a Peirce grading of a Jordan algebra J to come from a Peirce decomposition with respect to an idempotent of a Jordan algebra J̃ containing J as a subalgebra. We also show that the above condition holds automatically when J is nondegenerate.

2000 Mathematics Subject Classification: 17C27, 17C10.

Keywords: idempotent, Jordan algebra, Peirce grading

Introduction

Peirce gradings V = V_{2} ⊕ V_{1} ⊕ V_{0} of a Jordan system V are introduced in [4, 11] as a generalization of Peirce decompositions with respect to idempotents. As shown in [11], Peirce gradings are directly linked to involutive automorphisms and gradings.

In [4, 4.1], it is shown that the components V_{α}, α = 0, 1, 2 inherit nondegeneracy and Von Neumann regularity from V, among other properties. In [1], it is shown that V_{0} and V_{2} inherit strong primeness, primitivity, semiprimitivity and simplicity from V. These results extend those obtained by McCrimmon [6, 7] about inheritance of simplicity by the diagonal components of a Peirce decomposition with respect to an idempotent.

Neher [11] gives examples of Pierce gradings not coming from idempotents. Indeed, if we consider a Jordan system J̃ with an idempotent e and take any nonzero ideal J of J̃ such that e ∉ J, then the usual Peirce decomposition J̃ = J̃_{2}(e)⊕J̃_{1}(e)⊕ J̃_{0}(e) gives rise to a Peirce grading in J (with J_{α} = J̃_{α}(e) ∩ J for α ∈ {0, 1, 2}) which does not come, in general, from an idempotent of J. We will study, in the case of Jordan algebras, if every Peirce grading of J comes from a Peirce decomposition with respect of an idempotent in an algebra J̃ containing J as a subalgebra.

1 Partly supported by the Ministerio de Educación y Ciencia and Fondos FEDER, MTM2004-06580-C02-01, and by the Plan de Investigación del Principado de Asturias, FICYT IB05-017.

∗ Corresponding author.

The paper is divided into three parts, apart from a preliminary section in which we outline some known results and definitions. In the first section we study Peirce gradings of associative algebras and show that they always come from idempotents in bigger algebras. We also show that not all Peirce gradings in special Jordan algebras are induced by associative Peirce gradings. In the second section we give a necessary condition on Peirce gradings of a Jordan algebra J for them to come from idempotents in a bigger algebra, and build the natural extension J̃ of J where the idempotent element should be found. Finally, in Section 3, we prove that the necessary condition of the previous section is also sufficient for Jordan algebras without 2-torsion, by showing that then J̃ is indeed a Jordan algebra. We also prove that this additional condition holds automatically in sufficiently regular (for example nondegenerate) Jordan algebras.

0. Preliminaries

0.1 We will work with associative and Jordan algebras over an arbitrary ring of scalars Φ. The reader is referred to [3, 9, 10] for basic results, notation, and terminology, though we will stress here some definitions.

0.2 A quadratic algebra J in the sense of [5, Section 1] is given on a Φ-module by its squares x^{2} and products U_{x}y, for x, y ∈ J. They are quadratic in x and linear in y, so that we have their linearizations, given by x ∘ y = V_{x}y := (x + y)^{2} − x^{2} − y^{2}, and U_{x,z}y = V_{x,y}z = {x, y, z} := U_{x+z}y −U_{x}y −U_{z}y. A quadratic algebra is a Jordan algebra if it strictly satisfies

QJ1 V_{x,x} = V_{x^{2}},
QJ2 U_{x}V_{x} = V_{x}U_{x},
QJ3 U_{x}(x^{2}) = (x^{2})^{2},
QJ4 U_{x}U_{y}(x^{2}) = (U_{x}y)^{2},
QJ5 U_{x^{2}} = U_{x}^{2},
QJ6 U_{U_{x}y} = U_{x}U_{y}U_{x},

for any x, y ∈ J, i.e., the polynomials j_{1}(x, y) := V_{x,x}y − V_{x^{2}}y, j_{2}(x, y) := U_{x}V_{x}y − V_{x}U_{x}y, j_{3}(x) := U_{x}(x^{2}) − (x^{2})^{2}, j_{4}(x, y) := U_{x}U_{y}(x^{2}) − (U_{x}y)^{2}, j_{5}(x, y) := U_{x^{2}}y − U_{x}^{2}y, j_{6}(x, y, z) := U_{U_{x}y}z − U_{x}U_{y}U_{x}z vanish strictly on J, i.e., vanish on any scalar extension of J.

0.3 One can get Jordan algebras from associative algebras by symmetrization: Given an associative algebra R with products xy, for any x, y ∈ R, one can built a Jordan algebra, denoted R^{(+)}, on the same Φ-module and with the same squares as R (x^{2} = xx) and Jordan products U_{x}y = xyx. A Jordan algebra is said to be special when it is a subalgebra of R^{(+)}, for some associative algebra R.

0.4 A Jordan algebra J is said to be unital if there exists a unit element 1 ∈ J such that, U_{1}x = x, U_{x}1 = x^{2}, for any x ∈ J. The unit element is unique (when it exists) and idempotent (1^{2} = 1). Any Jordan algebra J has a unitization Ĵ = J ⊕Φ1, which is a Jordan algebra built on the direct sum of J and a free Φ-module with basis {1}, given by

(x+λ1)^{2} = x^{2}+2λx+λ^{2}1, U_{x+λ1}(y+μ1) = U_{x}y+μx^{2}+λ^{2}y+λx∘y+2λμx+λ^{2}μ1,

for any x, y ∈ J, λ, μ ∈ Φ. The algebra J is an ideal of Ĵ, and 1 is the unit element of Ĵ [5].

0.5 We will need the following identities of Jordan algebras, which are direct consequences of Macdonald’s Theorem [2]:

(i) {x, y, y} = x ∘ y^{2}, {x, y^{2}, y} = {x, y, y^{2}} = x ∘ y^{3},

(ii) (x ∘ y) ∘ z = {x, y, z} + {y, x, z},

(iii) 2x^{3} = x ∘ x^{2},

(iv) (x^{2} ∘ y) ∘ x = x^{2} ∘ (y ∘ x),

(v) (x ∘ y)^{2} = x ∘ U_{y}x + U_{x}y^{2} + U_{y}x^{2},

(vi) x^{2m+n} = U_{x^{m}}x^{n}, (x^{m})^{n} = x^{mn},

(vii) U_{x}U_{y}x = U_{x∘y}x − x ∘ U_{y}x^{2} − U_{y}x^{3},

(viii) U_{z}U_{x}U_{z}y = U_{z}U_{x∘z}y − {z^{2} ∘ x, y, U_{z}x} + {x, z^{2}, y} ∘ U_{z}x − {y, z, U_{z}U_{x}z} − U_{z^{2}}U_{x}y,

(ix) 2U_{x}y = x ∘ (x ∘ y) − x^{2} ∘ y.

