LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS Miguel Gómez

LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
Miguel Gómez Lozano & Mercedes Siles Molina
Departamento de Álgebra, Geometrı́a y Topologı́a
Universidad de Málaga
29071 Málaga (España)
e-mail: [email protected]; [email protected]
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 [FG1 ] introduced a notion of order in a ring which need not
have a unit, and gave [FG2 ] 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 [AM1 ] 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
1
2
M. GÓMEZ LOZANO AND M. SILES MOLINA
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 lanR (X) when it is necessary to emphasize the dependence on R. Similarly the right
annihilator of R, ran(X) = ranR (X), is defined by
ran(X) = {a ∈ R
|
xa = 0
for all
x ∈ X}.
We also write ann(X) = annR (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 → R
the canonical projection of R onto 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) R is a semiprime alternative ring.
(iv) I is an essential ideal of R if and only if ann(I) = 0.
(v) I = π(I) is an essential ideal of R.
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
3
If R is nondegenerate, then:
(vi) ann(I) = {x ∈ R | xIx = 0},
(vii) 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.
(ii).
(iii).
(iv).
(v).
(vi).
(vii).
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).
We have (I ∩ann(I))2 ⊆ I ann(I) = 0, and by semiprimeness of R we obtain I ∩ann(I) = 0.
Let J = π(J ) be an ideal of R, with J an ideal of R, such that (J)2 = 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 = J I, which implies J = 0.
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.
By (iii), R is semiprime, so, by (i), lanR (I) = annR (I). Hence and by (iv) it is enough to
prove that lanR (I) = 0. Let r be in lanR (I). Then rI ⊆ I ∩ ann(I), which is zero by (iii).
And by (i), r = 0.
By [CFGM, Lemma] the Jordan annihilator of I (see note after Lemma 1.3), annJ (I),
which coincides with the set {x ∈ R | xIx = 0}, is an ideal of R such that annJ (I) ∩ I = 0.
This implies (annJ (I) ∩ I)2 = 0 and since R is semiprime, annJ (I)I = 0. Hence annJ (I) ⊆
lan(I) = (by (i)) ann(I). The inclusion ann(I) ⊆ annJ (I) is obvious.
Assume that there exists a ∈ R such that aba = 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, R1 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]):
U = {x ∈ R | xR1 ⊆ N (R)} = {x ∈ R | R1 x ⊆ N (R)}.
(1.1)
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)
(ii)
(iii)
(iv)
N (L) = L ∩ N (R).
Z(L) = L ∩ Z(R).
U (L) = L ∩ U (R).
For every nonzero element y ∈ Z(L), y 2 6= 0 .
If R is nondegenerate then:
(v) N (L) 6= 0.
If R is nondegenerate and L * N (R) then:
(vi) Z(L) 6= 0.
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M. GÓMEZ LOZANO AND M. SILES MOLINA
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 R1 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 6= (R1 y)2 = (R1 )2 y 2 , hence y 2 6= 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) 6= 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) 6= 0 by [ZSSS, Corollary of Theorem 9.6]. Hence and by (i),
0 6= N (Ľ) = Ľ ∩ N (R) ⊆ L ∩ N (R) = N (L).
(vi). By the hypotheses, (L, L, R) 6= 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) 6= 0. Hence, and using twice (ii), 0 6= 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 (R1 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 6= (L, R, R) ⊆ L ∩ D. Following [Sl], the left socle of an alternative ring R, Socl (R), is the sum of all minimal left
ideals of R. The right socle, Socr (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 Socl (R) = 0
(Socr (R) = 0). If the ring R is semiprime then the left and the right socle coincide (in fact
Socl (R) = Socr (R) under a weaker condition than semiprimeness -see [Sl, Theorem F]-). In this
situation we define the socle of R, Soc(R), to be Socl (R) = Socr (R). If R is semiprime, by
[Sl, Proposition 4.15], Soc(R) = R1 ⊕ R2 where R1 and R2 are the following ideals of R: R1 is
the internal direct sum of all those ideals of R which are Cayley-Dickson algebras and R2 is the
internal direct sum of all ideals of R which are simple associative rings having minimal left (and
right) ideals (in fact R1 = Soc(D(R)) and R2 = Soc(U (R))).
Remark. Since each alternative ring R becomes a Jordan algebra with quadratic maps Ux y = xyx,
denote it by RJ , 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(RJ ) 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.
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
5
Lemma 1.4. Let R be a semiprime (nondegenerate) alternative ring. Then R = R/U is a
semiprime (nondegenerate) purely alternative ring.
1
Proof. We see first that R is purely alternative. Let x be in R such that (R x, R, R) = 0. This
implies (R1 x, R, R) ⊆ U and hence (R1 x, R, R) ⊆ U ∩ D = 0, by Lemma 1.3 (i), which implies
x ∈ U and, consequently, x = 0.
Now, since by Lemma 1.3 U = ann(D), R semiprime (nondegenerate) implies R semiprime
(nondegenerate), by Proposition 1.1 (iii) ((iv)). For any element a in an alternative ring R the left and right multiplications λa and ρa are
defined as follows:
λa b = ab = ρb a.
Lemma 1.5. Let R be a semiprime ring. Then for every element n ∈ N (R) the following identities
hold in R:
n[µx1 . . . µxr (y, z, t)] = µnx1 . . . µxr (y, z, t) = . . . = µx1 . . . µnxr (y, z, t) = µx1 . . . µxr (ny, z, t),
n(µx1 . . . µxr y, z, t) = (µnx1 . . . µxr y, z, t) = . . . = (µx1 . . . µnxr y, z, t) = (µx1 . . . µxr ny, z, t),
where µa denotes λa or ρa .
Proof. Since R = R/U is purely alternative, by Lemma 1.4, [ZSSS, Theorem 8.11] applies to get
N (R) = Z(R). Hence n[µx1 . . . µxr (y, z, t)] = µnx1 . . . µxr (y, z, t), and so n[µx1 . . . µxr (y, z, t)] =
µnx1 . . . µxr (y, z, t), which implies n[µx1 . . . µxr (y, z, t)] − µnx1 . . . µxr (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 R1 x ⊆ N and, consequently, x ∈ U ). Otherwise, 0 6= 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 6= u := λr1 . . . λrα t ∈ Z((t]). By condition (iv) in Lemma 1.2, 0 6= u2 = uλr1 . . . λrα (a, b, x) =
(by Lemma 1.5) λr1 . . . λrα (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
Zl (R) = {x ∈ U (R) | lan(x) is an essential left ideal of R}.
