Algebras of quotients of Lie algebras

Mercedes Siles Molina^{1} Departamento de Álgebra, Geometría y Topología, Universidad de Málaga, 29071 Málaga, España e-mail address: mercedes@agt.cie.uma.es

Abstract: In this paper we introduce the notion of algebra of quotients of a Lie algebra. Properties such as semiprimeness, primeness or nondegeneracy can be lifted from a Lie algebra to its algebras of quotients. We construct a maximal algebra of quotients for every semiprime Lie algebra and give a Passman-like characterization of this (unique) maximal algebra of quotients.

Introduction.

The notion of ring of quotients has played an important role in the development of the theories of associative and commutative rings. In the associative case, the notion of general left quotient ring was introduced by Utumi in 1956 (see [10]). In his paper Utumi proved that every associative ring R without nonzero elements annihilating R on the right hand side has a maximal left quotient ring Q_{max}^{l}(R) and constructed it. Maximal in the sense that every left quotient ring of R can be included in Q_{max}^{l}(R) via a monomorphism which is the identity when restricted to R.

Our aim is to introduce the notion of algebra of quotients of a Lie algebra, study its properties and construct a maximal algebra of quotients of a semiprime Lie algebra. We follow Utumi’s pattern but we cannot work as in the associative setting.

In the associative case, dense left ideals of the associative ring R and homomorphisms of left R-modules will provide the way to obtain Q_{max}^{l}(R). In the Lie case, essential ideals and partial derivations will do the work. The idea of considering partial derivations appears in Martinez’s paper [8], where the author uses Lie algebras to give an Ore condition for a Jordan algebra to have a classical algebra of quotients.

In the preliminaries, containing basic definitions and properties of Lie algebras, we study some facts concerning the center of an associative algebra and the center of a left quotient algebra of it. Section 2 is devoted to the notion of algebra of quotients of a Lie algebra, introduced by following the associative pattern, and characterized in terms of ideals with zero annihilator. We show that if L is a semiprime, prime or nondegenerate Lie algebra then every algebra of quotients is semiprime, prime or nondegenerate, respectively. Our proof of nondegeneracy needs two and three-torsion freedom. Associative algebras supply us with examples of Lie algebras of quotients. In the third section we construct the maximal algebra of quotients of every semiprime Lie algebra and characterized it. Finally we show that every semisimple Lie algebra is its own maximal algebra of quotients.

1 Partially supported by the MCYT, BFM2001-1938-C02-01 and the “Plan Andaluz de Investigación y Desarrollo Tecnológico”, FQM 264.

1. Preliminaries.

In this paper we will deal with Lie algebras over an arbitrary unital and commutative ring of scalars Φ. First of all we will give some definitions and basic notation.

A Φ-module L with a bilinear map [ , ] : L → L, denoted (x, y) ↦ [x, y] and called the bracket of x and y is called a Lie algebra over Φ if the following axioms are satisfied:

(i) [x, x] = 0,

(ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 (Jacobi identity),

for every x, y, z ∈ L.

Every associative algebra S gives rise to a Lie algebra S^{(−)} by considering the same module structure and product given by [x, y] = xy − yx.

A Lie algebra L is said to be semiprime if for every nonzero ideal I of L, I^{2} ≠ 0 (I^{2} := [I, I]). In other words, L has no abelian ideals. Any element x of a Lie algebra L determines a map ad x : L → L defined by ad x(y) = [x, y], which is a derivation of the Lie algebra L. An element x ∈ L is an absolute zero divisor if (ad x)^{2} = 0 (equivalently [[L, x], x] = 0). The element x is said to be a sandwich of L of thickness r if the identities ad x ady_{1} … ad y_{k} ad x = 0, with y_{1}, …, y_{k} ∈ L are satisfied for k = 0, …, r, but ad x adz_{1} … ad z_{r+1} ad x ≠ 0 for some z_{1}, …, z_{r+1} ∈ L. For r ≤ 1 we speak of a thin sandwich. By [5, Proposition 1.5.2], if Φ is two torsion free, the concepts of absolute zero divisor and thin sandwich are the same for a Lie algebra L. The algebra L is said to be nondegenerate (strongly nondegenerate in terms of trikin) if it does not contain nonzero absolute zero divisors. Semiprime Lie algebras may be degenerate. For an example of a semiprime degenerate Lie algebra see, for instance, [2, (0.3)].

For a subset X of a Lie algebra L recall that the set

Ann(X) = {a ∈ L | [a, x] = 0 for every x ∈ X}

is called the annihilator of X. This will be denoted by Ann_{L}(X) when it is necessary to emphasize the dependence on L. It is easy to check (by using the Jacobi identity) that Ann(X) is an ideal of L when X is an ideal of L.

1.1. Remark. For a Lie algebra L we have:

L nondegenerate ⇒ L semiprime ⇒ Ann(L) = 0.

The converses are not true.

Proof: Suppose L nondegenerate. Then, for every element y in an ideal I of L such that [I, I] = 0 we have [[L, y], y] ⊆ [I, I] = 0 and, by the nondegeneracy of L, y = 0. Hence, L must be semiprime.

