On solvable Z₃-graded alternative algebras
Let A=A₀⊕A₁⊕A₂ be an alternative Z₃-graded algebra. The main result of the paper is the following: if A0 is solvable and the characteristic of the ground field not equal 2,3 and 5, then A is solvable.
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Інститут прикладної математики і механіки НАН України
2015
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| Цитувати: | On solvable Z₃-graded alternative algebras / M. Goncharov // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 203-216. — Бібліогр.: 15 назв. — англ. |
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irk-123456789-1551402019-06-17T01:27:43Z On solvable Z₃-graded alternative algebras Goncharov, M. Let A=A₀⊕A₁⊕A₂ be an alternative Z₃-graded algebra. The main result of the paper is the following: if A0 is solvable and the characteristic of the ground field not equal 2,3 and 5, then A is solvable. 2015 Article On solvable Z₃-graded alternative algebras / M. Goncharov // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 203-216. — Бібліогр.: 15 назв. — англ. 1726-3255 http://dspace.nbuv.gov.ua/handle/123456789/155140 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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English |
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Let A=A₀⊕A₁⊕A₂ be an alternative Z₃-graded
algebra. The main result of the paper is the following: if A0 is
solvable and the characteristic of the ground field not equal 2,3
and 5, then A is solvable. |
| format |
Article |
| author |
Goncharov, M. |
| spellingShingle |
Goncharov, M. On solvable Z₃-graded alternative algebras Algebra and Discrete Mathematics |
| author_facet |
Goncharov, M. |
| author_sort |
Goncharov, M. |
| title |
On solvable Z₃-graded alternative algebras |
| title_short |
On solvable Z₃-graded alternative algebras |
| title_full |
On solvable Z₃-graded alternative algebras |
| title_fullStr |
On solvable Z₃-graded alternative algebras |
| title_full_unstemmed |
On solvable Z₃-graded alternative algebras |
| title_sort |
on solvable z₃-graded alternative algebras |
| publisher |
Інститут прикладної математики і механіки НАН України |
| publishDate |
2015 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/155140 |
| citation_txt |
On solvable Z₃-graded alternative algebras / M. Goncharov // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 203-216. — Бібліогр.: 15 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT goncharovm onsolvablez3gradedalternativealgebras |
| first_indexed |
2025-07-14T07:13:59Z |
| last_indexed |
2025-07-14T07:13:59Z |
| _version_ |
1837605574347325440 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 20 (2015). Number 2, pp. 203–216
© Journal “Algebra and Discrete Mathematics”
On solvable Z3-graded alternative algebras
Maxim Goncharov
Communicated by V. M. Futorny
Abstract. Let A = A0 ⊕ A1 ⊕ A2 be an alternative Z3-
graded algebra. The main result of the paper is the following: if A0
is solvable and the characteristic of the ground field not equal 2,3
and 5, then A is solvable.
1. Introduction
Let R be an algebra over a field F. Let G be a finite group of auto-
morphisms of R, and RG = {x ∈ R|φ(x) = x for all φ ∈ G} be a fixed
points subalgebra of R.
For Lie algebras there is a classical Higman result: if a Lie algebra L
has an automorphism φ of simple order p without fixed points (Lφ = 0),
then L is nilpotent [1]. Moreover, nil index h(p) in this case depends only
on the order p. The explicit estimate of the function h(p) was found in
the paper of Kreknin and Kostrikin [2]. At the same time Kreknin proved
that a Lie ring with a regular automorphism of an arbitrary finite order
is solvable [3]. It is also worth mentioning here a result of Makarenko [4]
who proved that if a Lie algebra L admits an automorphism of a prime
order p with a finite-dimensional fixed-point subalgebra of dimension t,
then L has a nilpotent ideal of nilpotency class bounded in terms of p
and of codimension bounded in terms of t and p.
If R is an associative algebra with a finite group of automorphisms G
then classical Bergman-Isaacs theorem says that if the subalgebra of fixed
Key words and phrases: alternative algebra, solvable algebra, Z3-graded algebra,
subalgebra of fixed points.
204 On solvable Z3-graded alternative algebras
points RG is nilpotent and R has no |G|-torsion, then R is nilpotent [5].
Kharchenko proved that under the same conditions, if RG is a PI-ring,
then R is a PI-ring [6]. For Jordan algebras the analogue of Kharchenko’s
result was proved by Semenov [7].