0.6 Following [1, 4, 9, 11], given a Jordan algebra J, a Peirce grading of J is a decomposition J = J_{2} ⊕ J_{1} ⊕ J_{0} (direct sum of Φ-modules) such that, for any α, β, γ ∈ {0, 1, 2},

(i) U_{J_{α}}J_{β} ⊆ J_{2α−β},
(ii) {J_{α}, J_{β}, J_{γ}} ⊆ J_{α−β+γ},
(iii) {J_{0}, J_{2}, J} = {J_{2}, J_{0}, J} = 0,
(iv) J_{2}^{2} ⊆ J_{2}, J_{0}^{2} ⊆ J_{0}, J_{1}^{2} ⊆ J_{0} + J_{2},
(v) J_{2} ∘ J_{1} + J_{0} ∘ J_{1} ⊆ J_{1}, J_{2} ∘ J_{0} = 0,

where J_{λ} = 0 if λ ∉ {0, 1, 2}. A Jordan algebra equipped with a Peirce grading will be said Peirce graded. We will consider the natural projections π_{α} : J ⟶ J_{α}, π_{α}(x) = x_{α}, α ∈ {0, 1, 2}. Notice that (i–v) involve at most degree two multiplications, so that any scalar extension of a Peirce graded Jordan algebra is naturally Peirce graded.

0.7 Notice that J_{0} y J_{2} can be exchanged, i.e., taking J′_{α} = J_{2−α} for α ∈ {0, 2}, we obtain a new Peirce grading J = J′_{2} ⊕ J′_{1} ⊕ J′_{0}.

0.8 Given an idempotent e in a Jordan algebra J, the usual Peirce decomposition J = J_{2}(e) ⊕ J_{1}(e) ⊕ J_{0}(e) of J with respect to e, where

J_{2}(e) = U_{e}J,  J_{1}(e) = (V_{e} − 2U_{e})J,  J_{0}(e) = (Id − V_{e} + U_{e})J  (1)

[3, Section I.5], is an example of Peirce grading of J which, following [11], will be called an idempotent Peirce grading. Notice the lack of “(0, 2)-symmetry” in idempotent Peirce gradings. On the other hand, we recall the basic fact that, for any x_{α} ∈ J_{α}(e),

e ∘ x_{α} = αx_{α}.  (2)

1. Peirce Gradings in Associative Algebras

1.1 Given an associative algebra R, a Peirce grading of R is a decomposition R = R_{11} ⊕ R_{10} ⊕ R_{01} ⊕ R_{00} (direct sum of Φ-modules) such that, for any α, β, γ, δ ∈ {0, 1},

(i) R_{αβ}R_{γδ} ⊆ R_{αδ}, if β = γ,
(ii) R_{αβ}R_{γδ} = 0, if β ≠ γ,

We will consider the natural projections π_{αβ} : R ⟶ R_{αβ}, π_{αβ}(x) = x_{αβ}, α, β ∈ {0, 1}.

Given an idempotent e in an associative algebra R, the usual Peirce decomposition R = R_{11}(e) ⊕ R_{10}(e) ⊕ R_{01}(e) ⊕ R_{00}(e) of R with respect to e is an example of Peirce grading of R.

1.2 Given a Peirce grading of an associative algebra R = R_{11} ⊕ R_{10} ⊕ R_{01} ⊕ R_{00}, let R̃ = R ⊕ Φe be the Φ-module obtained as a direct sum of R and a free Φ-module with basis {e}. Define a Φ-algebra structure on R̃ by the fact that R is a subalgebra of R̃ and

e^{2} = e,  xe = x_{11} + x_{01},  ex = x_{11} + x_{10},

for any x ∈ R, i.e., for any x, y ∈ R and λ, μ ∈ Φ,

(λe + x)(μe + y) = λμe + λ(y_{11} + y_{10}) + μ(x_{11} + x_{01}) + xy.

The proof of the following result is straightforward.

1.3 Theorem. Under the conditions of (1.2), R̃ is an associative algebra such that R is an ideal of R̃, and R_{αβ} = R̃_{αβ}(e) ∩ R, for any α, β ∈ {0, 1}.

1.4 Any Peirce grading R = R_{11} ⊕ R_{10} ⊕ R_{01} ⊕ R_{00} of an associative algebra R obviously induces a Peirce grading of J = R^{(+)} by taking J_{2} = R_{11}, J_{1} = R_{01} + R_{10}, J_{0} = R_{00}. The converse is false: Let R = G(e_{1}, e_{2}) be the nonunital Grassmann or exterior algebra in two generators over Φ (Φ is an arbitrary ring of scalars), and let J = R^{(+)}, which is a trivial algebra. Then, J_{0} = Φe_{1}, J_{2} = Φe_{2}, J_{1} = Φe_{1}e_{2} is a Peirce grading of J which does not come from any (associative) Peirce grading of R in the above sense, since the associative product J_{0}J_{2} is nonzero.

2. Peirce Gradings of Jordan Algebras

2.1 We will remark some basic multiplication properties of a Peirce graded Jordan algebra J = J_{2} ⊕J_{1} ⊕J_{0}, which are direct consequences of (0.5)(i)(ii) together with the definition (0.6). If x_{α}, y_{α}, z_{α} ∈ J_{α}, α = 0, 1, 2, then

(i) x_{2}^{2} ∘ x_{1} = (x_{1} ∘ x_{2}) ∘ x_{2}, x_{0}^{2} ∘ x_{1} = (x_{1} ∘ x_{0}) ∘ x_{0},

(x_{2} ∘ y_{2}) ∘ x_{1} = (x_{1} ∘ x_{2}) ∘ y_{2} + (x_{1} ∘ y_{2}) ∘ x_{2},
(x_{0} ∘ y_{0}) ∘ x_{1} = (x_{1} ∘ x_{0}) ∘ y_{0} + (x_{1} ∘ y_{0}) ∘ x_{0},