Now, suppose L semiprime. Then, since Ann(L) is an ideal of L which satisfies [Ann(L), Ann(L)] ⊆ [Ann(L), L] = 0, Ann(L) = 0.

To see that the converses are not true, consider the ℝ-algebra A^{(−)}, where

A = \{( \begin{matrix}a & 0 & 0 \\ 0 & b & 0 \\ c & d & 0 \end{matrix} ) | a, b, c, d ∈ ℝ\} .

Then Ann(A^{(−)}) = 0 while I = \{( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ c & d & 0 \end{matrix} ) | c, d ∈ ℝ\} is a nilpotent ideal of

A^{(−)}.

Finally, that semiprimeness does not implies nondegeneracy has been previously shown.

Every Lie algebra L (over Φ) satisfies that ad L = {ad x| x ∈ L} generates an associative algebra A(L) over Φ whose elements are Φ-linear combinations of all products

ad x_{1} ad x_{2} … ad x_{n},

where juxtaposition means the usual composition of endomorphisms. If Ann(L) = 0, then the map p ↦ ad p is a monomorphism of Lie algebras from L into A(L)^{(−)}. We shall make use of this fact.

An ideal I of a Lie algebra L is said to be essential if every nonzero ideal of L hits I (I ∩ J ≠ 0 for every nonzero ideal J of L). Essential ideals in semiprime Lie algebras can be characterized in terms of their annihilators.

1.2. Lemma. Let I be an ideal of a semiprime Lie algebra L. Then:

(i) I ∩ Ann(I) = 0.

(ii) I is an essential ideal of L if and only if Ann(I) = 0.

Proof: (i). We have (I ∩Ann(I))^{2} ⊆ [I, Ann(I)] = 0 and by the semiprimeness of L we obtain I ∩ Ann(I) = 0.

(ii). Consider an essential ideal I of L; then I ∩ Ann(I) = 0 by (i), and by the essentiallity, Ann(I) = 0. Conversely, suppose Ann(I) = 0. If J is an ideal of L satisfying J ∩ I = 0 then, since [I, J] ⊆ I ∩ J = 0, J ⊆ Ann(I) = 0 and hence I must be essential.

Recall that an associative algebra S is said to be a left quotient algebra of a subalgebra A if given p, q ∈ S, with p ≠ 0, there exists x ∈ A such that xp ≠ 0 and xq ∈ A. Right quotient algebras are defined similarly. An associative algebra A has a left quotient algebra if and only if it has no total right zero divisors different from zero (x in A is said to be a total right zero divisor if Ax = 0).

The center of an associative algebra A, denoted by Z(A), is just the annihilator of the Lie algebra A^{(−)}. A left ideal I of A is said to be dense in A if A is a left quotient algebra of I.

1.3. Lemma. Let A be a subalgebra of an associative algebra S, and suppose that S is a left quotient algebra of A.

(i) Z(A) = Z(S) ∩ A.

(ii) Z(S) = 0 implies Z(A) = 0.

(iii) [q, A] = 0 for some q ∈ S implies [q, S] = 0.

(iv) Let I be a dense left ideal of A, and q ∈ S. If [q, I] = 0, then [q, S] = 0.

Proof: (i). The containing Z(S)∩A ⊆ Z(A) is obvious. Now, consider x ∈ Z(A) and suppose x ∉ Z(S). Then [x, s] ≠ 0 for some s ∈ S. Apply that S is a left quotient algebra of A to find a ∈ A such that as ∈ A and 0 ≠ a[x, s] = axs−asx = (x ∈ Z(A)) xas − xas = 0, which is a contradiction.

(ii) follows by (i).

(iii). Suppose [q, A] = 0 but 0 ≠ [q, s] for some s ∈ S. Since S is a left algebra of quotients of A, there exists a ∈ A such that as ∈ A and 0 ≠ a[q, s] = aqs − asq = ([q, A] = 0 and as ∈ A) qas − qas = 0, a contradiction.

(iv) follows by (iii) taking into account that I being a dense left ideal of A implies that S is a left quotient algebra of I ([10, (1.5)]).

1.4. Remark. The converse of (ii) is not true: Let V be a left vector space over the field ℝ of infinite (countable) dimension, S = Endℝ(V) and R = Soc(S). Then S is a left quotient ring of R (in fact, S = Q_{max}^{l}(R), see, for example [6, (14.45)]) and Z(S) ≠ 0, while Z(R) = 0.

1.5. Lemma. Let A be a semiprime associative algebra such that Z(A) = 0. If I is an essential ideal of A, then I^{(−)} is an ideal of A^{(−)} such that Ann_{A^{(−)}}(I^{(−)}) = 0.

Proof: It is clear that I^{(−)} is an ideal of A^{(−)}. Suppose a ∈ Ann_{A^{(−)}}(I^{(−)}). Then [a, I] = 0 implies, by (1.3)(iv), [a, A] = 0. Since Z(A) = 0, a = 0.

2. Algebras of quotients of a Lie algebra.

Inspired by the notion of ring of quotients for associative rings given by Utumi in 1956 (see [10]) we introduce the notion of algebra of quotients of a Lie algebra.