The Bergman-Isaacs theorem was partially generalized by Martindale
and Montgomery to the case when G is a finite group of so called Jordan
automorphisms, that is a linear automorphisms that are automorphisms
of the adjoint Jordan algebra R(+) (note that in this case RG is not a
subalgebra in R, but a subalgebra in R(+)) [8].
Note, that in general for Jordan algebras Bergman-Isaacs theorem is
false - there is an example of a solvable non-nilpotent Jordan algebra J
with an automorphism of second order φ such that the ring of invariants
Jφ is nilpotent. However, Zhelyabin in [11] proved, that if a Jordan algebra
J over a field of characteristic not equal 2,3 admits an automorphism of
second order φ such that the algebra of invariants Jφ is solvable, then J
is solvable.
For alternative algebras in [12] it was proved that if A is an alternative
algebra over a field of characteristic not equal 2 with an automorphism
g of second order then the solvability of the algebra of fixed points Ag
implies the solvability of A. On the other hand, if the characteristic of
the ground field is zero and G is a finite group of automorphisms of an
alternative algebra A, then again the solvability of the algebra of fixed
points AG implies the solvability of A [7]. At the same time it is not
known if the similar result is true in positive characteristic.
In this work we study a special case of the problem for alternative
algebras: we consider a Z3-graded alternative algebra A = A0 ⊕ A1 ⊕ A2
and prove, that if the characteristic of the ground field not equal 2,3 and
5 and A0 is solvable, then A is solvable.
As a consequence we obtain the following result: if A is an alternative
algebra with an automorphism φ of order 2k3l, then under the same
conditions on the characteristic of the ground field, the solvability of the
subalgebra of fixed points Aφ implies the solvability of A.
2. Definitions and preliminary results
Let F be a field of characteristic not equal 2,3,5, A be an algebra
over F . If x, y, z ∈ A then (x, y, z) = (xy)z − x(yz) is the associator of
elements x, y, z, x ◦ y = xy + xy is a Jordan product of elements x and y
and [x, y] = xy − yx is a commutator of the elements x and y.
M. Goncharov 205
Definition. An algebra A is called Z3-graduated if A is a direct sum of
subspaces Ai, i ∈ Z3 : A = A0 ⊕ A1 ⊕ A2 and AiAj ⊆ Ai+j .
If A is a Z3-graded algebra, then for every i ∈ Z3 and x ∈ A by xi
we will denote the projection of the element x to the subspace Ai and if
M ⊂ A then Mi = {xi| x ∈ M}. An ideal I of A is called homogeneous
if Ij ⊂ I, j = 0, 1, 2 . If I is a homogeneous ideal of A, then the
factor-algebra A/I is also a Z3-graded algebra.
If φ is an automorphism of the algebra A, then by Aφ we denote the
subalgebra of fixed points of φ, that is Ag = {x ∈ A| φ(x) = x}.
Define subsets Ai, A<i> and A(i) as:
A2 = A<2> = A(1) = AA, An =
n−1∑
i=1
AiAn−i, A<n> = A<n−1>A
A(1) = A2, A(i) = A(i−1)A(i−1).
Definition. An algebra A is called nilpotent if Ai = 0 for some i. An al-
gebra is called solvable, if A(i) = 0 for some i.
It is clear that A(i) ⊂ A2i
, so every nilpotent algebra is solvable. If
A is an associative algebra then the inverse is also true: every solvable
associative algebra is nilpotent. But in general, a solvable algebra is not
necessary nilpotent. An example of an alternative solvable non-nilpotent
algebra was constructed by Dorofeev [15](can also be found in [13]).
Definition. An algebra A is called alternative, if for all x, y ∈ A:
(x, x, y) = (y, x, x) = 0. (1)
Let A be an alternative algebra. We will need the following identities
on A (that are the linearizations of the well-known Moufang identities):
(x1, x2y, z) + (x2, x1y, z) = (x1, y, z)x2 + (x2, y, z)x1. (2)
(x1, yx2, z) + (x2, yx1, z) = x1(x2, y, z) + x2(x1, y, z). (3)
(x1 ◦ x2, y, z) = (x1, x2y + yx2, z) + (x2, x1y + yx1, z). (4)
(x1 ◦ x2, y, z) = (x1, y, z) ◦ x2 + (x2, y, z) ◦ x1. (5)
Also, in A the following equalities hold([13]):
2[(a, b, c), d] = ([a, b], c, d) + ([b, c], a, d) + ([c, a], b, d), (6)
206 On solvable Z3-graded alternative algebras
(dx, y, z) + (d, x, [y, z]) = d(x, y, z) + (d, y, z)x. (7)
Let D(A) be the associator ideal of A, that is an ideal generated by
all associators (x, y, z), x, y, z ∈ A. In [13] it was shown that
D(A) = (A, A, A) + (A, A, A)A = (A, A, A) + A(A, A, A), (8)
where (A, A, A) = {
∑
i
(xi, yi, zi)| xi, yi, zi ∈ A}.
