Jordan Centers and Martindale-like Covers

José A. Anquela, Teresa Cortés ^{1} anque@pinon.ccu.uniovi.es, cortes@pinon.ccu.uniovi.es Departamento de Matemáticas, Universidad de Oviedo, C/ Calvo Sotelo s/n, 33007 Oviedo, Spain

Esther García ^{1} egarciag@mat.ucm.es Departamento de Álgebra, Universidad Complutense de Madrid, 28040 Madrid, Spain

Miguel Gómez-Lozano ^{2} magomez@agt.cie.uma.es Departamento de Álgebra, Geometría, y Topología, Facultad de Ciencias, Universidad de Málaga, Campus de Teatinos, 29071 Málaga, Spain

Abstract: In this paper we show that the scalar center of a nondegenerate quadratic Jordan algebra is contained in the scalar center of any of its Martindale-like covers.

Introduction

The notion of (weak) scalar center, introduced by Fulgham in [3], has revealed a central tool in the study of Martindale-like quotients [1, 4] of linear Jordan algebras mainly due to two facts:

(i) any nonzero ideal of a nondegenerate PI Jordan algebra contains nonzero central elements [2, 3.6], and

(ii) the scalar center of a nondegenerate linear Jordan algebra is contained in the scalar center of any of its Martindale-like algebras of quotients [1, 4.1].

MSC2000: 17C10.

Keywords: Jordan algebra, scalar center, Martindale-like cover.

1 Partially supported by the Ministerio de Ciencia y Tecnología and Fondos FEDER, BFM2001-1938-C02-02.

2 Partially supported by the Ministerio de Ciencia y Tecnología and Fondos FEDER, BFM2001-1938-C02-01, and the Junta de Andalucía, FQM-264.

Our aim in this paper is showing that the general (quadratic) version of (ii) holds. Indeed we will work at the slightly more general setting of what we call “Martindalelike covers”, defined in terms of natural “ideal absorption properties”. This result is basic in our forthcoming paper on polynomial identities and speciality of quadratic Martindale-like quotients, as well as we expect it to be useful in the description of Martindale-like quotients of strongly prime quadratic Jordan algebras satisfying a polynomial identity.

The proof of our main result is purely combinatorial, based on the fact that 2J + Ker 2Id_{J} is an essential ideal of any nondegenerate Jordan algebra J, which, with the use of annihilators, allows to split the problem into the 2-torsion free and the characteristic 2 cases.

The paper is divided into four sections. Section 0 is devoted to recalling basic facts and notions, including the essentialness of 2J +Ker 2Id_{J}, mentioned above, and the definition of the scalar center. In Section 1 we study characteristic 2 phenomena needed in the sequel, and their natural extensions to arbitrary Jordan algebras in terms of the annihilator Ann_{J}(Ker 2Id_{J}) of Ker 2Id_{J}. In the next section we establish the fundamental properties of Martindale-like covers. Finally, in Section 3, we prove our main theorem asserting the inheritance of the scalar center by Martindale-like covers of nondegenerate Jordan algebras. It turns out that for a central element z of J, and a cover Q of J, V_{z} is in the centroid of Q as soon as Q satisfies the natural outer ideal absorption properties, while for the fact that z is indeed central in Q, the inner ideal absorption property must be assumed too.

0. Preliminaries

0.1 We will deal with Jordan algebras over a ring of scalars Φ. The reader is referred to [5, 7, 11] for definitions and basic properties not explicitly mentioned or proved in this section. Given a Jordan algebra J, its products will be denoted x^{2}, U_{x}y, for x, y ∈ J. They are quadratic in x and linear in y and have linearizations denoted V_{x}y = x ∘ y, U_{x,z}y = {x, y, z} = V_{x,y}z, respectively. A Jordan algebra J is said to be unital if there is an element 1 ∈ J satisfying U_{1} = Id_{J} and U_{x}1 = x^{2}, for any x ∈ J (such an element can be shown to be unique and it is called the unit of J).

Every Jordan algebra J imbeds in a unital Jordan algebra Ĵ = J ⊕ Φ1 called its (free) unitization [11, 0.6].

A Jordan algebra J is said to be nondegenerate if zero is the only absolute zero divisor, i.e., zero is the only x ∈ J such that U_{x} = 0.

0.2 We will need the following identities valid for arbitrary Jordan algebras.

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

(ii) z ∘ U_{x}y = {z, x, y} ∘ x − y ∘ U_{x}z,

(iii) U_{U_{x}y} = U_{x}U_{y}U_{x}, U_{x^{2}} = (U_{x})^{2},

(iv) {x, U_{z}x, y} = {U_{x}z, z, y},

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

(vi) {z, x, U_{y_{1}}y_{2}} = {z, {x, y_{1}, y_{2}}, y_{1}} − {z, y_{2}, U_{y_{1}}x},

(vii) U_{x}(y ∘ z) = {x ∘ y, z, x} − y ∘ U_{x}z,

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

(ix) (U_{x}y)^{2} = U_{x}U_{y}x^{2},

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

(xi) U_{{a,x,b}}z = U_{a}U_{x}U_{b}z + U_{b}U_{x}U_{a}z + {a, x, U_{b}{x, a, z}} − {U_{a}U_{x}b, z, b},

(xii) {a, z, U_{a}U_{x}a} = {U_{a}z, x, U_{a}x},

(xiii) 2U_{a}U_{x}U_{a}z = {a, x, U_{a}{x, a, z}} − {U_{a}x, x, U_{a}z},

(xiv) U_{x}{a, b, c} = {x, a, {b, c, x}} − {U_{x}a, c, b}.

Indeed, (vi) is [7, JP10], (xi) is [7, JP21], (xiv) is [7, JP12], and the rest of them follow from Macdonald’s Theorem [6].