(ii) (x_{1} ∘ x_{2}) ∘ x_{0} = (x_{1} ∘ x_{0}) ∘ x_{2} = {x_{0}, x_{1}, x_{2}},

(iii) π_{2}((x_{1} ∘ x_{2}) ∘ y_{1}) = {x_{2}, x_{1}, y_{1}}, π_{0}((x_{1} ∘ x_{2}) ∘ y_{1}) = {x_{1}, x_{2}, y_{1}},

π_{2}((x_{1} ∘ x_{0}) ∘ y_{1}) = {x_{1}, x_{0}, y_{1}}, π_{0}((x_{1} ∘ x_{0}) ∘ y_{1}) = {x_{0}, x_{1}, y_{1}},

(iv) π_{2}({x_{1}^{2}, y_{1}, z_{1}}) = {π_{2}(x_{1}^{2}), y_{1}, z_{1}}, π_{0}({x_{1}^{2}, y_{1}, z_{1}}) = {π_{0}(x_{1}^{2}), y_{1}, z_{1}},

π_{2}({x_{1}, y_{1}^{2}, z_{1}}) = {x_{1}, π_{0}(y_{1}^{2}), z_{1}}, π_{0}({x_{1}, y_{1}^{2}, z_{1}}) = {x_{1}, π_{2}(y_{1}^{2})z_{1}},

(v) π_{2}(U_{x_{1}}y_{1}^{2}) = U_{x_{1}}(π_{0}(y_{1}^{2})), π_{0}(U_{x_{1}}y_{1}^{2}) = U_{x_{1}}(π_{2}(y_{1}^{2})),

(vi) π_{2}(x_{1}^{4}) = (π_{2}(x_{1}^{2}))^{2}, π_{0}(x_{1}^{4}) = (π_{0}(x_{1}^{2}))^{2}.

2.2 Let J be a subalgebra of a Jordan algebra J̃, and assume that J is equipped with a Peirce grading J = J_{2} ⊕ J_{1} ⊕ J_{0} induced by an idempotent e ∈ J̃, i.e., J_{α} = J̃_{α}(e) ∩ J. Then

x_{1}^{3} = π_{0}(x_{1}^{2}) ∘ x_{1} = π_{2}(x_{1}^{2}) ∘ x_{1}, for any x_{1} ∈ J_{1}.  (1)

Moreover, (1) holds strictly on J in the sense that any linearization of (1) holds in J, i.e., (1) holds in any (naturally Peirce graded) scalar extension of J. This is due to the fact that the hypotheses extend naturally to scalar extensions of J and J̃.

Indeed,

x_{1}^{3} = e ∘ x_{1}^{3} (by (0.8)(2), since x_{1}^{3} ∈ J_{1} by (0.6)(i))
= {e, x_{1}^{2}, x_{1}} (by (0.5)(i)) = (e ∘ x_{1}^{2}) ∘ x_{1} − {x_{1}^{2}, e, x_{1}} (by (0.5)(ii))

= 2π_{2}(x_{1}^{2}) ∘ x_{1} − {π_{0}(x_{1}^{2}), e, x_{1}} − {π_{2}(x_{1}^{2}), e, x_{1}} (by (0.6)(iv) and (0.8)(2))

= 2π_{2}(x_{1}^{2}) ∘ x_{1} − {π_{2}(x_{1}^{2}), e, x_{1}} (by (0.6)(iii) since e ∈ J̃_{2}(e))
= 2π_{2}(x_{1}^{2}) ∘ x_{1} − π_{2}(x_{1}^{2}) ∘ (e ∘ x_{1}) + {π_{2}(x_{1}^{2}), x_{1}, e} (by (0.5)(ii))
= 2π_{2}(x_{1}^{2}) ∘ x_{1} − π_{2}(x_{1}^{2}) ∘ x_{1} (by (0.8)(2) and (0.6)(ii)) = π_{2}(x_{1}^{2}) ∘ x_{1}.

The second equality follows from the first together with the following general fact:

2.3 If J = J_{2} ⊕ J_{1} ⊕ J_{0} is a Peirce graded Jordan algebra and x_{1} ∈ J_{1}, then

x_{1}^{3} = π_{0}(x_{1}^{2}) ∘ x_{1} ⟺ x_{1}^{3} = π_{2}(x_{1}^{2}) ∘ x_{1}.

Just notice that 2x_{1}^{3} = x_{1} ∘x_{1}^{2} = x_{1} ∘π_{0}(x_{1}^{2})+x_{1} ∘π_{2}(x_{1}^{2}) using (0.5)(iii) and (0.6)(iv).

2.4 There are examples of Peirce gradings not fulfilling (2.2)(1): Let S = Φ[X] be the free nonunital associative algebra in one generator X and let Φ be an arbitrary ring of scalars. Let L be the (associative) ideal of S generated by X^{4}, and A = S/L. Clearly, A is an associative algebra and a free Φ-module with basis

{a = X + L, b = X^{2} + L, c = X^{3} + L}.

In the special Jordan algebra J = A^{(+)} one can readily check that J_{2} = Φb, J_{1} = Φa+Φc, J_{0} = 0 is a Peirce grading, but a^{3} = c ≠ 0 and π_{0}(a^{2})∘a = 0 since π_{0}(a^{2}) = 0 because a^{2} = b ∈ J_{2}.

2.5 Let J = J_{2} ⊕ J_{1} ⊕ J_{0} be a Peirce graded Jordan algebra. Let J̃ = J ⊕ Φe be the Φ-module obtained as a direct sum of J and a free Φ-module with basis {e}. We can define a quadratic algebra structure (cf. [5, Section 1]) on J̃ by establishing, for any x, y ∈ J:

(i) J is a subalgebra of J̃,

(ii) e^{2} = e, which implies

e ∘ e = 2e,

(iii) e ∘ x = 2x_{2} + x_{1},

(iv) U_{e}e = e,

(v) U_{e}x = x_{2}, which implies

{e, x, e} = 2x_{2},

(vi) U_{x}e = x_{2}^{2} + x_{2} ∘ x_{1} + π_{0}(x_{1}^{2}), which implies

{x, e, y} = x_{2} ∘ y_{2} + x_{2} ∘ y_{1} + y_{2} ∘ x_{1} + π_{0}(x_{1} ∘ y_{1}),

(vii) {e, x, y} = x_{2} ∘ y_{2} + π_{2}(x_{1} ∘ y_{1}) + x_{2} ∘ y_{1} + x_{1} ∘ y_{0},