Proposition 2.1. For an alternative ring R, Zl (R) is an ideal of R.
Proof. For x, y ∈ Zl (R) we have x + y ∈ Zl (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 ∈ Zl (R) and r ∈ R.
Since lan(x) ⊆ lan(xr), Zl (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, R1 L) ⊆ (R, R, L) = 0. If Lr = 0, then L ⊆ lan(rx); otherwise there would exist
y ∈ L such that 0 6= yr ∈ U , and since lan(x) is an essential left ideal of R, we can take a
nonzero element z(yr) ∈ R1 (yr) ∩ lan(x), that is, 0 6= zy ∈ R1 y ∩ lan(rx) ⊆ L ∩ lan(rx). Finally,
suppose (R, R, L) 6= 0. Taking into account (R, R, L)(rx) ⊆ DU = 0, we obtain 0 6= (R, R, L) ⊆
L ∩ lan(rx). Definition 2.2. For an alternative ring R we call the ideal Zl (R) the left singular ideal of R.
Analogously, it is proved that
Zr (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.
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M. GÓMEZ LOZANO AND M. SILES MOLINA
Lemma 2.3. For every ideal I of a semiprime alternative ring R we have: Zl (I) = I ∩ Zl (R).
Proof. Assume I 6= 0. Take 0 6= x ∈ Zl (I). By condition (iii) in Lemma 1.2, x ∈ U (R). Now
we see that lanR (x) is an essential left ideal of R. Let L be a nonzero left ideal of R. If Lx = 0
then L ⊆ lanR (x). If Lx 6= 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 6=
IxL ∩ lanI (x) ⊆ L ∩ lanR (x).
Conversely, let 0 6= x ∈ I ∩ Zl (R). By Lemma 1.2 (iii), x ∈ U (I). Now we prove that lanI (x)
is an essential left ideal of I. Let L be a nonzero left ideal of I. If Lx = 0, then L ⊆ lanI (x). If
Lx 6= 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 lanR (x) is an essential left ideal
of R, 0 6= RxL ∩ lanR (x) and L ∩ lanI (x) is nonzero. Corollary 2.4. Let R be a semiprime alternative ring and I an essential ideal of R. Then Zl (R) =
0 if and only if Zl (I) = 0.
Proposition 2.5. For an alternative ring R the singular ideal Zl (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 ∈ Zl (R)
and let y be in R such that xyx = x. Then Rxy is a nonzero left ideal of R and 0 6= 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
Zl (R) = 0.
Proof. Since Soc(R) is a von Neumann regular ideal of R (by [L, Proposition 3 (2)]), Zl (Soc(R)) =
0 by Proposition 2.5, which implies Zl (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 6= 0.
(ii) ran(a) = ran(xa) for every x ∈ R such that xa 6= 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) Zl (R) and Zr (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 Zl (R) = Zr (R) = 0.
Proof. Since Zl (R) and Zr (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 Zl (R) = 0.
Similarly, we will say that R is right nonsingular if Zr (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 y = (xa)y
a
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
7
becomes an alternative ring, the (left) a-homotope ring, denoted by Ra , which has as an ideal
the set
Ker(a) = {x ∈ R | axa = 0}.
We define the (left) local ring of R at a as Ra /Ker(a) and denote it by Ra . The product of two
elements x, y ∈ Ra is designed by x y (:= (xa)y).
If ( , , )a and ( , , )a designate the associators in Ra and Ra , respectively, it is obvious that
for every x, y, z ∈ R, (x, y, z)a = (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)
(ii)
(iii)
(iv)
(v)
(vi)
x ∈ N (Ra ) if and only if ax or xa are in N (R). In this case ax and xa are in N (R).
x ∈ U (Ra ) if and only if axa ∈ U (R).
If x ∈ Zl (Ra ) then axa ∈ Zl (R).
If a ∈ Zl (R) then Zl (Ra ) = Ra .
Zl (R) = 0 if and only if Zl (Ra ) = 0 for all a ∈ U (R).
A nonzero element a ∈ U (R) is pseudo-uniform if and only if Ra has no nonzero divisors.
Proof.
(i). Suppose x ∈ N (Ra ). By (3.2), for every y, z ∈ R, 0 = (x, y, z)a = (xa, ya, z). This means
0 = a(xa, ya, z)a = (xa2 , ya2 , z) and by [ZSSS, Lemma 9.5] applied to the left ideal Ra2 we
obtain xa2 ∈ N (R). Hence 0 = (xa2 , 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), (x, y, z)a = 0, which implies x ∈ N (Ra ).
(ii). We have x ∈ U (Ra ) if and only if for every y ∈ (Ra )1 , (ya)x = y x ∈ N (Ra ) if and only
if (by (i)) [(ya)x]a = (by the right Moufang identity) y(axa) ∈ N (R) for every y ∈ R1 and
this is equivalent to axa ∈ U (R).
(iii). Take x ∈ Zl (Ra ). By (ii), axa ∈ U (R). Now let L be a nonzero left ideal of R. If Laxa = 0
then L ⊆ lanR (axa). Suppose Laxa 6= 0. Then (Laxa)2 6= 0, so L is a nonzero left ideal
of Ra . Since lanRa (x) is an essential left ideal of Ra we can choose an element y ∈ L such
that 0 6= y ∈ lanRa (x) ∩ L, that is, yax = 0 and we have 0 6= ay ∈ L ∩ lanR (axa). In any
case lanR (axa) ∩ L 6= 0, which implies that lanR (axa) is an essential left ideal of R.
(iv). Suppose a ∈ Zl (R) and take x ∈ Ra and a nonzero left ideal L of Ra . If L x = 0 then
L ⊆ lanRa (x). If L x 6= 0, take y ∈ L such that y x 6= 0, that is, 0 6= a[(ya)x]a =
(a ∈ U (R)) ayaxa. Since U (R) is nondegenerate, by [ZSSS, Theorem 8.4], U (R)ayaxa 6= 0
and hence Rayax is a nonzero left ideal of R. Now take 0 6= rayax ∈ lanR (a). Then
0 6= aRraya and aRrayaxa = 0, that is, 0 6= Rray = Rr y ⊆ lanRa (x) ∩ L.
