PEIRCE GRADINGS OF JORDAN SYSTEMS ^{1}

José A. Anquela and Teresa Cortés

Abstract: In this paper we prove that the diagonal components V_{0} and V_{2} of a Peirce grading of a Jordan pair or triple system V inherit strong primeness, primitivity and simplicity from V .

Dedicated to the memory of Eulalia García Rus.

Introduction

Peirce gradings V = V_{2} ⊕ V_{1} ⊕ V_{0} of a Jordan pair V are introduced in [8, 11] as a generalization of Peirce decompositions with respect to idempotents. In [8, 4.1] it is shown that the components V_{α}, α = 0, 1, 2 inherit nondegeneracy and regularity, among other conditions.

In the first section of this paper we show that V_{0} and V_{2} inherit strong primeness, primitivity, semiprimitivity and simplicity from V. The proof follows from the corresponding results on inheritance of regularity conditions by subquotients [3, 5, 6, 8], once that it is shown that V_{0} and V_{2} are isomorphic to certain subquotients of V.

In the second section we extend the above results to Peirce gradings of triple systems. Dealing with arbitrary Peirce gradings, instead of Peirce decompositions related to idempotents, has a cost, namely the existence of ideals which are not graded. This difficulty is overcome by introducing the notion of graded tight double pair which is used to link Jordan triple systems and Jordan pairs both with a Peirce grading. The properties of graded tight double pairs established in the second section of the paper are based on a study of ideals in Jordan pairs with a Peirce grading done in the previous section.

Since Peirce decompositions T = T_{2}(e) ⊕ T_{1}(e) ⊕ T_{0}(e) of a Jordan triple system T with respect to a tripotent e ∈ T are particular cases of Peirce gradings, the main result of the second section contains as a corollary the central result of McCrimmon [9] on the simplicity of T_{0}(e) and T_{2}(e) when T is simple. However, the proof given in this paper cannot be considered an alternative to the proof by McCrimmon in [9]. Indeed the results of this paper are based on the results of [5], which depend on the inheritance of regularity by local algebras proved in [6]. These latter results are based on the results by McCrimmon [9].

1 Partially supported by the DGES, PB97-1069-C02-02.

2000 Mathematics Subject Classification: 17C27, 17C10, 17C20.

Key words and phrases: Jordan system, Peirce grading, simple, primitive, strongly prime

0. Preliminaries

0.1 We will deal with Jordan pairs and triple systems over an arbitrary ring of scalars Φ. The reader is referred to [3, 4, 7, 10] for basic results, notation and terminology, though we will stress some definitions and basic notions.

—For a Jordan pair V = (V^{+}, V^{−}) we will denote the products by Q_{x}y, for any x ∈ V^{σ}, y ∈ V^{−σ}, σ = ±, with linearizations denoted by Q_{x,z}y = {x, y, z}.

—A Jordan triple system T is given by its products P_{x}y, for any x, y ∈ T, with linearizations denoted by P_{x,z}y = {x, y, z}.

0.2 By doubling any Jordan triple system T one obtains the double Jordan pair V (T) = (T, T) with products Q_{x}y = P_{x}y, for any x, y ∈ T. From a Jordan pair V = (V^{+}, V^{−}) one can get a (polarized) Jordan triple system T (V) = V^{+} ⊕ V^{−} by defining P_{x^{+}⊕x^{−}}(y^{+} ⊕ y^{−}) = Q_{x^{+}}y^{−} ⊕ Q_{x^{−}}y^{+} [7, 1.13, 1.14]

0.3 Regularity conditions through the functors V and T:

(i) For a Jordan triple system T, we have: (a) T is nondegenerate if and only if V (T) is nondegenerate,

(b) if V (T) is strongly prime, then T is strongly prime,

(c) if V (T) is simple, then T is simple. (ii) For a Jordan pair V, we have:

(a) V is nondegenerate if and only if T (V) is nondegenerate,

(b) V is strongly prime if and only if T (V) is strongly prime,

(c) V is simple if and only if T (V) is simple.