(viii) {e, e, x} = 2x_{2} + x_{1},

i.e., for any x, y ∈ J, α, β ∈ Φ,

(a) (x + αe)^{2} = x^{2} + α(2x_{2} + x_{1}) + α^{2}e, and
(b) \begin{matrix}U_{x+αe}(y + βe) = & U_{x}y + β(x_{2}^{2} + x_{2} ∘ x_{1} + π_{0}(x_{1}^{2})) + α^{2}y_{2} + αβ(2x_{2} + x_{1})+ \\ α(x_{2} ∘ y_{2} + π_{2}(x_{1} ∘ y_{1}) + x_{1} ∘ y_{2} + x_{0} ∘ y_{1}) + α^{2}βe. \end{matrix}

Under these conditions J is automatically an ideal of J̃ and the quotient J̃/J is isomorphic to the subalgebra Φe of J̃. Notice that Φe is a Jordan algebra isomorphic to Φ^{(+)}. We have the natural projections μ : J̃ ⟶ J and τ : J̃ ⟶ Φe, given by μ(x + αe) = x, τ(x + αe) = αe, for any x ∈ J, α ∈ Φ, and we remark that τ is a quadratic algebra epimorphism, whose kernel is J.

2.6 The “(0-2)-symmetry” of the Peirce grading in J disappears in J̃ (2.5). However, we will be able to make use of that symmetry by further extending J̃. Let ˆJ̃ be the unitization of J̃. Thus ˆ J̃ = J ⊕ Φe ⊕ Φ1. Notice that the subalgebra J ⊕ Φ1 of ˆ J̃ is simply the unitization Ĵ of J, hence a Jordan algebra. Let J′ = J Peirce graded by J′_{2} = J_{0}, J′_{1} = J_{1}, J′_{0} = J_{2}, build J̃′ as in (2.5), and let ˆ J̃′ be its unitization.

We can define a linear map φ : ˆ J̃ ⟶ ˆ J̃′ by

φ|Ĵ = IdĴ,  φ(e) = 1 − e,  (1)

which satisfies

φ(π_{α}(x)) = π_{α}(x) = π′_{2−α}(x) = π′_{2−α}(φ(x)),  α = 1, 2  (2)

for any x ∈ J.

The proof of the following result is straightforward.

2.7 Proposition. Under the conditions of (2.6), φ is a quadratic algebra isomorphism.

3. Main Results

3.1 In this section, when J is a Peirce graded Jordan algebra, J̃ will denote the quadratic algebra built in (2.5).

3.2 As in [5], we also consider the notion of commutative Jordan algebra: a linear algebra (C, ∘) over Φ, such that

x ∘ y = y ∘ x  and  (x^{2} ∘ y) ∘ x = x^{2} ∘ (y ∘ x)

hold strictly on J. Notice that, when 1/2 ∈ Φ, these are the usual linear Jordan algebras.

3.3 Proposition. Let (J, ( )^{2}, U) be a quadratic Φ-algebra such that

(i) 2U_{x}y = x ∘ (x ∘ y) − x^{2} ∘ y, for any x, y ∈ J and

(ii) the identity (x^{2} ∘ y) ∘ x = x^{2} ∘ (y ∘ x) holds strictly on J.

Then, the polynomials 4j_{1}, 8j_{2}, 8j_{3}, 32j_{4}, 16j_{5}, and 64j_{6} vanish strictly on J.

Proof: Let us consider the linear algebra (J, ∘), whose squares will be denoted

x˜2 := x ∘ x = 2x^{2}.  (1)

By (ii), (J, ∘) is a commutative Jordan algebra, so that (J, ( )˜2, Ũ) is a Jordan algebra [5, Cor. on p. 277], where

\begin{aligned}Ũ_{x}y & = 2x ∘ (x ∘ y) − (x ∘ x) ∘ y = 2(x ∘ (x ∘ y) − x^{2} ∘ y) = 4U_{x}y, \\ Ṽ_{x}y & = x ˜∘ y = (x + y)˜2 − x˜2 − y˜2 = 2x ∘ y = 2V_{x}y, \end{aligned}  (2)

hence it strictly satisfies QJ1 Ṽ_{x,x} = Ṽ_{x˜2}, i.e., 4V_{x,x} = 4V_{x^{2}},

QJ2 Ũ_{x}Ṽ_{x} = Ṽ_{x}Ũ_{x}, i.e., 8U_{x}V_{x} = 8V_{x}U_{x},
QJ3 Ũ_{x}(x˜2) = (x˜2)˜2, i.e., 8U_{x}(x^{2}) = 8(x^{2})^{2},
QJ4 Ũ_{x}Ũ_{y}(x˜2) = (Ũ_{x}y)˜2, i.e., 32U_{x}U_{y}(x^{2}) = 32(U_{x}y)^{2}
QJ5 Ũ_{x˜2} = Ũ_{x}^{2}, i.e., 16U_{x^{2}} = 16U_{x}^{2}
QJ6 Ũ_{Ũ_{x}y} = Ũ_{x}Ũ_{y}Ũ_{x}, i.e., 64U_{U_{x}y} = 64U_{x}U_{y}U_{x}.

We have shown that 4j_{1}, 8j_{2}, 8j_{3}, 32j_{4}, 16j_{5}, and 64j_{6} vanish strictly on J.

3.4 Lemma. Let J = J_{2} ⊕ J_{1} ⊕ J_{0} be a Peirce graded Jordan algebra. For any x, y ∈ J, we have the following equalities in J̃:

(i) (e ∘ e) ∘ x + ((x ∘ e) ∘ e) ∘ e = e ∘ (e ∘ x) + (x ∘ e) ∘ (e ∘ e),

(ii) (x^{2} ∘ e) ∘ e + ((x ∘ e) ∘ e) ∘ x = x^{2} ∘ (e ∘ e) + (x ∘ e) ∘ (e ∘ x),
(iii) (e ∘ y) ∘ x + ((x ∘ e) ∘ y) ∘ e = e ∘ (y ∘ x) + (x ∘ e) ∘ (y ∘ e).

If, in addition, the Peirce grading in J satisfies (2.2)(1), then also (iv) (x^{2} ∘ e) ∘ x = x^{2} ∘ (e ∘ x),

(v) (x^{2} ∘ y) ∘ e + ((x ∘ e) ∘ y) ∘ x = x^{2} ∘ (y ∘ e) + (x ∘ e) ∘ (y ∘ x).

