LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS

Miguel Gómez Lozano & Mercedes Siles Molina

Departamento de Álgebra, Geometría y Topología Universidad de Málaga 29071 Málaga (España) e-mail: miggl@ccuma.sci.uma.es; mercedes@agt.cie.uma.es

Abstract. In this paper we develop a Fountain-Gould-like Goldie theory for alternative rings. We characterize alternative rings which are Fountain-Gould left orders in semiprime alternative rings coinciding with their socle, and those which are Fountain-Gould left orders in semiprime artininian alternative rings.

Introduction.

Goldie’s Theorem is certainly one of the fundamental results of the theory of associative rings. Today this theorem is usually formulated as follows: A ring R is a classical left order in a semisimple (equivalently, semiprime artinian) ring Q if and only if R is semiprime, left nonsingular, and does not contain infinite direct sums of left ideals. Moreover, R is prime if and only if Q is simple.

In 1990 J. Fountain and V. Gould [FG_{1}] introduced a notion of order in a ring which need not have a unit, and gave [FG_{2}] a Goldie-like characterization of two-sided orders in semiprime rings with descending chain condition on principal one-sided ideals (equivalently, coinciding with their socle). In 1991 P.N. Ánh and L. Márki [AM_{1}] extended this result to one-sided orders.

In [EK] the authors give a version for alternative rings of the classical Goldie theory.

In our paper we extend to alternative rings the results and notions by Fountain-Gould, and Áhn-Márki, dealing with orders in nonnecessarily unital alternative rings. The main technique to overcome the absence of a unit element is the use of local rings at elements. This reflects in our results, so that many conditions are given in terms of local rings in the line of results like Proposition 3.2. Due to the generality of the notions we deal with, our results include those of the classical Goldie theory for alternative rings [EK] and even produce stronger results when specialized to the classical setting (cf. Theorem 7.3).

After a paragraph containing preliminary results, in the second section we introduce the notion of singular ideal for an alternative ring. We show its existence and establish some properties of the singular ideal of a semiprime ring.

In Section 3 we introduce local rings at elements of an alternative ring and give local characterizations of properties and notions like primeness, nonsingularity, left (right) Goldie dimension or coincidence with the socle. This technique of using local algebras to pass information back and forth between the ring and its local rings at elements, plays a fundamental role in our work.

Section 4 allows us to reduce the study of nondegenerate local Goldie alternative rings to the strongly prime case via the notions of uniform ideal and essential subdirect product.

The following section is devoted to the introduction of left quotient rings of an alternative ring, extending the given by Utumi in [U]. We also relate left quotient rings with classical left quotient rings (via local rings).

Typeset by AMS-TEX

In Section 6, Fountain-Gould left orders in alternative rings are introduced, and a common denominator property is given. We also study the relationship among classical, Fountain-Gould, weak Fountain-Gould left orders, and left quotient rings.

The main results are given in Section 7: We characterize Fountain-Gould left orders in nondegenerate alternative rings coinciding with its socle, and specialize this result to the classical case.

§1. Preliminaries.

The defining axioms for an alternative ring R are the left and right alternative laws:

(x, x, y) = 0 = (y, x, x),

where (x, y, z) = (xy)z−x(yz) is the associator of the three elements x, y, z ∈ R. As a consequence we have the fact that the associator is an alternating function of its arguments. The standard reference for alternative rings is [ZSSS].

If X is a nonempty set of an alternative ring R, then the left annihilator of X is define to be the set

lan(X) = {a ∈ R | ax = 0 for all x ∈ X},

written lan_{R}(X) when it is necessary to emphasize the dependence on R. Similarly the right annihilator of R, ran(X) = ran_{R}(X), is defined by

ran(X) = {a ∈ R | xa = 0 for all x ∈ X}.

We also write ann(X) = ann_{R}(X) := lan(X) ∩ ran(X) to denote the annihilator of X. It is easily seen that if X is a left (right) ideal of R, then ran(X) (lan(X)) is a right (left) ideal of R, and if X is an ideal, then lan(X), ran(X) and ann(X) are ideals of R.

For every subset X of R we have the third annihilator property:

lan(ran(lan(X))) = lan(X) and ran(lan(ran(X))) = ran(X).

The following three basic central subsets can be considered in the ring R: the associative center N = N(R), the commutative center K = K(R), and the center Z = Z(R), defined by:

N = {x ∈ R | (x, R, R) = (R, x, R) = (R, R, x) = 0},
K = {x ∈ R | [x, R] = 0},
Z = N ∩ K,

where [x, y] = xy − yx is the commutator of two elements x, y ∈ R.

If a ∈ N(R), then it is obvious that lan(a) (ran(a)) is a left (right) ideal of R.

A ring without nonzero trivial ideals (i.e., ideals with zero multiplication) is called semiprime. By [ZSSS, Exercise 8, pg. 182], every semiprime alternative ring does not contain nonzero trivial left (right) ideals. An element a of an alternative ring R is called an absolute zero divisor if aRa = (0). The ring R is called nondegenerate (or strongly semiprime) if R does not contain nonzero absolute zero divisors.

Proposition 1.1. Let I be an ideal of a semiprime alternative ring R and denote by π : R → \overline{R} the canonical projection of R onto \overline{R} = R/ann(I). Then:

(i) lan(I) = ran(I) = ann(I) is a two-sided ideal of R.

(ii) I ∩ ann(I) = 0.

(iii) \overline{R} is a semiprime alternative ring.

(iv) I is an essential ideal of R if and only if ann(I) = 0.

(v) π(I) is an essential ideal of \overline{R}.

If R is nondegenerate, then:

(vi) ann(I) = {x ∈ R | xIx = 0}, (vii) \overline{R} is a nondegenerate alternative ring.

Proof.

(i). Let x be in lan(I). We will see that x ∈ ran(I). For every r ∈ R and y ∈ I,

(1) r(yx) = −(r, y, x) + (ry)x = (x, y, r) + (ry)x = (ry)x

and

(2) (rx)y = (r, x, y) + r(xy) = (x, y, r) = 0.

By (1), Ix is a left ideal of R and by (2), (Ix)^{2} = 0. Since R is semiprime, Ix = 0. Similarly we prove ran(I) ⊆ lan(I).

(ii). We have (I∩ann(I))^{2} ⊆ I ann(I) = 0, and by semiprimeness of R we obtain I∩ann(I) = 0. (iii). Let \overline{J} = π(J) be an ideal of \overline{R}, with J an ideal of R, such that (\overline{J})^{2} = \overline{0}. Since J^{2} ⊆ ann(I),

(I ∩ J)^{3} ⊆ J^{2}I + IJ^{2} ⊆ ann(I) I = 0 and semiprimeness of R implies I ∩ J = 0. In consequence IJ = 0 = JI, which implies \overline{J} = \overline{0}.

(iv) . If I is an essential ideal of R, then (by (ii)) I ∩ann(I) = 0 implies ann(I) = 0. Conversely, supposse ann(I) = 0. If J is an ideal of R satisfying J ∩ I = 0 then, since IJ ⊆ I ∩ J = 0, BY (I), J ⊆ ann(I) = 0 and hence I must be essential

(v). By (iii), \overline{R} is semiprime, so, by (i), lan\overline{R}(\overline{I}) = ann\overline{R}(\overline{I}). Hence and by (iv) it is enough to prove that lan\overline{R}(\overline{I}) = \overline{0}. Let \overline{r} be in lan\overline{R}(\overline{I}). Then rI ⊆ I ∩ ann(I), which is zero by (iii). And by (i), \overline{r} = \overline{0}.

(vi). By [CFGM, Lemma] the Jordan annihilator of I (see note after Lemma 1.3), ann^{J}(I), which coincides with the set {x ∈ R | xIx = 0}, is an ideal of R such that ann^{J}(I) ∩ I = 0. This implies (ann^{J}(I)∩I)^{2} = 0 and since R is semiprime, ann^{J}(I)I = 0. Hence ann^{J}(I) ⊆ lan(I) = (by (i)) ann(I). The inclusion ann(I) ⊆ ann^{J}(I) is obvious.

(vii). Assume that there exists \overline{a} ∈ \overline{R} such that \overline{a}\overline{b}\overline{a} = \overline{0} for every b ∈ R. In particular, for every b ∈ I we have aba ∈ I ∩ ann(I), which is zero by (ii), and therefore a ∈ ann(I), by (v). □

From now on, for a ring R, R^{1} will denote its unitization, that is, R if the ring is unital, or R × ℤ with product (x, m)(y, n) := (xy + nx + my, mn) if R has no unity. Following [ZSSS], given a ring R we call every ideal contained in the associative center N(R) a nuclear ideal, and the largest nuclear ideal the associative nucleus of the ring R. We denote the latter by U = U(R). This ideal of R can be characterized as follows (see [ZSSS, Proposition 8.9]):

(1.1)  U = {x ∈ R | xR^{1} ⊆ N(R)} = {x ∈ R | R^{1}x ⊆ N(R)}.

For a left ideal L of an alternative ring R, denote by Ľ the largest ideal of R contained in L.

Lemma 1.2. Let L be a nonzero left ideal of a semiprime alternative ring R. Then: (i) N(L) = L ∩ N(R). (ii) Z(L) = L ∩ Z(R). (iii) U(L) = L ∩ U(R). (iv) For every nonzero element y ∈ Z(L), y^{2} ≠ 0 .

If R is nondegenerate then:

(v) N(L) ≠ 0. If R is nondegenerate and L ⊈ N(R) then:

(vi) Z(L) ≠ 0.

Proof.

(i) and (ii) are [ZSSS, Theorem 9.2 and Theorem 9.3]. (iii). Consider u ∈ U(L), x, y ∈ L and r ∈ R. By (i), u ∈ N(R) and since U(L) is an ideal of L,

xu ∈ U(L) ⊆ N(L) = L ∩ N(R) (by (i)). Hence and by [ZSSS, Corollary 2 of Lemma 7.3], (ru, x, y) = (r, xu, y) = 0. This implies R^{1}u ⊆ N(L) = L ∩ N(R) (by (i)) and by (1.1), u ∈ U(R). The reverse inclusion is obvious.

(iv). By semiprimeness of R and since y ∈ Z(L), 0 ≠ (R^{1}y)^{2} = (R^{1})^{2}y^{2}, hence y^{2} ≠ 0.

(v). Let y be a nonzero element of L. If (y, L, R) = 0 then y ∈ N(R) by [ZSSS, Lemma 9.5]. Suppose on the contrary (y, L, R) ≠ 0. By [ZSSS, Lemma 9.3], (L, L, R) ⊆ Ľ and we have that the nonzero ideal Ľ of R is a nondegenerate alternative ring, by [ZSSS, Theorem 9.7], with N(Ľ) ≠ 0 by [ZSSS, Corollary of Theorem 9.6]. Hence and by (i), 0 ≠ N(Ľ) = Ľ ∩ N(R) ⊆ L ∩ N(R) = N(L).

(vi). By the hypotheses, (L, L, R) ≠ 0. By [ZSSS, Lemma 9.3], (L, L, R) ⊆ Ľ and by [ZSSS, Lemma 9.7], Ľ is a nondegenerate alternative ring. Since Ľ is not associative then by [ZSSS, Theorem 9.7], Z(Ľ) ≠ 0. Hence, and using twice (ii), 0 ≠ Z(Ľ) = Ľ ∩ Z(R) ⊆ L ∩ Z(R) = Z(L). □

The associator ideal D = D(R) of an alternative ring R is the ideal generated by the set (R, R, R) of all associators.

Lemma 1.3. Let R be a semiprime alternative ring. Then:

(i) U ∩ D = 0.

(ii) U = ann(D).

(iii) U ⊕ D is essential as a left (right) ideal of R.

Proof. (i) is [BM, Remark 2.3].

(ii). Let a be in U. Then aD ⊆ U ∩ D = 0 (by (i)) and by Proposition 1.1 (i), a ∈ ann(D). Conversely, a ∈ ann(D) implies (R^{1}a, R, R) ⊆ D ∩ ann(D) = 0 (by Proposition 1.1 (ii)), which implies a ∈ U.

(iii). Now, let L be a nonzero left (right) ideal of R. If (L, R, R) = 0 then L ⊆ U. Otherwise 0 ≠ (L, R, R) ⊆ L ∩ D. □

Following [Sl], the left socle of an alternative ring R, Soc_{l}(R), is the sum of all minimal left ideals of R. The right socle, Soc_{r}(R), is defined analogously. We adopt the usual convention for the sum of an empty collection, thus if R has no minimal left (right) ideals then Soc_{l}(R) = 0 (Soc_{r}(R) = 0). If the ring R is semiprime then the left and the right socle coincide (in fact Soc_{l}(R) = Soc_{r}(R) under a weaker condition than semiprimeness -see [Sl, Theorem F]-). In this situation we define the socle of R, Soc(R), to be Soc_{l}(R) = Soc_{r}(R). If R is semiprime, by [Sl, Proposition 4.15], Soc(R) = R_{1} ⊕ R_{2} where R_{1} and R_{2} are the following ideals of R: R_{1} is the internal direct sum of all those ideals of R which are Cayley-Dickson algebras and R_{2} is the internal direct sum of all ideals of R which are simple associative rings having minimal left (and right) ideals (in fact R_{1} = Soc(D(R)) and R_{2} = Soc(U(R))).