Let L and Q be Lie algebras, with L ⊆ Q. For any q ∈ Q we denote by _{L}(q) the linear span in Q of q and the elements of the form ad x_{1} … ad x_{n}q, with n ∈ ℕ and x_{1}, …, x_{n} ∈ L. In particular, if q ∈ L, _{L}(q) denotes the ideal of L generated by q.

2.1. Definition. Let L be a subalgebra of a Lie algebra Q. We will say that Q is an algebra of quotients of L if given p, q ∈ Q, with p ≠ 0, there exists x ∈ L such that [x, p] ≠ 0 and [x, _{L}(q)] ⊆ L. The algebra L will be called a subalgebra of

quotients of Q.

2.2. Definition. An element x in a Lie algebra L is said to be a total zero divisor if x ∈ Ann(L).

2.3. Remark. Every Lie algebra L without nonzero total zero divisors (equivalently Ann(L) = 0) is an algebra of quotients of itself and, conversely, if L has an algebra of quotients then Ann(L) = 0.

Since every associative algebra S gives rise to a Lie algebra S^{(−)}, it is natural to ask about the relation between S being a left (or right) algebra of quotients of A and S^{(−)} being an algebra of quotients of A^{(−)}. We will return to this question at the end of this section.

2.4. Examples. There are associative algebras A which have left algebras of quotients but A^{(−)} has not an algebra of quotients. Take, for example, a field K, which is a left and a right algebra of quotients of itself, while Ann(K^{(−)}) = K^{(−)}.

Conversely there are examples of Lie algebras of the form A^{(−)}, for A associative, such that Ann(A^{(−)}) = 0 (that is, A^{(−)} is an algebra of quotients of itself) and A contains total right zero divisors (which means that A has not a left algebra of quotients): Consider the associative algebra A given in (1.1). Then Ann(A^{(−)}) = 0

and (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ c & d & 0 \end{matrix} ) is a total right zero divisor of A, for every c, d ∈ ℝ.

The following result shows how certain properties of a Lie algebra L are inherited by each algebra of quotients Q. In fact, we just need a weaker condition on Q.

2.5. Definition. An overalgebra Q of a Lie algebra L is said to be a weak algebra of quotients of L if for every nonzero element q ∈ Q there exists x ∈ L such that 0 ≠ [x, q] ∈ L.

2.6. Remark. Every algebra of quotients of a Lie algebra L is a weak algebra of quotients. The converse is not true.

Indeed, consider the ℂ-module P of all polinomials ∑_{r=0}^{m} α_{r}x^{r}, with α_{i} ∈ ℂ and m ∈ ℕ. Denote by σ : ℂ → ℂ the complex conjugation. Then the following product makes P into a Lie algebra:

[∑_{r=0}^{m} α_{r}x^{r}, ∑_{s=0}^{n} β_{s}x^{s}] = ∑_{r,s}(α_{r}β_{s}^{σ} − β_{s}α_{r}^{σ})x^{r+s}.

Let Q be the Lie algebra P/I, where I denotes the ideal of P consisting of all polynomials whose first nonzero term has degree at least 4, and let L be the following subalgebra of Q:

L = \{\overline{α}_{0}+ \overline{α}_{2}\overline{x} ^{2} + \overline{α}_{3}\overline{x} ^{3} | α_{0}, α_{2}, α_{3} ∈ ℂ\},

where \overline{y} denotes the class of an element y ∈ P in P/I. Then Q is a weak algebra of quotients of L, but Q is not a quotient algebra of L since no l ∈ L satisfies [l, \overline{x}] ∈ L and [l, \overline{x}^{3}] ≠ 0.

2.7. Proposition. Let Q be a weak algebra of quotients of a subalgebra L.

(i) For every nonzero ideal I of Q, I ∩ L is a nonzero ideal of L (i.e., L is tight in Q).

(ii) L semiprime (prime) implies Q semiprime (prime).

(iii) If Φ is two and three-torsion free and L is nondegenerate, then Q is nondegenerate.

Proof: (i). Take 0 ≠ y ∈ I and choose x ∈ L such that 0 ≠ [x, y] ∈ I ∩ L.

(ii). Suppose L prime. If I, J are nonzero ideals of Q then, by (i), Ĩ = I ∩ L and J̃ = J ∩ L are nonzero ideals of L. By the primeness of L, [Ĩ, J̃] ≠ 0, so [I, J] ≠ 0, which proves that Q is prime. The semiprimeness of Q follows from that of L in an analogous way.

(iii). Let q be in Q such that (F.1) ad q ad q = 0. Since L is semiprime, Q is semiprime too (by (ii)), so that Ann(Q) = 0 and

p ↦ ad p is a monomorphism from Q into A(Q)^{(−)}, and we can identify L and Q with subalgebras of A(Q)^{(−)}. This fact will be used in what follows. Inside A(Q)^{(−)}, (F.1) reads:

(F.2) q q = 0. The element q is a thin sandwich in Q, that is, (F.3) q p q = 0 for every p ∈ Q. This implies (F.4) qztq = qtzq for every z, t ∈ Q. Denote x := [a, q], for a ∈ Q. (F.5) x^{2} = −qa^{2}q. x^{2} = (aq − qa)(aq − qa) = aqaq − aq^{2}a − qa^{2}q + qaqa = −qa^{2}q (by (F.2) and

(F.3)).