Proof: Using (0.6), for any x, y ∈ J,

x^{2} = \underbrace{x_{2}^{2} + π_{2}(x_{1}^{2})}_{=π_{2}(x^{2})} + \underbrace{x_{1} ∘ x_{2} + x_{1} ∘ x_{0}}_{=π_{1}(x^{2})} + \underbrace{x_{0}^{2} + π_{0}(x_{1}^{2})}_{=π_{0}(x^{2})},  (1)

x ∘ y = \underbrace{x_{2} ∘ y_{2} + π_{2}(x_{1} ∘ y_{1})}_{=π_{2}(x∘y)} + \underbrace{x_{1} ∘ y_{2} + x_{1} ∘ y_{0} + x_{2} ∘ y_{1} + x_{0} ∘ y_{1}}_{=π_{1}(x∘y)} +

\underbrace{x_{0} ∘ y_{0} + π_{0}(x_{1} ∘ y_{1})}_{=π_{0}(x∘y)} .  (2)

Using (1), (2), (0.6), and (2.5), one can readily check:

(i) (e ∘ e) ∘ x + ((x ∘ e) ∘ e) ∘ e = e ∘ (e ∘ x) + (x ∘ e) ∘ (e ∘ e) = 12x_{2} + 3x_{1}, (ii) (x^{2} ∘ e) ∘ e + ((x ∘ e) ∘ e) ∘ x = x^{2} ∘ (e ∘ e) + (x ∘ e) ∘ (e ∘ x) = 12x_{2}^{2} + 4π_{2}(x_{1}^{2}) +

2x_{1}^{2} + 6x_{1} ∘ x_{2} + 2x_{1} ∘ x_{0}, (iii) (e ∘ y) ∘ x + ((x ∘ e) ∘ y) ∘ e = e ∘ (y ∘ x) + (x ∘ e) ∘ (y ∘ e) = 6x_{2} ∘ y_{2} + 3y_{2} ∘ x_{1} +

3y_{1} ∘ x_{2} + x_{1} ∘ y_{1} + 2π_{2}(x_{1} ∘ y_{1}) + x_{0} ∘ y_{1} + x_{1} ∘ y_{0}. Now, let us assume that the Peirce grading in J satisfies (2.2)(1). Using (1), (2),

(0.6), and (2.5), we obtain:

\begin{matrix}(x^{2} ∘ e) ∘ x = & 4x_{2}^{3} + 2π_{2}(x_{1}^{2}) ∘ x_{2} + 2x_{2}^{2} ∘ x_{1} + 2π_{2}(x_{1}^{2}) ∘ x_{1}+ \\ (x_{1} ∘ x_{2}) ∘ x_{2} + (x_{1} ∘ x_{2}) ∘ x_{1} + (x_{1} ∘ x_{2}) ∘ x_{0}+ \\ (x_{1} ∘ x_{0}) ∘ x_{2} + (x_{1} ∘ x_{0}) ∘ x_{1} + (x_{1} ∘ x_{0}) ∘ x_{0} \end{matrix}  (3)

and

\begin{matrix}x^{2} ∘ (e ∘ x) = & 4x_{2}^{3} + 2π_{2}(x_{1}^{2}) ∘ x_{2} + 2(x_{1} ∘ x_{2}) ∘ x_{2} + 2(x_{1} ∘ x_{0}) ∘ x_{2}+ \\ x_{2}^{2} ∘ x_{1} + π_{2}(x_{1}^{2}) ∘ x_{1} + (x_{1} ∘ x_{2}) ∘ x_{1}+ \\ (x_{1} ∘ x_{0}) ∘ x_{1} + x_{0}^{2} ∘ x_{1} + π_{0}(x_{1}^{2}) ∘ x_{1}. \end{matrix}  (4)

By (2.1)(i)(ii), and (2.2)(1), (3) and (4) coincide, i.e., (iv) holds.

Let a = (x^{2} ∘ y) ∘ e + ((x ∘ e) ∘ y) ∘ x, b = x^{2} ∘ (y ∘ e) + (x ∘ e) ∘ (y ∘ x). Using (1), (2), (0.6), and (2.5), it can be checked that

π_{2}(a) = 2x_{2}^{2} ∘ y_{2} + 2x_{1}^{2} ∘ y_{2} + 2(x_{2} ∘ y_{2}) ∘ x_{2} + (x_{1} ∘ y_{1}) ∘ x_{2}+
π_{2}(2(x_{1} ∘ x_{2}) ∘ y_{1} + 2(x_{1} ∘ x_{0}) ∘ y_{1} + 2(x_{2} ∘ y_{1}) ∘ x_{1}+
(x_{1} ∘ y_{2}) ∘ x_{1} + (x_{1} ∘ y_{0}) ∘ x_{1}),
π_{2}(b) = 2x_{2}^{2} ∘ y_{2} + 2x_{1}^{2} ∘ y_{2} + 2x_{2} ∘ (x_{2} ∘ y_{2}) + 2x_{2} ∘ (x_{1} ∘ y_{1})+
π_{2}((x_{1} ∘ x_{2}) ∘ y_{1} + (x_{1} ∘ x_{0}) ∘ y_{1} + x_{1} ∘ (x_{1} ∘ y_{2})+
x_{1} ∘ (x_{1} ∘ y_{0}) + x_{1} ∘ (x_{2} ∘ y_{1}) + x_{1} ∘ (x_{0} ∘ y_{1})),

and π_{2}(a) = π_{2}(b) because

π_{2}((x_{1} ∘ x_{2}) ∘ y_{1} + (x_{1} ∘ x_{0}) ∘ y_{1} + (x_{2} ∘ y_{1}) ∘ x_{1}) = x_{2} ∘ (x_{1} ∘ y_{1}) + π_{2}(x_{1} ∘ (x_{0} ∘ y_{1}))

by (2.1)(iii) and (0.5)(ii).