Remark. Since each alternative ring R becomes a Jordan algebra with quadratic maps U_{x}y = xyx, denote it by R^{J}, we are allowed to apply Jordan-theoretic results to alternative rings.

We can transfer the Jordan notion of inner ideal to alternative rings: an inner ideal K of R is an additive subgroup of R such that xRx ⊆ K for any x ∈ K. We have that Soc(R) = Soc(R^{J}) is the sum of all minimal inner ideals of R (see [CFGM]) and Soc(R) is a von Neumann regular ring, that is, every element of Soc(R) is von Neumann regular (an element a of an alternative ring R is said to be von Neumann regular if there exists b ∈ R satisfying aba = a).

An alternative ring R for which U(R) = 0 is called purely alternative.

The case nondegenerate in the following lemma was established in [BM, Lemma 2.4 (2)], althought with a different proof.

Lemma 1.4. Let R be a semiprime (nondegenerate) alternative ring. Then \overline{R} = R/U is a semiprime (nondegenerate) purely alternative ring.

Proof. We see first that \overline{R} is purely alternative. Let \overline{x} be in \overline{R} such that (\overline{R}^{1}\overline{x}, \overline{R}, \overline{R}) = \overline{0}. This implies (R^{1}x, R, R) ⊆ U and hence (R^{1}x, R, R) ⊆ U ∩ D = 0, by Lemma 1.3 (i), which implies x ∈ U and, consequently, \overline{x} = \overline{0}.

Now, since by Lemma 1.3 U = ann(D), R semiprime (nondegenerate) implies \overline{R} semiprime (nondegenerate), by Proposition 1.1 (iii) ((iv)). □

For any element a in an alternative ring R the left and right multiplications λ_{a} and ρ_{a} are defined as follows:

λ_{a}b = ab = ρ_{b}a.

Lemma 1.5. Let R be a semiprime ring. Then for every element n ∈ N(R) the following identities hold in R:

\begin{matrix}n[μ_{x_{1}} … μ_{x_{r}}(y, z, t)] = μ_{nx_{1}} … μ_{x_{r}}(y, z, t) = … = μ_{x_{1}} … μ_{nx_{r}}(y, z, t) = μ_{x_{1}} … μ_{x_{r}}(ny, z, t), \\ n(μ_{x_{1}} … μ_{x_{r}}y, z, t) = (μ_{nx_{1}} … μ_{x_{r}}y, z, t) = … = (μ_{x_{1}} … μ_{nx_{r}}y, z, t) = (μ_{x_{1}} … μ_{x_{r}}ny, z, t), \end{matrix}

where μ_{a} denotes λ_{a} or ρ_{a}.

Proof. Since \overline{R} = R/U is purely alternative, by Lemma 1.4, [ZSSS, Theorem 8.11] applies to get N(\overline{R}) = Z(\overline{R}). Hence \overline{n}[μ_{\overline{x}_{1}} … μ_{\overline{x}_{r}}(\overline{y}, \overline{z}, \overline{t})] = μ_{\overline{nx}_{1}} … μ_{\overline{x}_{r}}(\overline{y}, \overline{z}, \overline{t}), and so \overline{n[μ_{x_{1}} … μ_{x_{r}}(y, z, t)]} = \overline{μ_{nx_{1}} … μ_{x_{r}}(y, z, t)}, which implies n[μ_{x_{1}} … μ_{x_{r}}(y, z, t)] − μ_{nx_{1}} … μ_{x_{r}}(y, z, t) ∈ U ∩ D = 0 (by Lemma 1.3 (i)). The other equalities can be proved similarly. □

Lemma 1.6. If R is a nondegenerate alternative ring, then

U(R) = {x ∈ R | Rx ⊆ N(R)} = {x ∈ R | xR ⊆ N(R)}.

Proof. Clearly, U ⊆ {x ∈ R | Rx ⊆ N}. Conversely, suppose x ∈ R such that Rx ⊆ N and prove x ∈ N (whence R^{1}x ⊆ N and, consequently, x ∈ U). Otherwise, 0 ≠ t := (a, b, x) for some a, b ∈ R. Since (t] ⊆ D and, by Lemma 1.3 (i), U ∩ D = 0, (t] ⊊ N, hence we can apply Lemma 1.2 (vi) to get 0 ≠ u := λ_{r_{1}} … λ_{r_{α}}t ∈ Z((t]). By condition (iv) in Lemma 1.2, 0 ≠ u^{2} = uλ_{r_{1}} … λ_{r_{α}}(a, b, x) = (by Lemma 1.5) λ_{r_{1}} … λ_{r_{α}}(a, b, ux) = 0 (since Rx ⊆ N), which leads to contradiction. The other equality follows analogously. □

§2. The singular ideal of an alternative ring.

The notion of nonsingularity proved particularly useful in the general theory of quotient rings initiated by Y. Utumi in 1956 [U]. Here we show the existence of a singular ideal in any alternative ring R and study its properties when the ring is semiprime. We define

Z_{l}(R) = {x ∈ U(R) | lan(x) is an essential left ideal of R}.

Proposition 2.1. For an alternative ring R, Z_{l}(R) is an ideal of R.

Proof. For x, y ∈ Z_{l}(R) we have x + y ∈ Z_{l}(R) because U is an ideal of R and lan(x) ∩ lan(y) ⊆ lan(x + y) implies that lan(x + y) is an essential left ideal of R. Take x ∈ Z_{l}(R) and r ∈ R. Since lan(x) ⊆ lan(xr), Z_{l}(R) is a right ideal. Now we prove that lan(rx) is an essential left ideal of R. Let L be a nonzero left ideal of R. Suppose (R, R, L) = 0. In this case, L ⊆ U because (R, R, R^{1}L) ⊆ (R, R, L) = 0. If Lr = 0, then L ⊆ lan(rx); otherwise there would exist y ∈ L such that 0 ≠ yr ∈ U, and since lan(x) is an essential left ideal of R, we can take a nonzero element z(yr) ∈ R^{1}(yr) ∩ lan(x), that is, 0 ≠ zy ∈ R^{1}y ∩ lan(rx) ⊆ L ∩ lan(rx). Finally, suppose (R, R, L) ≠ 0. Taking into account (R, R, L)(rx) ⊆ DU = 0, we obtain 0 ≠ (R, R, L) ⊆ L ∩ lan(rx). □

Definition 2.2. For an alternative ring R we call the ideal Z_{l}(R) the left singular ideal of R. Analogously, it is proved that

Z_{r}(R) = {x ∈ U(R) | lan(x) is an essential right ideal of R}

is an ideal of R, which will be called the right singular ideal.

Lemma 2.3. For every ideal I of a semiprime alternative ring R we have: Z_{l}(I) = I ∩ Z_{l}(R).

Proof. Assume I ≠ 0. Take 0 ≠ x ∈ Z_{l}(I). By condition (iii) in Lemma 1.2, x ∈ U(R). Now we see that lan_{R}(x) is an essential left ideal of R. Let L be a nonzero left ideal of R. If Lx = 0 then L ⊆ lan_{R}(x). If Lx ≠ 0 then IxL would be a nonzero left ideal of I (otherwise IxL = 0 would imply LxLxLx ⊆ IxLx = 0, which is not possible by the semiprimeness of R) and 0 ≠ IxL ∩ lan_{I}(x) ⊆ L ∩ lan_{R}(x).

Conversely, let 0 ≠ x ∈ I ∩ Z_{l}(R). By Lemma 1.2 (iii), x ∈ U(I). Now we prove that lan_{I}(x) is an essential left ideal of I. Let L be a nonzero left ideal of I. If Lx = 0, then L ⊆ lan_{I}(x). If Lx ≠ 0 then RxL is a nonzero left ideal of R (RxL = 0 would imply LxLx ⊆ RxLx = 0 so Lx would be a nonzero trivial left ideal in a semiprime ring). Since lan_{R}(x) is an essential left ideal of R, 0 ≠ RxL ∩ lan_{R}(x) and L ∩ lan_{I}(x) is nonzero. □

Corollary 2.4. Let R be a semiprime alternative ring and I an essential ideal of R. Then Z_{l}(R) = 0 if and only if Z_{l}(I) = 0.

Proposition 2.5. For an alternative ring R the singular ideal Z_{l}(R) does not contain any nonzero von Neumann regular element.

Proof. Suppose on the contrary that there is a nonzero von Neumann regular element x ∈ Z_{l}(R) and let y be in R such that xyx = x. Then Rxy is a nonzero left ideal of R and 0 ≠ rxy ∈ Rxy ∩ lan(x) for some r ∈ R, which leads to contradiction since rxy = rxyxy = 0. □

Corollary 2.6. Let R be a semiprime alternative ring whose socle is an essential ideal. Then Z_{l}(R) = 0.

Proof. Since Soc(R) is a von Neumann regular ideal of R (by [L, Proposition 3 (2)]), Z_{l}(Soc(R)) = 0 by Proposition 2.5, which implies Z_{l}(R) = 0 (Corollary 2.4). □

Lemma 2.7. Let R be an alternative ring. For a nonzero element a ∈ U(R), the following conditions are equivalent:

(i) lan(a) = lan(ax) for every x ∈ R such that ax ≠ 0.

(ii) ran(a) = ran(xa) for every x ∈ R such that xa ≠ 0.

Definition 2.8. An element a in an alternative ring R satisfying the equivalent conditions in the previous lemma is said to be pseudo-uniform.

Corollary 2.9. Let R be a semiprime alternative ring.

(i) Z_{l}(R) and Z_{r}(R) have not nonzero pseudo-uniform elements.

(ii) If R satisfies the ascending chain condition (acc) on the left annihilators of the form lan(x), with x ∈ U(R), then Z_{l}(R) = Z_{r}(R) = 0.

Proof. Since Z_{l}(R) and Z_{r}(R) are contained in U(R), the proof of this result is analogue to the proof of [GS, Corollary 1.2] □

Definition 2.10. A nondegenerate alternative ring R will be called left nonsingular if Z_{l}(R) = 0. Similarly, we will say that R is right nonsingular if Z_{r}(R) = 0, while nonsingular means that R is both left and right nonsingular.

§3. The local rings of an alternative ring.

Local algebras at elements were introduced by K. Meyberg. Usually they are presented as follows: recall that for an alternative ring R and an element a ∈ R, the abelian group of R endowed with the (left) a-homotope product

x ▪_{a} y = (xa)y

becomes an alternative ring, the (left) a-homotope ring, denoted by R^{a}, which has as an ideal the set

Ker(a) = {x ∈ R | axa = 0}.

We define the (left) local ring of R at a as R^{a}/Ker(a) and denote it by R_{a}. The product of two elements \overline{x}, \overline{y} ∈ R_{a} is designed by \overline{x} ▪ \overline{y} (:= \overline{(xa)y}).

If ( , , )^{a} and ( , , )_{a} designate the associators in R^{a} and R_{a}, respectively, it is obvious that for every x, y, z ∈ R, (\overline{x}, \overline{y}, \overline{z})_{a} = \overline{(x, y, z)^{a}}. Moreover, it is not difficult to prove:

(3.1)  (x, y, z)^{a} = (xa, ya, z) + (xa, y, a)z.

In particular, if a ∈ N(R) then:

(3.2)  (x, y, z)^{a} = (xa, ya, z).

In what follows we will use [ZSSS, Corollary 2 of Lemma 7.3] without mention it.

Proposition 3.1. Let R be a semiprime alternative ring and a ∈ N(R). We have: (i) \overline{x} ∈ N(R_{a}) if and only if ax or xa are in N(R). In this case ax and xa are in N(R). (ii) \overline{x} ∈ U(R_{a}) if and only if axa ∈ U(R). (iii) If \overline{x} ∈ Z_{l}(R_{a}) then axa ∈ Z_{l}(R). (iv) If a ∈ Z_{l}(R) then Z_{l}(R_{a}) = R_{a}.

(v) Z_{l}(R) = 0 if and only if Z_{l}(R_{a}) = 0 for all a ∈ U(R). (vi) A nonzero element a ∈ U(R) is pseudo-uniform if and only if R_{a} has no nonzero divisors.

Proof.

(i). Suppose \overline{x} ∈ N(R_{a}). By (3.2), for every y, z ∈ R, \overline{0} = (\overline{x}, \overline{y}, \overline{z})_{a} = \overline{(xa, ya, z)}. This means 0 = a(xa, ya, z)a = (xa^{2}, ya^{2}, z) and by [ZSSS, Lemma 9.5] applied to the left ideal Ra^{2} we obtain xa^{2} ∈ N(R). Hence 0 = (xa^{2}, y, z) = (xa, ya, z) for every y, z ∈ R, and by [ZSSS, Lemma 9.5] again, xa ∈ N(R). Analogously we prove ax ∈ N(R).