Let J2(A)={
∑
i
αia
2
i |αi ∈F, a∈A} and J6(A)={
∑
i
αia
6
i |αi ∈F , a∈A}.
Suppose A is an alternative algebra, then if char(F ) 6= 2 then J2(A) is
an ideal of A and if char(F ) 6= 2, 3, 5 then J6(A) is also an ideal in A(see,
for example, [13]).
3. Properties of Z3-graded alternative algebras
In this section we will get some technical results that we will need.
Throughout this section A = A0 ⊕ A1 ⊕ A2 is an arbitrary alternative
Z3-graded algebra.
Lemma 1.
1)
(A2
0, A1, A2) ⊂ A2
0 (9)
2) For every x ∈ A1, y ∈ A2, a1, a2 ∈ A0:
(x(a1a2))y = x((a1a2)y)+a′, (y(a1a2))x = y((a1a2)x)+a′′ (10)
for some a′, a′′ ∈ A2
0.
3)
(A0A1)(A2
0A2) ⊂ A2
0, (A0A2)(A2
0A1) ⊂ A2
0 (11)
(A1A2
0)(A2A0) ⊂ A2
0, (A2A2
0)(A1A0) ⊂ A2
0. (12)
Proof. Let x ∈ A1, y ∈ A2 and a1, a2 ∈ A0. Then using (7) we get:
(a1a2, x, y) = −(a1, a2, [x, y]) + a1(a2, x, y) + (a1, x, y)a2 ⊂ A2
0.
And (9) is proved. It is easy to see that (10) follows from (9).
Let us prove (11) and (12). It is easy to see that they are similar and it
is enough to prove one of these inclusions. Using (7) and (9) we compute:
(A0A1)(A2
0A2) ⊂ A0(A1(A2
0A2)) + (A0, A1, A2
0A2)
⊂ A2
0 + (A0, A1, A2
0)A2 + A2
0(A0, A1, A2) + (A1, A2
0, A2)
⊂ A2
0 + ((A2
0)A1)A2 ⊂ A2
0.
M. Goncharov 207
Remark. From (10) it is follows that A2
0 + (A1A2
0)A2 = A2
0 + A1(A2
0A2)
and A2
0 + (A2A2
0)A1 = A2
0 + A2(A2
0A1). This allows us to omit brackets
in such a sentences without ambiguity.
Lemma 2.
(D(A))1 ⊆ A0A1 + (A1, A2, A1) + (A2, A0, A2). (13)
(D(A))2 ⊆ A0A2 + (A2, A1, A2) + (A1, A0, A1). (14)
Proof. It is enough to prove one of these equations. Let us prove (13).
Using (8) we have:
(D(A))1 ⊆ (A1, A1, A1)A1 + (A0, A0, A0)A1 + (A2, A2, A2)A1
+ (A1, A0, A0)A0 + (A1, A2, A1)A0 + (A2, A2, A0)A0
+ (A1, A1, A0)A2 + (A2, A2, A1)A2 + (A0, A2, A0)A2
+ (A0, A0, A1) + (A1, A2, A1) + (A2, A0, A2).
Using (6) we get:
(A1, A0, A0)A0 ⊆ A0A1 + [A0, (A1, A0, A0)]
⊆ A0A1 + (A0, A1, A0) ⊆ A0A1.
Similarly, we obtain that
(A1, A2, A1)A0 + (A2, A2, A0)A0 ⊆ A0A1 + (A2, A0, A2).
By (2) we compute:
(A1, A1, A0)A2 ⊆ (A2, A1, A0)A1 + (A1, A0, A0) + (A1, A2, A1)
⊆ A0A1 + (A1, A2, A1),
(A2, A2, A1)A2 ⊆ A0A1 + (A2, A0, A2) + (A1, A2, A1).
And, finally, using (1) we obtain the following inclusion:
(A0, A2, A0)A2 ⊆ A0A1 + (A2, A0, A2).
Summing up the obtained inclusions we finally have that:
(D(A))1 ⊆ A0A1 + (A1, A2, A1) + (A2, A0, A2).
Lemma 3.