(i)(a) and (ii)(a) are straightforward; (i)(b)(c) follow from the fact that any nonzero ideal I of T gives rise to the nonzero ideal V (I) = (I, I) of V (T); (ii)(b)(c) are proved in [1, page 230].

0.4 Let V be a Jordan pair, M ⊆ V^{+} be an inner ideal of V. The set

Ker_{V} M = Ker M = {x ∈ V^{−} | Q_{M}x = Q_{M}Q_{x}M = 0}

gives rise to the ideal (0, Ker M) of the subpair (M, V^{−}) of V. The quotient

(M, V^{−})/(0, Ker M) = (M, V^{−}/ Ker M)

is called the subquotient of V determined by M. Similarly, one can consider subquotients of V determined by an inner ideal M ⊆ V^{−} of V (see [8]).

If V is nondegenerate, then Ker M = {x ∈ V^{−} | Q_{M}x = 0} [8, 2.4].

0.5 Most regularity conditions are inherited by subquotients: Let V be a Jordan pair, 0 ≠ M ⊆ V^{σ} be an inner ideal of V, and S be the subquotient of V determined by M.

(i) If V is nondegenerate (resp. semiprimitive), then S is nondegenerate (resp. semiprimitive) [8, 2.2].

(ii) If V is strongly prime, then S is strongly prime [6, 3.2(v)].

(iii) If V is primitive, then S is primitive [3, 3.10; 4, 2.2]. Simplicity is not inherited by arbitrary subquotients (cf. [5, 5.2]). However, information about the heart Heart S of S can be obtained when V is simple:

(iv) If V is simple, then H = Heart S of S is simple. Moreover H^{σ} = Q_{M}V^{−σ} =

Q_{S^{σ}}S^{−σ} and H^{−σ} = S^{−σ} [5, 5.1].

1. Peirce Gradings of Jordan Pairs

1.1 Following [8, 3.6, 4.1], given a Jordan pair V = (V^{+}, V^{−}), a Peirce grading of V is a decomposition V^{σ} = V_{2}^{σ} ⊕ V_{1}^{σ} ⊕ V_{0}^{σ}, σ = ±, (direct sum of Φ-modules) such that, for any α, β, γ ∈ {0, 1, 2},

(i) Q_{V_{α}^{σ}}V_{β}^{−σ} ⊆ V_{2α−β}^{σ},

(ii) {V_{α}^{σ}, V_{β}^{−σ}, V_{γ}^{σ}} ⊆ V_{α−β+γ}^{σ},
(iii) {V_{0}^{σ}, V_{2}^{−σ}, V^{σ}} = {V_{2}^{σ}, V_{0}^{−σ}, V^{σ}} = 0,

with the convention that V_{λ}^{σ} = 0 if λ ∉ {0, 1, 2}. The pairs (V_{α}^{+}, V_{α}^{−}) for α ∈ {0, 1, 2} will be denoted V_{α}, so that V = V_{2} ⊕ V_{1} ⊕ V_{0}.

Notice that the roles of V_{0} and V_{2} can be exchanged in the above definition, i.e., by taking Ṽ_{α}^{σ} = V_{2−α}^{σ}, we obtain a new Peirce grading V = Ṽ_{2} ⊕ Ṽ_{1} ⊕ Ṽ_{0}.

Given an idempotent e in a Jordan pair V the usual Peirce decomposition of V = V_{2}(e) ⊕ V_{1}(e) ⊕ V_{0}(e) with respect to e [7, Section I.5] is an example of Peirce grading of V. We remark the lack of “(0, 2)-symmetry” in Peirce decompositions given by idempotents, so that the studies of V_{2}(e) and V_{0}(e) require different arguments (cf. [7, 9])

1.2 From the multiplication rules given in the above definition it is clear that V_{α} is a subpair of V for α = 0, 1, 2 and, moreover, V_{0}^{σ}, V_{2}^{σ} are inner ideals of V whose subquotients will be deeply related to the pairs V_{0}, V_{2}, respectively. This relationship, on which the results of this paper are based, is obtained in the following lemma due to Ottmar Loos.