(v). Zl (R) = 0 implies, by (iii), that for every a ∈ U (R), Zl (Ra ) = 0. Conversely, suppose
Zl (Ra ) = 0 for every a ∈ U (R). By (iv), a ∈ Zl (R) would imply Ra = Zl (Ra ) = 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)]). 8
M. GÓMEZ LOZANO AND M. SILES MOLINA
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 Ra , then it is not difficult to
prove
(3.3)
Uxa y = Ux Ua 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 6= 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 = λa or ϕ = IdR
(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-dimR (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-dimR (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)
(ii)
(iii)
(iv)
All the local rings of R at nonzero elements are nondegenerate.
R is prime if and only if all the local rings of R at nonzero elements are prime.
If R is simple, then all the local rings of R at nonzero elements of N (R) are simple.
For any element a ∈ R, u − dimR (a) ≤ u − dim(Ra ). If a is in N(R), then a has left Goldie
dimension equal to n if and only if Ra has left Goldie dimension equal to n.
(v) For every a ∈ R, x ∈ Soc(Ra ) if and only if axa ∈ Soc(R). Hence R coincides with its socle
if and only if Ra 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 x ∈ Ra such that x y x = 0 for every y ∈ Ra .
Then 0 = Ux Ua y = Uxa y by (3.3) and hence UUa x y = (by Macdonald’s identity -see [ZSSS]-)
Ua Ux Ua y = 0 for every y ∈ R. By nondegeneracy of R we have Ua x = 0, that is, x = 0.
(ii). Assume that R is prime. Take a ∈ R and let I, J be two ideals of Ra such that I J = 0. If
x and y were nonzero elements of I and J respectively, then for every r ∈ R, 0 = x (r y),
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
9
which implies 0 = Ua (x [r y]) = (axa){[(ra)y]a} = (axa)[r(aya)] and by [ZSSS, Exercise
a
a
1, pg. 197], axa = 0 or aya = 0, i.e., x = 0 or y = 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 I and J of Rx+y
are nonzero because, by (i), (x + y)I(x + y) = xIx 6= 0 and (x + y)J (x + y) = yJ y 6= 0.
Moreover I J = 0, which contradicts the primeness of Rx+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
ϕ:
Ra
x
→ Ra
7
→
x
and
ψ:
Ra
x
→ R
7→ xa
are ring isomorphisms because a ∈ N (R) = Z(R) and Z(R) is a field. Hence Ra , 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 6= 0:
If U L 6= 0, then there exist u ∈ U , l ∈ L such that 0 6= ul ∈ L. Since L ⊆ R (a] there
Pm
Pm
exist xji in R1 such that l = i=1 λxi1 . . . λxin a and ul = va with v = i=1 uρxi1 . . . ρxin
i
i
∈ U . By nondegeneracy of R, 0 6= (va)t(va) for some t ∈ R. Now, v, va ∈ N implies
0 6= (at)(va) = a(tv)a ∈ L ∩ aRa.
If U L = 0, then (L, R, R) 6= 0 (otherwise L ⊆ U and L2 = 0, which is not possible
because R is nondegenerate) and D∩L (which contains (L, R, R)) is nonzero.
PmBy Lemma 1.2
(ii) and (vi), we can choose 0 6= l ∈ (L∩D)∩Z(R) = Z(L∩D). Write l = i=1 λxi1 . . . λxin a
i
Pm
as before. By Lemma 1.2 (v), 0 6= l2 = ( i=1 λxi1 . . . λxin a)2 , which implies al 6= 0, and by
i
nondegeneracy of R, 0 6= (al)t(al) = a(l2 t)a ∈ L ∩ aRa.
(2) Let {Lα } be a family of nonzero left ideals of R contained in R (a] and whose sum is direct.
By (1), Lα ∩ aRa is nonzero and it is clear that if we define Lα = {x ∈ Ra | axa ∈ Lα },
then {Lα } is a family of nonzero left ideals of Ra whose sum is direct, which proves that
u-dimR (a) ≤ u-dim(Ra ).
Now, suppose a ∈ N (R) and let {Li }ni=1 be a direct sum of nonzero left ideals of Ra .
Choose 0 6= li ∈ N (Li ) = Li ∩ N (Ra ) (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 ti := ali a is in N (R)
for every i = 1, . . . , n. Hence and using the nondegeneracy of R we have that {Rti }ni=1 is a
family of nonzero left ideals of R contained in Ra. We claim that the sum of these ideals
is direct because if r1 t1 + · · · + rn tn = 0 with, for example, r1 t1 6= 0,
nondegeneracy
of
Pby
Pn
n
R, there exists s ∈ R such that asr1 al1 a 6= 0, that is, 0 6= sr1 l1 ∈ i=2 R li ⊆ i=2 Li ,
which is a contradiction.
(v). Since x y x = Uxa y = Ux Ua y (by (3.3)), I is an inner ideal of Ra if and only if I =
{axa | x ∈ I} is an inner ideal of R, hence x ∈ Soc(Ra ) 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] Ra is an artinian alternative ring
with capacity equal to the rank of a.
Conversely if Ra is artinian then Ra has bounded length for the chains of inner ideals
of the form I with I an inner ideal of RJ contained in aRa, hence for chains of principal
inner ideals of the form bRb with b in the inner ideal of RJ 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. 10
M. GÓMEZ LOZANO AND M. SILES MOLINA
§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 Zl (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 6= 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 6= 0. Denote by L and L′ the nonzero left
ideals of R generated by Bu and Cu respectively. Then 0 6= 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. Q
A subdirect product of alternative rings R ≤
Rα will Q
be called an essential subdirect
product if R contains an essential ideal of the full product Rα . If R is actually contained in
the direct sum of the Rα , 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 Rα .
(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 Rα = R/ann(Mα ), where {Mα } is the family of all maximal uniform ideals
of R.