208 On solvable Z3-graded alternative algebras
1)
(A2
0A1, A0, A2) ⊂ A2
0, (A2
0A2, A0, A1) ⊂ A2
0. (15)
2)
((A1, A2, A1) + (A2, A0, A2))A2
0(A1 ◦ A1) ⊂ A2
0. (16)
((A2, A1, A2) + (A1, A0, A1))A2
0(A2 ◦ A2) ⊂ A2
0. (17)
3) For all n > 2:
(A1A<n>
0 )(A0A2) ⊂ A1(A<n+1>
0 )A2 + A2
0, (18)
(A2A<n>
0 )(A0A1) ⊂ A2(A<n+1>
0 )A1 + A2
0. (19)
Proof. It is easy to see that it is enough to prove only one inclusion in
every statement. We will prove the first inclusion in all cases.
By (7) we have:
(A2
0A1, A0, A2) ⊂ A2
0(A1, A0, A2) + (A2
0, A0, A2)A1 + (A2
0, A1, A2)
⊂ A2
0 + (A2
0A2)A1 ⊂ A2
0.
And (15) is proved.
Using (6), (9) and (15) we compute:
((A1, A2, A1) + (A2, A0, A2))A2
0(A1 ◦ A1)
⊂ A2
0 + [(A1, A2, A1) + (A2, A0, A2), A2
0(A1 ◦ A1)]
⊂ A2
0 + (A2
0(A1 ◦ A1), A0, A1) + (A2
0(A1 ◦ A1), A2, A2)
⊂ A2
0 + (A2
0(A1 ◦ A1), A2, A2).
Using (2) and (4) we have:
(A2
0(A1 ◦ A1), A2, A2) ⊂ (A1, A2
0, A2) + A2
0 + (A1 ◦ A1, A2
0, A2)A2
⊂ A2
0 + (A1, A2
0, A0)A2 ⊂ A2
0.
Thus, ((A1, A2, A1) + (A2, A0, A2))A2
0(A1 ◦ A1) ⊂ A2
0.
Let us prove (18). Using (9) and (15) we get:
(A1A<n>
0 )(A0A2) ⊂ ((A1A<n>
0 )A0)A2 + (A1A<n>
0 , A0, A2)
⊂ A2
0 + (A1A<n+1>
0 )A2 + (A1, A<n>
0 , A0)A2
⊂ A2
0 + A1(An+1
0 A2).
M. Goncharov 209
Lemma 4.
1) Let char(F ) 6= 2. Then A is solvable if and only if J2(A) is solvable.
2) Let char(F ) 6= 2, 3, 5. Then A is solvable if and only if J6(A) is
solvable.
Proof. The proof is similar for both cases. Let us prove 2.
If A is solvable then clearly J6(A) is solvable.
Suppose J6(A) is solvable. Consider the factor algebra A = A/J6(A).
Then for every x in A: x6 = 0, that is A is a nil algebra of nil-index 6.
Since the characteristic of the ground field F not equal 2,3 or 5, then by
Zhevlakov’s theorem A is solvable ([14], the proof can also be found in
[13]). Thus, A is solvable.
Lemma 5. Let A be a Z3-graded alternative algebra over a field F of
characteristic not equal 2,3,5. Then we have the following inclusions:
1) (J2(A))1 ⊂ A0 ◦ A1 + A2 ◦ A2, (J2(A))2 ⊂ A0 ◦ A2 + A1 ◦ A1.
2) (J6(A))0 ⊂ A2
0 + A1A2
0A2 + A2A2
0A1.
Proof. The first assertion is obvious.
Let us prove 2. We will use the following notation: if u, v ∈ A then
u ≡ v means that u − v ∈ A2
0 + A1A2
0A2 + A2A2
0A1
Let x∈A1, y ∈A2, a∈A0. It is sufficient to prove that ((x+y+a)6)0 ≡0.
First we will proof the following inclusion:
x(y, x, a)x2 + x2(y, x, a)x ∈ A2
0 (20)
Indeed, using (5) and (2) we have
A2
0 ∋ 2(xy, x3, a) = (xy, x2, a) ◦ x + (xy, x, a) ◦ x2
= x(xy, x, a)x + (xy, x, a)x2 + (xy, x, a) ◦ x2
= x(y, x, a)x2 + 2(y, x, a)x3 + x2(y, x, a)x.