1.3 Lemma [O. Loos]. Let V = V_{2} ⊕ V_{1} ⊕ V_{0} be a Peirce grading of a nondegenerate Jordan pair V. The subquotient S of V determined by the inner ideal M = V_{α}^{σ} (σ = ±, α = 0, 2) is isomorphic to V_{α}. Indeed, an isomorphism φ : V_{α} ⟶ S is given by φ^{σ} = Id_{V_{α}^{σ}}, φ^{−σ}(x) = x + Ker M, for any x ∈ V_{α}^{−σ}.

Proof: By (0, 2)-symmetry and (+, −)-symmetry, we can take M = V_{2}^{+}. We will prove the equality Ker_{V} M = V_{1}^{−} + V_{0}^{−} from which everything else readily follows.

Notice that Ker_{V} M = {x ∈ V^{−} | Q_{M}x = 0} by nondegeneracy of V, so that V_{1}^{−} + V_{0}^{−} ⊆ Ker_{V} M by (1.1)(i). Conversely, let x ∈ Ker_{V} M and write x = x_{0} + x_{1} + x_{2} with x_{i} ∈ V_{i}^{−} (i = 0, 1, 2). We just need to show that x_{2} = 0. Indeed x_{2} ∈ Ker_{V} M since V_{1}^{−} + V_{0}^{−} ⊆ Ker_{V} M, hence x_{2} ∈ Ker_{V} M ∩ V_{2}^{−} = Ker_{V_{2}} V_{2}^{+} = 0 by [8, 1.4] since V_{2} is nondegenerate by [8, 4.1(ii)].

Our main result for Peirce gradings of pairs follows from the application of the previous lemma to known results of inheritance of regularity by subquotients (0.5). However, this is not the case for the inheritance of nondegeneracy by Peirce components [8, 4.1(ii)]. That is indeed our starting point since it is needed in the proof of (1.3).

1.4 Theorem. Let V = V_{2} ⊕ V_{1} ⊕ V_{0} be a Peirce grading of a Jordan pair V. Let α ∈ {0, 2} and assume V_{α} ≠ 0.

(i) If V is strongly prime then V_{α} is strongly prime.

(ii) If V is primitive (resp. semiprimitive) then V_{α} is primitive (resp. semiprim-

itive). (iii) If V is simple then V_{α} is simple. Proof: In all cases V is nondegenerate (strongly prime ≡ prime + nondegenerate,

primitive ⟹ strongly prime [3, 3.9], semiprimitive ⟹ nondegenerate [7, 4.15], simple ⟹ nondegenerate [5, 2.4 (ii)]), thus V_{α} is nondegenerate by [8, 4.1(ii)], and both V_{α}^{+} and V_{α}^{−} are nonzero. Moreover, (1.3) applies, so that V_{α} is isomorphic to the subquotient S determined by V_{α}^{+}.

(i) follows from (0.5)(ii).

(ii) follows from (0.5)(iii) (resp. (0.5)(i)).

(iii) If V is simple then the heart of S is simple by (0.5)(iv), hence the heart H of V_{2} is simple. Moreover, using the isomorphism φ : V_{2} ⟶ S given in (1.3), H^{−} = φ^{−1}((Heart S)^{−}) = φ^{−1}(S^{−}) = V_{α}^{−}. By (+, −)-symmetry, also H^{+} = V_{α}^{+}, and we have shown H = V_{α}, hence V_{α} is simple.

1.5 One of the difficulties when dealing with arbitrary Peirce gradings is the existence of ideals which are not graded. Let us consider an arbitrary grading V = V_{2} ⊕ V_{1} ⊕ V_{0} of a Jordan pair V. An ideal I of V is said to be graded if I = (I ∩ V_{2}) + (I ∩ V_{1}) + (I ∩ V_{0}). Equivalently, if π_{α} : V ⟶ V_{α} denote the natural projections, π_{α}(I) ⊆ I for all α = 0, 1, 2. When the Peirce grading considered is a Peirce decomposition with respect to an idempotent e of V, every ideal is graded since the projections π_{α} are built out of elements of the multiplication algebra of V (cf. [7, Section I.5]).