Proof. (i) ⇒ (ii). In general any subdirect product R of a family {Rα } of nondegenerate
Q alternative
rings is nondegenerate. Let M ⊆ R be an essential ideal of the full direct product Rα and set
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
11
Q
Mα := M ∩ Rα , where we are regarding Rα as an
Q ideal of Rα . Then Mα is a nonzero ideal of Rα
contained in R since M is an essential ideal of Rα . In fact Mα is a uniform ideal of R since Mα
is uniform in Rα because Rα is strongly prime and any ideal of R contained in Mα is an ideal of
Rα . Now if I is a nonzero ideal of R then πα (I) is a nonzero ideal of Rα for some index α. Hence
by strongly primeness of Rα we have 0 6= πα (I) ∗ Mα ⊆ I ∩ Mα . Therefore I contains the nonzero
ideal I ∩ Mα , which is uniform since it is contained in Mα .
(ii) ⇒ (iii) is trivial.
P
(iii) ⇒ (i). Let
Mα be the
P sum of all maximal uniform ideals of R, which is direct
P by
Proposition 4.2 (iii). Since ann( Mα ) is a closed ideal, it must be zero: otherwise ann( Mα )
would contain a uniform ideal, and therefore a maximal uniform ideal because
it is closed, which
P
leads to contradiction. Hence, by a standard argument, ∩ann(Mα ) = ann( Mα ) = 0 implies that
R is a subdirect product of the alternative rings Rα := R/ann(Mα ) each of which is a strongly
prime alternative ring by Proposition 4.2 (i). Q
Q
Finally, the homomorphic image of ⊕Mα in
Rα is an essential ideal of Rα since Mα can
be regarded as an essential ideal of Rα , 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 R the quotient
ring R/ann(I). We have:
(i) Any direct sum of nonzero left ideals of 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
R has also finite left Goldie dimension.
(ii) If a ∈ R has finite left Goldie dimension in R, then a := a + ann(I) has finite left Goldie
dimension in R.
(iii) If Zl (R) = 0 then Zl (R) = 0 . Moreover, if R is nondegenerate and I is a uniform ideal,
then R is a strongly prime left nonsingular alternative ring.
Proof.
P
(i). Let
Lα be a direct sum of nonzero left ideals of R. Denote by π : R → R the canonical
projection of R onto R. By Proposition 1.1 (v), π(I) is an essential left ideal of R.PHence
Lα := π −1 (Lα )∩I is a nonzero leftP
ideal of R, for each index α. Now,
P we show that Lα is
a direct sum. Indeed, if x ∈ Lβ ∩ ( α6=β Lα ), then π(x) ∈ Lβ ∩ ( α6=β Lα ) = 0. Therefore
x ∈ IP∩ ann(I) = 0 (by Proposition 1.1 (ii)).
(ii). Let
Lα be a direct sum of nonzero left ideals of R inside the principal left ideal R (a].
By taking Lα := I(π −1 (Lα )) we obtain a direct sum of nonzero left ideals of R contained
in R (a].
(iii). By Lemma 2.3, Zl (R) = 0 implies Zl (I) = 0. But I can be regarded as an ideal of R via
the isomorphism x 7→ x + ann(I), for x ∈ I and by Proposition 1.1 (v) we have that I is
an essential ideal of R. By Proposition 1.1 (iii) and Corollary 2.4, Zl (R) = 0.
If R is nondegenerate and I is uniform, then 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,
⊕Mα ⊳ R ≤ ⊕R/ann(Mα ),
where Mα 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 Rα = R/ann(Mα ), with Mα a maximal
uniform ideal of R, each of which is a strongly prime left local Goldie alternative ring by Proposition
12
M. GÓMEZ LOZANO AND M. SILES MOLINA
4.2 (i) and conditions (ii) and (iii) in Lemma 4.4. Let us see that R ⊆ ⊕Rα . Otherwise, there
would exist x ∈ R such that x ∈
/ ann(Mα ) for an infinite number of α′ s. Say x ∈
/ ann(Mα ) for
every α ∈ Λ. Denote by IαPthe left ideal of R generated by Mα x. Then 0 6= Mα x ⊆ Iα ⊆ Mα for
every α ∈ Λ, and the sum α∈Λ Iα 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 Rα . Moreover, each Rα 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 6= 0, there exists r ∈ N (R) such that rp 6= 0 and rq ∈ R.
It is not difficult to see that for any finite familly {q1 , . . . , qn } ⊆ Q, with q1 6= 0, there exists
r ∈ N (R) such that rq1 6= 0 and rqi ∈ 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 U p 6= 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 6= t := (x, y, zp). Since (t] ⊆ D, we can apply Lemma 1.2 to find 0 6= u := λr1 . . . λrα t ∈ Z((t]). By
condition (iv) in Lemma 1.2, 0 6= u2 = uλr1 . . . λrα (x, y, zp) = (by Lemma 1.5) λr1 . . . λrα (x, y, zup).
In particular, up 6= 0.
In any case, N p 6= 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 6= 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 6= q ∈ Q such that qQq = 0. Since Q is a left quotient ring
of R there exists r ∈ N (R) such that 0 6= 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 lanR (X) ⊆ lanR (Y ) if and only if lanQ (X) ⊆ lanQ (Y ).
(iii) U (R) = U (Q) ∩ R.
If R is semiprime then:
(iv) Zl (R) = Zl (Q) ∩ R.
(v) Zl (R) = 0 if and only if Zl (Q) = 0.
(vi) Any direct sum {Li } of nonzero left ideals of R gives rise to a direct sum {Li } 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.
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
13
If R is nondegenerate then:
(viii) For every nonzero element a ∈ R the local ring Qa of Q at a is a left quotient ring of Ra .
Proof.
(i). The containement Z(Q) ∩ R ⊆ Z(R) is obvious. Conversely, take x ∈ Z(R) and suppose
x ∈
/ Z(Q). Then [p, x] 6= 0 for some p ∈ Q. Apply that Q is a left quotient ring of R
to find n ∈ N (R) such that 0 6= 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 lanR (X) = lanQ (X) ∩ R, lanQ (X) ⊆ lanQ (Y ) implies lanR (X) ⊆ lanR (Y ). Suppose
lanR (X) ⊆ lanR (Y ). If there were an element q ∈ lanQ (X) such that qy 6= 0 for some y ∈ Y
then there would exist an element r ∈ N (R) such that 0 6= rqy and rq ∈ R (because Q is
a left quotient ring of R). Then rq ∈ lanR (X) but rq ∈
/ lanR (Y ), which is a contradiction.