Thus, x(y, x, a)x2 + x2(y, x, a)x ∈ A2
0. Similarly, one can prove the follow-
ing inclusion:
y(x, y, a)y2 + y2(x, y, a)y ∈ A2
0. (21)
Consider p = (x + y + a)3. Then we have:
p0 = x3 + y3 + a3 + (x ◦ a)y + (y ◦ a)x + (x ◦ y)a,
p1 = x2y + (x ◦ y)x + y2a + (a ◦ y)y + a2x + (a ◦ x)a,
p2 = y2x + (y ◦ x)y + x2a + (a ◦ x)x + a2y + (a ◦ y)a.
210 On solvable Z3-graded alternative algebras
Since ((x + y + a)6)0 = p2
0 + p1 ◦ p2, then it is enough to proof that:
(x2y + (x ◦ y)x) ◦ (y2x + (y ◦ x)y) ≡ 0, (22)
(y2a + (a ◦ y)y) ◦ (y2x + (y ◦ x)y) ≡ 0, (23)
(a2x + (a ◦ x)a) ◦ (y2x + (y ◦ x)y) ≡ 0, (24)
(x2y + (x ◦ y)x) ◦ (x2a + (a ◦ x)x) ≡ 0, (25)
(y2a + (a ◦ y)y) ◦ (x2a + (a ◦ x)x) ≡ 0, (26)
(a2x + (a ◦ x)a) ◦ (x2a + (a ◦ x)x) ≡ 0, (27)
(x2y + (x ◦ y)x) ◦ (a2y + (a ◦ y)a) ≡ 0, (28)
(y2a + (a ◦ y)y) ◦ (a2y + (a ◦ y)a) ≡ 0, (29)
(a2x + (a ◦ x)a) ◦ (a2y + (a ◦ y)a) ≡ 0. (30)
The equivalences (22),(27) and (29) are obvious. Let us prove (23).
We have:
(y2a)(y2x) + (y2x)(y2a)
= (y2ay)(yx) − (y2a, y, yx) + ((y2x)y2)a − (y2x, y2, a)
≡ −(a, y, x)y3 − y(y2x, y, a) − (y2x, y, a)y ≡ y(y, x, a)y2,
(y2a)((y ◦ x)y) + (y ◦ x)y)(y2a)
= ((y2a)(y ◦ x)))y + (y2a, y ◦ x, y) + (y ◦ x)(y3a) + (y ◦ x, y, y2a)
≡ (y2(a(y ◦ x)))y + (y2, a, y ◦ x)y + (y2a, y ◦ x, y) + (y ◦ x, y, y2a)
≡ y(y, a, x)y2 + 2(y ◦ x, y, a)y2 = 3y(y, a, x)y2,
((a ◦ y)y)(y2x) + (y2x)((ay + ya)y)
= ((a ◦ y)y2)(yx) + ((a ◦ y)y, y, yx) + y((yx)(ay2)) + (y, yx, ay2)
+ (y2xy)(ay) − (y2x, y, ay)
≡ y((yx)a)y2 − y(yx, a, y2) + y2(y, x, a)y + ((y2xy)a)y
− (y2xy, a, y) − y(x, y, a)y2
≡ y2(y, x, a)y + y2((xy)a)y + (y2, xy, a)y−y(x, a, y)y2−y(x, y, a)y2
≡ y2(y, x, a)y,
((a ◦ y)y)((y ◦ x)y)) + ((y ◦ x)y))((a ◦ y)y)
= (ay2)((y ◦ x)y) + (yay)(yxy + xy2) + (y ◦ x)(y(a ◦ y)y)
+ (y ◦ x, y, (a ◦ y)y)
M. Goncharov 211
≡ a(y2((y ◦ x)y)) + (a, y2, (y ◦ x)y) + (yay2)(xy) − (yay, y, xy)
+ ((yay)x)y2 − (yay, x, y2) + y2(x, y, a)y
≡ 3y2(x, y, a)y + y((ay)x)y2 + (y, ay, x)y2
≡ 3y2(x, y, a)y + y(a(yx))y2 + y(a, y, x)y2 + y(y, a, x)y2
≡ 3y2(x, y, a)y.
Summing up the obtained equations we have:
(y2a + (a ◦ y)y) ◦ (y2x + (y ◦ x)y)
≡ y(y, x, a)y2 + 3y(y, a, x)y2 + y2(y, x, a)y + 3y2(x, y, a)y ≡ 0.
That proves (23). Using similar arguments one can obtain (25).