In general, if I is graded with respect to a Peirce grading V = V_{2} ⊕ V_{1} ⊕ V_{0}, then a Peirce grading is induced in the quotient V/I. Namely V/I = (V_{2} + I)/I ⊕ (V_{1} + I)/I ⊕ (V_{0} + I)/I.

To extend (1.4) to triple systems we will need to construct pairs V associated to a triple system T (quotients of V (T)) which, on the one hand, are tight in the sense that they do not have essentially more ideals than those coming from ideals of T, and, on the other hand, behave nicely with respect to a given Peirce grading of T. The next result is aimed to find suitable graded ideals of V (T) which will be used to test the behaviour of such V.

1.6 Lemma. If V = V_{2} ⊕ V_{1} ⊕ V_{0} is a Peirce grading of a Jordan pair V and I is an ideal of V, then the ideal M = Q_{I}I = (Q_{I^{+}}I^{−}, Q_{I^{−}}I^{+}) of I satisfies M ⊆ (I ∩ V_{2}) + (I ∩ V_{1}) + (I ∩ V_{0}). As a consequence, if V is semiprime, then every nonzero ideal I of V contains a nonzero graded ideal of V.

Proof: Let, x ∈ I^{σ}, y ∈ I^{−σ}, we can write x = x_{2} + x_{1} + x_{0}, y = y_{2} + y_{1} + y_{0}, where x_{α} = π_{α}(x), y_{α} = π_{α}(y), α = 0, 1, 2. By the Peirce multiplication rules (1.1),

π_{0}(Q_{x}y) = Q_{x_{0}}y_{0} + Q_{x_{1}}y_{2} + {x_{0}, y_{1}, x_{1}} =
(Q_{x_{0}}y − Q_{x_{0}}(y_{1} + y_{2})) + (Q_{x}y_{2} − Q_{x_{0}+x_{2}}y_{2} − {x_{0} + x_{2}, y_{2}, x_{1}})+
({x_{0}, y_{1}, x} − {x_{0}, y_{1}, x_{0} + x_{2}}) =
(Q_{x_{0}}y − Q_{x_{0}}y_{1} − Q_{x_{0}}y_{2})+
(Q_{x}y_{2} − Q_{x_{0}}y_{2} − Q_{x_{2}}y_{2} − {x_{0}, y_{2}, x_{2}} − {x_{0}, y_{2}, x_{1}} − {x_{2}, y_{2}, x_{1}})+
({x_{0}, y_{1}, x} − {x_{0}, y_{1}, x_{0}} − {x_{0}, y_{1}, x_{2}}) =_{(by(1.1))}
Q_{x_{0}}y + Q_{x}y_{2} − Q_{x_{2}}y_{2} − {x_{2}, y_{2}, x_{1}} + {x_{0}, y_{1}, x} − {x_{0}, y_{1}, x_{2}} =
Q_{x_{0}}y + Q_{x}y_{2} − Q_{x_{2}}y_{2} − ({x_{2}, y_{2}, x} − {x_{2}, y_{2}, x_{0}} − {x_{2}, y_{2}, x_{2}})+
{x_{0}, y_{1}, x} − ({x_{0}, y, x_{2}} − {x_{0}, y_{0}, x_{2}} − {x_{0}, y_{2}, x_{2}}) =_{(by(1.1))}
Q_{x_{0}}y + Q_{x}y_{2} − Q_{x_{2}}y_{2} − {x_{2}, y_{2}, x} + 2Q_{x_{2}}y_{2} + {x_{0}, y_{1}, x} − {x_{0}, y, x_{2}} =

Q_{x_{0}}y+Q_{x}y_{2}−{x_{2}, y_{2}, x}+(Q_{x_{2}}y−Q_{x_{2}}y_{0}−Q_{x_{2}}y_{1})+{x_{0}, y_{1}, x}−{x_{0}, y, x_{2}} =_{(by(1.1))}

Q_{x_{0}}y + Q_{x}y_{2} − {x_{2}, y_{2}, x} + Q_{x_{2}}y + {x_{0}, y_{1}, x} − {x_{0}, y, x_{2}} ∈ I^{σ}

since x ∈ I^{σ} and y ∈ I^{−σ}.