(F.6) x^{3} = 0. x^{3} = (by (F.5)) −qa^{2}q(aq − qa) = 0 by (F.2) and (F.3). (F.7) xzx = aqzaq − qazaq + qazqa for every z ∈ Q. xzx = (aq − qa)z(aq − qa) = aqzaq − aqzqa − qazaq + qazqa = (by (F.3))

aqzaq − qazaq + qazqa.

By the proof of [5, Lemma 1.5.6] the following relations hold: (F.8) x^{2}yx^{2} = 0 for every y ∈ Q. (F.9) xyx^{2} = x^{2}yx for every y ∈ Q. Denote u := [x, [x, y]], for y ∈ Q. Then (F.10) u = −qa^{2}qy − yqa^{2}q − 2aqyaq + 2qayaq − 2qayqa. u = x^{2}y+yx^{2}−2xyx = (by (F.5) and (F.7)) −qa^{2}qy−yqa^{2}q−2aqyaq+2qayaq−

2qayqa.

(F.11) u^{2} = x^{2}y^{2}x^{2}. u^{2} = (x^{2}y + yx^{2} − 2xyx)^{2} = x^{2}yx^{2}y + x^{2}y^{2}x^{2} − 2x^{2}yxyx + yx^{4}y + yx^{2}yx^{2} −

2yx^{3}yx − 2xyx^{3}y − 2xyxyx^{2} + 4xyx^{2}yx = (by (F.6) and (F.8)) x^{2}y^{2}x^{2} − 2x^{2}yxyx − 2xyxyx^{2} + 4xyx^{2}yx = (by (F.9)) x^{2}y^{2}x^{2}.

Now, we prove (F.12) 0 = qulqtpq + qupqtlq + qutqplq + qlpqtuq + qltqpuq + qptqluq for every u, l, t, p ∈ Q. Take [u, [l, [p, [t, q]]]] instead of p in (F.3). Since [l, [p, [t, q]]] = [l, p(tq − qt) − (tq − qt)p] = [l, ptq − pqt − tqp + qtp] =

lptq − lpqt − ltqp + lqtp − ptql + pqtl + tqpl − qtpl,

0 = q[u, [l, [p, [t, q]]]]q = qu(lptq − lpqt − ltqp + lqtp − ptql + pqtl + tqpl − qtpl)q − q(lptq − lpqt − ltqp + lqtp − ptql + pqtl + tqpl − qtpl)uq = qulqtpq + qupqtlq + qutqplq + qlpqtuq + qltqpuq + qptqluq (by (F.2) and (F.3)).

Moreover (F.13) 0 = qa^{2}ql^{2}q + ql^{2}qa^{2}q + 4qalqalq for every a, l ∈ Q. Substitute in (F.12) u and l by a, and t and p by l. 0 = qa^{2}ql^{2}q + qalqlaq + qalqlaq + qalqlaq + qalqlaq + ql^{2}qa^{2}q = (by (F.4))

qa^{2}ql^{2}q + ql^{2}qa^{2}q + 4qalqalq.

Substitute in (F.12) u, l, t by a, and p by l. 0 = qa^{2}qalq + qalqa^{2}q + qa^{2}qlaq + qalqa^{2}q + qa^{2}qlaq + qlaqa^{2}q = (by (F.4))

3qa^{2}qalq + 3qalqa^{2}q.

Since Q is three-torsion free, (F.14) qa^{2}qalq + qalqa^{2}q = 0 for every a, l ∈ Q. Take l = a in (F.14) and use that Q is two torsion-free. (F.15) qa^{2}qa^{2}q = 0 for every a ∈ Q. Multiply (F.13) by qa^{2} on the left hand side and apply (F.15). (F.16) 0 = qa^{2}ql^{2}qa^{2}q + 4qa^{2}qalqalq for every a, l ∈ Q. Apply that Q is a weak algebra of quotients of L and consider a ∈ L such that

x := [a, q] ∈ L. The element α = [x, [x, a]] ∈ L satisfies:

α^{2} = (by (F.10))(−3qa^{2}qa − 3aqa^{2}q + 2qa^{3}q)^{2} = 0 by (F.2), (F.3) and (F.15). Since L has no absolute zero divisors, α = 0, that is,

−3qa^{2}qa − 3aqa^{2}q + 2qa^{3}q = 0, and if we multiply by yq on the right hand side, for every y ∈ Q,

0 = −3qa^{2}qayq − 3aqa^{2}qyq + 2qa^{3}qyq = (by (F.3)) −3qa^{2}qayq.

Since Φ is three-torsion free,

(F.17) qa^{2}qayq = 0 for every y ∈ Q.

For every y ∈ L we have that v = [x, [x, y]] ∈ L satisfies

v^{2} = (by (F.11)) x^{2}y^{2}x^{2} = (by (F.5)) qa^{2}qy^{2}qa^{2}q = (by (F.16))

= −4qa^{2}qayqayq = 0 by (F.17),

but L has no absolute zero divisors. This implies v = 0. Apply again the absence of sandwich elements in L to obtain x = 0. Since Q is a weak algebra of quotients of L, q = 0 and hence Q must be nondegenerate.