Conversely, if xa or ax are in N(R) then for every y, z ∈ R, (ax, y, z) = (xa, y, z) = 0 and by (3.2), \overline{(x, y, z)^{a}} = \overline{0}, which implies \overline{x} ∈ N(R_{a}).

(ii). We have \overline{x} ∈ U(R_{a}) if and only if for every \overline{y} ∈ (R_{a})^{1}, \overline{(ya)x} = \overline{y} ▪ \overline{x} ∈ N(R_{a}) if and only if (by (i)) [(ya)x]a = (by the right Moufang identity) y(axa) ∈ N(R) for every y ∈ R^{1} and this is equivalent to axa ∈ U(R).

(iii). Take \overline{x} ∈ Z_{l}(R_{a}). By (ii), axa ∈ U(R). Now let L be a nonzero left ideal of R. If Laxa = 0 then L ⊆ lan_{R}(axa). Suppose Laxa ≠ 0. Then (Laxa)^{2} ≠ 0, so \overline{L} is a nonzero left ideal of R_{a}. Since lan_{R_{a}}(\overline{x}) is an essential left ideal of R_{a} we can choose an element y ∈ L such that \overline{0} ≠ \overline{y} ∈ lan_{R_{a}}(\overline{x}) ∩ \overline{L}, that is, \overline{yax} = \overline{0} and we have 0 ≠ ay ∈ L ∩ lan_{R}(axa). In any case lan_{R}(axa) ∩ L ≠ 0, which implies that lan_{R}(axa) is an essential left ideal of R.

(iv). Suppose a ∈ Z_{l}(R) and take \overline{x} ∈ R_{a} and a nonzero left ideal L of R_{a}. If L ▪ \overline{x} = \overline{0} then L ⊆ lan_{R_{a}}(\overline{x}). If L ▪ \overline{x} ≠ \overline{0}, take \overline{y} ∈ L such that \overline{y} ▪ \overline{x} ≠ \overline{0}, that is, 0 ≠ a[(ya)x]a = (a ∈ U(R)) ayaxa. Since U(R) is nondegenerate, by [ZSSS, Theorem 8.4], U(R)ayaxa ≠ 0 and hence Rayax is a nonzero left ideal of R. Now take 0 ≠ rayax ∈ lan_{R}(a). Then 0 ≠ aRraya and aRrayaxa = 0, that is, \overline{0} ≠ \overline{Rray} = \overline{Rr} ▪ \overline{y} ⊆ lan_{R_{a}}(\overline{x}) ∩ L.

(v). Z_{l}(R) = 0 implies, by (iii), that for every a ∈ U(R), Z_{l}(R_{a}) = 0. Conversely, suppose Z_{l}(R_{a}) = 0 for every a ∈ U(R). By (iv), a ∈ Z_{l}(R) would imply R_{a} = Z_{l}(R_{a}) = \overline{0}, so aU(R)a ⊆ aRa = 0 and hence a = 0 because U(R) is nondegenerate by [ZSSS, Theorem 9.4].

(vi) follows analogously to the associative case (see [GS, Proposition 3.2 (v)]). □

Let R be an alternative ring. We say that R is a prime ring if for any two of its ideals I and J it follows from the equality IJ = (0) that either I = (0) or J = (0), and R is said to be strongly prime if it is prime and nondegenerate. By [ZSSS, Exercise 1, pg. 197], this is equivalent to: (aR)b = 0 or a(Rb) = 0 implies a = 0 or b = 0, with a, b ∈ R.

If a ∈ R and we denote by U^{a} the quadratic operator defined on R^{a}, then it is not difficult to prove

(3.3)  U_{x}^{a}y = U_{x}U_{a}y.

Let R be an alternative ring. Given an element a in R we denote by (a] (or by _{R}(a] if it is necessary to specify the ring R) the principal left ideal of R generated by a in R (we notice that a ∈ (a]). A nonzero left ideal I of R will be called uniform if for any nonzero left ideals B and C of R inside I we have B ∩ C ≠ 0. An element a ∈ R is said to be l-uniform if (a] is uniform. A nonzero ideal I of R will be called uniform if the intersection of two nonzero ideals of R contained in I is nonzero.

It is clear that the linear span of all operators φ : R → R of the form φ = λ_{a} or φ = Id_{R} (a ∈ R) is a unital associative algebra, denote it by Λ(R), and R is clearly a left Λ(R)-module. Similarly, Π(R) is defined as the linear span of all the right multiplications plus the identity on R. Then R becomes a right Π(R)-module. This allows us to apply results of modules to alternative rings. The left ideals of R are precisely the Λ(R)-submodules of R, and the right ideals are the Π(R)-submodules of R.

Let L be a left ideal of R which does not contain infinite direct sums of nonzero left ideals. By [G, Proposition 3.19] there exists a nonnegative integer n called the left Goldie (or uniform) dimension of L, denoted by u-dim_{R}(L) or simply by u-dim(L), such that L contains a direct sum of n nonzero left ideals and any direct sum of nonzero left ideals contained in L has at most n summands (notice that any direct sum of n nonzero left ideals is essential in the sense that it intersects any nonzero left ideal contained in L and its summands are necessarily uniform). Now a uniform nonzero left ideal is just a nonzero left ideal of left Goldie dimension one.

The left Goldie (or uniform) dimension of an element a ∈ R, denoted by u-dim_{R}(a) or simply by u-dim(a), is the left Goldie dimension of _{R}(a]. If any element of R has finite left Goldie dimension, we will say that R has finite left local Goldie dimension.

With the following result we prove that, besides the left nonsingularity, there are other notions, concerning a nondegenerate alternative ring, which can be characterized in terms of its local rings at elements.

Proposition 3.2. Let R be a nondegenerate alternative ring. Then:

(i) All the local rings of R at nonzero elements are nondegenerate.

(ii) R is prime if and only if all the local rings of R at nonzero elements are prime.

(iii) If R is simple, then all the local rings of R at nonzero elements of N(R) are simple.

(iv) For any element a ∈ R, u − dim_{R}(a) ≤ u − dim(R_{a}). If a is in N(R), then a has left Goldie

dimension equal to n if and only if R_{a} has left Goldie dimension equal to n.

(v) For every a ∈ R, \overline{x} ∈ Soc(R_{a}) if and only if axa ∈ Soc(R). Hence R coincides with its socle if and only if R_{a} is artinian for each a ∈ R.

(vi) If R coincides with its socle, then R has finite both left and right local Goldie dimension.

Proof.

(i). Let a be an element in R and suppose \overline{x} ∈ R_{a} such that \overline{x} ▪ \overline{y} ▪ \overline{x} = \overline{0} for every \overline{y} ∈ R_{a}. Then \overline{0} = \overline{U_{x}U_{a}y} = \overline{U_{x}^{a}y} by (3.3) and hence U_{Uax}y = (by Macdonald’s identity -see [ZSSS]-) U_{a}U_{x}U_{a}y = 0 for every y ∈ R. By nondegeneracy of R we have U_{a}x = 0, that is, \overline{x} = \overline{0}.

(ii). Assume that R is prime. Take a ∈ R and let I, J be two ideals of R_{a} such that I ▪J = \overline{0}. If \overline{x} and \overline{y} were nonzero elements of I and J respectively, then for every r ∈ R, \overline{0} = \overline{x} ▪ (\overline{r} ▪ \overline{y}), which implies 0 = U_{a}(x ▪_{a} [r ▪_{a} y]) = (axa){[(ra)y]a} = (axa)[r(aya)] and by [ZSSS, Exercise

1, pg. 197], axa = 0 or aya = 0, i.e., \overline{x} = \overline{0} or \overline{y} = \overline{0}.

Conversely, let I and J be two nonzero ideals of R of zero product. Choose nonzero elements x and y in I ∩ N(R) and J ∩ N(R) respectively. Then the ideals \overline{I} and \overline{J} of R_{x+y} are nonzero because, by (i), (x + y)I(x + y) = xIx ≠ 0 and (x + y)J(x + y) = yJy ≠ 0. Moreover \overline{I} ▪ \overline{J} = \overline{0}, which contradicts the primeness of R_{x+y}.

(iii). By Kleinfeld’s Theorem (see [ZSSS, Corollary 1 of Theorem 7.5]) R is associative or a Cayley-Dickson ring over Z(R). In the first case the result follows from [FGGS, Proposition 5.2 (iii)]. In the second one the maps

\begin{aligned}φ : & R^{a} & → & R_{a} & and & ψ : & R^{a} & → & R \\ & x & ↦ & \overline{x} & & & x & ↦ & xa \end{aligned}

are ring isomorphisms because a ∈ N(R) = Z(R) and Z(R) is a field. Hence R_{a}, which is isomorphic to R, is a simple ring.

(iv).

(1) For every nonzero left ideal L of R contained in _{R}(a] we have L ∩ aRa ≠ 0:

If UL ≠ 0, then there exist u ∈ U, l ∈ L such that 0 ≠ ul ∈ L. Since L ⊆ _{R}(a] there exist x_{i}^{j} in R^{1} such that l = ∑_{i=1}^{m} λ_{x_{1}^{i}} … λ_{x_{n_{i}}^{i}} a and ul = va with v = ∑_{i=1}^{m} uρ_{x_{1}^{i}} … ρ_{x_{n_{i}}^{i}} ∈ U. By nondegeneracy of R, 0 ≠ (va)t(va) for some t ∈ R. Now, v, va ∈ N implies 0 ≠ (at)(va) = a(tv)a ∈ L ∩ aRa.

If UL = 0, then (L, R, R) ≠ 0 (otherwise L ⊆ U and L^{2} = 0, which is not possible because R is nondegenerate) and D∩L (which contains (L, R, R)) is nonzero. By Lemma 1.2 (ii) and (vi), we can choose 0 ≠ l ∈ (L∩D)∩Z(R) = Z(L∩D). Write l = ∑_{i=1}^{m} λ_{x_{1}^{i}} … λ_{x_{n_{i}}^{i}} a as before. By Lemma 1.2 (v), 0 ≠ l^{2} = (∑_{i=1}^{m} λ_{x_{1}^{i}} … λ_{x_{n_{i}}^{i}} a)^{2}, which implies al ≠ 0, and by nondegeneracy of R, 0 ≠ (al)t(al) = a(l^{2}t)a ∈ L ∩ aRa.

(2) Let {L_{α}} be a family of nonzero left ideals of R contained in _{R}(a] and whose sum is direct. By (1), L_{α} ∩ aRa is nonzero and it is clear that if we define L_{α} = \{\overline{x} ∈ R_{a} | axa ∈ L_{α}\}, then {L_{α}} is a family of nonzero left ideals of R_{a} whose sum is direct, which proves that u-dim_{R}(a) ≤ u-dim(R_{a}).

Now, suppose a ∈ N(R) and let {L_{i}}_{i=1}^{n} be a direct sum of nonzero left ideals of R_{a}. Choose \overline{0} ≠ \overline{l}_{i} ∈ N(L_{i}) = L_{i} ∩ N(R_{a}) (which is possible by virtue of (i) and Lemma 1.2 (iv)). By Proposition 3.1 (i) and since a ∈ N(R), the nonzero element t_{i} := al_{i}a is in N(R) for every i = 1, …, n. Hence and using the nondegeneracy of R we have that {Rt_{i}}_{i=1}^{n} is a family of nonzero left ideals of R contained in Ra. We claim that the sum of these ideals is direct because if r_{1}t_{1} + ⋯ + r_{n}t_{n} = 0 with, for example, r_{1}t_{1} ≠ 0, by nondegeneracy of R, there exists s ∈ R such that asr_{1}al_{1}a ≠ 0, that is, \overline{0} ≠ \overline{sr}_{1} ▪ \overline{l}_{1} ∈ ∑_{i=2}^{n} \overline{R} ▪ \overline{l}_{i} ⊆ ∑_{i=2}^{n} L_{i}, which is a contradiction.

(v). Since \overline{x} ▪ \overline{y} ▪ \overline{x} = \overline{U_{x}^{a}y} = \overline{U_{x}U_{a}y} (by (3.3)), I is an inner ideal of R_{a} if and only if I =

\{axa | \overline{x} ∈ I\} is an inner ideal of R, hence \overline{x} ∈ Soc(R_{a}) if and only if axa ∈ Soc(R).

Now, by [L, Proposition 3 (2)] it follows that if a ∈ Soc(R) then a has finite rank and it is von Neumann regular. Hence and by [L, Corollary 1] R_{a} is an artinian alternative ring with capacity equal to the rank of a.

Conversely if R_{a} is artinian then R_{a} has bounded length for the chains of inner ideals of the form \overline{I} with I an inner ideal of R^{J} contained in aRa, hence for chains of principal inner ideals of the form \overline{bRb} with b in the inner ideal of R^{J} generated by a. Thus by [L, Proposition 3 (2)] again, a ∈ Soc(R).

(vi). By (v) any local ring of R at an element of R is artinian and hence it has finite both left and right Goldie dimension. Then, by (iv), R has finite both left and right local Goldie dimension. □

§4. Left local Goldie alternative rings.