We have shown π_{0}(M^{σ}) ⊆ I^{σ}, σ = ±, i.e., π_{0}(M) ⊆ I. By (0, 2)-symmetry, π_{2}(M) ⊆ I and, hence, π_{1}(M) ⊆ I: for any m ∈ M^{σ}, π_{1}(m) = m − π_{0}(m) − π_{2}(m) ∈ M + I ⊆ I since M ⊆ I. Therefore, M ⊆ π_{2}(M) + π_{1}(M) + π_{0}(M) ⊆ (I ∩ V_{2}) + (I ∩ V_{1}) + (I ∩ V_{0}).

If V is semiprime and I ≠ 0, then M ≠ 0, hence (I ∩ V_{2}) + (I ∩ V_{1}) + (I ∩ V_{0}) ≠ 0 and the ideal of V generated by (I ∩ V_{2}) + (I ∩ V_{1}) + (I ∩ V_{0}) is a nonzero graded ideal of V contained in I.

2. Peirce Gradings of Jordan Triple Systems

2.1 Given a Jordan triple system T, a Peirce grading of T is a decomposition T = T_{2} ⊕ T_{1} ⊕ T_{0} (direct sum of Φ-modules) such that, for any α, β, γ ∈ {0, 1, 2},

(i) P_{T_{α}}T_{β} ⊆ T_{2α−β},

(ii) {T_{α}, T_{β}, T_{γ}} ⊆ T_{α−β+γ},
(iii) {T_{0}, T_{2}, T } = {T_{2}, T_{0}, T } = 0,

with the convention that T_{λ} = 0 if λ ∉ {0, 1, 2}.

As for pairs, the roles of T_{0} and T_{2} can be exchanged in the above definition, so that, by taking T̃_{α} = T_{2−α}, we obtain a new Peirce grading T = T̃_{2} ⊕ T̃_{1} ⊕ T̃_{0}.

Given a tripotent e in a Jordan triple system T the usual Peirce decomposition of T = T_{2}(e) ⊕ T_{1}(e) ⊕ T_{0}(e) with respect to e [9, 1.1] is an example of Peirce grading of T [9, 1.4].

The formal similarity between Peirce gradings of Jordan pairs and Peirce gradings of triple systems comes from the following obvious though fundamental fact:

2.2 Let T = T_{2} + T_{1} + T_{0} be a decomposition of a Jordan triple system T as a sum of Φ-submodules of T. Then T = T_{2} + T_{1} + T_{0} is a Peirce grading of T if and only if V (T) = V (T_{2}) + V (T_{1}) + V (T_{0}) is a Peirce grading of V (T).

2.3 As for pairs, from the multiplication rules given in (2.1) (or from (1.2) with (2.2)), it is clear that T_{α} is a subtriple of T for α = 0, 1, 2 and, moreover, T_{0} and T_{2} are inner ideals of T.

2.4 Let T = T_{2} ⊕ T_{1} ⊕ T_{0} be a Peirce grading of a Jordan triple system T so that V (T) = V (T_{2}) ⊕ V (T_{1}) ⊕ V (T_{0}) is a Peirce grading of V (T), according to (2.2). Let I a graded ideal of V (T) which is maximal among all graded ideals of V (T) such that I^{+} ∩ I^{−} = 0. The quotient

V = V (T)/I = (T/I^{+}, T/I^{−})

will be called a graded tight double pair of T, associated to the grading T = T_{2} ⊕ T_{1} ⊕ T_{0}. By Zorn’s lemma, an ideal I satisfying the above conditions does always exist. Moreover, given any graded ideal L of V (T) such that L^{+} ∩ L^{−} = 0, there is an ideal I containing L which is maximal among all graded ideals of V (T) with I^{+} ∩ I^{−} = 0.