Let us prove (24):
(a2x) ◦ (y2x + (y ◦ x)y)
≡ a2(xy2x + (y ◦ x)y) + (a2, x, y2x + (y ◦ x)y) ≡ 0,
((a ◦ x)a) ◦ (y2x + (y ◦ x)y)
= (xa2) ◦ (y2x + (y ◦ x)y) + (axa)(y2x + (y ◦ x)y)
+ (y2x + (y ◦ x)y)(axa)
≡ a((xa)(y2x + (y ◦ x)y)) + (a, xa, (y2x + (y ◦ x)y))
+ ((y2x + (y ◦ x)y)(ax))a − ((y2x + (y ◦ x)y), ax, a)
≡ a(a, x, (y2x + (y ◦ x)y)) − ((y2x + (y ◦ x)y), x, a)a ≡ 0.
Thus, (a2x) ◦ (y2x + (y ◦ x)y) + ((a ◦ x)a) ◦ (y2x + (y ◦ x)y) ≡ 0 and
(24) is proved. Similarly, one can prove (28).
Consider (26). We have:
(y2a) ◦ (x2a) = ((y2a)x2)a − (y2a, x2, a) + ((x2a)y2)a − (x2a, y2, a)
≡ −a(y2a, x2, a) − a(x2, y2, a) ≡ 0.
Similarly, (y2a) ◦ (ax2) + (ay2) ◦ (ax2) + (ay2) ◦ (x2a) ≡ 0. Further, we
compute:
(y2a)(xax)+(ay2)(xax) = y2(axax)+(y2, a, xax)+a(y2(xax))+(a, y2, xax)
≡ y(((y(ax))a)x)−y(y(ax), a, x)≡−y(y(ax), a, x).
Using (7) and (9) we get:
−y(y(ax), a, x) = (y(ax), a, y)x − (y(ax), a, yx) − ([y(ax), a], y, x)
≡ (y(ax), a, y)x = ((ax, a, y)y)x = (((x, a, y)a)y)x ≡ 0.
212 On solvable Z3-graded alternative algebras
Using similar computations one can prove that (xax)(y2a)+(xax)(ay2)≡0.
Finally,
(yay)(xax) + (xax)(yay)
= ((yay)x)(ax) − (yay, x, ax) + x((ax)(yay)) + (x, ax, yay)
≡ ((ya)(yx))(ax) + (ya, y, x)(ax) + x(((ax)y)(ay)) − x(ax, y, ay)
≡ (y(a(yx)))(ax) + (y, a, yx)(ax) + (ya, y, x)(ax) + x((((ax)y)a)y)
− x((ax)y, a, y) − x(ax, y, ay)
≡ ((y, a, x)y)(ax) + ((a, y, x)y)(ax) − x(y(ax, a, y))
− x(y(ax, y, a)) = 0.
And (26) is proved. The equality (30) can be proved in a similar
way.
4. The main part
Recall that if A is an algebra, then by D(A) we denote the ideal
generated by associators. Define subalgebras Ki and Ti as
K1 := J2(A), T1 := D(K1), Ki := J2(Ti−1), Ti := D(Ki).
It is easy to see that:
A ⊇ K1 ⊇ T1 ⊇ K2 ⊇ ... ⊇ Ki ⊇ Ti ⊇ ...
Lemma 6. If for some i > 1 Ti or Ki is solvable, then A is solvable.
Proof. By lemma 4 A is solvable if and only if J2(A) = K1 is solvable.
Since D(K1) is a homogeneous ideal, then K1/D(K1) - is an associative
Z3-graded algebra with a solvable even part. Thus, by Bergman-Isaacs
theorem K1/D(K1) is nilpotent and if D(K1) is solvable, then A is
solvable.
Similar arguments show that Ki and Ti are solvable if and only if Ti−1
is solvable.
Lemma 7. Let A be a Z3-graded algebra and A0 = 0. If char F 6= 2, 3,
then A is solvable.
Proof. Consider J3(A) = {
∑
i
x3
i |xi ∈ A}. Using similar arguments as in
lemma 4 we get, that A is solvable if and only if J3(A) is solvable. For all
x ∈ A1 and y ∈ A2 we have:
(x + y)3 = x3 + y3 + x2y + yx2 + xyx + y2x + xy2 + yxy.
M. Goncharov 213
But x3 ∈ A0, y3 ∈ A0 and xy ∈ A0. Thus (x+y)3 = 0 and J3(A) = 0.
Theorem 1. Let A be a Z3-graded alternative algebra over a field F .
If A0 is solvable and char F 6= 2, 3, 5, then A is solvable.