In this section we show how to reduce the study of nondegenerate local Goldie alternative rings to the strongly prime case, via the notions of uniform ideal and essential subdirect product.

Every alternative ring R with Z_{l}(R) = 0 and such that every element has finite left Goldie dimension will be called a left local Goldie alternative ring. If additionally R has finite left (global) Goldie dimension then R will be called a left Goldie alternative ring. Right and two-sided corresponding notions are defined dually.

Proposition 4.1. Let R be a semiprime alternative ring. Then every l-uniform element generates a uniform ideal.

Proof. Let I be the ideal generated in R by a l-uniform element u and let B and C be nonzero ideals of R contained in I. We note that Bu ≠ 0. Otherwise u ∈ ran(B) = ann(B) (by Proposition 1.1 (i)), which implies I ⊆ ann(B), and since B ⊆ I, we would have B = B ∩ ann(B) = 0 (by Proposition 1.1 (ii)), a contradiction. Analogously Cu ≠ 0. Denote by L and L′ the nonzero left ideals of R generated by Bu and Cu respectively. Then 0 ≠ L ∩ L′ ⊆ B ∩ C by the l-uniformity of u. □

Since given an alternative ring R, the lattice L(R) of all ideals of R is an algebraic lattice relative to the ∗-product B ∗ C := (BC), where (X) denotes the ideal of R generated by X, and XY the linear span of all products xy with x ∈ X and y ∈ Y, we can apply the results of [FeGa] to obtain the following results. We notice that B ∗ C = 0 if and only if BC = 0. Hence R is semiprime (prime) if and only if the algebraic lattice (L(R), ∗) is semiprime (prime). Note that since R is alternative, B ∗ C is merely BC.

Proposition 4.2. Let R be a semiprime alternative ring. Then:

(i) A nonzero ideal I of R is uniform if and only if the annihilator ideal ann(I) is maximal among all annihilator ideals ann(B) with B being a nonzero ideal of R, equivalently, R/ann(I) is a prime alternative ring. Moreover if R is nondegenerate then R/ann(I) is a strongly prime ring.

(ii) For each uniform ideal I of R there exists a unique maximal uniform ideal U of R containing

I; actually U = ann(ann(I)). (iii) The sum of all maximal uniform ideals of R is direct.

Proof. Suppose that R is nondegenerate and that R/ann(I) is a prime alternative ring. By Proposition 1.1 (vii), the ring R/ann(I) is nondegenerate and so it is strongly prime.

The rest of the statements follows as a particular case from [FeGa, Proposition 3.1], using that L(R) is a modular lattice. □

A subdirect product of alternative rings R ≤ ∏ R_{α} will be called an essential subdirect product if R contains an essential ideal of the full product ∏ R_{α}. If R is actually contained in the direct sum of the R_{α}, then R will be called an essential subdirect sum. An ideal I of a nondegenerate alternative ring R is called a closed ideal if I = ann(ann(I)). By the third annihilator property an ideal is closed if and only if it is the annihilator of an ideal. Note that by Proposition 4.2 (ii), maximal uniform ideals are closed.

Theorem 4.3. For an alternative ring R the following conditions are equivalent:

(i) R is an essential subdirect product of strongly prime alternative rings R_{α}.

(ii) R is nondegenerate and every nonzero ideal of R contains a uniform ideal.

(iii) R is nondegenerate and every nonzero closed ideal of R contains a uniform ideal. Actually we can take R_{α} = R/ann(M_{α}), where {M_{α}} is the family of all maximal uniform ideals of R.

Proof. (i) ⇒ (ii). In general any subdirect product R of a family {R_{α}} of nondegenerate alternative rings is nondegenerate. Let M ⊆ R be an essential ideal of the full direct product ∏ R_{α} and set M_{α} := M ∩ R_{α}, where we are regarding R_{α} as an ideal of ∏ R_{α}. Then M_{α} is a nonzero ideal of R_{α} contained in R since M is an essential ideal of ∏ R_{α}. In fact M_{α} is a uniform ideal of R since M_{α} is uniform in R_{α} because R_{α} is strongly prime and any ideal of R contained in M_{α} is an ideal of R_{α}. Now if I is a nonzero ideal of R then π_{α}(I) is a nonzero ideal of R_{α} for some index α. Hence by strongly primeness of R_{α} we have 0 ≠ π_{α}(I) ∗ M_{α} ⊆ I ∩ M_{α}. Therefore I contains the nonzero ideal I ∩ M_{α}, which is uniform since it is contained in M_{α}.

(ii) ⇒ (iii) is trivial.

(iii) ⇒ (i). Let ∑ M_{α} be the sum of all maximal uniform ideals of R, which is direct by Proposition 4.2 (iii). Since ann(∑ M_{α}) is a closed ideal, it must be zero: otherwise ann(∑ M_{α}) would contain a uniform ideal, and therefore a maximal uniform ideal because it is closed, which leads to contradiction. Hence, by a standard argument, ∩ann(M_{α}) = ann(∑ M_{α}) = 0 implies that R is a subdirect product of the alternative rings R_{α} := R/ann(M_{α}) each of which is a strongly prime alternative ring by Proposition 4.2 (i).

Finally, the homomorphic image of ⊕M_{α} in ∏ R_{α} is an essential ideal of ∏ R_{α} since M_{α} can be regarded as an essential ideal of R_{α}, by condition (v) in Proposition 1.1. □

Lemma 4.4. Let R be a semiprime alternative ring and I an ideal of R. Denote by \overline{R} the quotient ring R/ann(I). We have:

(i) Any direct sum of nonzero left ideals of \overline{R} gives rise to a direct sum of nonzero left ideals of R with the same number of summands. Hence if R has finite left Goldie dimension then \overline{R} has also finite left Goldie dimension.

(ii) If a ∈ R has finite left Goldie dimension in R, then \overline{a} := a + ann(I) has finite left Goldie

dimension in \overline{R}. (iii) If Z_{l}(R) = 0 then Z_{l}(\overline{R}) = \overline{0} . Moreover, if R is nondegenerate and I is a uniform ideal,

then \overline{R} is a strongly prime left nonsingular alternative ring.

Proof.

(i). Let ∑ \overline{L}_{α} be a direct sum of nonzero left ideals of \overline{R}. Denote by π : R → \overline{R} the canonical projection of R onto \overline{R}. By Proposition 1.1 (v), π(I) is an essential left ideal of \overline{R}. Hence L_{α} := π^{−1}(\overline{L}_{α})∩I is a nonzero left ideal of R, for each index α. Now, we show that ∑ L_{α} is a direct sum. Indeed, if x ∈ L_{β} ∩ (∑_{α≠β} L_{α}), then π(x) ∈ \overline{L}_{β} ∩ (∑_{α≠β} \overline{L}_{α}) = \overline{0}. Therefore x ∈ I ∩ ann(I) = 0 (by Proposition 1.1 (ii)).

(ii). Let ∑ \overline{L}_{α} be a direct sum of nonzero left ideals of \overline{R} inside the principal left ideal \overline{R}(\overline{a}]. By taking L_{α} := I(π^{−1}(\overline{L}_{α})) we obtain a direct sum of nonzero left ideals of R contained

in _{R}(a].

(iii). By Lemma 2.3, Z_{l}(R) = 0 implies Z_{l}(I) = 0. But I can be regarded as an ideal of \overline{R} via the isomorphism x ↦ x + ann(I), for x ∈ I and by Proposition 1.1 (v) we have that \overline{I} is an essential ideal of \overline{R}. By Proposition 1.1 (iii) and Corollary 2.4, Z_{l}(\overline{R}) = \overline{0}.

If R is nondegenerate and I is uniform, then \overline{R} is strongly prime by condition (i) in Proposition 4.2. □

Theorem 4.5. Let R be a nondegenerate left local Goldie alternative ring. Then R is an essential subdirect sum of strongly prime left local Goldie alternative rings. More precisely,

⊕M_{α} ◁ R ≤ ⊕R/ann(M_{α}),

where M_{α} ranges over all maximal uniform ideals of R. If R is actually left Goldie, then R is an essential subdirect sum of finitely many strongly prime left Goldie alternative rings.

Proof. Since R has finite left local Goldie dimension, any nonzero ideal of R contains an l-uniform element and hence a uniform ideal, by Proposition 4.1. Then, by Theorem 4.3, R is an essential subdirect product of the strongly prime alternative rings R_{α} = R/ann(M_{α}), with M_{α} a maximal uniform ideal of R, each of which is a strongly prime left local Goldie alternative ring by Proposition 4.2 (i) and conditions (ii) and (iii) in Lemma 4.4. Let us see that R ⊆ ⊕R_{α}. Otherwise, there would exist x ∈ R such that x ∉ ann(M_{α}) for an infinite number of α′s. Say x ∉ ann(M_{α}) for every α ∈ Λ. Denote by I_{α} the left ideal of R generated by M_{α}x. Then 0 ≠ M_{α}x ⊆ I_{α} ⊆ M_{α} for every α ∈ Λ, and the sum ∑_{α∈Λ}I_{α} is direct. This implies that x has infinite left Goldie dimension, a contradiction.

Suppose additionally that R has finite left Goldie dimension. Then it follows, from Proposition 4.2 (iii), that R contains only a finite number of maximal uniform ideals, and hence R is an essential subdirect sum of a finite number of R_{α}. Moreover, each R_{α} has now finite left Goldie dimension by Lemma 4.4 (i). □

§5. Left quotient rings and classical left quotient rings.

In this section we introduce the notion of left quotient ring of an alternative ring, which extends that of Utumi given for associative rings in [U]. We also establish the relationship between classical left quotient rings and the new left quotient rings.

Definition 5.1. Let R be a subring of a (non necessarily unital) alternative ring Q. We will say that Q is a left quotient ring of R if:

(1) N(R) ⊆ N(Q) and

(2) given p, q ∈ Q, with p ≠ 0, there exists r ∈ N(R) such that rp ≠ 0 and rq ∈ R.

It is not difficult to see that for any finite familly {q_{1}, …, q_{n}} ⊆ Q, with q_{1} ≠ 0, there exists r ∈ N(R) such that rq_{1} ≠ 0 and rq_{i} ∈ R, for i ∈ {1, …, n}.

Example 5.2. Every nondegenerate alternative ring R is a left quotient ring of itself.

Proof. Let p be a nonzero element in R. If p ∈ U, then Up ≠ 0 (because U is nondegenerate, by [ZSSS, Lemma 9.7]). Suppose p ∉ U. By Lemma 1.6, there exist x, y, z ∈ R such that 0 ≠ t := (x, y, zp). Since (t] ⊆ D, we can apply Lemma 1.2 to find 0 ≠ u := λ_{r_{1}} … λ_{r_{α}}t ∈ Z((t]). By condition (iv) in Lemma 1.2, 0 ≠ u^{2} = uλ_{r_{1}} … λ_{r_{α}}(x, y, zp) = (by Lemma 1.5) λ_{r_{1}} … λ_{r_{α}}(x, y, zup). In particular, up ≠ 0.

In any case, Np ≠ 0, which proves the claim. □

Proposition 5.3. Let Q be an alternative ring which is a left quotient ring of an alternative ring R. Then:

(i) L ∩ R ≠ 0 for every nonzero left ideal L of Q.

(ii) If R is semiprime, prime, or nondegenerate, then Q is semiprime, prime, or nondegenerate too.

Proof. (i) follows from the definition.

(ii). If R is semiprime (or prime) then Q is semiprime (or prime) too because every nonzero ideal of Q meets R. Now, suppose 0 ≠ q ∈ Q such that qQq = 0. Since Q is a left quotient ring of R there exists r ∈ N(R) such that 0 ≠ rq ∈ R. Taking into account N(R) ⊆ N(Q) we have (rq)R(rq) = r[q(Rr)q] ⊆ r(qQq) = 0 and by nondegeneracy of R, rq = 0, a contradiction. □

Proposition 5.4. Let Q be an alternative ring which is a left quotient ring of an alternative ring R. Then:

(i) Z(R) = Z(Q) ∩ R.

(ii) For X, Y ⊆ R we have lan_{R}(X) ⊆ lan_{R}(Y) if and only if lan_{Q}(X) ⊆ lan_{Q}(Y).

(iii) U(R) = U(Q) ∩ R. If R is semiprime then:

(iv) Z_{l}(R) = Z_{l}(Q) ∩ R.

(v) Z_{l}(R) = 0 if and only if Z_{l}(Q) = 0.

(vi) Any direct sum {L_{i}} of nonzero left ideals of R gives rise to a direct sum {L_{i}} of nonzero

left ideals of Q, with the same number of sumands. (vii) If Q has finite left (local) Goldie dimension, then R has finite left (local) Goldie dimension

as well.

If R is nondegenerate then: (viii) For every nonzero element a ∈ R the local ring Q_{a} of Q at a is a left quotient ring of R_{a}.

Proof.