Using (1.5), we have a Peirce grading

V = (V (T_{2}) + I)/I ⊕ (V (T_{1}) + I)/I ⊕ (V (T_{0}) + I)/I

induced by the grading of T.

The behaviour of graded tight double pairs is formally similar to usual tight double pairs [3, 5.1] replacing sometimes ideals by graded ideals. Namely, under the above conditions:

(i) If 0 ≠ x ∈ T then either 0 ≠ x + I^{+} ∈ V^{+} or 0 ≠ x + I^{−} ∈ V^{−}.

(ii) If L is a nonzero ideal (resp. graded ideal) of T, then ((L + I^{+})/I^{+}, (L + I^{−})/I^{−}) is a nonzero ideal (resp. graded ideal) of V.

(iii) If M = (M^{+}/I^{+}, M^{−}/I^{−}) is a nonzero graded ideal of V, then M^{+} ∩ M^{−} is a nonzero graded ideal of T.

We remark that, unlike (ii), we need in (iii) that the ideal be graded to obtain a nonzero ideal of T since I is maximal only among the graded ideals of V (T) not hitting T. However, we are dealing with properties defined in terms of ideals which are not necessarily graded. This difficulty is the motivation of (1.6), which provides the way to obtain graded ideals inside arbitrary ideals.

2.5 Lemma. Let T = T_{2} ⊕T_{1} ⊕T_{0} be a Peirce grading of a Jordan triple system T, V = V (T)/I be a graded tight double pair of T. Then:

(i) T is nondegenerate if and only if V is nondegenerate.

(ii) T is strongly prime if and only if V is strongly prime.

(iii) T is primitive if and only if V is primitive.

(iv) T is simple if and only if V is simple.

Proof: (i) If V is nondegenerate, then T is nondegenerate using (2.4)(i). Conversely, if T is nondegenerate, then V is nondegenerate, as in the proof of 3.4 of [6], where only the fact that I^{+} ∩ I^{−} = 0 is used.

(ii) If T is strongly prime, then V is nondegenerate by (i). Moreover, if A and B are nonzero orthogonal ideals of V, then, by (1.6), they contain nonzero orthogonal graded ideals Ã and B̃ of V which give rise to nonzero orthogonal ideals of T by (2.4)(iii), which is impossible. Conversely, if V is strongly prime, then T is nondegenerate by (i) and it is also prime since nonzero orthogonal ideals of T would give rise to nonzero orthogonal ideals of V by (2.4)(ii).

(iii) If T is primitive, let K be a primitizer of T at b with modulus c (b, c ∈ T). The proof of [2, 5.8.2] applies here to show that either K + I^{+}/I^{+} ≠ V^{+} or K + I^{−}/I^{−} ≠ V^{−} and is a modular inner ideal of V. Assume that, for example K̃ = K + I^{+}/I^{+} ≠ V^{+} Now, by (1.6), given any nonzero ideal A of V, there exists a nonzero graded ideal L = (L^{+}/I^{+}, L^{−}/I^{−}) of V such that L ⊆ A. Thus L^{+} ∩ L^{−} is a nonzero ideal of T by (2.4)(iii) so that K + (L^{+} ∩ L^{−}) = T and K̃ complements L, hence it also complements A. We have shown that K̃ is a primitizer of V, hence V is primitive.

Conversely, if, for example, K/I^{+} is a primitizer of V at b + I^{−} ∈ V^{−}, then it is readily checked that K is a proper inner ideal of T which is modular at b. Moreover, K complements nonzero ideals of T using (2.4)(ii).