Proof. Let A
(m)
0 = 0 and n = 2m. Consider Tn and define I = J6(Tn).
By lemmas 4 and 6 it is enough to prove that I is solvable. By lemma 5
we have I0 ⊂ A2
0 + (Tn)2(A2
0)(Tn)1 + (Tn)1(A2
0)(Tn)2.
Our aim now is to prove that
(Tn)1(A2
0)(Tn)2 ⊂ A2
0 + (Tn−1)1A<3>
0 (Tn−1)2. (31)
Indeed, since Tn ⊂ Kn = J2(Tn−1) then by lemma 5:
(Tn)2 ⊂ A0 ◦ (Tn−1)2 + (Tn−1)1 ◦ (Tn−1)1.
By (12) and (18) we have that
(Tn)1A2
0(A0 ◦ (Tn−1)2) ⊂ (Tn−1)1A<3>
0 (Tn−1)2 + A2
0.
Using inclusion (13) we get:
(Tn)1 = (D(Kn))1
⊆ A0(Kn)1 + ((Kn)1, (Kn)2, (Kn)1) + ((Kn)2, A0, (Kn)2).
And now it is left to use inclusions (11) and (16) to prove (31). Similar
reasons shows us that (Tn)2(A2
0)(Tn)1 ⊂ A2
0 + (Tn−1)2A<3>
0 (Tn−1)1 and
we may conclude that
I0 ⊂ A2
0 + (Tn−1)1A<3>
0 (Tn−1)2 + (Tn−1)2A<3>
0 (Tn−1)1.
Now we can continue to use similar arguments and get that
I0 ⊂ A2
0 + (Tn−2)1A<4>
0 (Tn−2)2 + (Tn−2)2A<4>
0 (Tn−2)1.
And finally, we will get that
I0 ⊂ A2
0 + A1A<n>
0 A2 + A2A<n>
0 A1. (32)
Let us prove that A1A<n>
0 A2 ⊂ A2
0. For this we will prove that for all
k > 2:
A1(Ak
0, A0, A0)A2 ⊂ A2
0. (33)
214 On solvable Z3-graded alternative algebras
Indeed, using (2) and (9) we have:
A1(Ak
0, A0, A0)A2 ⊂ A1((A2, A0, A0)Ak
0) + A1(Ak
0, A2, A0) ⊂ A2
0.
Moreover, from (18) and (33) we see that
A1(((...((A2
0, A0, A0)A0)A0)...A0)A2 ⊂ A2
0. (34)
Now we can use (34) to obtain the following inclusions:
A1A<n>
0 A2 ⊂ A1(((A2
0A2
0)A0)...A0)A2 + A2
0
⊂ A1(((A2
0A2
0)A2
0)...A2
0)A2 + A2
0
⊂ A1(((A
(2)
0 A
(2)
0 )...)A
(2)
0 )A2 + A2
0
⊂ ... ⊂ A1A
(m)
0 A2 + A2
0 = A2
0.
Similarly, A2A<n>
0 A1 ⊂ A2
0. Thus, I0 ⊂ A2
0 = A
(1)
0 .
Now we can start from the beginning with the ideal I and construct an
ideal I ′ such that I (and, thus, A) is solvable if and only if I ′ is solvable and
I ′
0 ⊂ I2
0 ⊂ A
(2)
0 . Repeating this construction, in the end we will construct
an sublagebra Ĩ such that A is solvable if and only if Ĩ is solvable and
Ĩ0 ⊂ A
(m)
0 = 0. But by lemma 7 Ĩ is solvable, so A is also solvable.
Corollary 1. Let A be an alternative algebra with an automorphism φ
of order 3. If char F 6= 2, 3, 5 and the subalgebra Aφ of fixed points with
respect to φ is solvable, then A is solvable.
Proof. If the ground field F is algebraically closed, then we can consider
subspaces Aξ = {x ∈ A| φ(x) = ξx} and Aξ2 = {x ∈ A| φ(x) = ξ2x},
where ξ is a primitive cube root of unity. It is easy to see that A =
Aξ ⊕ Aξ2 ⊕ Aφ and A is a Z3-graded algebra. Since Aφ is solvable, then
by theorem 1 A is solvable.
If F is not algebraically closed we can consider it’s algebraic closure F
and an algebra A = A ⊗F F . Then A is an alternative algebra over F and
A is solvable if and only if A is solvable. We can define an automorphisms
φ on A by putting: φ(a ⊗ α) = φ(a) ⊗ α for all a ∈ A, α ∈ F . Then φ is an
automorphism of order 3 and the subalgebra of fixed points A
φ
= Aφ ⊗ F
is solvable. Thus, A is solvable and, finally, A is solvable.