0.3 We recall that an ideal I of a Jordan algebra J is just a Φ-submodule of J satisfying U_{I}J + I^{2} + U_{J}I + I ∘ J ⊆ I, equivalently, U_{I} Ĵ + UĴI ⊆ I, which implies {I, J, J} ⊆ I using (0.2)(i). An ideal I of J is said to be essential if it hits every nonzero ideal of J, i.e., I ∩ L ≠ 0 for any nonzero ideal L of J.

0.4 In a Jordan algebra J, the annihilator Ann_{J}(I) of an ideal I of J is an ideal of J which, when J is nondegenerate, is given by

Ann_{J}(I) = {x ∈ J | U_{x}I = 0} = {x ∈ J | U_{I}x = 0}

[8, 1.3, 1.7; 12, 1.3]. An ideal I of J will be said sturdy if Ann_{J}(I) = 0. It is easy to prove that essential ideals coincide with sturdy ideals in any semiprime Jordan algebra.

0.5 The centroid Γ(J) of a Jordan algebra J is the set of linear maps acting “scalarly” in Jordan products [10]:

Γ(J) = {T ∈ End_{Φ}(J) | T U_{x} = U_{x}T, T V_{x} = V_{x}T,
T^{2}(x^{2}) = (T (x))^{2}, T^{2}U_{x} = U_{T (x)}, for any x ∈ J}.

It is immediate that T V_{x,y} = V_{x,y}T, T U_{x,y} = U_{x,y}T for any T ∈ Γ(J), and any x, y ∈ J. Clearly, ΦId_{J} ⊆ Γ(J). By [10, 2.5], when J has no nonzero extreme elements (for example when J is nondegenerate), Γ(J) is a unital associative commutative Φalgebra and J is a Jordan algebra over Γ(J).

0.6 Lemma. If J is a nondegenerate Jordan algebra and T ∈ Γ(J), then

(i) the sum T (J)+Ker T is direct and, indeed, Ker T = Ker T^{n} for any positive integer n,

(ii) T (J) and Ker T are ideals of J,

(iii) T (J) + Ker T is an essential ideal of J.

Proof: (i) Let x ∈ J such that T^{2}(x) = 0. Then, for any y ∈ J, we have

U_{U_{T (x)}y} = U_{T (x)}U_{y}U_{T (x)} = U_{T (x)}U_{y}T^{2}U_{x} = T^{2}U_{T (x)}U_{y}U_{x} = U_{T^{2}(x)}U_{y}U_{x} = 0,

hence U_{T (x)}y = 0 by nondegeneracy. This shows U_{T (x)} = 0, hence T (x) = 0 again by nondegeneracy. We have proved Ker T = Ker T^{2}, which readily implies our assertion.

(ii) By [10, 2.6] we already know that T (J) is an ideal of J and Ker T is an outer ideal of J. But, under nondegeneracy, Ker T is also an inner ideal of J: for any x ∈ Ker T, y ∈ Ĵ, T^{2}(U_{x}y) = U_{T (x)}y = 0, hence U_{x}y ∈ Ker T^{2} = Ker T by (i).

(iii) Given a nonzero ideal L of J, if T (L) = 0, then 0 ≠ L ⊆ L ∩ Ker T ⊆ L ∩ (T (J)+Ker T). Otherwise, there exists x ∈ L such that T (x) ≠ 0. By nondegeneracy, 0 ≠ U_{T (x)}J = T^{2}U_{x}J = U_{x}T^{2}(J) ⊆ U_{L}J ∩ T (J) ⊆ L ∩ (T (J) + Ker T).

0.7 Following [3], the (weak) center of J is the set C(J) of all elements z ∈ J such that U_{z}, V_{z} ∈ Γ(J), which is a subalgebra of J when J is nondegenerate [3, Th. 1, 2]. More explicitly, z ∈ J lies in C(J) if and only if

c_{i}(z, J, J) = 0, for i = 1, 2, 3, 5, 6, and c_{i}(z, J) = 0 for i = 4, 7,

where

c_{1}(z, x, y) = V_{z}U_{x}y − U_{x}V_{z}y = z ∘ U_{x}y − U_{x}(z ∘ y),
c_{2}(z, x, y) = V_{z}V_{x}y − V_{x}V_{z}y = z ∘ (x ∘ y) − x ∘ (z ∘ y),
c_{3}(z, x, y) = U_{V_{z}x}y − V_{z}^{2}U_{x}y = U_{z∘x}y − z ∘ (z ∘ U_{x}y),
c_{4}(z, x) = (V_{z}x)^{2} − V_{z}^{2}x^{2} = (z ∘ x)^{2} − z ∘ (z ∘ x^{2}),
c_{5}(z, x, y) = U_{z}U_{x}y − U_{x}U_{z}y,
c_{6}(z, x, y) = U_{z}V_{x}y − V_{x}U_{z}y = U_{z}(x ∘ y) − x ∘ U_{z}y,
c_{7}(z, x) = (U_{z}x)^{2} − U_{z}^{2}x^{2},

since c_{5}(z, J, J) = 0 and (0.2)(iii) imply U_{U_{z}x} = U_{z}U_{x}U_{z} = U_{z}^{2}U_{x}, for any x ∈ J.

We claim that z ∈ C(J) also satisfies c_{8}(z, J, J) = 0, where c_{8}(z, x, y) = {U_{z}x, y, x}−2U_{z}U_{x}y, which readily follows from the fact that U_{z} ∈ Γ(J). If J is nondegerate then also c_{9}(z, J) = 0 for c_{9}(z, x) = U_{z}x^{2} − U_{x}z^{2}, since c_{9}(z, x) = c_{5}(z, x, 1) and C(J) ⊆ C(Ĵ) by [3, Cor. 1].