(iv) If T is simple, then it is nondegenerate by [5, 2.4 (ii)], hence V is nondegenerate by (i) and, since it is clearly nonzero, it is not trivial. If A is a nonzero ideal of V, then there exists a nonzero graded ideal L = (L^{+}/I^{+}, L^{−}/I^{−}) of V such that L ⊆ A. Now L^{+} ∩ L^{−} is a nonzero ideal of T by (2.4)(iii), hence T = L^{+} ∩ L^{−}, which implies L = A = V, and we have shown that V is simple.

Conversely, assume that V is simple, so that V and hence T are nontrivial. We have that L = I^{+} + I^{−} is an ideal of T. If L = 0 then I = 0 and V = V (T) is simple, which implies T is simple by (0.3)(i)(c). If L ≠ 0, then ((L+I^{+})/I^{+}, (L+I^{−})/I^{−}) = (L/I^{+}, L/I^{−}) is a nonzero ideal of V. By simplicity of V, V = (L/I^{+}, L/I^{−}) = ((I^{+} ⊕ I^{−})/I^{+}, (I^{+} ⊕ I^{−})/I^{−}) and we have T = I^{+} ⊕ I^{−} = T (I) is a polarized triple system. Moreover V = ((I^{+} ⊕ I^{−})/I^{+}, (I^{+} ⊕ I^{−})/I^{−}) is clearly isomorphic to (I^{−}, I^{+}), hence (I^{−}, I^{+}) is a simple Jordan pair, its opposite I is also simple, and T = T (I) is simple by (0.3)(ii)(c).

2.6 Remark: Notice that, when T_{1} = T_{0} = 0 and T_{2} = T (which is trivially a Peirce grading of T) the graded ideals of V (T) are exactly the ideals of V (T), so that a graded tight double pair of T with respect to this grading is a usual tight double pair [3, 5.1]. As a consequence, (2.5) is a generalization of [3, 5.2(ii)] and also yields the equivalence between the simplicity of a Jordan triple system and any of its tight double pairs.

Now we can prove our main result concerning Peirce gradings of Jordan triple systems.

2.7 Theorem. Let T = T_{2} ⊕ T_{1} ⊕ T_{0} be a Peirce grading of a Jordan triple system T. Let α ∈ {0, 2} and assume T_{α} ≠ 0.

(i) If T is strongly prime (resp. nondegenerate) then T_{α} is strongly prime (resp. nondegenerate).

(ii) If T is primitive (resp. semiprimitive) then T_{α} is primitive (resp. semiprim-

itive). (iii) If T is simple then T_{α} is simple. Proof: Let V = V (T)/I be a graded tight double pair of T and V = V_{2} ⊕V_{1} ⊕V_{0}

be the Pierce grading induced in V by the grading of T as in (2.4). Since T_{α} ≠ 0, and I^{+} ∩ I^{−} = 0, either (T_{α} + I^{+})/I^{+} ≠ 0 or (T_{α} + I^{−})/I^{−} ≠ 0, i.e., V_{α} ≠ 0.

(i) If T is nondegenerate, then so is V (T) by (0.3)(i)(a), hence V (T_{α}) is nondegenerate by (2.2) and [8, 4.1(ii)]. Thus T_{α} is nondegenerate by (0.3)(i)(a).

If T is strongly prime, then V is strongly prime by (2.5)(ii), hence V_{α} is strongly prime by (1.4)(i). If A, B are nonzero orthogonal ideals of T_{α}, then ((A+I^{+})/I^{+}, (A+ I^{−})/I^{−}), ((B + I^{+})/I^{+}, (B + I^{−})/I^{−}) are nonzero orthogonal ideals of V_{α}, which contradicts primeness of V_{α}. We have shown that T_{α} is prime and we already know that it is nondegenerate.

(ii) Notice that a Jordan triple system T is semiprimitive if and only if V (T) is a semiprimitive Jordan pair: if Rad( ) denotes the Jacobson radical, Rad V (T) = (Rad T, Rad T) (cf. [7, Section 4]). Thus, if T is semiprimitive, then so is V (T), hence V (T_{α}) is semiprimitive by (2.2) and (1.4)(ii), and T_{α} is semiprimitive.