(iii). It is clear that U (R) ⊇ U (Q) ∩ R. Conversely take x ∈ U (R) and suppose (px, q1 , q2 ) 6= 0
for some p ∈ Q1 , q1 , q2 ∈ Q. Since Q is a left quotient ring of R there exists a ∈ N (R) such
that 0 6= a(px, q1 , q2 ) (= ((ap)x, q1 , q2 )) and ap ∈ R. Now ap ∈ R and x ∈ U (R) implies
(ap)x ∈ N (R) ⊆ N (Q). Hence a(px, q1 , q2 ) must be zero, a contradiction.
(iv). In this item, (iii) will be in our mind. Take x ∈ Zl (R) and let L be a nonzero left ideal of
Q. Then R ∩ L is a nonzero left ideal of R and hence 0 6= lanR (x) ∩ R ∩ L ⊆ lanQ (x) ∩ L.
Conversely, let x ∈ Zl (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 ∩ lanR (x). Suppose Kx 6= 0 and let l = u + d be a nonzero element of K, with
u ∈ U (R) and d ∈ D(R), satisfying 0 6= lx = ux + dx = ux. By [ZSSS, Theorem 9.4], U (R)
is nondegenerate, so there exists 0 6= v ∈ U (R) with 0 6= vlx. Then 0 6= vl ∈ U (R) ∩ L.
Apply that lanQ (x) is an essential left ideal of Q and choose 0 6= q(vl) ∈ Q(vl) ∩ lanQ (x).
Notice that since 0 6= 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 6= q(vl), with
q ∈ Q, there exists a ∈ N (R) satisfying 0 6= (aq)(vl) and aq ∈ R. Then (aq)(vl) is a
nonzero element in L ∩ lanR (x).
(v) follows from (iv) and Proposition 5.3 (i).
(vi). Let {Li } be a collection of nonzero left ideals of R whose sum is direct. Take an Li of this
family. If (R, R, Li ) = 0 then Li ⊆ U (R) and since R is semiprime, L2i 6= 0, hence Li := Qli
is a nonzero left ideal of Q, for some li ∈ Li . Suppose (R, R, Li ) 6= 0 and choose a nonzero
element ti ∈ (R, R, Li ). Define in this case Li :=Q (ti ]. We claim that the sum of all the
(nonzero) Li ’s is direct. Suppose on the contrary that there is a sum
X
(1)
i1
i1 t1 + . . . +
λqi1 . . . λqm
1
1
X
in
in tn + qn+1 ln+1 + . . . + qm lm = 0
λqin . . . λqm
n
1
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 6= 0 and ax ∈ R for every x ∈ X. Multiplying (1)
by a on the left side and applying Lemma 1.5 we have
X
i1
i1 t1 + . . . +
λqi1 . . . λaqm
1
1
X
in tn + (aqn+1 )ln+1 + . . . + (aqm )lm = 0
λqin . . . λaqm
n
1
in
with some summand (at) nonzero. Since X is a finite set, in a finite number of steps we
obtain
X
X
i1 t1 + . . . +
in tn + rn+1 ln+1 + . . . + rm lm = 0
λri1 . . . λrm
λrin . . . λrm
i1
1
1
in
1
n
14
M. GÓMEZ LOZANO AND M. SILES MOLINA
with all the appearing elements in R and some summand nonzero. But this contradicts
that the sum of the Li ’s is direct.
(vii) follows analogously to (vi).
(viii).
(1) Given x, y, p, q 1 , q 2 ∈ Qa such that (q 1 , q 2 , p x)a 6= 0, there exist n1 , n2 , n3 ∈ N (R) such
that [(n1 p)a]x, [(n1 p)a]y, n1 p, n2 q1 , n3 q2 , n1 y, n1 ay ∈ R and 0 6= (n1 p x, n2 q1 , n3 q2 )a .
Indeed, 0 6= (q1 , q 2 , p x)a = (q1 , q2 , (pa)x)a = ((pa)x, q1 , q2 )a = (by (3.1) and the right
Moufang identity) (p(axa), q1 a, q2 ) + (p(axa), q1 , a)q2 implies
0 6= t := a(p(axa), q1 a, q2 )a + a[(p(axa), q1 , a)q2 ]a.
Apply three times that Q is a left quotient ring of R to find n1 , n2 , n3 ∈ N (R) such that
n1 (axa), [(n1 p)a]x, n1 p, n2 q1 , n3 q2 , n1 y, n1 ay ∈ R and n1 n2 n3 t 6= 0. By Lemma 1.5,
n1 n2 n3 t = ρa λa [((n1 p)(axa), n2 q1 a, n3 q2 ) + ((n1 p)(axa), n2 q1 , a)(n3 q2 )]
= ρa λa ((n1 p) x, n2 q1 , n3 q2 )a
(by (3.1))
and this implies 0 6= (n1 p x, n2 q1 , n3 q2 )a .
(2) N (Ra ) ⊆ N (Qa ).
/ N (Qa ). Then, there exist q 1 , q 2 ∈ Qa such that (q 1 , q 2 , x)a 6=
Suppose x ∈ N (Ra ) but x ∈
0. Reasoning as in (1), we can find n1 , n2 ∈ N (R) satisfying: n1 q1 , n2 q2 ∈ R and
0 6= (x, n1 q1 , n2 q2 )a , which is a contradiction since x ∈ N (Ra ).
(3) Qa is a left quotient ring of Ra .
Let x, y be in Qa with x 6= 0. Suppose first x ∈ U (Qa ). Since Q is a left quotient
ring of R, there exists n ∈ N (R) such that n(axa) 6= 0 and nx, ny, n(axa), nax, nay,
n(aya) ∈ R. By nondegeneracy of R, [n(axa)]r[n(axa)] 6= 0 for some r ∈ R. Hence,
Ua Uxa (rn) = Ua Ux Ua (rn) 6= 0, which implies x rn x = Uxa (rn) 6= 0. In particular, rn x 6= 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 rn x ∈ U (Qa ) ∩ Ra , which implies rn x ∈ U (Ra ). Since Ra is
nondegenerate, U (Ra ) is nondegenerate too. Let z be in U (Ra ) such that rnxz rnx 6= 0.