1. Characteristic 2 Phenomena

1.1 We remark that, by applying (0.6) to T = 2Id_{J} in a nondegenerate Jordan algebra J, 2x = 0 if and only if 2^{n}x = 0 for a positive integer n.

On the other hand, if 2x = 0 and x ∈ Ann(Ker 2Id_{J}), then x = 0: x ∈ Ker 2Id_{J} ∩ Ann(Ker 2Id_{J}) = 0 since J is semiprime and Ker 2Id_{J} is an ideal of J by (0.6)(ii).

1.2 Remark: In a nondegenerate Jordan algebra J, U_{y}x = U_{y}(−x), i.e., U_{y}2x = 0, for any x ∈ J, y ∈ Ker 2Id_{J}: U_{y}2x = 2U_{y}x = 0 since U_{y}x ∈ Ker 2Id_{J} by (0.6)(ii).

1.3 Lemma. Let J be a nondegenerate Jordan algebra, and let a, b ∈ J. If (U_{a} − U_{b})J ⊆ Ann_{J}(Ker 2Id_{J}), then a − b ∈ Ann_{J}(Ker 2Id_{J}).

Proof: (I) U_{{a,J,b}}J ⊆ Ann_{J}(Ker 2Id_{J}): for any y ∈ Ker 2Id_{J}, x, z ∈ J, using (0.2)(xi),

U_{y}U_{{a,x,b}}z = U_{y}[U_{a}U_{x}U_{b}z + U_{b}U_{x}U_{a}z + {a, x, U_{b}{x, a, z}} − {U_{a}U_{x}b, z, b}] =
= U_{y}[U_{a}U_{x}U_{a}z + U_{a}U_{x}U_{a}z + {a, x, U_{a}{x, a, z}} − {U_{b}U_{x}b, z, b}]
(for t ∈ J, U_{a}t − U_{b}t ∈ Ann_{J}(Ker 2Id_{J}), which is an ideal of J) =
= U_{y}[2U_{a}U_{x}U_{a}z + {a, x, U_{a}{x, a, z}} − {U_{b}x, x, U_{b}z}] (by (0.2)(xii)) =
= U_{y}[2U_{a}U_{x}U_{a}z + {a, x, U_{a}{x, a, z}} − {U_{a}x, x, U_{a}z}]
(for t ∈ J, U_{a}t − U_{b}t ∈ Ann_{J}(Ker 2Id_{J}), which is an ideal of J) =
= U_{y}[4U_{a}U_{x}U_{a}z] (by (0.2)(xiii)) = 0

by (1.2).

(II) {a, J, b} ⊆ Ann_{J}(Ker 2Id_{J}): using (0.2)(iii), for any y ∈ Ker 2Id_{J}, x ∈ J, U_{U_{y}{a,x,b}} = U_{y}U_{{a,x,b}}U_{y} = 0 by (I), hence U_{y}{a, x, b} = 0 by nondegeneracy of J.

(III) U_{a−b}J ⊆ Ann_{J}(Ker 2Id_{J}): for any y ∈ Ker 2Id_{J}, x ∈ J,

\begin{aligned}U_{y}U_{a−b}x & = U_{y}[U_{a}x + U_{b}x − \{a, x, b\}] = U_{y}[U_{a}x + U_{a}x − \{a, x, b\}] (as above) = \\ & = U_{y}[2U_{a}x − \{a, x, b\}] = 0 \end{aligned}

by (1.2) and (II).

Finally, for any y ∈ Ker 2Id_{J}, U_{U_{a−b}y} = U_{a−b}U_{y}U_{a−b} (by (0.2)(iii)) = 0, by (III), hence U_{a−b}y = 0 by nondegeneracy, and a − b ∈ Ann_{J}(Ker 2Id_{J}) (0.4).

Under the assumption of characteristic 2, (1.3) turns into the following result of independent interest, though it is not explicitly needed in the sequel.

1.4 Corollary. Let J be a nondegenerate Jordan algebra of characteristic two (2J = 0), a, b ∈ J. If U_{a} = U_{b}, then a = b.

Proof: Use (1.3) and the fact that Ann_{J}(Ker 2Id_{J}) = Ann_{J}(J) = 0 by nondegeneracy.

2. Martindale-like Covers

2.1 When J and Q are Jordan algebras such that J is a subalgebra of Q, we will say that Q is a cover of J. We will consider the following ideal absorption properties for a cover Q of J: the outer ideal absorption properties

(IA1) For any 0 ≠ q ∈ Q there exists an essential ideal I of J such that

0 ≠ U_{I}q ⊆ J, (IA2) For any q ∈ Q there exists an essential ideal I of J such that I ∘ q ⊆ J,

and the inner ideal absorption property

(IA3) For any q ∈ Q there exists an essential ideal I of J such that U_{q}I ⊆ J. A cover Q of J will be said a Martindale-like cover if it satisfies (IA1–3).

2.2 Remark: Assuming (IA1), condition (IA2) can be replaced by (IA2′) For any q ∈ Q there exists an essential ideal I of J such that {q, I, I} ⊆ J.

Indeed, (0.2)(i) implies that {q, I, I} ⊆ (q ∘ I) ∘ I + {I, q, I} ⊆ J when I is the intersection of the ideals in (IA1) and (IA2) for the element q. Conversely, if I and L are essential ideals satisfying U_{I}q + {q, L, L} ⊆ J, then K := U_{I∩L}(I ∩ L) is an essential ideal of J by [12, 1.2(a)], and (0.2)(ii) yields

\begin{aligned}q ∘ K & = q ∘ U_{I∩L}(I ∩ L) ⊆ \{q, I ∩ L, I ∩ L\} ∘ (I ∩ L) + (I ∩ L) ∘ U_{I∩L}q ⊆ \\ & ⊆ \{q, L, L\} ∘ J + J ∘ U_{I}q ⊆ J. \end{aligned}

2.3 Remark: Notice that any cover Q of J satisfying (IA1) is tight over J, i.e., any nonzero ideal of Q hits J. As a consequence, if J is nondegenerate then Q is also nondegenerate (cf. [9, 2.9(iii)]). Similarly, J is free of 2-torsion if and only if Q is free of 2-torsion, using tightness, (0.6)(ii), and the obvious fact that Ker 2Id_{J} = J ∩ Ker 2Id_{Q}.