2.8. Definition. Let L be a subalgebra of a Lie algebra Q. For any q ∈ Q we define

(L : q) := {x ∈ L | [x, _{L}(q)] ⊆ L}.

The following definition has been inspired by that of [1, (0.4)] .

2.9. Definition. Let L be a subalgebra of a Lie algebra Q. We say that Q is ideally absorbed into L if for every nonzero element q ∈ Q there exists an ideal I of L with Ann_{L}(I) = 0 such that 0 ≠ [I, q] ⊆ L.

It is immediate to see that if Q is ideally absorbed into L then Q is a weak algebra of quotients of L.

2.10. Lemma. Let L be a subalgebra of a Lie algebra Q, with L ⊆ Q, and consider q ∈ Q.

(i) (L : q) is an ideal of L.

(ii) If Q is an algebra of quotients of L then Ann_{L}((L : q)) = 0. In particular, (L : q) is an essential ideal of L.

(iii) If Q is ideally absorbed into L, then Ann_{L}((L : q)) = 0. In particular, (L : q) is an essential ideal of L.

Proof: (i) follows immediately by using the Jacobi identity.

(ii). First, we will prove that (L : q) is an essential ideal of L. Let I be a nonzero ideal of L. Take 0 ≠ y ∈ I. Apply that Q is an algebra of quotients of L to find x ∈ L such that 0 ≠ [x, y] and [x, _{L}(q)] ⊆ L. Since for every q̃∈ _{L}(q) we have [[x, y], q̃] = [x, [y, q̃]] − [y, [x, q̃]] ∈ L, then [x, y] is a nonzero element of I ∩ (L : q). This shows the essentiallity of (L : q). Now, suppose Ann_{L}((L : q)) ≠ 0. Since (L : q) is an essential ideal of L, there exists 0 ≠ u ∈ Ann_{L}((L : q))∩(L : q). Apply that Q is an algebra of quotients of L to find x ∈ L such that [x, u] ≠ 0 and [x,_{L}(q)] ⊆ L. This last containment means x ∈ (L : q), and so [x, u] = 0 (because u ∈ Ann_{L}((L : q))), which is a contradiction.

(iii). Consider an ideal I of L with Ann_{L}(I) = 0 such that [I, q] ⊆ L. We are going to see I ⊆ (L : q), in which case Ann_{L}((L : q)) ⊆ Ann_{L}(I) = 0 and our proof will be complete.

We claim [I, ad x_{1} … ad x_{n}q] ⊆ L for every x_{1}, …, x_{n} ∈ L, with n ∈ ℕ. We will prove it by induction on n. For n = 1, [I, [x_{1}, q]] ⊆ [[I, x_{1}], q] + [x_{1}, [I, q]] ⊆ [I, q] + [L, [I, q]] ⊆ L. Suppose the claim true for n − 1 elements in L and consider x_{1}, … x_{n} ∈ L. Then, [I, ad x_{1} … ad x_{n}q] ⊆ [[I, x_{1}], ad x_{2} … ad x_{n}q]+ [x_{1}, [I, ad x_{2} … ad x_{n}q]] ⊆ [I, ad x_{2} … ad x_{n}q]+ [L, [I, ad x_{2} … ad x_{n}q]] ⊆ L by the induction hypothesis. Now it follows I ⊆ (L : q).

2.11. Lemma. Let L be a subalgebra of a Lie algebra Q and suppose that Q is a weak algebra of quotients of L. Then for every ideal I of L, Ann_{L}(I) = 0 implies Ann_{Q}(I) = 0.

Proof: Suppose 0 ≠ q ∈ Ann_{Q}(I). By the hypothesis there exists an element x in L such that 0 ≠ [x, q] ∈ L. Since Ann_{L}(I) = 0, there exists y ∈ I satisfying 0 ≠ [y, [x, q]] = [[y, x], q] + [x, [y, q]] ⊆ [I, q] + [L, [I, q]] = 0, which is a contradiction.

2.12. Notation. Denote by I_{e}(L) the set of all essential ideals of a Lie algebra L.

Clearly, given I, J ∈ I_{e}(L) we have I ∩ J ∈ I_{e}(L). If L is semiprime, then I^{2} ∈ I_{e}(L).

2.13. Lemma. Let L be a semiprime Lie algebra and Q an algebra of quotients of L. Then for every essential ideal I of L we have that Q is an algebra of quotients of I.

Proof: Consider p, q ∈ Q, with p ≠ 0. By (2.10)(ii), (L : q) ∈ I_{e}(L), so that (L : q)∩I ∈ I_{e}(L) and, consequently, ((L : q)∩I)^{2} ∈ I_{e}(L). Apply (1.2) and (2.11) to obtain Ann_{Q}((L : q)∩I)^{2} = 0. So there exist y, z ∈ (L : q)∩I such that [[y, z], p] ≠ 0. Moreover, [[y, z], _{I}(q)] ⊆ [y, [z, _{I}(q)]]+[z, [y, _{I}(q)]] ⊆ [y, [z, _{L}(q)]]+[z, [y, _{L}(q)]] ⊆ [I, L] ⊆ I.