(i). The containement Z(Q) ∩ R ⊆ Z(R) is obvious. Conversely, take x ∈ Z(R) and suppose x ∉ Z(Q). Then [p, x] ≠ 0 for some p ∈ Q. Apply that Q is a left quotient ring of R to find n ∈ N(R) such that 0 ≠ n[p, x] and np ∈ R. By [ZSSS, Lemma 7.1], n[p, x] = [np, x] − [n, x]p = 0 since x ∈ Z(R) and np ∈ R, which is a contradiction.

(ii). Since lan_{R}(X) = lan_{Q}(X) ∩ R, lan_{Q}(X) ⊆ lan_{Q}(Y) implies lan_{R}(X) ⊆ lan_{R}(Y). Suppose lan_{R}(X) ⊆ lan_{R}(Y). If there were an element q ∈ lan_{Q}(X) such that qy ≠ 0 for some y ∈ Y then there would exist an element r ∈ N(R) such that 0 ≠ rqy and rq ∈ R (because Q is a left quotient ring of R). Then rq ∈ lan_{R}(X) but rq ∉ lan_{R}(Y), which is a contradiction.

(iii). It is clear that U(R) ⊇ U(Q) ∩ R. Conversely take x ∈ U(R) and suppose (px, q_{1}, q_{2}) ≠ 0 for some p ∈ Q^{1}, q_{1}, q_{2} ∈ Q. Since Q is a left quotient ring of R there exists a ∈ N(R) such that 0 ≠ a(px, q_{1}, q_{2}) (= ((ap)x, q_{1}, q_{2})) and ap ∈ R. Now ap ∈ R and x ∈ U(R) implies (ap)x ∈ N(R) ⊆ N(Q). Hence a(px, q_{1}, q_{2}) must be zero, a contradiction.

(iv). In this item, (iii) will be in our mind. Take x ∈ Z_{l}(R) and let L be a nonzero left ideal of Q. Then R ∩ L is a nonzero left ideal of R and hence 0 ≠ lan_{R}(x) ∩ R ∩ L ⊆ lan_{Q}(x) ∩ L. Conversely, let x ∈ Z_{l}(Q) ∩ R and take a nonzero left ideal L of R. Denote by K the left ideal (of R) (U(R) ⊕ D(R)) ∩ L, which is nonzero by Lemma 1.3 (iii). If Kx = 0 then K ⊆ L ∩ lan_{R}(x). Suppose Kx ≠ 0 and let l = u + d be a nonzero element of K, with u ∈ U(R) and d ∈ D(R), satisfying 0 ≠ lx = ux + dx = ux. By [ZSSS, Theorem 9.4], U(R) is nondegenerate, so there exists 0 ≠ v ∈ U(R) with 0 ≠ vlx. Then 0 ≠ vl ∈ U(R) ∩ L. Apply that lan_{Q}(x) is an essential left ideal of Q and choose 0 ≠ q(vl) ∈ Q(vl) ∩ lan_{Q}(x). Notice that since 0 ≠ vl ∈ U(R) ⊆ U(Q) and U(Q) is nondegenerate (by Proposition 5.3, Q is semiprime, and by [ZSSS, Theorem 8.4], U(Q) is semiprime, equivalently nondegenerate), Q(vl) is a nonzero left ideal of Q. As Q is a left quotient ring of R, given 0 ≠ q(vl), with q ∈ Q, there exists a ∈ N(R) satisfying 0 ≠ (aq)(vl) and aq ∈ R. Then (aq)(vl) is a nonzero element in L ∩ lan_{R}(x).

(v) follows from (iv) and Proposition 5.3 (i). (vi). Let {L_{i}} be a collection of nonzero left ideals of R whose sum is direct. Take an L_{i} of this

family. If (R, R, L_{i}) = 0 then L_{i} ⊆ U(R) and since R is semiprime, L_{i}^{2} ≠ 0, hence L_{i} := Ql_{i} is a nonzero left ideal of Q, for some l_{i} ∈ L_{i}. Suppose (R, R, L_{i}) ≠ 0 and choose a nonzero element t_{i} ∈ (R, R, L_{i}). Define in this case L_{i} :=_{Q} (t_{i}]. We claim that the sum of all the (nonzero) L_{i}’s is direct. Suppose on the contrary that there is a sum

(1) ∑_{i_{1}} λ_{q_{1}^{i_{1}}} … λ_{q_{m_{1}}^{i_{1}}} t_{1} + … + ∑_{i_{n}} λ_{q_{1}^{in}} … λ_{q_{mn}^{in}} t_{n} + q_{n+1}l_{n+1} + … + q_{m}l_{m} = 0

with some summand nonzero, say t. Denote by X the set of all the elements of Q which appears in (1). Since Q is a left quotient ring of R, given the nonzero element t and given X, there exists a ∈ N(R) such that at ≠ 0 and ax ∈ R for every x ∈ X. Multiplying (1) by a on the left side and applying Lemma 1.5 we have

∑_{i_{1}} λ_{q_{1}^{i_{1}}} … λ_{aq_{m_{1}}^{i_{1}}} t_{1} + … + ∑_{i_{n}} λ_{q_{1}^{in}} … λ_{aq_{mn}^{in}} t_{n} + (aq_{n+1})l_{n+1} + … + (aq_{m})l_{m} = 0

with some summand (at) nonzero. Since X is a finite set, in a finite number of steps we obtain

∑_{i_{1}} λ_{r_{1}^{i_{1}}} … λ_{r_{m_{1}}^{i_{1}}} t_{1} + … + ∑_{i_{n}} λ_{r_{1}^{in}} … λ_{r_{mn}^{in}} t_{n} + r_{n+1}l_{n+1} + … + r_{m}l_{m} = 0

with all the appearing elements in R and some summand nonzero. But this contradicts that the sum of the L_{i}’s is direct.

(vii) follows analogously to (vi). (viii).

(1) Given \overline{x}, \overline{y}, \overline{p}, \overline{q}_{1}, \overline{q}_{2} ∈ Q_{a} such that (\overline{q}_{1}, \overline{q}_{2}, \overline{p} ▪ \overline{x})_{a} ≠ \overline{0}, there exist n_{1}, n_{2}, n_{3} ∈ N(R) such

that [(n_{1}p)a]x, [(n_{1}p)a]y, n_{1}p, n_{2}q_{1}, n_{3}q_{2}, n_{1}y, n_{1}ay ∈ R and \overline{0} ≠ (\overline{n_{1}p} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a}.

Indeed, \overline{0} ≠ (\overline{q}_{1}, \overline{q}_{2}, \overline{p} ▪ \overline{x})_{a} = \overline{(q_{1}, q_{2}, (pa)x)^{a}} = \overline{((pa)x, q_{1}, q_{2})^{a}} = (by (3.1) and the right Moufang identity) \overline{(p(axa), q_{1}a, q_{2})} + \overline{(p(axa), q_{1}, a)q_{2}} implies

0 ≠ t := a(p(axa), q_{1}a, q_{2})a + a[(p(axa), q_{1}, a)q_{2}]a.

Apply three times that Q is a left quotient ring of R to find n_{1}, n_{2}, n_{3} ∈ N(R) such that n_{1}(axa), [(n_{1}p)a]x, n_{1}p, n_{2}q_{1}, n_{3}q_{2}, n_{1}y, n_{1}ay ∈ R and n_{1}n_{2}n_{3}t ≠ 0. By Lemma 1.5,

n_{1}n_{2}n_{3}t = ρ_{a}λ_{a}[((n_{1}p)(axa), n_{2}q_{1}a, n_{3}q_{2}) + ((n_{1}p)(axa), n_{2}q_{1}, a)(n_{3}q_{2})]
= ρ_{a}λ_{a}((n_{1}p) ▪ x, n_{2}q_{1}, n_{3}q_{2})^{a} (by (3.1))

and this implies \overline{0} ≠ (\overline{n_{1}p} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a}.

(2) N(R_{a}) ⊆ N(Q_{a}).

Suppose \overline{x} ∈ N(R_{a}) but \overline{x} ∉ N(Q_{a}). Then, there exist \overline{q}_{1}, \overline{q}_{2} ∈ Q_{a} such that (\overline{q}_{1}, \overline{q}_{2}, \overline{x})_{a} ≠ \overline{0}. Reasoning as in (1), we can find n_{1}, n_{2} ∈ N(R) satisfying: n_{1}q_{1}, n_{2}q_{2} ∈ R and \overline{0} ≠ (\overline{x}, \overline{n_{1}q_{1}}, \overline{n_{2}q_{2}})_{a}, which is a contradiction since \overline{x} ∈ N(R_{a}).

(3) Q_{a} is a left quotient ring of R_{a}.

Let \overline{x}, \overline{y} be in Q_{a} with \overline{x} ≠ \overline{0}. Suppose first \overline{x} ∈ U(Q_{a}). Since Q is a left quotient ring of R, there exists n ∈ N(R) such that n(axa) ≠ 0 and nx, ny, n(axa), nax, nay, n(aya) ∈ R. By nondegeneracy of R, [n(axa)]r[n(axa)] ≠ 0 for some r ∈ R. Hence, U_{a}U_{x}^{a}(rn) = U_{a}U_{x}U_{a}(rn) ≠ 0, which implies \overline{x}▪\overline{rn}▪\overline{x} = \overline{U_{x}^{a}(rn)} ≠ \overline{0}. In particular, \overline{rn}▪\overline{x} ≠ \overline{0} and [(rn)a]x = [r(na)]x = (r, na, x) + r[(na)x] = (r, a, nx) + r(nax) ∈ R. Analogously, [(rn)a]y ∈ R. Consequently \overline{rn} ▪ \overline{x} ∈ U(Q_{a}) ∩ R_{a}, which implies \overline{rn} ▪ \overline{x} ∈ U(R_{a}). Since R_{a} is nondegenerate, U(R_{a}) is nondegenerate too. Let \overline{z} be in U(R_{a}) such that \overline{rn}▪\overline{x}▪\overline{z}▪\overline{rn}▪\overline{x} ≠ \overline{0}. Then \overline{z} ▪ \overline{rn} ∈ U(R_{a}), \overline{0} ≠ \overline{z} ▪ \overline{rn} ▪ \overline{x} and \overline{z} ▪ \overline{rn} ▪ \overline{y} ∈ R_{a}.

Suppose now \overline{x} ∉ U(Q_{a}). Since Q_{a} is nondegenerate (Proposition 3.2 (i)), by Lemma 1.6 there exist \overline{p}, \overline{q}_{1}, \overline{q}_{2} ∈ Q_{a} such that \overline{0} ≠ (\overline{q}_{1}, \overline{q}_{2}, \overline{p} ▪ \overline{x})_{a}. By (1) we can find n_{1}, n_{2}, n_{3} ∈ N(R) satisfying \overline{0} ≠ \overline{r} := (\overline{n_{1}p} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a} and [(n_{1}p)a]x, n_{2}q_{1}, n_{3}q_{2}, [(n_{1}p)a]y, n_{1}y, n_{1}ay ∈ R. This implies \overline{r} ∈ D(R_{a}). Denote by \overline{L} := _{R_{a}}(\overline{r}] ⊆ D(R_{a}) and let \overline{l} be a nonzero element in Z(\overline{L}), which exists by Lemma 1.2 (vi). By Lemma 1.2 (iv), \overline{l}^{2} ≠ \overline{0}, hence \overline{0} ≠ \overline{l} ▪ (\overline{n_{1}p} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a} = (\overline{n_{1}p} ▪ \overline{l} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a} = (\overline{p} ▪ \overline{ln_{1}} ▪ \overline{x}, \overline{n_{2}q_{1}}, \overline{n_{3}q_{2}})_{a} (since n_{1} ∈ N(R) ⊆ N(Q) and by Lemma 1.5 and (3.1)).

Finally we note that the element \overline{ln_{1}} ∈ N(R_{a}) because for \overline{u}, \overline{v} ∈ R_{a}, (\overline{n_{1}l}, \overline{u}, \overline{v})_{a} = (since n_{1} ∈ N(R) ⊆ N(Q) and by (3.1) and Lemma 1.5) (\overline{l}, \overline{n_{1}u}, \overline{v})_{a} = \overline{0} because \overline{l} ∈ Z(R_{a}). Moreover, \overline{ln_{1}} ▪ \overline{x} ≠ \overline{0} and \overline{ln_{1}} ▪ \overline{y} = \overline{[(ln_{1})a]y} = \overline{[(l(n_{1}a)]y} = \overline{(l, n_{1}a, y)} + \overline{l[(n_{1}a)y]} = \overline{(l, a, n_{1}y)} + \overline{l(n_{1}ay)} ∈ R_{a}. □

Let R be an alternative ring and S′ = S′(R) the subset of R defined by:

S′ = {a ∈ N(R) | a is regular in R}.

Suppose that S′ is nonempty. An alternative ring Q with 1 is called a classical left quotient ring of a subring R (relative to S′ ⊆ R), or R is said to be a classical left order in Q if:

(1) every element of S′ is invertible in Q, and

(2) each element q ∈ Q can be written in the form q = a^{−1}x, where a ∈ S′ and x ∈ R.

The proof of the following Proposition is analogue to that of [St, Lemma 1.10, pg. 54].