In the next result we go further in the tightness of Martindale-like covers, in fact of covers just satisfying (IA1).

2.4 Proposition. Let J be a nondegenerate Jordan algebra and Q be a cover of J satisfying (IA1). Then, for any 0 ≠ q ∈ Q, and any essential ideal L of J, U_{L}q ≠ 0 and U_{q}L ≠ 0. If J has not 2-torsion, then also L ∘ q ≠ 0.

Proof: Given 0 ≠ q ∈ Q, let I be an essential ideal of J such that 0 ≠ U_{I}q ⊆ J, so that we can take x ∈ I such that 0 ≠ U_{x}q. For any essential ideal L of J, 0 ≠ U_{U_{x}q}L since Ann_{J}(L) = 0. But U_{U_{x}q}L = U_{x}U_{q}U_{x}L (by (0.2)(iii)) ⊆ U_{x}U_{q}L, which implies U_{q}L ≠ 0.

If U_{L}q = 0, then U_{L[t]}q = 0 in the algebra Q[t] of polynomials over Q. Notice that Q is nondegenerate by (2.3), which readily implies that Q[t] is also nondegenerate. For any h ∈ L[t], let a := U_{h}U_{q}h ∈ Q[t]. By (0.2)(iii),

U_{a}Q[t] = U_{h}U_{q}U_{h}U_{q}U_{h}Q[t] = U_{U_{h}q}U_{q}U_{h}Q[t] = 0

since U_{h}q = 0, hence a = 0 by nondegeneracy. For x, y ∈ L, the coefficient of t in U_{x+ty}U_{q}(x + ty) is U_{x}U_{q}y + U_{x,y}U_{q}x, which is then zero. But, on the other hand, U_{x,y}U_{q}x = {U_{x}q, q, y} (by (0.2)(iv)) = 0, hence we obtain U_{L}U_{q}L = 0. Fixing any x ∈ L such that U_{q}x ≠ 0, we then have 0 ≠ U_{U_{q}x}L = U_{q}U_{x}U_{q}L ⊆ U_{q}U_{L}U_{q}L, which contradicts U_{L}U_{q}L = 0. This shows U_{L}q ≠ 0.

Finally, in case J has not 2-torsion, 0 ≠ 2U_{L}q ⊆ L∘(L∘q)+L^{2}∘q (by (0.2)(v)) ⊆ L ∘ (L ∘ q) + L ∘ q implies L ∘ q ≠ 0.

As a consequence, we can choose a single ideal to nontrivially absorb any given finite set of elements in the cover.

2.5 Corollary. Let J be a nondegenerate Jordan algebra and Q be a cover of J satisfying (IA1). Given a finite set q_{1}, …, q_{n} of nonzero elements in Q, there exists an essential ideal I of J such that 0 ≠ U_{I}q_{i} ⊆ J, for all i = 1, …, n.

If Q also satisfies (IA2) and/or (IA3), then the ideal I above can also be assumed to satisfy I ∘ q_{i} + {q_{i}, I, I} ⊆ J (with 0 ≠ I ∘ q_{i} in case J has not 2-torsion), and/or 0 ≠ U_{q_{i}}I ⊆ J, respectively, for all i = 1, …, n.

Proof: Apply (2.4) and (2.2) together with the fact that the finite intersection of essential ideals is also essential.

2.6 If J is a nondegenerate Jordan algebra without 2-torsion, a cover Q of J is a Martindale-like cover of J if and only if for any 0 ≠ q ∈ Q there exists an essential ideal I of J such that 0 ≠ I ∘ q ⊆ J (when 1/2 ∈ Φ, this just amounts to saying that Q is a Jordan algebra of Martindale-like quotients of J with respect to the filter of all essential ideals of J in the sense of [4, 5.1]).

Indeed, a Martindale-like cover of J satisfies (IA2) and, moreover, I ∘ q ≠ 0 for any 0 ≠ q ∈ Q by (2.4) in the absence of 2-torsion. Conversely, assume that, for any 0 ≠ q ∈ Q, there exists an essential ideal I of J such that 0 ≠ I ∘ q ⊆ J. Clearly, M := 2I is an essential ideal of J and

U_{M}q = 2(2U_{I}q) ⊆ 2(I ∘ (I ∘ q) + I^{2} ∘ q) (by (0.2)(v)) ⊆ I ∘ J + I ∘ q ⊆ J.

Moreover, for x ∈ I such that x ∘ q ≠ 0, we have, by sturdiness of I (cf. (0.4)),

0 ≠ U_{I}(x ∘ q) ⊆ {I ∘ x, q, I} + x ∘ U_{I}q (by (0.2)(vii)) ⊆ U_{I}q + x ∘ U_{I}q,

which implies U_{I}q ≠ 0, hence 0 ≠ 4U_{I}q = U_{M}q, and we have established (IA1).