Being a left quotient algebra of an associative algebra A can be characterized by using left ideals of A with zero right annihilator.

2.14. Lemma. Let A be a subalgebra of an associative algebra Q. Then Q is a left quotient algebra of A if and only if for every nonzero element q ∈ Q there exists a left ideal I of A with ran_{A}(I) = 0 such that 0 ≠ Iq ⊆ A.

Proof: If Q is a left quotient algebra of A, given 0 ≠ q, by [7, Lemma 4.3.2] (A : q) := {x ∈ A | xq ∈ A} is a dense left ideal of A. This means ran_{A}((A : q)) = 0 and 0 ≠ (A : q)q ⊆ A.

Conversely, consider p, q ∈ Q with p ≠ 0. By the hypothesis there exists a left ideal I of A with ran_{A}(I) = 0 such that 0 ≠ Ip ⊆ A. Consider yp ≠ 0, with y ∈ I. If Iq = 0 we have found y ∈ A satisfying yp ≠ 0 and yq ∈ A. If yq ≠ 0, apply again the hypothesis and consider a left ideal J of A with ran_{A}(J) = 0 and such that 0 ≠ Jyq ⊆ A. Then zyp ≠ 0 for some z ∈ J (because 0 ≠ yp ∈ A and ran_{A}(J) = 0) and zyq ∈ Jyq ⊆ A. This proves that Q is a left quotient algebra of A.

For a Lie algebra L, the notion of algebra of quotients can be expressed in terms of ideals of L with zero annihilator.

2.15. Proposition. Let L be a subalgebra of a Lie algebra Q. Then Q is an algebra of quotients of L if and only if Q is ideally absorbed into L.

Proof: Suppose Q an algebra of quotients of L and consider 0 ≠ q ∈ Q. By (2.10) (i) and (ii), (L : q) is an ideal of L with zero annihilator in L. By (2.11) this implies Ann_{Q}(L : q) = 0, so that 0 ≠ [(L : q), q] ⊆ L.

Conversely, suppose that Q is ideally absorbed into L. Take p, q ∈ Q, with p ≠ 0. By (2.10) (iii), Ann_{L}((L : q)) = 0 and by (2.11), Ann_{Q}((L : q)) = 0. Hence [(L : q), p] ≠ 0, which shows that Q is an algebra of quotients of L.

The following result provides some examples of algebras of quotients of Lie algebras.

Let A be a semiprime associative algebra and denote by Q_{s}(A) its Martindale symmetric algebra of quotients.

2.16. Proposition. Let A and S be associative algebras such that A is semiprime, Z(S) = 0 and A ⊆ S ⊆ Q_{s}(A). Then S^{(−)} is an algebra of quotients of A^{(−)}.

Proof: Take 0 ≠ s ∈ S. Since S ⊆ Q_{s}(A) there exists an essential ideal I of A such that Is + sI ⊆ A (see, for example [6, (14.25)]). By (1.5), I^{(−)} is an ideal of A^{(−)} such that Ann_{A^{(−)}}(I^{(−)}) = 0 and by (1.3) (iv), Z(S) = 0 implies [I^{(−)}, s] ≠ 0. The result follows by (2.15) because 0 ≠ [I^{(−)}, s] ⊆ A.

2.17. Remark. The converse of the previous result is not true. There are examples of associative algebras A ⊆ S (A a subalgebra of S) with Z(S) = 0 such that S^{(−)} is an algebra of quotients of A^{(−)} but S is not an algebra of quotients of A. Consider, for example, S = A the associative algebra given in (1.1).

3. The maximal algebra of quotients of a semiprime Lie algebra.

In this section we will construct a maximal algebra of quotients for every semiprime Lie algebra L. Maximal in the sense that every algebra of quotients of L can be considered inside this maximal algebra of quotients via a monomorphism which restricted to L is the identity.

3.1. Definition. Given an ideal I of a Lie algebra L, we say that a homomorphism of Φ-modules δ : I → L is a partial derivation if for every x, y ∈ I it is satisfied:

δ[x, y] = [δx, y] + [x, δy].

Denote by PDer(I, L) the set of all partial derivations of I in L.

3.2. Lemma. Let L be a semiprime Lie algebra and consider the set

D := {(δ, I) | I ∈ I_{e}(L), δ ∈ PDer(I, L)}.

Define on D the following relation:

(δ, I) ≡ (μ, J) if and only if there exists K ∈ I_{e}(L), K ⊆ I ∩ J such that δ_{|_{K}} = μ_{|_{K}} .

Then ≡ is an equivalence relation.

Proof: Straightforward.