Then z rn ∈ U (Ra ), 0 6= z rn x and z rn y ∈ Ra .
Suppose now x ∈
/ U (Qa ). Since Qa is nondegenerate (Proposition 3.2 (i)), by Lemma 1.6
there exist p, q 1 , q 2 ∈ Qa such that 0 6= (q 1 , q 2 , p x)a . By (1) we can find n1 , n2 , n3 ∈ N (R)
satisfying 0 6= r := (n1 p x, n2 q1 , n3 q2 )a and [(n1 p)a]x, n2 q1 , n3 q2 , [(n1 p)a]y, n1 y, n1 ay ∈
R. This implies r ∈ D(Ra ). Denote by L := Ra (r] ⊆ D(Ra ) and let l be a nonzero
2
element in Z(L), which exists by Lemma 1.2 (vi). By Lemma 1.2 (iv), l 6= 0, hence
0 6= l (n1 p x, n2 q1 , n3 q2 )a = (n1 p l x, n2 q1 , n3 q2 )a = (p ln1 x, n2 q1 , n3 q2 )a (since
n1 ∈ N (R) ⊆ N (Q) and by Lemma 1.5 and (3.1)).
Finally we note that the element ln1 ∈ N (Ra ) because for u, v ∈ Ra , (n1 l, u, v)a = (since
n1 ∈ N (R) ⊆ N (Q) and by (3.1) and Lemma 1.5) (l, n1 u, v)a = 0 because l ∈ Z(Ra ).
Moreover, ln1 x 6= 0 and ln1 y = [(ln1 )a]y = [(l(n1 a)]y = (l, n1 a, y) + l[(n1 a)y] =
(l, a, n1 y) + l(n1 ay) ∈ Ra . 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].
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
15
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 6= y ∈ L ∩ N (R)
such that lan(y) = 0.
Proof. In this proof we will use Lemma 1.2 (v) without mention it. Let a1 be a nonnilpotent element
of L ∩ N (R) with lan(a1 ) as large as possible. We have lan(a1 ) ⊆ lan(a21 ) and a21 is a nonnilpotent
element of L ∩ N (R), hence lan(a1 ) = lan(a21 ). If lan(a1 ) = 0 we would have finished. Otherwise
lan(a1 ) ∩ L 6= 0. Let a2 be a nonnilpotent element of lan(a1 ) ∩ L ∩ N (R) with lan(a2 ) as large as
possible. As before lan(a2 ) = lan(a22 ). It is easy to verify that the sum Ra1 + Ra2 is direct and
that lan(a1 + a2 ) = lan(a1 ) ∩ lan(a2 ). If lan(a1 + a2 ) = 0 we stop. If not, let a3 be a nonnilpotent
element of lan(a1 + a2 ) ∩ L ∩ N (R) with lan(a3 ) as large as possible. Again lan(a3 ) = lan(a23 ) and
Ra1 + Ra2 + Ra3 is a direct sum. Thus lan(a1 ) + lan(a2 ) + lan(a3 ) = lan(a1 ) ∩ lan(a2 ) ∩ lan(a3 ).
This process must stop because R does not contain infinite direct sums of nonzero left ideals, so
there are elements a1 , . . . , an in L ∩ N (R) such that lan(a1 + · · · + an ) = 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 6= 0. Then, given q there exist
a ∈ S ′ , b ∈ R such that q = a−1 b, hence, aq ∈ R and ap 6= 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 ρa1 . . . ρan q with ai ∈ 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 a1 , . . . , an ∈ R,
(ax)(ρa1 . . . ρan q) = (a, x, ρa1 . . . ρan q) + a[x(ρa1 . . . ρan q)]
= (x, ρa1 . . . ρan q, a) + a[x(ρa1 . . . ρan q)]
= [x(ρa1 . . . ρan q)]a − x(ρa ρa1 . . . ρan q) + a[x(ρa1 . . . ρan 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 lanR (s) = 0 if and only if s ∈ Inv(Q).
(iii) Every essential ideal I of R is a classical left order in Q.
16
M. GÓMEZ LOZANO AND M. SILES MOLINA
Proof.
(i). Consider a nonzero left ideal L of R and take an element 0 6= 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 a1 , . . . , an ∈ R, x(ρa1 . . . ρan q) = ρa1 . . . ρan (xq).
Now yq = 0 implies y[q)R = 0. Hence 0 6= y ∈ (R : q)∩L. If yq 6= 0, since Q is a left quotient
ring of R, there exists a ∈ N (R) such that 0 6= a(yq) ∈ R and therefore 0 6= (ay)[q)R ⊆ R.
So ay is a nonzero element in (R : q) ∩ L.
(ii). Take s ∈ N (R) such that lanR (s) = 0. By Lemma 5.5 , s ∈ Inv(Q) if and only if lanQ (s) =
0. Suppose 0 6= q ∈ lanQ (s). Given q, since Q is a left quotient ring of R, there exists an
element a ∈ N (R) such that 0 6= aq ∈ R. Then, (aq)s = a(qs) = 0 and hence 0 6= aq ∈
lanR (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 6= IL ⊆ I ∩ L.
(2) Reg(I) ∩ N (I) ⊆ Inv(Q).
Let y be in Reg(I) ∩ N (I). Then lanR (y) ∩ I = lanI (y) = 0 implies, by (1), lanR (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 6= s ∈ N (R) ∩ (R : q) ∩ I = N (I) ∩ (R : q) with
lanR (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 Ra of R at a is a classical left order in the
nondegenerate artinian alternative ring Qa .
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), Ra and Qa are nondegenerate and by Proposition 3.2
(v), Qa is artinian. Finally, since Qa is a left quotient ring for Ra (by Proposition 5.4 (viii))
we can apply Proposition 5.9 (iii) to obtain that Ra is a classical left order in Qa .
(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 letPaRb = 0 for some nonzero
n
elements a, b ∈ R. By simplicity of Q, QbQ = Q so a =
i=1 pi bqi with pi , qi ∈ Q.