Furthermore, M ∘ q ⊆ J, and {q, M, M} ⊆ J as in the proof of (2.2). We now just need to show (IA3). Let L be an essential ideal of J such that q^{2} ∘ L ⊆ J, and let K := U_{M}M ∩ L, which is an essential ideal of J by [12, 1.2(a)], and we will show U_{q}2K ⊆ J. First, q ∘ U_{M}M ⊆ M: for any x, y ∈ M,

q∘U_{x}y = {q, x, y}∘x−y∘U_{x}q (by (0.2)(ii)) ⊆ {q, M, M}∘M +M ∘U_{M}q ⊆ J ∘M ⊆ M.

Thus, by (0.2)(v), U_{q}2K = 2U_{q}K ⊆ q ∘ (q ∘ K) + q^{2} ∘ K ⊆ q ∘ (q ∘ U_{M}M) + q^{2} ∘ L ⊆ q ∘ M + q^{2} ∘ L ⊆ J.

3. Center Inheritance in Martindale-like Covers

The proof of the next result is just the quadratic version of the proof of [1, 4.1]. In the generalization a factor 2 comes out.

3.1 Lemma. Let J be a nondegenerate Jordan algebra, Q be a cover of J satisfying (IA1) and (IA2), and z ∈ C(J). Then, 2z ∘ (p ∘ q) = 2(z ∘ p) ∘ q, for any p, q ∈ Q, i.e., 2V_{z}V_{q} = 2V_{q}V_{z}, for any q ∈ Q.

Proof: (I) For any q ∈ Q and any x ∈ J such that x∘q ∈ J, z∘(x∘q) = (z∘x)∘q:

Use (2.5) to find an essential ideal I of J such that U_{I}q + {q, I, I} ⊆ J. For any y_{1}, y_{2} ∈ I, and t ∈ Ĵ,

{z ∘ t, q, U_{y_{1}}y_{2}} = {z ∘ t, {q, y_{1}, y_{2}}, y_{1}} − {z ∘ t, y_{2}, U_{y_{1}}q} (by (0.2)(vi)) =
= z ∘ {t, {q, y_{1}, y_{2}}, y_{1}} − z ∘ {t, y_{2}, U_{y_{1}}q} =
= z ∘ {t, q, U_{y_{1}}y_{2}} (by (0.2)(vi))

(1) since {q, y_{1}, y_{2}}, U_{y_{1}}q ∈ J, z ∈ C(J), and C(J) ⊆ C(Ĵ) [3, Cor. 1]. Now, if K := U_{I}I and y ∈ K,

U_{y}((z ∘ x) ∘ q) = {y ∘ (z ∘ x), q, y} − (z ∘ x) ∘ U_{y}q (by (0.2)(vii)) =
= {z ∘ (y ∘ x), q, y} − z ∘ (x ∘ U_{y}q) (since x, y, U_{y}q ∈ J, z ∈ C(J)) =
= z ∘ {y ∘ x, q, y} − z ∘ (x ∘ U_{y}q) (by (1)) =
= z ∘ (U_{y}(x ∘ q)) (by (0.2)(vii)) = U_{y}(z ∘ (x ∘ q))

since y, x ∘ q ∈ J and z ∈ C(J). We have shown that U_{K}((z ∘ x) ∘ q − z ∘ (x ∘ q)) = 0, which implies (z ∘ x) ∘ q − z ∘ (x ∘ q) = 0 by (2.4) since K is an essential ideal of J by [12, 1.2(a)].

(II) Let q ∈ Q, and I be an essential ideal of J satisfying I ∘ q + U_{I}q + {q, I, I} ⊆ J, that can be found by (2.5). Then (z ∘ q) ∘ x = z ∘ (q ∘ x) for any x ∈ U_{I}I:

(z ∘ q) ∘ x = 2{z, q, x} − z ∘ (q ∘ x) + (z ∘ x) ∘ q (by linearized (0.2)(v)) =
= 2{z, q, x} (by (I)) = {z ∘ 1, q, x} =
= z ∘ {1, q, x} (by (1)) = z ∘ (q ∘ x).

(III) For any p, q ∈ Q, 2(z ∘ p) ∘ q = 2z ∘ (p ∘ q):

By (2.5), we can find an essential ideal I of J such that I ∘ p + U_{I}p + {p, I, I} + I ∘ q + U_{I}q + {q, I, I} + I ∘ (p ∘ q) + U_{I}(p ∘ q) + {p ∘ q, I, I} ⊆ J. Let K := U_{I}I and L := U_{K}K. Notice that

U_{L}q ⊆ K.  (2)

Indeed, U_{L}q is spanned by elements of the form U_{U_{a}b}q and {U_{a′}b′, q, U_{a}b}, where a, b, a′, b′ ∈ K, and

U_{U_{a}b}q = U_{a}U_{b}U_{a}q (by (0.2)(iii)) ⊆ U_{K}U_{K}U_{I}q ⊆ U_{K}U_{K}J ⊆ K,

whereas

\begin{aligned}\{U_{a′}b′, q, U_{a}b\} & ⊆ \{K, q, U_{a}b\} ⊆ \{K, \{q, a, b\}, a\} + \{K, b, U_{a}q\} (by (0.2)(vi)) ⊆ \\ & ⊆ \{K, \{q, I, I\}, K\} + \{K, K, U_{I}q\} ⊆ \{K, J, K\} + \{K, K, J\} ⊆ K. \end{aligned}

Now, for any y ∈ L,

U_{y}(2(z ∘ p) ∘ q) = 2[{y ∘ (z ∘ p), q, y} − (z ∘ p) ∘ U_{y}q] (by (0.2)(vii)) =

= 2[{z ∘ (y ∘ p), q, y} − z ∘ (p ∘ U_{y}q)] (by (II) since y, U_{y}q ∈ K by (2)) =

= 2[z ∘ {y ∘ p, q, y} − z ∘ (p ∘ U_{y}q)] (by (1) since y ∘ p ∈ J and y ∈ K) =
= 2z ∘ U_{y}(p ∘ q) (by (0.2)(vii)) =
= z ∘ [(y ∘ (p ∘ q)) ∘ y − y^{2} ∘ (p ∘ q)] (by (0.2)(v)) =
= [(y ∘ (z ∘ (p ∘ q))) ∘ y − y^{2} ∘ (z ∘ (p ∘ q))] (by (II)) =
= 2U_{y}(z ∘ (p ∘ q)) (by (0.2)(v)) = U_{y}(2z ∘ (p ∘ q)).