3.3. Notation. Denote by Q(L) the quotient set D/ ≡. Let δ_{I} denote the equivalence class of (δ, I) in Q(L).

3.4. Theorem. Let L be a semiprime Lie algebra, and let Q := Q(L) be as in (3.3). Define the following maps:

\begin{aligned}· : & Φ × Q & → & Q \\ & (α, δ_{I}) & ↦ & αδ_{I} \end{aligned} where \begin{aligned}αδ : & I & → & L \\ & y & ↦ & αδ(y) = δ(αy) \end{aligned}

\begin{aligned}+ : & Q × Q & → & Q \\ & (δ_{I}, μ_{J}) & ↦ & (δ + μ)_{I∩J} \end{aligned} where \begin{aligned}δ + μ : & I ∩ J & → & L \\ & x & ↦ & δ(x) + μ(x) \end{aligned}

\begin{aligned}[ , ] : & Q × Q & → & Q \\ & (δ_{I}, μ_{J}) & ↦ & [δ, μ]_{(I∩J)^{2}} \end{aligned} where \begin{aligned}[δ, μ] : & (I ∩ J)^{2} & → & L \\ & x & ↦ & δμ(x) − μδ(x) \end{aligned}

Algebras of quotients of Lie algebras 13

Then Q, with these operations, is a Lie algebra containing L as a subalgebra, via the Lie monomorphism

\begin{aligned}φ : & L & → & Q \\ & x & ↦ & ad x_{L} \end{aligned}

Proof: The more “difficult” point in proving that Q is a Lie algebra is to see that the Lie bracket is well defined on Q. Indeed, given δ_{I}, μ_{J} ∈ Q, for every z, t ∈ (I ∩ J) we have [δ, μ]([z, t]) = δμ([z, t]) − μδ([z, t]) = δ([μ(z), t] + [z, μ(t)]) − μ([δ(z), t] + [z, δ(t)]), which makes sense since [μ(z), t], [z, μ(t)], [δ(z), t], [z, δ(t)] ∈ I∩J. Moreover, δ([z, t]), μ([z, t]) ∈ I ∩ J. Since δ and μ are partial derivations, [δ, μ] : (I ∩ J)^{2} → L is a partial derivation.

Finally, it is easy to check that φ defines a homomorphism of Lie algebras. The injectivity of φ follows since ad x_{L} = 0 for some x ∈ L means [x, L] = 0 and hence x ∈ Ann(L), which is zero by the semiprimeness of L and (1.1).

For any X ⊆ L, write X^{φ} to denote the image of X inside Q(L) via the monomorphism φ defined in (3.4).

3.5. Remark. For every δ_{I} ∈ Q(L), and ad x_{L} ∈ I^{φ} (x ∈ I) we have [δ_{I}, ad x_{L}] = ad δx_{L} ∈ L^{φ}.

Proof: For any y ∈ I, [δ, adx]y = δ[x, y]−[x, δy] = [δx, y]+[x, δy]−[x, δy] = [δx, y] = (ad δx)y and so [δ, adx]_{I} = (ad δx)_{I} = (ad δx)_{L} ∈ L^{φ}.

The maximallity of the algebra of quotients constructed in (3.4) is shown in the next result.

3.6. Proposition. Let L be a semiprime Lie algebra. Then Q(L) is semiprime and an algebra of quotients of L. Moreover, Q(L) is maximal among the algebras of quotients of L, in the sense that if S is an algebra of quotients of L, then there exists a Lie monomorphism ψ : S → Q(L) which is the identity on L. In particular, the map

\begin{aligned}ψ : & S & → & Q(L) \\ & s & ↦ & ad s_{(L:s)} \end{aligned}

is a monomorphism of Lie algebras which is the identity when restricted to L.

Proof: Take δ_{I}, μ_{I} ∈ Q(L), with δ_{I} ≠ 0 (we can consider the same I for δ and μ because if J, K ∈ I_{e}(L) were such that δ : J → L, μ : K → L, then I := J∩K ∈ I_{e}(L) and so δ_{J} = δ_{I} and μ_{K} = μ_{I}). Choose a ∈ I such that δa ≠ 0. Then ad a_{L} ∈ L^{φ} satisfies: [δ_{I}, ada_{L}] = (by (3.5)) ad δa_{L} ≠ 0 (for every ideal P ∈ I_{e}(L), [δa, P] = 0 implies δa ∈ Ann(P) = 0 by the semiprimeness of L and (1.2) (ii)) and for every λ_{I} ∈ _{L^{φ}}(μ_{I}), [λ_{I}, ad a_{L}] = (by (3.5)) ad λa_{L} ∈ L^{φ}. This proves that Q(L) is an algebra of quotients of L^{φ}. The semiprimeness follows by condition (ii) in (2.7).

Now suppose S is an algebra of quotients of L and consider the map

\begin{aligned}ψ : & S & → & Q(L) \\ & s & ↦ & ad s_{(L:s)} \end{aligned}

By (2.10) (ii), ψ is well defined. Moreover, ψ is a Lie homomorphism. To prove the injectivity, suppose s ∈ S such that ψ(s) = 0, that is, [s, K] = 0 for some ideal K ∈ I_{e}(L), K ⊆ (L : s). By the semiprimeness of L and (1.2) (ii), Ann_{L}(K) = 0 = (by (2.11)) Ann_{S}(K). This implies s = 0.

3.7. Definition. For a semiprime Lie algebra L, the algebra constructed in (3.4) will be called the maximal algebra of quotients of L. Denote it by Q_{m}(L).

The axiomatic characterization of the Martindale ring of quotients given by D. Passman in [9] has inspired us to give the following description of the maximal algebra of quotients of a semiprime Lie algebra.