Corollary 2. Let A =
n−1∑
i=0
Ai be a Zn-graded alternative algebra, where
n = 2k3l and k + l > 1. If char F 6= 2, 3, 5 and the subalgebra A0 is
solvable, then A is solvable.
M. Goncharov 215
Proof. If k = 0 then by corollary 1 A is solvable. Suppose k > 1. We will
use an induction on l. If l = 0 then the result follows from the paper of
Smirnov [12]. Let l > 1. Then we can consider subspaces Â0 =
∑
i
A3i,
Â1 =
∑
i
A1+3i, Â2 =
∑
i
A2+3i. Then A = Â0 ⊕Â1 ⊕Â2 - is a Z3-gradation
of A. By theorem 1 A is solvable if and only if Â0 is solvable. On the other
hand it is easy to see that Â0 is a Zn′-graded algebra, where n′ = 2k3l−1
and (Â0)0 = A0 is solvable. Now we may use the induction and get that
Â0 is solvable. Hence, A is solvable.
Corollary 3. Let A be an alternative algebra with an automorphism φ
of order 2k3l. If char F 6= 2, 3, 5 and the subalgebra Aφ of fixed points with
respect to φ is solvable, then A is solvable.
Acknowledgements
The author is grateful to Prof. I. Shestakov for his attention to this
work. The author would also like to gratitude the University of Sao Paulo
and Sao Paulo research foundation (FAPESP) for hostility and funds that
enabled the author to make this article.
The author was supported by FAPESP(2013/02039-1) and by RFFI
(12-01-33031).
References
[1] G. Higman, Groups and rings which have automorphisms without non-trivial fixed
elements, J. London Math. Soc. 32 2 (1957) 321-334.
[2] V.A. Kreknin, Kostrikin A.I., Lie algebras with a regular automorphism, Sov.
Math. Dokl. 4 (1963) 355-358.
[3] V.A. Kreknin, Solvability of a Lie algebra containing a regular automorphism,
Sib. Math. J. 8 (1967) 536-537.
[4] N.Yu. Makarenko, A nilpotent ideal in the Lie rings with automorphism of prime
order, Sib. Mat. Zh. 46, 6 (2005) 1360-1373.
[5] G.M. Bergman, I. M. Isaacs, Rings with fixed-point-free group actions, Proc.
London Math. Soc. 27 (1973) 69-87.
[6] V.K. Kharchenko, Galois extensions and quotient rings, Algebra and Logic 13 4
(1974) 265-281.
[7] A.P. Semenov, Subrings of invariants of a finite group of automorphisms of a
Jordan ring, Sib. Math. J. 32 1 (1991) 169-172.
[8] W.S. Martindale, S. Montgomery, Fixed elements of Jordan automorphisms of
associative rings Pacific J. Math 72 1 (1977) 181-196.
216 On solvable Z3-graded alternative algebras
[9] M. Nagata, On the nilpotancy of nil-algebras, J. Math. Soc. Japan 4 (1952)
296-301.
[10] G. Higman, On a conjecture of Nagata, Proc. Camb. Phil. Soc. 52 (1956) 1-4.
[11] V.N. Zhelyabin, Jordan superalgebras with a solvable even part, Algebra and Logic
34 1 (1995) 25-34.
[12] O.N. Smirnov, Solvability of alternative Z2-graded algebras and alternative super-
algebras, Sib. Math. J. 32 6 (1991) 1030-1034.
[13] K.A. Zhevlakov, A.M. Slinko, I.P. Shestakov, A.I. Shirshov, Rings that are Nearly
Associative, translated from the Russian by Harry F. Smith, Academic Press, New
York, 1982.
[14] K.A. Zhevlakov, Solvability of alternative nil-rings, Sib. Math. J. 3 (1962) 368-377
(in Russian).
[15] G.V. Dorofeev, An instance of a solvable, though nonnilpotent, alternative ring,
Uspehi Mat. Nauk 15 3 (1960) 147-150 (in Russian).
Contact information
M. E. Goncharov Institute of Mathematic and Statistic,
University of Sao Paulo,
Sao Paulo, SP, 05508-090, Brazil,
Sobolev Institute of Mathematics
Novosibirsk, 630058, Russia
E-Mail(s): goncharov.gme@gmail.com
Received by the editors: 21.09.2014
and in final form 21.09.2014.
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