If T is primitive, then V is primitive by (2.5)(iii), hence V_{α} is primitive by (1.4)(ii). Moreover, W_{α} = (T_{α}/(T_{α} ∩ I^{+}), T_{α}/(T_{α} ∩ I^{−}) is also primitive since it is isomorphic to V_{α} . Let us take, for example, a primitizer K̃ = K/(T_{α} ∩ I^{+}) of W_{α} at b + (T_{α} ∩ I^{−}) ∈ W_{α}^{−} (b ∈ T_{α}). It is very easy to check that K is a proper inner ideal of T_{α} modular at b. Moreover, for any nonzero ideal A of T_{α}, ((A + (T_{α} ∩ I^{+}))/(T_{α} ∩ I^{+}), (A + (T_{α} ∩ I^{−}))/(T_{α} ∩ I^{−})) is a nonzero ideal of W_{α},

hence K̃ + (A + (T_{α} ∩ I^{+}))/(T_{α} ∩ I^{+}) = W_{α}, and K + A = T_{α}, i.e., K is a primitizer for T_{α}.

(iii) If T is simple, then V is simple by (2.5)(iv), hence V_{α} is simple by (1.4)(iii). As above, W_{α} = (T_{α}/(T_{α} ∩ I^{+}), T_{α}/(T_{α} ∩ I^{−}) is isomorphic to V_{α}, hence W_{α} is simple. Now, since M = I ∩ V (T_{α}) = (T_{α} ∩ I^{+}, T_{α} ∩ I^{−}) is an ideal of V (T_{α}), (T_{α} ∩ I^{+}) + (T_{α} ∩ I^{−}) is an ideal of T_{α} and

L = ([(T_{α} ∩ I^{+}) + (T_{α} ∩ I^{−})]/(T_{α} ∩ I^{+}), [(T_{α} ∩ I^{+}) + (T_{α} ∩ I^{−})]/(T_{α} ∩ I^{−}))

is an ideal of W_{α}. By simplicity of W_{α}, either L = 0 or L = W_{α}.

If L = 0, then T_{α} ∩ I^{+} = T_{α} ∩ I^{−} = 0, W_{α} is isomorphic to V (T_{α}), which is then simple, and T_{α} is simple by (0.3)(i)(c).

If L = W_{α}, then W_{α} is isomorphic to (T_{α} ∩I^{−}, T_{α} ∩I^{+}), hence (T_{α} ∩I^{−}, T_{α} ∩I^{+}) and its opposite M are simple. Moreover, we also have that T_{α} = (T_{α} ∩ I^{+}) + (T_{α} ∩ I^{−}) = (T_{α} ∩ I^{+}) ⊕ (T_{α} ∩ I^{−}) = T (M), and T_{α} is simple by (0.3)(ii)(c).

2.8 Remark: We will not explicitly deal with Jordan algebras since every reasonable definition of Peirce grading for algebras (formally reproducing the multiplication rules of Peirce decompositions with respect to idempotents) would be in particular a Peirce grading of the underlying triple system, giving rise, by duplication, to a Peirce grading of the corresponding double pair. Since all regularity conditions we have considered are equivalent for a Jordan algebra J and the pair V (J) (nondegeneracy with obvious proof, semiprimitivity by [7, 4.17], strong primeness by [6, 1.12], primitivity by [4, 0.11], and simplicity by [5, 3.4]) they are inherited by the diagonal components of a Peirce grading of J, applying [8, 4.1(ii)] and (1.4) to V (J).

Acknowledgments: We want to thank Prof. Ottmar Loos for allowing us to include his result (Lemma 1.3) in our paper, as well as for bringing the problem of the inheritance of regularity by the components of Peirce gradings to our attention.

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

E-mail address: cortes@pinon.ccu.uniovi.es