Since Q is a left quotient ring of R, given a, pi , there exists c ∈ R such that ca 6= 0 and
cpi ∈ R. By nondegeneracy of R there exists r ∈ R such that arca 6= 0. Then 0 6= arca =
P
n
i=1 arcpi bqi ⊆ 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
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
17
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.
2
Proof. Take in [ZSSS, (7.18)] z = x, t = a, y = a♯ , x = a♯ . Since {a♯ a♯ a} = 2a♯ a = 2a♯
3
then (x, a, a♯ ) = 2(x, a♯ , a♯ ) + (x, a♯ aa♯ , a♯ ) = 3(x, a♯ , a♯ ) = 0 which implies, by the Generalized
3
Theorem of Artin, that the subring generated by {a, a♯ , 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 a2 x = a2 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 Q1 a ⊇ Q1 a2 ⊇ . . .
must terminate, Q1 an = Q1 an+1 = . . . for some natural number n. In particular, Q1 an = Q1 an+4
implies an = qan+3 for some q ∈ Q. Apply that a is right semiregular in Q to obtain a = qa4 .
Hence a3 = a2 qa4 and, again by right semiregularity of a, a = a2 qa2 ∈ a2 Qa2 . 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 Q1 a ⊇ Q1 a2 ⊇ . . . 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.
18
M. GÓMEZ LOZANO AND M. SILES MOLINA
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.
2
Proof. Take q ∈ Q and b ∈ S, y ∈ R such that q = b♯ y. Then q = b♯ (by). By Lemma 6.1 the
2
subring generated by {b, b♯ , y} is associative, so, if we take a = b♯ 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.
2
2
2
2
Take x = a, y = a♯ , z, t ∈ R in (2). Then (aa♯ , z, t) = a♯ (a, z, t) + (a♯ , z, t)a, implies
2
(a♯ , z, t) = (a♯ , z, t)a
(3)
for
2
z, t ∈ R.
2
2
2
Take x = a3 , y = a♯ , z, t ∈ R in (2). Then (a3 a♯ , z, t) = a♯ (a3 , z, t) + (a♯ , z, t)a3 implies
2
2
(a♯ , z, t)a3 = 0. Multiply by a♯ on the right side and take into account Lemma 6.1. Then
2
(a♯ , z, t)a = 0 and by (3),
(a♯ , z, t) = 0
(4)
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,
2
q ∈ Q.
2
2
2
For x = a ∈ S, y = a♯ , z = x ∈ R, t = q ∈ Q in (2), (aa♯ , x, q) = a♯ (a, x, q) + (a♯ , x, q)a and
taking into account (6),
2
(a♯ , x, q) = (a♯ , x, q)a.
(7)
2
2
Take x = a3 ∈ S, y = a♯ , z = x ∈ R, t = q ∈ Q in (2). Then, for x ∈ R, q ∈ Q, (a3 a♯ , x, q) =
2
2
2
2
a♯ (a3 , x, q) + (a♯ , x, q)a3 ; by (6) and multiplying by a♯ on the right side (a♯ , x, q)a = 0. Hence,
by (7)
(8)
(a♯ , x, q) = 0
for
x ∈ R,
q ∈ Q.
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
19
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,
(n, q, q ′ ) = 0
(9)
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.
2
Then, for every p, q ∈ Q, 0 = (a, p, q) = (a3 a♯ , p, q) = (since a ∈ N (R) ⊆ N (Q) and by [ZSSS,
2
2
2
Lemma 7.1]) a3 (a♯ , p, q). Multiply by a♯ on the left side and apply Lemma 6.1: 0 = a(a♯ , p, q) =
2
(a ∈ N (R) ⊆ N (Q) and [ZSSS, Lemma 7.1]) (aa♯ , 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 [FG1 , 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 q1 , . . . , qn ∈ Q there exist a ∈ S, b1 , . . . , bn ∈ R
such that for every i = 1, . . . , n, qi = a♯ bi and aa♯ bi = bi .
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 qi = x♯i yi , with xi ∈ N (R) and yi ∈ R. By Lemma 6.4, x♯i ∈ N (Q), and by
[AM2 , Theorem 5] (which is valid for weak Fountain-Gould left orders) and (1), there exist
a, ui ∈ N (R) satisfying x♯i = a♯ ui and aa♯ ui = ui for i = 1, . . . , n. Define bi = ui yi . Then
for every i ∈ {1, . . . , n}, qi = a♯ bi and aa♯ bi = bi . 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 SemiRegr (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 ps2 = qs2 but ps − qs 6= 0. Apply that Q is a left quotient ring of
R to find a ∈ N (R) satisfying a(ps − qs) 6= 0 and ap, aq ∈ R. Now, ps2 − qs2 = 0 implies
0 = a(ps2 −qs2 ) = (ap−aq)s2 . Since ap, aq ∈ R, and s ∈ SemiRegr (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.
20
M. GÓMEZ LOZANO AND M. SILES MOLINA
Proof.
(i). By Lemma 6.4, N (R) ⊆ N (Q). Given p, q ∈ Q with p 6= 0, by the common denominator
property there exist b ∈ S, a1 , a2 ∈ R satisfying p = b♯ a1 , q = b♯ a2 and bb♯ ai = ai . Then
bp 6= 0 and bq ∈ R.
(ii). By Lemma 6.6, S ⊆ SemiRegr (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 = va2 .
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) = ⊕Qα where Qα is a simple associative ring coinciding
with its socle or a Cayley-Dickson algebra.
Denote by πα : Q → Qα the canonical projection.
By Lemma 6.8, for every α, πα (R) is a Fountain-Gould left order in Qα . If Qα is associative,
then πα (R) is strongly prime by [FG2 , Proposition 3.6]. If Qα is a Cayley-Dickson algebra then
πα (R) is a classical left order in Qα , by Proposition 6.7 (iii). Moreover πα (R) is strongly prime:
suppose a, b ∈ πα (R) such that aπα (R)b = 0 and take an arbitrary element q = u−1 v ∈ Qα with
u ∈ N (πα (R)) ⊆ N (Q) = Z(Q) and v ∈ πα (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 πα (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 Ra of R at a is
a classical left order in the nondegenerate artinian alternative ring Qa .
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, Qa , is a left quotient ring of Ra . 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 FountainGould 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, Ra is a classical left
order in Qa . Denote by I the set { y ∈ Ra | y ∈ I}. It is easy to see that I is an ideal of Ra .