We have shown U_{L}(2(z∘p)∘q−2z∘(p∘q)) = 0, which implies 2(z∘p)∘q−2z∘(p∘q) = 0 by (2.4), since L is an essential ideal of J by [12, 1.2(a)].

3.2 Theorem. Let J be a nondegenerate Jordan algebra, Q be a cover of J satisfying (IA1) and (IA2), and z ∈ C(J). Then,

2c_{i}(z, Q, Q) = 0, for i = 1, 2, 3, 5, 6, 8 and 2c_{i}(z, Q) = 0 for i = 4, 7, 9.

Proof: By (1.1), it is enough to prove 2^{n}c_{i}(z, Q, …) = 0 for some positive integer n. On the other hand, we claim that, for any c_{i}, i = 1, …, 9, there exists a positive integer n such that 2^{n}c_{i} can be expressed in terms of 2 times “∘-products”. As an example, for p, q ∈ Q, using (0.2)(v) yields

8c_{3}(z, p, q) = 8[U_{z∘p}q − z ∘ (z ∘ U_{p}q)] =
= 4[(z ∘ p) ∘ ((z ∘ p) ∘ q) − (z ∘ p)^{2} ∘ q − z ∘ (z ∘ [p ∘ (p ∘ q) − p^{2} ∘ q])] =
= 2[2(z ∘ p) ∘ ((z ∘ p) ∘ q) − [(z ∘ p) ∘ (z ∘ p)] ∘ q
− z ∘ (z ∘ [2p ∘ (p ∘ q) − (p ∘ p) ∘ q])].

Now, our result follows from (3.1).

The above result is enough to obtain a generalization of [1, 4.1] for 2-torsion free Jordan algebras.

3.3 Corollary. Let J be a nondegenerate Jordan algebra without 2-torsion, Q be a cover of J satisfying (IA1) and (IA2). Then, C(J) ⊆ C(Q).

Proof: Use (0.7), (3.2), and the fact that Q has not 2-torsion by (2.3).

3.4 Corollary. Let J be a nondegenerate Jordan algebra, Q be a cover of J satisfying (IA1) and (IA2). Then, 2C(J) ⊆ C(Q).

Proof: For any z ∈ C(J), and any i = 1, …, 7, c_{i}(2z, Q, …) = 2^{k}c_{i}(z, Q, …) (for some positive integer k) = 0 by (3.2), hence 2z ∈ C(Q) by (0.7).

In order to extend (3.3) to the general quadratic case we will proceed in two steps. In the first one we will study the centrality in Q of the operator V_{z} for a central element of J, and show that only conditions (IA1) and (IA2) are needed. Our first result is the natural generalization of [3, Cor.2].

3.5 Lemma. In a nondegenerate Jordan algebra J, C(J) ∘ Ker 2Id_{J} = 0.

Proof: Let z ∈ C(J), x ∈ Ker 2Id_{J}, and y ∈ J. By (0.2)(viii),

\begin{aligned}U_{z∘x}y & = U_{z}U_{x}y + U_{x}U_{z}y + z ∘ U_{x}(y ∘ z) − \{U_{z}x, y, x\} = \\ & = 2U_{z}U_{x}y + (z ∘ z) ∘ U_{x}y − 2U_{z}U_{x}y = 2z^{2} ∘ U_{x}y ∈ 2J. \end{aligned}

But 4U_{z∘x}y = U_{z∘2x}y = 0, hence U_{z∘x}y = 0 by (1.1). We have shown U_{z∘x}J = 0, hence z ∘ x = 0 by nondegeneracy.

The next two results are meant to “lift” (3.5) to covers satisfying (IA1).

3.6 Lemma. If J is a nondegenerate Jordan algebra and Q is a cover of J satisfying (IA1), then C(J) ∘ Ker 2Id_{Q} = 0.

Proof: Let z ∈ C(J), q ∈ Ker 2Id_{Q}, and let I be an essential ideal of J such that U_{I}q ⊆ J. Notice that L := I ∩ (2J + Ker 2Id_{J}) is an essential ideal of J by (0.6)(iii). For any y ∈ L, using (0.2)(vii),

U_{y}(z ∘ q) = {y ∘ z, q, y} − z ∘ U_{y}q.

But writing y = 2a+b for a ∈ J, b ∈ Ker 2Id_{J}, {y∘z, q, y} = {2a∘z, q, y}+{b∘z, q, y} = {a ∘ z, 2q, y} (since b ∘ z = 0 by (3.5)) = 0 since q ∈ Ker 2Id_{Q}. On the other hand, 2U_{y}q = U_{y}2q = 0, hence U_{y}q ∈ J ∩Ker 2Id_{Q} = Ker 2Id_{J}, so that z ∘U_{y}q = 0 by (3.5). We have shown U_{L}(z ∘ q) = 0, which implies z ∘ q = 0 by (2.4).

3.7 Lemma. If J is a nondegenerate Jordan algebra and Q is a cover of J satisfying (IA1), then C(J) ∘ Q ⊆ Ann_{Q}(Ker 2Id_{Q}).