Lemma 5.5. Let Q be a unital alternative ring which satisfies the descending chain condition (dcc) on principal left ideals. Then the following conditions on an element x ∈ N(Q) are equivalent:

(i) x is invertible,

(ii) x is regular,

(iii) lan(x) = 0.

The proof of the following proposition follows partially the proof of [EK, Theorem 1].

Proposition 5.6. Let R be a nondegenerate alternative ring which satisfies the acc for the left annihilators of the form lan(a) with a ∈ N(R) and does not contain infinite direct sums of nonzero left ideals. Then for every essential left ideal L of R there exists an element 0 ≠ y ∈ L ∩ N(R) such that lan(y) = 0.

Proof. In this proof we will use Lemma 1.2 (v) without mention it. Let a_{1} be a nonnilpotent element of L ∩ N(R) with lan(a_{1}) as large as possible. We have lan(a_{1}) ⊆ lan(a_{1}^{2}) and a_{1}^{2} is a nonnilpotent element of L ∩ N(R), hence lan(a_{1}) = lan(a_{1}^{2}). If lan(a_{1}) = 0 we would have finished. Otherwise lan(a_{1}) ∩ L ≠ 0. Let a_{2} be a nonnilpotent element of lan(a_{1}) ∩ L ∩ N(R) with lan(a_{2}) as large as possible. As before lan(a_{2}) = lan(a_{2}^{2}). It is easy to verify that the sum Ra_{1} + Ra_{2} is direct and that lan(a_{1} + a_{2}) = lan(a_{1}) ∩ lan(a_{2}). If lan(a_{1} + a_{2}) = 0 we stop. If not, let a_{3} be a nonnilpotent element of lan(a_{1} + a_{2}) ∩ L ∩ N(R) with lan(a_{3}) as large as possible. Again lan(a_{3}) = lan(a_{3}^{2}) and Ra_{1} + Ra_{2} + Ra_{3} is a direct sum. Thus lan(a_{1}) + lan(a_{2}) + lan(a_{3}) = lan(a_{1}) ∩ lan(a_{2}) ∩ lan(a_{3}). This process must stop because R does not contain infinite direct sums of nonzero left ideals, so there are elements a_{1}, …, a_{n} in L ∩ N(R) such that lan(a_{1} + ⋯ + a_{n}) = 0. □

Lemma 5.7. Let R be a classical left order in a (unital) alternative ring Q. Then Q is a left quotient ring of R.

Proof. By [EK, Lemma 5], N(R) ⊆ N(Q). Take p, q ∈ Q with p ≠ 0. Then, given q there exist a ∈ S′, b ∈ R such that q = a^{−1}b, hence, aq ∈ R and ap ≠ 0 because a is invertible in Q. □

Given a subring R of an alternative ring Q, and an element q ∈ Q we define the set

(R : q) = {x ∈ R | x[q)_{R} ⊆ R},

where [q)_{R} denotes the linear span of q and the elements of the form ρ_{a_{1}} … ρ_{a_{n}}q with a_{i} ∈ R.

Lemma 5.8. Let R be a subring of an alternative ring Q and q ∈ Q. Then (R : q) is a left ideal of R.

Proof. Take x ∈ (R : q) and a ∈ R. Then for n ∈ ℕ and a_{1}, …, a_{n} ∈ R,

(ax)(ρ_{a_{1}} … ρ_{a_{n}}q) = (a, x, ρ_{a_{1}} … ρ_{a_{n}}q) + a[x(ρ_{a_{1}} … ρ_{a_{n}}q)]
= (x, ρ_{a_{1}} … ρ_{a_{n}}q, a) + a[x(ρ_{a_{1}} … ρ_{a_{n}}q)]
= [x(ρ_{a_{1}} … ρ_{a_{n}}q)]a − x(ρ_{a}ρ_{a_{1}} … ρ_{a_{n}}q) + a[x(ρ_{a_{1}} … ρ_{a_{n}}q)],

which is in R because x ∈ (R : q). Hence ax ∈ (R : q). □

Proposition 5.9. Let Q be a (nondegenerate) left quotient ring of a nondegenerate alternative ring R. Then:

(i) (R : q) is an essential left ideal of R for every nonzero q ∈ Q. If Q is artinian, then:

(ii) For every element s ∈ N(R) we have lan_{R}(s) = 0 if and only if s ∈ Inv(Q).

(iii) Every essential ideal I of R is a classical left order in Q.

Proof.

(i). Consider a nonzero left ideal L of R and take an element 0 ≠ y ∈ L∩N(R) (what is possible by virtue of Lemma 1.2 (v)). By induction, and taking into account N(R) ⊆ N(Q), it is easy to see that for any n ∈ ℕ, x ∈ N(R) and a_{1}, …, a_{n} ∈ R, x(ρ_{a_{1}} … ρ_{a_{n}}q) = ρ_{a_{1}} … ρ_{a_{n}}(xq). Now yq = 0 implies y[q)_{R} = 0. Hence 0 ≠ y ∈ (R : q)∩L. If yq ≠ 0, since Q is a left quotient ring of R, there exists a ∈ N(R) such that 0 ≠ a(yq) ∈ R and therefore 0 ≠ (ay)[q)_{R} ⊆ R. So ay is a nonzero element in (R : q) ∩ L.

(ii). Take s ∈ N(R) such that lan_{R}(s) = 0. By Lemma 5.5 , s ∈ Inv(Q) if and only if lan_{Q}(s) = 0. Suppose 0 ≠ q ∈ lan_{Q}(s). Given q, since Q is a left quotient ring of R, there exists an element a ∈ N(R) such that 0 ≠ aq ∈ R. Then, (aq)s = a(qs) = 0 and hence 0 ≠ aq ∈ lan_{R}(s) = 0, which is a contradiction.

(iii).

(1) I is essential as a left (right) ideal of R:

Let L be a nonzero left ideal of R. If IL = 0 then L ⊆ ran(I) = ann(I) = 0 (by Proposition 1.1 (i) and (iv)), a contradiction. Therefore 0 ≠ IL ⊆ I ∩ L.

(2) Reg(I) ∩ N(I) ⊆ Inv(Q).

Let y be in Reg(I) ∩ N(I). Then lan_{R}(y) ∩ I = lan_{I}(y) = 0 implies, by (1), lan_{R}(y) = 0 and by (ii), y ∈ Inv(Q).

(3) Take q ∈ Q. By conditions (ii) and (vii) in Proposition 5.4, R satisfies the acc for the left annihilators of the form lan(a) with a ∈ N(R) and does not contain infinite direct sums of nonzero left ideals, so we can apply Proposition 5.6 to (R : q) ∩ I, which by (i) and (1) is an essential left ideal of R, and find 0 ≠ s ∈ N(R) ∩ (R : q) ∩ I = N(I) ∩ (R : q) with lan_{R}(s) = 0. By (ii), s ∈ Inv(Q), and sq = t ∈ R implies q = s^{−1}t, which completes the proof. □

Theorem 5.10. Let Q be a (nondegenerate) alternative ring coinciding with its socle which is a left quotient ring of a nondegenerate alternative ring R. Then:

(i) R is left local Goldie. (ii) For each nonzero element a ∈ R the local ring R_{a} of R at a is a classical left order in the

nondegenerate artinian alternative ring Q_{a}.

Moreover

(1) R is prime if and only if Q is simple.

(2) R has finite left Goldie dimension if and only if Q is artinian.

Proof.

(i). By Corollary 2.6, Q is nonsingular and by Proposition 5.4 (v), R is left nonsingular. Proposition 3.2 (vi) says that Q has finite left local Goldie dimension. Hence and using condition (vii) in Proposition 5.4 we obtain that R has finite left local Goldie dimension.

(ii). Let a ∈ R. By Proposition 3.2 (i), R_{a} and Q_{a} are nondegenerate and by Proposition 3.2 (v), Q_{a} is artinian. Finally, since Q_{a} is a left quotient ring for R_{a} (by Proposition 5.4 (viii)) we can apply Proposition 5.9 (iii) to obtain that R_{a} is a classical left order in Q_{a}.

(1) If R is prime then Q is simple by Proposition 5.3 (i) and by the structure of the socle (if Q

were not simple then it would contain two orthogonal nonzero ideals).

Conversely, let Q be simple. Suppose Q associative and let aRb = 0 for some nonzero elements a, b ∈ R. By simplicity of Q, QbQ = Q so a = ∑_{i=1}^{n} p_{i}bq_{i} with p_{i}, q_{i} ∈ Q. Since Q is a left quotient ring of R, given a, p_{i}, there exists c ∈ R such that ca ≠ 0 and cp_{i} ∈ R. By nondegeneracy of R there exists r ∈ R such that arca ≠ 0. Then 0 ≠ arca = ∑_{i=1}^{n} arcp_{i}bq_{i} ⊆ aRbQ = 0, a contradiction.

Finally, suppose Q a Cayley-Dickson algebra (over Z(Q) = N(Q)). By Lemma 1.2 (v), every nonzero ideal of R contains a nonzero element of N(R). Since Q is a left quotient ring of R, N(R) ⊆ N(Q) = Z(Q), so every nonzero element of N(R) is regular in Q and can not be a zero divisor in R.

(2) Suppose Q artinian. Since Q satisfy the acc on all left ideals, from condition (vii) in Proposition 5.4 it follows that R has finite left Goldie dimension. Conversely, if Q were not artinian it would have infinite left Goldie dimension, hence R would have infinite left Goldie dimension too (applying Proposition 5.3 (i)). □

§6. Fountain-Gould left orders in alternative rings.

Let a be an element of an alternative ring R. We say that b in R is the group inverse of a if the following conditions hold:

aba = a, bab = b, ab = ba.

It is easy to see that a has a group inverse b in R if and only if there exists a unique idempotent e (e = ab) in R such that a is invertible in the alternative ring eRe (with inverse b), hence the group inverse is unique and a is said to be locally invertible. Denote by a^{♯} the group inverse of a and by LocInv(R) the set of all locally invertible elements of R.

A subring S of a ring R is called strongly associative (see [ZSSS]) if for any elements a, b ∈ S and r ∈ R, the equalities (a, b, r) = (a, r, b) = (r, a, b) = 0 holds.

Althought the following result follows as a consequence of [FGSS, Theorem 4 (2)], we included here a direct proof.

Lemma 6.1. Let a be an element in an alternative ring R and suppose that there exists the group inverse a^{♯} of a in R. Then the subring of R generated by a and a^{♯} is strongly associative.

Proof. Take in [ZSSS, (7.18)] z = x, t = a, y = a^{♯}, x = a^{♯}. Since {a^{♯}a^{♯}a} = 2a^{♯}^{2}a = 2a^{♯} then (x, a, a^{♯}^{3}) = 2(x, a^{♯}, a^{♯}) + (x, a^{♯}aa^{♯}, a^{♯}) = 3(x, a^{♯}, a^{♯}) = 0 which implies, by the Generalized Theorem of Artin, that the subring generated by {a, a^{♯}^{3}, x} (which is the same that the generated by {a, a^{♯}, x}) is associative. □

An element a ∈ R is left semiregular (or left square cancellable) if a^{2}x = a^{2}y implies ax = ay, for x, y ∈ R ∪ {1}.

Analogously is defined right semiregular (right square cancellable) element and semiregular (square cancellable) means both left and right semiregular. We denote by SemiReg_{l}(R), SemiReg_{r}(R) and SemiReg(R) the sets of all left, right and two-sided semiregular elements of R, respectively.

Proposition 6.2. Let Q be an alternative ring which satisfies dcc on principal left ideals generated by elements of N(Q). Then every right semiregular element of N(Q) has a group inverse in Q.

Proof. Let a be a right semiregular element in N(Q). Since the descending chain Q^{1}a ⊇ Q^{1}a^{2} ⊇ … must terminate, Q^{1}a^{n} = Q^{1}a^{n+1} = … for some natural number n. In particular, Q^{1}a^{n} = Q^{1}a^{n+4} implies a^{n} = qa^{n+3} for some q ∈ Q. Apply that a is right semiregular in Q to obtain a = qa^{4}. Hence a^{3} = a^{2}qa^{4} and, again by right semiregularity of a, a = a^{2}qa^{2} ∈ a^{2}Qa^{2}. By [FGSS, Theorem 1 (i)], a has a group invere in Q. □

Remark. This last result remains valid if in Q every chain of the form Q^{1}a ⊇ Q^{1}a^{2} ⊇ … terminates. We define the following subset S in R :

S = {a ∈ N(R) | a is semiregular in R}.

We now define a subring R to be a Fountain-Gould left order in an alternative ring Q if: (1) Every element of S has a group inverse in Q, and

(2) every element q ∈ Q can be written in the form q = a^{♯}x, where a ∈ S and x ∈ R.

We also say that Q is a ring of Fountain-Gould left quotients of R. Similarly we define Fountain-Gould right order and ring of Fountain-Gould right quotients. If R is both a Fountain-Gould left and right order in Q, then we say that R is an Fountain-Gould order in Q and that Q is a Fountain-Gould ring of quotients of R. When condition (2) is satisfied we speak about weak Fountain-Gould left order.