We also have, by (1), (2), (0.6) and (2.5)

π_{1}(a) = (x_{1} ∘ x_{2}) ∘ y_{2} + (x_{1} ∘ x_{0}) ∘ y_{2} + (x_{1} ∘ x_{2}) ∘ y_{0} + (x_{1} ∘ x_{0}) ∘ y_{0}+
x_{2}^{2} ∘ y_{1} + x_{1}^{2} ∘ y_{1} + x_{0}^{2} ∘ y_{1} + 2(x_{2} ∘ y_{2}) ∘ x_{1} + (x_{1} ∘ y_{1}) ∘ x_{1}+
2(x_{2} ∘ y_{1}) ∘ x_{2} + 2(x_{2} ∘ y_{1}) ∘ x_{0} + (x_{1} ∘ y_{2}) ∘ x_{2} + (x_{1} ∘ y_{2}) ∘ x_{0}+
(x_{1} ∘ y_{0}) ∘ x_{2} + (x_{1} ∘ y_{0}) ∘ x_{0},
π_{1}(b) = 2(x_{1} ∘ x_{2}) ∘ y_{2} + 2(x_{1} ∘ x_{0}) ∘ y_{2} + x_{2}^{2} ∘ y_{1} + x_{1}^{2} ∘ y_{1} + x_{0}^{2} ∘ y_{1}+
2x_{2} ∘ (x_{1} ∘ y_{2}) + 2x_{2} ∘ (x_{1} ∘ y_{0}) + 2x_{2} ∘ (x_{2} ∘ y_{1}) + 2x_{2} ∘ (x_{0} ∘ y_{1})+
x_{1} ∘ (x_{2} ∘ y_{2}) + x_{1} ∘ (x_{1} ∘ y_{1}) + x_{1} ∘ (x_{0} ∘ y_{0}),

and π_{1}(a) = π_{1}(b) because (x_{1} ∘x_{2})∘y_{0} +(x_{1} ∘x_{0})∘y_{0} +(x_{2} ∘y_{2})∘x_{1} +2(x_{2} ∘y_{1})∘x_{0} +(x_{1} ∘y_{2})∘x_{0} +(x_{1} ∘y_{0})∘x_{0} = (x_{1} ∘x_{2})∘y_{2} +(x_{1} ∘x_{0})∘y_{2} +x_{2} ∘(x_{1} ∘y_{2})+x_{2} ∘(x_{1} ∘y_{0})+2x_{2} ∘(x_{0} ∘y_{1})+x_{1} ∘(x_{0} ∘y_{0}) using (2.1)(i)(ii).

Let φ be the map defined in (2.6). By (2.6)(1)(2), and the fact that φ is a quadratic algebra homomorphism (2.7),

φ(π_{0}(a − b)) = π′_{2}(φ(a − b)) =

π′_{2}((x^{2′}∘′y)∘′(1−e)+((x∘′(1−e))∘′y)∘′x−(x^{2′}∘′(y∘′(1−e))+(x∘′(1−e))∘′(y∘′x))) = π′_{2}((x^{2′} ∘′ y) ∘′ 1 + ((x ∘′ 1) ∘′ y) ∘′ x − (x^{2′} ∘′ (y ∘′ 1) + (x ∘′ 1) ∘′ (y ∘′ x)))− π′_{2}((x^{2′} ∘′ y) ∘′ e + ((x ∘′ e) ∘′ y) ∘′ x − (x^{2′} ∘′ (y ∘′ e) + (x ∘′ e) ∘′ (y ∘′ x))) =

π′_{2}(2x^{2′} ∘′ y + (2x ∘′ y) ∘′ x − (x^{2′} ∘′ 2y + 2x ∘′ (y ∘′ x)))−

π′_{2}((x^{2′} ∘′ y) ∘′ e + ((x ∘′ e) ∘′ y) ∘′ x − (x^{2′} ∘′ (y ∘′ e) + (x ∘′ e) ∘′ (y ∘′ x))) = π′_{2}(0) − π′_{2}((x^{2′} ∘′ y) ∘′ e + ((x ∘′ e) ∘′ y) ∘′ x − (x^{2′} ∘′ (y ∘′ e) + (x ∘′ e) ∘′ (y ∘′ x))) = 0 since J̃′ satisfies the same conditions as J̃.

By injectivity of φ (2.7), π_{0}(a − b) = 0, and we have shown π_{α}(a − b) = 0 for any α = 0, 1, 2, i.e., a − b = 0, that is to say, (v) holds.

3.5 Theorem. If J = J_{2} ⊕ J_{1} ⊕ J_{0} is a Peirce graded Jordan algebra, then J̃ satisfies (3.3)(i). If the Peirce grading of J satisfies (2.2)(1) strictly, then J̃ satisfies (3.3)(ii). If, in addition, J does not have 2-torsion, then J̃ is a Jordan algebra, and J_{α} = J̃_{α}(e) ∩ J, for any α ∈ {0, 1, 2}.

Proof: By the definition (2.5) of J̃, for any α, β ∈ Φ, x, y ∈ J,

((x + αe) ∘ (y + βe)) ∘ (x + αe) − (x + αe)^{2} ∘ (y + βe) =

(x∘y+β(2x_{2}+x_{1})+α(2y_{2}+y_{1})+2αβe)∘(x+αe)−(x^{2}+α(2x_{2}+x_{1})+α^{2}e)∘(y+βe) =

(x ∘ y) ∘ x − x^{2} ∘ y + α((2y_{2} + y_{1}) ∘ x + e ∘ (x ∘ y) − (2x_{2} + x_{1}) ∘ y)+
α^{2}(e ∘ (2y_{2} + y_{1}) − e ∘ y) + αβ(e ∘ (2x_{2} + x_{1}) + 2e ∘ x − e ∘ (2x_{2} + x_{1}))+
α^{2}β(2e ∘ e − e ∘ e) + β((2x_{2} + x_{1}) ∘ x − e ∘ x^{2}) =
(x ∘ y) ∘ x − x^{2} ∘ y + 2α(y_{2} ∘ x_{1} + y_{1} ∘ x_{0} + x_{2} ∘ y_{2} + π_{2}(x_{1} ∘ y_{1}))+
2α^{2}y_{2} + 2αβ(2x_{2} + x_{1}) + 2α^{2}βe + 2β(x_{2}^{2} + x_{2} ∘ x_{1} + π_{0}(x_{1}^{2})),  (1)

using (3.4)(1)(2), (0.6), and (2.5). Now, (2.5)(b) and (1) with (0.5)(ix) for J imply (3.3)(i) for J̃.