3.8. Theorem. Let L be a semiprime Lie algebra and consider an overalgebra S of L. Then S is isomorphic to Q_{m}(L), under an isomorphism which is the identity on L, if and only if S satisfies the following properties:

(i) For any s ∈ S there exists I ∈ I_{e}(L) such that [I, s] ⊆ L.

(ii) For s ∈ S and I ∈ I_{e}(L), [I, s] = 0 implies s = 0.

(iii) For any I ∈ I_{e}(L), δ ∈ PDer(I, L), δ : I → L, there exists s ∈ S such that δx = [s, x] for every x ∈ I.

Proof: Define

\begin{aligned}ψ : & S & → & Q_{m}(L) \\ & s & ↦ & ad s_{I} \end{aligned}

where I is a nonzero essential ideal of L satisfying [I, s] ⊆ L; this exists by (i), and so the map is well defined. Moreover it is a Lie homomorphism: given s, t ∈ S, by (i) there exist I, I′, I′′ ∈ I_{e}(L) such that [I, s] ⊆ L, [I′, t] ⊆ L, [I′′, [s, t]] ⊆ L. Take J = I ∩ I′ ∩ I′′. Then J^{2} ∈ I_{e}(L) and ψ([s, t]) = ad[s, t]_{J^{2}} = [ad s, adt]_{J^{2}} = [ψ(s), ψ(t)]. The other conditions to be a Lie homomorphism follow immediately.

The map ψ is injective (if ad s_{I} = 0 for some s ∈ S, by (ii), s = 0) and surjective (consider δ_{I} ∈ Q(L); by (iii), there exists s ∈ S such that δx = ad s(x) for every x ∈ I, hence δ_{I} = ψ(s)). Finally, ψ is the identity on L, by identifying L with L^{φ}, where φ is the map defined in (3.4).

Conversely, we will prove that Q(L) satisfies the three conditions.

(i). Consider q ∈ Q(L). By (3.6) and (2.10) (ii), (L : q) ∈ I_{e}(L) and, by the definition, [(L : q), q] ⊆ L.

(ii). Take q ∈ Q(L) and I ∈ I_{e}(L) such that [I, q] = 0. Then q ∈ Ann_{Q(L)}(I) = 0 by (3.6) and (2.11).

(iii). Given I^{φ} ∈ I_{e}(L^{φ}) and \overline{δ} ∈ PDer(I^{φ}, L^{φ}), \overline{δ} : I^{φ} → L^{φ}, we have to find q ∈ Q(L) such that for every y^{φ} ∈ I^{φ}, \overline{δ}y^{φ} = [q, y^{φ}]. Define

\begin{aligned}δ : & I & → & L \\ & x & ↦ & φ^{−1}(\overline{δ}(ad x_{L})) \end{aligned}

Then q := δ_{I} ∈ PDer(I, L) satisfies [q, x^{φ}] = [δ_{I}, ad x_{L}] = (by (3.5)) (ad δx)_{L} = (δ x)^{φ} = \overline{δ}(ad x_{L}) = \overline{δ}x^{φ}.

Finally, we compute the maximal algebra of quotients of a semisimple Lie algebra, where by a semisimple Lie algebra we understand a Lie algebra with zero radical (see, for example, [3]).

Denote by Der(L) the Lie algebra of all derivations of a Lie algebra L. 3.9. Lemma. If L is a semisimple Lie algebra, then Q_{m}(L) = L. Proof: Let’s write L = ⊕_{i=1}^{n}L_{i}, with L_{i} an ideal of L, simple as a Lie algebra

([3, Theorem 5.2]). By [3, Corollary 5.2 and Theorem 5.2], every ideal of L is a sum of the L_{i}’s. Hence, the only essential ideal of L is L itself, and so Q_{m}(L) ≅ Der L. By [3, Theorem 5.3], Der L ≅ad L ≅L (Z(L) = 0 by the semisimplicity of L).

ACKNOWLEDGMENTS

The author would like to thanks Professors J. A. Anquela and T. Cortés for their comments on a preliminary version of the manuscript.

REFERENCES

[1] J. A. Anquela, T. Cortés, K. McCrimmon, F. Montaner, “Strong primeness of Hermitian Jordan Systems”. J. Algebra 198 (1997), 311–326.

[2] E. García, “Inheritance of primeness by ideals in Lie algebras”. (Preprint).

[3] J. E. Humphreys, ‘Introduction to Lie Algebras and Representation Theory. Springer-Verlag New York Inc., 1972.

[4] N. Jacobson, Lie algebras. Dover Publications Inc., 1962.

[5] A. I. Kostrikin, Around Burnside. Springer-Verlag Berlin Heidelberg, 1990.

[6] T. Y. Lam Lectures on Modules and Rings. Graduate texts in Mathematics vol. 189, Springer-Verlag (New York, 1999).

[7] J. Lambek Lectures on Rings and Modules Chelsea Publishing Company (New York, 1976).

[8] C. Martínez, “The Ring of Fractions of a Jordan Algebra”. J. Algebra 237 (2001), 798–812.

[9] D. S. Passman “Computing the Symmetric Ring of Quotients”. J. Algebra 105 (1987), 207–235.

[10] Y. Utumi, “On quotient rings”. Osaka J. Math. 8 (1956), 1–18.