The result of the following lemma will be very used although we will not mention it.

Lemma 6.3. Let R be a weak Fountain-Gould left order in an alternative ring Q. Then every element q ∈ Q can be written as q = a^{♯}x, with a ∈ S, x ∈ R and aa^{♯}x = x.

Proof. Take q ∈ Q and b ∈ S, y ∈ R such that q = b^{♯}y. Then q = b^{♯}^{2}(by). By Lemma 6.1 the subring generated by {b, b^{♯}, y} is associative, so, if we take a = b^{♯}^{2} and x = by, the result is proved. □

Lemma 6.4. If R is a weak Fountain-Gould left order in an alternative ring Q, then N(R) ⊆ N(Q) and a^{♯} ∈ N(Q) for every a ∈ S ∩ LocInv(Q).

Proof. In every alternative ring the following identity is valid (Kleinfeld’s function):

(1)  (xy, z, t) − y(x, z, t) − (y, z, t)x = ([x, y], z, t) + (x, y, [z, t]).

Suppose x and y in the subring of Q generated by an element and its group inverse. By Lemma 6.1,

(2)  (xy, z, t) = y(x, z, t) + (y, z, t)x for every z, t ∈ Q.

Let a be an element in S.

Take x = a, y = a^{♯}^{2}, z, t ∈ R in (2). Then (aa^{♯}^{2}, z, t) = a^{♯}^{2}(a, z, t) + (a^{♯}^{2}, z, t)a, implies

(3)  (a^{♯}, z, t) = (a^{♯}^{2}, z, t)a for z, t ∈ R.

Take x = a^{3}, y = a^{♯}^{2}, z, t ∈ R in (2). Then (a^{3}a^{♯}^{2}, z, t) = a^{♯}^{2}(a^{3}, z, t) + (a^{♯}^{2}, z, t)a^{3} implies (a^{♯}^{2}, z, t)a^{3} = 0. Multiply by a^{♯}^{2} on the right side and take into account Lemma 6.1. Then (a^{♯}^{2}, z, t)a = 0 and by (3),

(4)  (a^{♯}, z, t) = 0 for z, t ∈ R.

The following identity is valid in any ring:

(5)  (xy, z, t) = x(y, z, t) + (x, y, z)t + (x, yz, t) − (x, y, zt).

Now, suppose n ∈ N(R), a ∈ S, x, y ∈ R and take in (5) x = a^{♯}, y = y, z = n, t = x. Then (a^{♯}y, n, x) = a^{♯}(y, n, x) + (a^{♯}, y, n)x + (a^{♯}, yn, x) − (a^{♯}, y, nx). If we call q = a^{♯}b and apply (4),

(6)  (q, n, x) = 0 for n ∈ N(R), x ∈ R, q ∈ Q.

For x = a ∈ S, y = a^{♯}^{2}, z = x ∈ R, t = q ∈ Q in (2), (aa^{♯}^{2}, x, q) = a^{♯}^{2}(a, x, q) + (a^{♯}^{2}, x, q)a and taking into account (6),

(7)  (a^{♯}, x, q) = (a^{♯}^{2}, x, q)a.

Take x = a^{3} ∈ S, y = a^{♯}^{2}, z = x ∈ R, t = q ∈ Q in (2). Then, for x ∈ R, q ∈ Q, (a^{3}a^{♯}^{2}, x, q) = a^{♯}^{2}(a^{3}, x, q) + (a^{♯}^{2}, x, q)a^{3}; by (6) and multiplying by a^{♯}^{2} on the right side (a^{♯}^{2}, x, q)a = 0. Hence, by (7)

(8)  (a^{♯}, x, q) = 0 for x ∈ R, q ∈ Q.

Now, consider in (5) x = n ∈ N(R), y = q ∈ Q, z = a^{♯}, t = x ∈ R. Then (nq, a^{♯}, x) = n(q, a^{♯}, x)+(n, q, a^{♯})x−(nq, a^{♯}, x)+(n, qa^{♯}, x). By (6), (8) and since for every element q′ ∈ Q there exists a ∈ S, x ∈ Q such that q′ = a^{♯}x,

(9)  (n, q, q′) = 0 for n ∈ N(R) and q, q′ ∈ Q,

which implies N(R) ⊆ N(Q). Finally, take a ∈ N(R) and suppose that there exists a^{♯} ∈ Q. Then, for every p, q ∈ Q, 0 = (a, p, q) = (a^{3}a^{♯}^{2}, p, q) = (since a ∈ N(R) ⊆ N(Q) and by [ZSSS, Lemma 7.1]) a^{3}(a^{♯}^{2}, p, q). Multiply by a^{♯}^{2} on the left side and apply Lemma 6.1: 0 = a(a^{♯}^{2}, p, q) = (a ∈ N(R) ⊆ N(Q) and [ZSSS, Lemma 7.1]) (aa^{♯}^{2}, p, q) = (a^{♯}, p, q), which completes the proof. □

Clearly, if R is a classical left order in a unital alternative ring Q (relative to S′ = Reg(R) ∩ N(R)), then R is a weak Fountain-Gould left order in Q. The converse is not true in general, as it was shown by Fountain and Gould in [FG_{1}, Example 3.1] for the associative case.

Theorem 6.5 (Common denominator property). Suppose that R is a weak Fountain-Gould left order in an alternative ring Q. Then, given q_{1}, …, q_{n} ∈ Q there exist a ∈ S, b_{1}, …, b_{n} ∈ R such that for every i = 1, …, n, q_{i} = a^{♯}b_{i} and aa^{♯}b_{i} = b_{i}.

Proof.

(1) N(R) is a weak Fountain-Gould left order in N(Q):

By Lemma 6.4, N(R) ⊆ N(Q). If q ∈ N(Q) then there exist a ∈ S, b ∈ R such that q = a^{♯}b with b = aa^{♯}b = aq ∈ N(Q) ∩ R = N(R).

(2) Write q_{i} = x_{i}^{♯}y_{i}, with x_{i} ∈ N(R) and y_{i} ∈ R. By Lemma 6.4, x_{i}^{♯} ∈ N(Q), and by [AM_{2}, Theorem 5] (which is valid for weak Fountain-Gould left orders) and (1), there exist a, u_{i} ∈ N(R) satisfying x_{i}^{♯} = a^{♯}u_{i} and aa^{♯}u_{i} = u_{i} for i = 1, …, n. Define b_{i} = u_{i}y_{i}. Then for every i ∈ {1, …, n}, q_{i} = a^{♯}b_{i} and aa^{♯}b_{i} = b_{i}. □

We notice that by the structure of the socle of a semiprime alternative ring, since for associative rings nondegeneracy and semiprimeness are equivalent, and since every Cayley-Dickson algebra is nondegenerate, for an alternative ring Q the following statements are equivalent:

(i) Q is nondegenerate and coincides with its socle,

(ii) Q is semiprime and coincides with its socle.

Lemma 6.6. Let Q be a left quotient ring of an alternative ring R. Then SemiReg_{r}(R) ⊆ SemiReg_{r}(Q).

Proof. Let s be in SemiReg_{r}(R) and suppose s ∉ SemiReg_{r}(Q). This means that there exist p, q ∈ Q ∪ {1} such that ps^{2} = qs^{2} but ps − qs ≠ 0. Apply that Q is a left quotient ring of R to find a ∈ N(R) satisfying a(ps − qs) ≠ 0 and ap, aq ∈ R. Now, ps^{2} − qs^{2} = 0 implies 0 = a(ps^{2}−qs^{2}) = (ap−aq)s^{2}. Since ap, aq ∈ R, and s ∈ SemiReg_{r}(R), this implies (ap−aq)s = 0, a contradiction. □

We stand out the following relations among classical, Fountain-Gould and weak Fountain-Gould left orders, and left quotient rings:

Proposition 6.7. Let R be a subring of an alternative ring Q.

(i) If Q is a weak Fountain-Gould left quotient ring of R, then Q is a left quotient ring of R. Suppose Q nondegenerate and coinciding with its socle.

(ii) If R is a weak Fountain-Gould left order in Q then R is a Fountain-Gould left order in Q. If Q is also artinian then

(iii) R is a classical left order in Q if and only if R is a Fountain-Gould left order in Q.

Proof.

(i). By Lemma 6.4, N(R) ⊆ N(Q). Given p, q ∈ Q with p ≠ 0, by the common denominator property there exist b ∈ S, a_{1}, a_{2} ∈ R satisfying p = b^{♯}a_{1}, q = b^{♯}a_{2} and bb^{♯}a_{i} = a_{i}. Then

bp ≠ 0 and bq ∈ R.

(ii). By Lemma 6.6, S ⊆ SemiReg_{r}(Q) and by Proposition 6.2, S ⊆ LocInv(Q).

(iii). Suppose that R is a Fountain-Gould left order in Q.

(1) Let q ∈ Q. By the common denominator property, given q and 1 there exist u ∈ S, v, w ∈ R such that q = u^{♯}v, 1 = u^{♯}w and uu^{♯}v = v, uu^{♯}w = w. This implies w = u and u^{♯} = u^{−1}, whence q = u^{−1}v.

(2) Now, consider a ∈ Reg(R) ∩ N(R). It is clear that a ∈ S, so there exists a^{♯} ∈ Q. By (1) we can write a^{♯} = u^{−1}v with u ∈ N(R) and v ∈ R. Therefore ua^{♯} = v and hence ua = va^{2}. Since a ∈ Reg(R), u = va. Now, a^{♯} = u^{−1}v implies a^{♯}a = u^{−1}va = u^{−1}u = 1.

Conversely, since every classical left order is a weak Fountain-Gould left order, the result follows by (ii). □

Lemma 6.8. Let R be an alternative ring which is a Fountain-Gould left order in a nondegenerate alternative ring Q coinciding with its socle, and let f : Q → Q′ be a ring epimorphism. Then f(R) is a Fountain-Gould left order in Q′.

Proof. It is clear that if a ∈ N(R) then f(a) ∈ N(Q′) and that if a^{♯} is the group inverse of a in Q then f(a^{♯}) is the group inverse of f(a) in Q′. Now, given q ∈ Q, let a ∈ S(R) and b ∈ R be such that q = a^{♯}b. Then f(q) = f(a)^{♯}f(b) and we have proved that f(R) is a weak Fountain-Gould left order in Q′. Since Q′ is nondegenerate and coinciding with its socle, by Proposition 6.7 (ii) we obtain the conclussion. □

Lemma 6.9. Let R be an alternative ring which is a Fountain-Gould left order in a nondegenerate alternative ring Q which coincides with its socle. Then R is nondegenerate.

Proof. By [Sl, Proposition 4.15], Soc(Q) = ⊕Q_{α} where Q_{α} is a simple associative ring coinciding with its socle or a Cayley-Dickson algebra.

Denote by π_{α} : Q → Q_{α} the canonical projection.

By Lemma 6.8, for every α, π_{α}(R) is a Fountain-Gould left order in Q_{α}. If Q_{α} is associative, then π_{α}(R) is strongly prime by [FG_{2}, Proposition 3.6]. If Q_{α} is a Cayley-Dickson algebra then π_{α}(R) is a classical left order in Q_{α}, by Proposition 6.7 (iii). Moreover π_{α}(R) is strongly prime: suppose a, b ∈ π_{α}(R) such that aπ_{α}(R)b = 0 and take an arbitrary element q = u^{−1}v ∈ Q_{α} with u ∈ N(π_{α}(R)) ⊆ N(Q) = Z(Q) and v ∈ π_{α}(R). Then (aq)b = u^{−1}[(av)b] = 0. Hence (aq)b = 0 and by [ZSSS, Exercise 1, pg 197], a = 0 or b = 0. In any case π_{α}(R) is strongly prime for every α, which implies nondegeneracy of R. □

Proposition 6.10. Let R be an alternative ring which is a Fountain-Gould left order in a nondegenerate alternative ring Q equal to its socle. Then for every a ∈ R, the local ring R_{a} of R at a is a classical left order in the nondegenerate artinian alternative ring Q_{a}.

Proof. We notice that R is nondegenerate by Lemma 6.9. By Proposition 5.4 (viii), for every a ∈ R the local ring of Q at a, Q_{a}, is a left quotient ring of R_{a}. Apply Proposition 5.9 (iii) to finish the proof. □

Proposition 6.11. Let R be an alternative ring which is a Fountain-Gould left order in a nondegenerate alternative ring Q equal to its socle. Then any essential ideal I of R is also a Fountain-Gould left order in Q.

Proof. First we point out that the ring R is nondegenerate by Lemma 6.9.

Since Q is nondegenerate and coincides with its socle, by Proposition 6.7 (ii) it is enough to prove that I is a weak Fountain-Gould left order in Q.

Take q ∈ Q and write q = a^{♯}b with a ∈ S and b ∈ R. By Proposition 6.10, R_{a} is a classical left order in Q_{a}. Denote by \overline{I} the set \{ \overline{y} ∈ R_{a} | y ∈ I\}. It is easy to see that \overline{I} is an ideal of R_{a}.