Proof: Let z ∈ C(J), p ∈ Ker 2Id_{Q}, and q ∈ Q. By (0.2)(vii),
U_{p}(z ∘ q) = {p ∘ z, q, p} − z ∘ U_{p}q = 0

by (3.6) since p, U_{p}q ∈ Ker 2Id_{Q}. This shows z ∘ q ∈ Ann_{Q}(Ker 2Id_{Q}) (cf. (0.4) since Q is nondegenerate by (2.3)).

3.8 Theorem. If J is a nondegenerate Jordan algebra and Q is a cover of J satisfying (IA1) and (IA2), then V_{z} ∈ Γ(Q) for any z ∈ C(J), equivalently,

c_{1}(z, Q, Q) = c_{2}(z, Q, Q) = c_{3}(z, Q, Q) = c_{4}(z, Q) = 0.

Proof: Notice that, c_{1}(z, Q, Q), c_{2}(z, Q, Q), c_{3}(z, Q, Q), c_{4}(z, Q) are contained in Ann_{Q}(Ker 2Id_{Q}) by (3.7), since they lie in the ideal of Q generated by z ∘ Q. Now, the result follows by using (3.2) and (1.1).

3.9 Theorem. Let J be a nondegenerate Jordan algebra, Q be a cover of J satisfying (IA1) and (IA2), and z ∈ C(J). Then

(i) {z, p, q} = {z, q, p} = {p, z, q}, for any p, q ∈ Q,

(ii) c_{6}(z, Q, Q) = c_{7}(z, Q) = c_{9}(z, Q) = 0.

Proof: (i) By (0.2)(i) and (3.8),

\begin{matrix}\{z, p, q\} = −\{p, z, q\} + (p ∘ z) ∘ q = −\{p, z, q\} + p ∘ (z ∘ q) = \{z, q, p\}, and \\ \{z, p, q\} = −\{z, q, p\} + (p ∘ q) ∘ z = −\{z, q, p\} + p ∘ (q ∘ z) = \{p, z, q\}. \end{matrix}

(ii) If c_{9}(z, Q) = 0 then, for any p ∈ Q,

c_{7}(z, p) = (U_{z}p)^{2} − U_{z}^{2}p^{2} = (U_{z}p)^{2} − U_{z}U_{p}z^{2} = 0

by (0.2)(ix). Thus we will show c_{6}(z, Q, Q) = c_{9}(z, Q) = 0, and we just need to prove that c_{6}(z, Q, Q), c_{9}(z, Q) ⊆ Ann_{Q}(Ker 2Id_{Q}) by (3.2) and (1.1). For any p, q ∈ Q, y ∈ Ker 2Id_{Q},

U_{y}c_{6}(z, p, q) = U_{y}(U_{z}(p ∘ q) − p ∘ U_{z}q) = U_{y}(U_{z}(p ∘ q) + p ∘ U_{z}q) (by (1.2) since Q

is nondegenerate by (2.3)) = U_{y}({z ∘ p, q, z}) (by (0.2)(vii)) = 0

since z ∘ p ∈ Ann_{Q}(Ker 2Id_{Q}) (by (3.7)) implies {z ∘ p, q, z} ∈ Ann_{Q}(Ker 2Id_{Q}). Also

U_{y}c_{9}(z, p) = U_{y}(U_{z}p^{2} − U_{p}z^{2}) = U_{y}(U_{z}p^{2} + U_{p}z^{2}) (by (1.2) since Q

is nondegenerate by (2.3)) = U_{y}((z ∘ p)^{2} − z ∘ U_{p}z) (by (0.2)(x)) = 0

since z ∘ Q ⊆ Ann_{Q}(Ker 2Id_{Q}) by (3.7).

3.10 Lemma. Let J be a nondegenerate Jordan algebra, Q be a cover of J satisfying (IA1) and (IA2), and z ∈ C(J). Then

U_{z}U_{p}q = −U_{p}U_{z}q + {U_{z}p, q, p}, for any p, q ∈ Q.

Proof: Notice that c_{10}(z, p, q) := U_{z}U_{p}q + U_{p}U_{z}q − {U_{z}p, q, p} = −c_{5}(z, p, q) − c_{8}(z, p, q), hence 2c_{10}(z, p, q) = 0 by (3.2). Using (1.1), we just need to prove c_{10}(z, Q, Q) ⊆ Ann_{Q}(Ker 2Id_{Q}). But using (0.2)(viii) yields c_{10}(z, p, q) = U_{z∘p}q − z ∘ U_{p}(q ∘ z) ∈ Ann_{Q}(Ker 2Id_{Q}) by (3.7).

Notice that, up to now, only the outer ideal absorption properties have been needed. The next results, aimed at studying the centrality of U_{z}, will make explicit use of inner ideal absorption.

3.11 Lemma. Let J be a nondegenerate Jordan algebra, Q be a Martindale-like cover of J, and z ∈ C(J). Then c_{5}(z, J, Q) = 0.

Proof: Let x ∈ J, q ∈ Q, and I be an essential ideal of J such that U_{q}I ⊆ J, U_{I}(U_{z}U_{x}q) ⊆ J, and U_{I}(U_{x}U_{z}q) ⊆ J, which exists by (2.5). For any y ∈ I, a ∈ J,

U_{U_{y}U_{z}U_{x}q}a = U_{y}U_{z}U_{x}U_{q}U_{x}U_{z}U_{y}a (by (0.2)(iii)) =

= U_{y}U_{x}U_{z}U_{q}U_{x}U_{z}U_{y}a (since U_{q}U_{x}U_{z}U_{y}a ⊆ U_{q}I ⊆ J, and U_{z} ∈ Γ(J)) =

= U_{y}U_{x}U_{z}U_{q}U_{z}U_{x}U_{y}a (since U_{y}a ∈ J, and U_{z} ∈ Γ(J)) =
= U_{U_{y}U_{x}U_{z}q}a (by (0.2)(iii)).