Let us now assume that the Peirce grading of J satisfies (2.2)(1). Using that e is an idempotent of J̃ (2.5)(ii), we have

((x + αe)^{2} ∘ (y + βe)) ∘ (x + αe) = ((x^{2} + αx ∘ e + α^{2}e) ∘ (y + βe)) ∘ (x + αe) =

(x^{2} ∘ y) ∘ x + α((x^{2} ∘ y) ∘ e + ((x ∘ e) ∘ y) ∘ x) + α^{2}((e ∘ y) ∘ x + ((x ∘ e) ∘ y) ∘ e)+

α^{3}(e ∘ y) ∘ e + β(x^{2} ∘ e) ∘ x + αβ((x^{2} ∘ e) ∘ e + ((x ∘ e) ∘ e) ∘ x)+
α^{2}β((e ∘ e) ∘ x + ((x ∘ e) ∘ e) ∘ e) + α^{3}β(e ∘ e) ∘ e,  (2)

and

(x + αe)^{2} ∘ ((y + βe) ∘ (x + αe)) = (x^{2} + αx ∘ e + α^{2}e) ∘ ((y + βe) ∘ (x + αe)) =

x^{2} ∘ (y ∘ x) + α(x^{2} ∘ (y ∘ e) + (x ∘ e) ∘ (y ∘ x)) + α^{2}(e ∘ (y ∘ x) + (x ∘ e) ∘ (y ∘ e))+

α^{3}e ∘ (y ∘ e) + βx^{2} ∘ (e ∘ x) + αβ(x^{2} ∘ (e ∘ e) + (x ∘ e) ∘ (e ∘ x))+
α^{2}β(e ∘ (e ∘ x) + (x ∘ e) ∘ (e ∘ e)) + α^{3}βe ∘ (e ∘ e).  (3)

The equality between (2) and (3) follows from (0.5)(iv) applied to J, and (3.4). Moreover, any scalar extension of J̃ comes from the corresponding scalar extension of J (naturally Peirce graded, and satisfying (2.2)(1) since J satisfies it strictly) through the construction (2.5), hence it also satisfies the equality (2) = (3), i.e., J̃ satisfies (3.3)(ii).

By (3.3), the polynomials 4j_{1}, 8j_{2}, 8j_{3}, 32j_{4}, 16j_{5}, and 64j_{6} vanish strictly on J̃, hence, for any linearization p of j_{i}, i = 1, …, 6, 2^{k}p vanishes on J̃, for some k. Now let us assume that J does not have 2-torsion, and recall the natural projections μ, τ defined in (2.5). We have that 0 = μ(2^{k}p(J̃)) = 2^{k}μ(p(J̃)) implies μ(p(J̃)) = 0 since μ(p(J̃)) ⊆ J. On the other hand, since τ is a quadratic algebra homomorphism (2.5), τ(p(J̃)) ⊆ p(τ(J̃)) ⊆ p(Φe) = 0 because Φe is a Jordan algebra. Thus p(J̃) = 0, and we have shown that the polynomials j_{i}, i = 1, …, 6, vanish strictly on J̃, i.e., J̃ is a Jordan algebra. Finally, the equalities J_{α} = J̃_{α}(e) ∩ J, for any α ∈ {0, 1, 2}, are immediate consequences of (0.8) and (2.5).

3.6 The unitization Ĵ of J (0.4) is a particular case of J̃ when we consider in J the trivial Peirce grading given by J_{2} = J, J_{1} = J_{0} = 0, which obviously satisfies (2.2)(1) strictly, and write e = 1. In this sense, (3.5) can be viewed as a generalization of [5, Theorem 5] in absence of 2-torsion.

3.7 Notice that if J does not have 2-torsion and satisfies (2.2)(1), then it automatically satisfies it strictly, which happens with any identity of degree at most three in each of its variables.

3.8 Corollary. A Peirce grading J = J_{2} ⊕ J_{1} ⊕ J_{0} of a Jordan algebra without 2-torsion J satisfies (2.2)(1) if and only if there exists a Jordan algebra J̃ with an idempotent e such that J is a subalgebra of J̃, and J_{α} = J̃_{α}(e) ∩ J, for any α ∈ {0, 1, 2}.

Proof: The “only if” follows from (3.5) and (3.7), while the “if” is proved in (2.2).

In the following results, we will show that basic conditions of regularity on a Peirce graded algebra automatically imply (2.2)(1).

3.9 Lemma. Let J = J_{2} ⊕ J_{1} ⊕ J_{0} be a Peirce graded Jordan algebra. Then z ∘ J = z^{2} = 0, for any z = x_{1}^{3} − π_{2}(x_{1}^{2}) ∘ x_{1}, with x_{1} ∈ J_{1}.

Proof: Using (0.6)(iv), x_{1}^{2} = a_{2}+a_{0}, where a_{2} = π_{2}(x_{1}^{2}) ∈ J_{2}, a_{0} = π_{0}(x_{1}^{2}) ∈ J_{0}. If y_{0} ∈ J_{0}, then {x_{1}, a_{2}, y_{0}} = {a_{0}, x_{1}, y_{0}} = 0 by (0.6)(ii)(iii), hence

\begin{aligned}(π_{2}(x_{1}^{2}) ∘ x_{1}) ∘ y_{0} & = (a_{2} ∘ x_{1}) ∘ y_{0} = \{a_{2}, x_{1}, y_{0}\} + \{x_{1}, a_{2}, y_{0}\} (by (0.5)(ii)) \\ & = \{a_{2}, x_{1}, y_{0}\} = \{a_{2}, x_{1}, y_{0}\} + \{a_{0}, x_{1}, y_{0}\} = \{x_{1}^{2}, x_{1}, y_{0}\} = x_{1}^{3} ∘ y_{0} \end{aligned}

by (0.5)(i), which implies z ∘ J_{0} = 0.

By (0.7), also z ∘ J_{2} = 0. Let y_{1} ∈ J_{1}

(π_{2}(x_{1}^{2}) ∘ x_{1}) ∘ y_{1} = (a_{2} ∘ x_{1}) ∘ y_{1} = {a_{2}, x_{1}, y_{1}} + {x_{1}, a_{2}, y_{1}} (by (0.5)(ii))
= π_{2}({x_{1}^{2}, x_{1}, y_{1}}) + π_{0}({x_{1}, x_{1}^{2}, y_{1}}) (by (2.1)(iv))
= π_{2}(x_{1}^{3} ∘ y_{1}) + π_{0}(x_{1}^{3} ∘ y_{1}) (by (0.5)(i)) = x_{1}^{3} ∘ y_{1},

which implies z ∘ J_{1} = 0, and we have shown z ∘ J = 0. Now,

(π_{2}(x_{1}^{2}) ∘ x_{1})^{2} = (a_{2} ∘ x_{1})^{2} = a_{2} ∘ U_{x_{1}}a_{2} + U_{a_{2}}x_{1}^{2} + U_{x_{1}}a_{2}^{2} (by (0.5)(v))