Moreover, it is an essential ideal. Indeed, let L be a nonzero ideal of R_{a} and choose \overline{0} ≠ \overline{l} ∈ N(L), which is possible by virtue of Lemma 1.2 (v), taking into account that R_{a} is nondegenerate by Proposition 3.2 (i). Then, since 0 ≠ ala and I is an essential ideal of R, by Proposition 1.1 (iv) and (vi), alaIala ≠ 0. In particular, there exists 0 ≠ y ∈ I such that ayala ≠ 0 and hence we have found a nonzero element \overline{yal} = \overline{y} ▪ \overline{l} in \overline{I} ∩ L.

By condition (iii) in Proposition 5.9, \overline{I} is a classical left order in the ring Q_{a}. Hence \overline{I} contains an invertible element of Q_{a}, say \overline{u}, with \overline{u} ∈ N(\overline{I}). Then \overline{p} = \overline{u}^{−1} for some \overline{p} ∈ N(Q_{a}), and apaua = auapa = a, which implies (multiplying by a^{♯}^{2} on the left hand side and taking into account a = a^{♯}a^{2}), a^{♯} = a^{♯}pa^{♯}a^{2}ua. An easy computation shows that a^{♯}pa^{♯} = (a^{2}ua^{2})^{♯}. Finally, q = a^{♯}b = a^{♯}pa^{♯}a^{2}uab = (a^{2}ua^{2})^{♯}(a^{2}uab), with a^{2}ua^{2} ∈ N(I) and a^{2}uab ∈ I (because u ∈ I), proves our claim.

We notice that the use of parentheses is not necessary because a ∈ N(R) ⊆ N(Q) and by Proposition 3.1 (i), al, la, au, ua, ap, pa ∈ N(Q). □

§7. The main Theorem.

Theorem 7.1. For an alternative ring R the following conditions are equivalent:

(i) R is a Fountain-Gould left order in a nondegenerate alternative ring Q which coincides with its socle,

(ii) R is nondegenerate, the set {lan(a) | a ∈ U(R)} satisfies the maximum condition and for

each a ∈ R, (a] has finite Goldie dimension, (iii) R is a nondegenerate left local Goldie ring. (iv) R is nondegenerate and for every a ∈ R, the local ring of R at a, R_{a}, is a classical left order

in a semisimple artinian alternative ring. In this case,

(1) R is prime if and only if Q is simple, and

(2) R is left Goldie if and only if Q is artinian.

Proof. (i) ⇒ (ii). By Lemma 6.9, R is nondegenerate. Moreover, since Q is a left quotient ring of R (Proposition 6.7 (i)) it follows from Theorem 5.10 (i) that R has finite left local Goldie dimension, and from Proposition 5.4 (ii) and (iii) that R satisfies the acc on the left annihilators of a single element in U(R) (since Q satisfies the acc on the left annihilators of a single element in U(Q)).

(ii) ⇒ (iii) is a consequence of condition (ii) in Corollary 2.9.

(iii) ⇒ (i). By Theorem 4.5, R is an essential subdirect sum of strongly prime left local Goldie alternative rings. More precisely, ⊕M_{α} ◁ R ≤ ⊕R/ann(M_{α}), where M_{α} ranges over all maximal uniform ideals of R. This allows us to reduce the question to the case that R is strongly prime because if we prove that each R_{α} := R/ann(M_{α}) is a Fountain-Gould left order in a (simple) alternative ring Q_{α} coinciding with its socle then, by Proposition 6.11, M_{α}, regarded as an ideal of R_{α}, is also a left order in Q_{α}. Since direct sums preserves Fountain-Gould left orders, ⊕M_{α} is a Fountain-Gould left order in the nondegenerate alternative ring coinciding with its socle ⊕Q_{α}, and hence R is also a Fountain-Gould left order in ⊕Q_{α}.

Suppose then that R is a strongly prime left local Goldie alternative ring. By [ZSSS, Theorem 9.9], R is associative or a Cayley-Dickson ring. In the first case the result follows from [AM_{1}, Theorem 1] and in the second one from the definition of a Cayley-Dickson ring.

(i) ⇒ (iv) follows from Proposition 6.10.

(iv) ⇒ (iii). We notice that R_{a} is nondegenerate by Proposition 3.2 (i) Since Q_{a} is left nonsingular and a left quotient ring of R_{a} (Lemma 5.5), R_{a} is left nonsingular too by Proposition 5.4 (vi), and applying Proposition 3.1 (v) we obtain the nonsingularity of R. On the other hand, (a] has finite Goldie dimension by Proposition 3.2 (iv).

Finally (1) and (2) had been proved in Theorem 5.10 in view of Proposition 6.7 (i). □

Besides we can give more information involving the two rings.

Theorem 7.2. Let R be a subring of a nondegenerate alternative ring Q coinciding with its socle. The following conditions are equivalent:

(i) R is a weak Fountain-Gould left order in Q, (ii) R is a Fountain-Gould left order in Q, (iii) (R is nondegenerate) N(R) ⊆ N(Q), Q = N(R)QR and for every nonzero element a ∈ R

the local ring of R at a, R_{a}, is a classical left order in the semisimple artinian alternative ring Q_{a},

(iv) for every finite subset Y of Q there exists an element a ∈ S such that Y ⊆ aQa and R_{a} is

a classical left order in the semisimple artinian alternative ring Q_{a},

(v) R is nondegenerate, Q is a left quotient ring of R and Q = N(R)QR.

Proof. Write Q as a sum of ideals, Q = ⊕Q_{α}, where the Q_{α}’s are simple alternative rings and Q_{α} = Soc(Q_{α}). Let π_{α} be the canonical projection π_{α} : Q → Q_{α} and define I_{α} := R ∩ Q_{α}.

(i) ⇔ (ii) is Proposition 6.7 (ii).

(ii) ⇒ (iii). Nondegeneracy of R follows from Lemma 6.9 and N(R) ⊆ N(Q) from Lemma 6.4. Given an element q ∈ Q there exist a ∈ S, b ∈ R satisfying q = a^{♯}b and aa^{♯}b = b, hence q = a^{♯}b = aa^{♯}^{2}b ∈ N(R)QR. The other assertion is Proposition 6.10.

(ii) ⇒ (iv). By Lemma 6.8, π_{α}(R) is a Fountain-Gould left order in Q_{α} and by Theorem 7.1 (1), π_{α}(R) is a prime ring. Hence every nonzero ideal of π_{α}(R), in particular I_{α}, is essential in π_{α}(R) (notice that I_{α} ≠ 0 since Q is a left quotient ring of R, by Proposition 5.3 (i)) and by Proposition 6.11, I_{α} is a Fountain-Gould left order in Q_{α}.

If Q_{α} is associative then (iv) is satisfied for I_{α} and Q_{α} by [GS, Theorem 4.12] and if Q_{α} is a Cayley-Dickson algebra over its center then Q_{α} = xQ_{α}x for every nonzero element x ∈ N(I_{α}) and by Proposition 6.10, (I_{α})_{x} is a classical left order in (Q_{α})_{x}.

Take q ∈ Q and write q = q_{1} + … q_{n} with q_{i} ∈ Q_{α_{i}}. As we have proved, there exist x_{i} ∈ S(I_{α_{i}}) ⊆ S such that q_{i} ∈ x_{i}Q_{α_{i}}x_{i} = x_{i}Qx_{i} and (I_{α_{i}})_{x_{i}} = R_{x_{i}} is a classical left order in Q_{x_{i}}. Write x = ∑_{i=1}^{n} x_{i}. Then q ∈ Q_{x} and R_{x} is a classical left order in Q_{x}. For more than one element the prove is similar.

(iv) ⇒ (i). Is analogue to (iii) ⇒ (i) in [GS, Theorem 4.12].

(ii) ⇒ (v). By Proposition 6.7, Q is a left quotient ring of R. Nondegeneracy of R is obtained by applying Lemma 6.9, and Q = N(R)QR can be proved as in (ii) ⇒ (iii).

(iii) ⇒ (i).

(1) π_{α}(N(R)) ≠ 0 and I_{α} ≠ 0 for every α.

Let 0 ≠ q_{α} ∈ Q_{α}. Since Q = N(R)QR, q_{α} = ∑ r^{i}p^{i}s^{i}, and applying π_{α}, 0 ≠ q_{α} = π_{α}(q_{α}) = ∑ π_{α}(r^{i})π_{α}(p^{i})π_{α}(s^{i}), which implies π_{α}(r^{i}) ≠ 0 for some r^{i} ∈ N(R).

Take r ∈ N(R), with π_{α}(r) ≠ 0. Since Q is nondegenerate, 0 ≠ π_{α}(r)Qπ_{α}(r) = rQ_{α}r. Let 0 ≠ rp_{α}r ∈ rQ_{α}r. Apply that R_{r} is a classical left order in Q_{r} to find \overline{a} ∈ Reg(R_{r}) ∩ N(R_{r}), \overline{b} ∈ R_{r} such that \overline{p}_{α} = (\overline{a})^{−1} ▪ \overline{b}. Then \overline{0} ≠ \overline{b} = \overline{a} ▪ \overline{p}_{α} = \underline{arp_{α}}, that is, 0 ≠ rbr = rarp_{α}r ∈ R ∩ Q_{α} = I_{α}.

(2) For every nonzero u ∈ π_{α}(R) the ring (π_{α}(R))_{u} is a classical left order in (Q_{α})_{u}.

Take r ∈ R satisfying π_{α}(r) = u. By hypothesis, R_{r} is a classical left order in Q_{r}. Notice that the following is a ring epimorphism:

\begin{aligned}φ : & Q_{r} & ⟶ & Q_{u} \\ & \overline{q} & ↦ & { \atop π_{α}(q)} \end{aligned}

Hence, by the analogue result to Lemma 6.8 for classical left orders, φ(R_{r}) = (π_{α}(R))_{u} is a classical left order in Q_{u}.

(3) In order to simplify the notation, we will write R = π_{α}(R) and Q = Q_{α}. For every α, Q is

a Fountain-Gould left quotient ring of R.

Suppose Q associative. We note that Q = RQR and by (2), R is nondegenerate. Hence, by the [GS, Theorem 4.12], R is a Fountain-Gould left order in Q.

Suppose Q a Cayley-Dickson algebra over its center. Then N(Q) = Z(Q) is a field. By (1) there exists 0 ≠ u ∈ π_{α}(N(R)) ⊆ π_{α}(N(Q)) ⊆ N(Q). By (2), given q ∈ Q = uQu, there exist \overline{a} ∈ Reg(R_{u}) ∩ N(R_{u}) ⊆ N(R) ⊆ N(Q), \overline{b} ∈ R_{u}, such that \overline{u^{−1}q} = \overline{a}^{−1} ▪ \overline{b}. It is not difficult to see that the map

\begin{aligned}ψ : & Q & ⟶ & Q_{u} \\ & q & ↦ & \overline{u^{−1}q} \end{aligned}

is a ring isomorphism, so \overline{u^{−1}q} = \overline{a}^{−1} ▪ \overline{b} implies q = (ua)^{−1}(ub) = a^{−1}b.

(4) R is a weak Fountain-Gould left order in Q.

Reasoning as in (ii) ⇒ (iv) we obtain that I_{α} is a Fountain-Gould left order in Q. This implies that ⊕I_{α} is a weak Fountain-Gould left order in ⊕Q_{α} and since ⊕I_{α} ⊆ R ⊆ Q = ⊕Q_{α} we have that R is a weak Fountain-Gould left order in Q, as required.

(v) ⇒ (iii) follows by Theorem 5.10 (ii). □

Notice that, unlike the usual Goldie-type results, the previous theorem deals with conditions on the ring of quotients Q rather than on the original ring R. The associative version of Theorem 7.2 ([GS, Theorem 5.13]) is used to identify Q inside bigger rings like the maximal ring of quotients of R, so it is natural to expect a similar use of Theorem 7.2 in the alternative setting once the appropriate maximal ring of quotients notion is introduced.

As a consequence of Theorems 7.1 and 7.2 we obtain the following result for classical left orders in artinian alternative rings coinciding with its socle, which improves the main theorem in [EK].

Theorem 7.3. For an alternative ring R the following conditions are equivalent:

(i) R is a classical left order in a nondegenerate artinian alternative ring Q coinciding with its socle,

(ii) there is a nondegenerate artinian alternative ring coinciding with its socle Q such that Q

is a left quotient ring of R, (iii) R is nondegenerate and a subring of a nondegenerate artinian alternative ring coinciding

with its socle Q such that N(R) ⊆ N(Q), Q = N(R)QR, and for every element a ∈ R the ring R_{a} is a classical left order in the nondegenerate artinian alternative ring Q_{a},

(iv) R is nondegenerate, satisfies the acc for the left annihilators lan(a) with a ∈ U(R) and has

finite left Goldie dimension,

(v) R is nondegenerate, left nonsingular and has finite left Goldie dimension. In this case, R is prime if and only if Q is simple.

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