By (1.3), we have U_{y}c_{5}(z, x, q) = U_{y}U_{z}U_{x}q − U_{y}U_{x}U_{z}q ∈ Ann_{J}(Ker 2Id_{J}). But 2U_{y}c_{5}(z, x, q) = U_{y}2c_{5}(z, x, q) = 0 by (3.2), hence U_{y}c_{5}(z, x, q) = 0 by (1.1).

We have shown that U_{I}c_{5}(z, x, q) = 0, which implies c_{5}(z, x, q) = 0 by (2.4).

3.12 Lemma. Let J be a nondegenerate Jordan algebra, Q be a Martindale-like cover of J, and z ∈ C(J). Then, for any x, y ∈ J, q ∈ Q,

(i) {U_{z}x, q, x} = 2U_{z}U_{x}q ∈ 2Q, so that {U_{z}x, q, x} ∈ Ann_{Q}(Ker 2Id_{Q}),

(ii) {U_{z}x, q, y} + {x, q, U_{z}y} ∈ Ann_{Q}(Ker 2Id_{Q}).

Proof: By (3.10), U_{z}U_{x}q = −U_{x}U_{z}q + {U_{z}x, q, x}, which implies (i) using (3.11), (2.3), and (1.2), whereas (ii) follows by linearizing (i).

3.13 Lemma. Let J be a nondegenerate Jordan algebra, Q be a Martindale-like cover of J, z ∈ C(J), q ∈ Q, and I be an essential ideal of J such that U_{q}I +U_{I}q ⊆ J. Then,

{U_{z}q, y, q} = 2U_{z}U_{q}y ∈ 2Q, so that {U_{z}q, y, q} ∈ Ann_{Q}(Ker 2Id_{Q}),

for any y ∈ I.

Proof: For any x ∈ I, u ∈ Ker 2Id_{Q},

U_{u}U_{x}{U_{z}q, y, q} = U_{u}[{x, U_{z}q, {y, q, x}} − {U_{x}U_{z}q, q, y}] (by (0.2)(xiv)) =
= U_{u}[U_{z}{x, q, {y, q, x}} − {U_{z}U_{x}q, q, y}]

(by applying (3.11) to both terms since {y, q, x} ∈ U_{I}q ⊆ J) =

= U_{u}[U_{z}{x, q, {y, q, x}} − {U_{x}q, q, U_{z}y}]
(by (3.12)(ii) since U_{x}q ∈ U_{I}q ⊆ J, and (1.2)) =
= U_{u}[U_{z}{x, q, {y, q, x}} − {x, U_{q}x, U_{z}y}] (by (0.2)(iv)) =

= U_{u}[U_{z}{x, q, {y, q, x}} − U_{z}{x, U_{q}x, y}]
(since U_{q}x ∈ U_{q}I ⊆ J and U_{z} ∈ Γ(J)) =
= U_{u}[U_{z}{x, q, {y, q, x}} − U_{z}{U_{x}q, q, y}] (by (0.2)(iv)) =
= U_{u}U_{z}U_{x}{q, y, q} (by (0.2)(xiv)) = U_{u}2U_{z}U_{x}U_{q}y = 0

by (1.2). Also U_{u}U_{x}2U_{z}U_{q}y = 0 by (1.2), hence we have shown

U_{I}c_{8}(z, q, y) = U_{I}[{U_{z}q, y, q} − 2U_{z}U_{q}y] ⊆ Ann_{Q}(Ker 2Id_{Q}).

But 2U_{I}c_{8}(z, q, y) = U_{I}2c_{8}(z, q, y) = 0 by (3.2), so that U_{I}c_{8}(z, q, y) = 0 by (1.1). Therefore c_{8}(z, q, y) = 0 by (2.4), i.e., {U_{z}q, y, q} = 2U_{z}U_{q}y.

3.14 Lemma. Let J be a nondegenerate Jordan algebra, Q be a Martindale-like cover of J, z ∈ C(J), q ∈ Q, and I be an essential ideal of J such that U_{q}I +U_{I}q ⊆ J. Then, for any y ∈ I, c_{5}(z, q, y) = 0.

Proof: By (3.10), U_{q}U_{z}y = −U_{z}U_{q}y + {U_{z}q, y, q} = U_{z}U_{q}y using (3.13).

3.15 Proposition. Let J be a nondegenerate Jordan algebra, Q be a Martindale-like cover of J, and z ∈ C(J). Then c_{5}(z, Q, Q) = 0.

Proof: Let p, q ∈ Q and I be an essential ideal of J such that U_{I}q + {q, I, I} + U_{I}p + U_{p}I ⊆ J, which exists by (2.5). If we take K := U_{I}I, we also have that K is an essential ideal of J [12, 1.2(a)], and U_{K}q ⊆ I, as in (III)(2) of the proof of (3.1).

For any x ∈ K,
U_{x}U_{z}U_{p}q = U_{{p,z,x}}q − U_{p}U_{z}U_{x}q − {p, z, U_{x}{z, p, q}} + {U_{p}U_{z}x, q, x} ((0.2)(xi)) =
= U_{{z,p,x}}q − U_{z}U_{p}U_{x}q − {z, p, U_{x}{p, z, q}} + {U_{z}U_{p}x, q, x}
(by (3.9)(i), and (3.14) since U_{x}q ∈ I) = U_{x}U_{p}U_{z}q

using again (0.2)(xi). We have shown that U_{K}c_{5}(z, p, q) = 0, which implies that c_{5}(z, p, q) = 0 by (2.4).

3.16 Theorem. Let J be a nondegenerate Jordan algebra, Q be a Martindalelike cover of J. Then C(J) ⊆ C(Q).

Proof: Put together (3.8), (3.9)(ii), and (3.15), and use (0.7).

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