= U_{a_{2}}x_{1}^{2} + U_{x_{1}}a_{2}^{2} (by (0.6)(i)(v)) = U_{a_{2}}(a_{2} + a_{0}) + U_{x_{1}}a_{2}^{2} = U_{a_{2}}a_{2} + U_{x_{1}}a_{2}^{2} (by (0.6)(i)) = U_{a_{2}+a_{0}}a_{2} + U_{x_{1}}a_{2}^{2} (by (0.6)(i)(iii)) = U_{x_{1}^{2}}a_{2} + U_{x_{1}}a_{2}^{2} = U_{x_{1}}U_{x_{1}}a_{2} + U_{x_{1}}a_{2}^{2} (by QJ5) = U_{x_{1}}(U_{x_{1}}a_{2} + a_{2}^{2}) = U_{x_{1}}(π_{0}(U_{x_{1}}x_{1}^{2}) + π_{2}(x_{1}^{4})) (by (2.1)(v)(vi))

= U_{x_{1}}(π_{0}(x_{1}^{4}) + π_{2}(x_{1}^{4})) (by (0.5)(vi)) = U_{x_{1}}x_{1}^{4} (by (0.6)(iv)(v)) = x_{1}^{6},

(1)

hence

z^{2} = (x_{1}^{3} − π_{2}(x_{1}^{2}) ∘ x_{1})^{2} = (x_{1}^{3})^{2} + (π_{2}(x_{1}^{2}) ∘ x_{1})^{2} − x_{1}^{3} ∘ (π_{2}(x_{1}^{2}) ∘ x_{1})

= x_{1}^{6} + x_{1}^{6} − (z + (π_{2}(x_{1}^{2}) ∘ x_{1})) ∘ (π_{2}(x_{1}^{2}) ∘ x_{1}) (by (0.5)(vi) and (1)) = x_{1}^{6} + x_{1}^{6} − (π_{2}(x_{1}^{2}) ∘ x_{1}) ∘ (π_{2}(x_{1}^{2}) ∘ x_{1}) (since z ∘ J = 0) = x_{1}^{6} + x_{1}^{6} − 2(π_{2}(x_{1}^{2}) ∘ x_{1})^{2} = 0

by (1).

3.10 Lemma. Let J be a Jordan algebra, z ∈ J.

(i) z ∘ J = z^{2} = 0 ⟹ z^{3} ∘ J = (z^{3})^{2} = (z^{3})^{3} = 0. (ii) z ∘ J = z^{3} = 0 ⟹ U_{z}U_{J}z = 0. (iii) z ∘ J = z^{2} = z^{3} = 0 ⟹ U_{U_{z}x}J = 0, for any x ∈ J. Proof: (i) For any x ∈ J,

z^{3} ∘ x = {z^{2}, z, x} (by (0.5)(i)) = 0,
(z^{3})^{2} = z^{6} = U_{z^{2}}z^{2} (by (0.5)(vi)) = 0,
(z^{3})^{3} = z^{9} = U_{z^{2}}z^{5} (by (0.5)(vi)) = 0.

(ii) For any x ∈ J,

U_{z}U_{x}z = U_{z∘x}z − z ∘ U_{x}z^{2} − U_{x}z^{3} (by (0.5)(vii)) ⊆ U_{z∘J}z − z ∘ J − U_{J}z^{3} = 0.

(iii) For any x, y ∈ J,

U_{U_{z}x}y = U_{z}U_{x}U_{z}y (by QJ6) = U_{z}U_{x∘z}y − {z^{2} ∘ x, y, U_{z}x} + {x, z^{2}, y} ∘ U_{z}x−
{y, z, U_{z}U_{x}z} − U_{z^{2}}U_{x}y (by (0.5)(viii)) = 0

by the hypotheses and (ii).

3.11 Theorem. If J = J_{2} ⊕ J_{1} ⊕ J_{0} is a Peirce graded Jordan algebra, then (2.2)(1) holds strictly when J is under any of the following circumstances:

(i) J does not have 2-torsion and does not contain nonzero invisible elements,

(ii) J is special and semiprime,

(iii) J is nondegenerate.

Proof: We remark that any of the properties (i–iii) on J is inherited by the scalar extension of J ⊗_{Φ} Φ̃ of J when Φ̃ is the unital, associative, commutative ring of Φ-polynomials in an infinite set of variables. Then, if we show that (2.2)(1) holds on J, it will also hold on J ⊗_{Φ} Φ̃, which implies (2.2)(1) holds strictly on J.

Let x_{1} ∈ J_{1}, and z = x_{1}^{3} − π_{2}(x_{1}^{2}) ∘ x_{1}. We have to show that, assuming (i), (ii) or (iii), z = 0.

(i) By (3.9), z ∘ J = 0, which implies that z is invisible (cf. [8]) since J does not have 2-torsion. Hence z = 0.

(ii) Since J is special, we can find an associative algebra A such that it is a envelope of J, i.e., J ≤ A^{(+)} and A is generated as an associative algebra by J. Let I be the ideal of A generated by z. Since z^{2} = z ∘ J = 0, it can be readily seen that II = 0. Then L = I ∩ J is an ideal of J such that U_{L}L ⊆ III = 0, which, by semiprimeness of J, implies L = 0. Hence z ∈ I ∩ J = L = 0.

(iii) Notice that t := z^{3} satisfies t ∘ J = t^{2} = t^{3} = 0 by (3.10)(i). Hence, for any x ∈ J, U_{t}x is an absolute zero divisor of J by (3.10)(iii). By nondegeneracy of J, U_{t}x = 0 for any x ∈ J, i.e., t itself is an absolute zero divisor of J, hence t = 0 again by nondegeneracy. Thus, z^{3} = 0, and (3.10)(iii) implies that U_{z}x is an absolute zero divisor of J, for any x ∈ J, but this implies, as above, that z = 0.

3.12 Corolary. If J = J_{2} ⊕ J_{1} ⊕ J_{0} is a Peirce graded Jordan algebra without 2-torsion in any of the situations (3.11)(i–iii), then J̃ is a Jordan algebra with an idempotent e such that J is a subalgebra of J̃, and J_{α} = J̃_{α}(e) ∩ J, for any α ∈ {0, 1, 2}.

3.13 Further Comments: The condition of absence of 2-torsion can be removed in (3.5) (hence in (3.8) and (3.12)). In that general setting, the proof is much more involved and lengthy and will be the subject of a forthcoming paper.

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