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
21
Moreover, it is an essential ideal. Indeed, let L be a nonzero ideal of Ra and choose 0 6= l ∈ N (L),
which is possible by virtue of Lemma 1.2 (v), taking into account that Ra is nondegenerate by
Proposition 3.2 (i). Then, since 0 6= ala and I is an essential ideal of R, by Proposition 1.1 (iv)
and (vi), alaIala 6= 0. In particular, there exists 0 6= y ∈ I such that ayala 6= 0 and hence we have
found a nonzero element yal = y l in I ∩ L.
By condition (iii) in Proposition 5.9, I is a classical left order in the ring Qa . Hence I contains
an invertible element of Qa , say u, with u ∈ N (I). Then p = u−1 for some p ∈ N (Qa ), and
2
apaua = auapa = a, which implies (multiplying by a♯ on the left hand side and taking into
account a = a♯ a2 ), a♯ = a♯ pa♯ a2 ua. An easy computation shows that a♯ pa♯ = (a2 ua2 )♯ . Finally,
q = a♯ b = a♯ pa♯ a2 uab = (a2 ua2 )♯ (a2 uab), with a2 ua2 ∈ N (I) and a2 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, Ra , 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, ⊕Mα ⊳ R ≤ ⊕R/ann(Mα ), where Mα 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 Rα := R/ann(Mα ) is a Fountain-Gould left order in a (simple)
alternative ring Qα coinciding with its socle then, by Proposition 6.11, Mα , regarded as an ideal
of Rα , is also a left order in Qα . Since direct sums preserves Fountain-Gould left orders, ⊕Mα is
a Fountain-Gould left order in the nondegenerate alternative ring coinciding with its socle ⊕Qα ,
and hence R is also a Fountain-Gould left order in ⊕Qα .
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 [AM1 ,
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 Ra is nondegenerate by Proposition 3.2 (i) Since Qa is left nonsingular and a left quotient ring of Ra (Lemma 5.5), Ra 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.
22
M. GÓMEZ LOZANO AND M. SILES MOLINA
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, Ra , is a classical left order in the semisimple artinian alternative
ring Qa ,
(iv) for every finite subset Y of Q there exists an element a ∈ S such that Y ⊆ aQa and Ra is
a classical left order in the semisimple artinian alternative ring Qa ,
(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 = ⊕Qα , where the Qα ’s are simple alternative rings and
Qα = Soc(Qα ). Let πα be the canonical projection πα : Q → Qα and define Iα := R ∩ Qα .
(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
2
q = a♯ b = aa♯ b ∈ N (R)QR. The other assertion is Proposition 6.10.
(ii) ⇒ (iv). By Lemma 6.8, πα (R) is a Fountain-Gould left order in Qα and by Theorem 7.1 (1),
πα (R) is a prime ring. Hence every nonzero ideal of πα (R), in particular Iα , is essential in πα (R)
(notice that Iα 6= 0 since Q is a left quotient ring of R, by Proposition 5.3 (i)) and by Proposition
6.11, Iα is a Fountain-Gould left order in Qα .
If Qα is associative then (iv) is satisfied for Iα and Qα by [GS, Theorem 4.12] and if Qα is a
Cayley-Dickson algebra over its center then Qα = xQα x for every nonzero element x ∈ N (Iα ) and
by Proposition 6.10, (Iα )x is a classical left order in (Qα )x .
Take q ∈ Q and write q = q1 + . . . qn with qi ∈ Qαi . As we have proved, there exist xi ∈
S(Iαi ) ⊆ SPsuch that qi ∈ xi Qαi xi = xi Qxi and (Iαi )xi = Rxi is a classical left order in Qxi .
n
Write x = i=1 xi . Then q ∈ Qx and Rx is a classical left order in Qx . 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) πα (N (R)) 6= 0 and Iα 6= 0 for every α.
P
Let 0 6= qα ∈ Qα . Since Q = N (R)QR, qα = r i pi si , and applying πα , 0 6= qα = πα (qα )
P
= πα (r i )πα (pi )πα (si ), which implies πα (r i ) 6= 0 for some r i ∈ N (R).
Take r ∈ N (R), with πα (r) 6= 0. Since Q is nondegenerate, 0 6= πα (r)Qπα (r) = rQα r.
Let 0 6= rpα r ∈ rQα r. Apply that Rr is a classical left order in Qr to find a ∈ Reg(Rr ) ∩
N (Rr ), b ∈ Rr such that pα = (a)−1 b. Then 0 6= b = a pα = arpα , that is, 0 6= rbr =
rarpα r ∈ R ∩ Qα = Iα .
(2) For every nonzero u ∈ πα (R) the ring (πα (R))u is a classical left order in (Qα )u .
Take r ∈ R satisfying πα (r) = u. By hypothesis, Rr is a classical left order in Qr . Notice
that the following is a ring epimorphism:
ϕ:
Qr
q
−→
7→
Qu
^
π
α (q)
Hence, by the analogue result to Lemma 6.8 for classical left orders, ϕ(Rr ) = (πα (R))u is
a classical left order in Qu .
(3) In order to simplify the notation, we will write R = πα (R) and Q = Qα . 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.
LEFT QUOTIENT RINGS OF ALTERNATIVE RINGS
23
Suppose Q a Cayley-Dickson algebra over its center. Then N (Q) = Z(Q) is a field. By
(1) there exists 0 6= u ∈ πα (N (R)) ⊆ πα (N (Q)) ⊆ N (Q). By (2), given q ∈ Q = uQu,
there exist a ∈ Reg(Ru ) ∩ N (Ru ) ⊆ N (R) ⊆ N (Q), b ∈ Ru , such that u−1 q = a−1 b. It
is not difficult to see that the map
ψ:
Q
q
−→ Qu
7→ u−1 q
is a ring isomorphism, so u−1 q = a−1 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 Iα is a Fountain-Gould left order in Q. This
implies that ⊕Iα is a weak Fountain-Gould left order in ⊕Qα and since ⊕Iα ⊆ R ⊆ Q =
⊕Qα 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 Ra is a classical left order in the nondegenerate artinian alternative ring Qa ,
(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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