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Chapter 6 Solvability and Nilpotency of Alternative Algebras
CHAPTER
6
Solvability and Nilpotency of Alternative Algebras
1.
THE NAGATA-HIGMAN THEOREM
In connection with Shirshov’s theorem on the local nilpotency of an alternative nil-algebra of bounded index, there arises a natural question: To what extent is the localness condition in that theorem essential, that is, is not every alternative nil-algebra of bounded index nilpotent? In this section we show that in the case of associative algebras, with a sufficiently nice ring of operators, the answer to this question turns out to be yes. We shall need the following
LEMMA 1. Let A be an alternative algebra. We denote by Z,(A) the set z,,(A)=
1aial)aiE 0,ai E A
( i
Then (n!)2Z,(A)A+ (n!)’AZ,(A)E Z,,(A).
I
.
PROOF. We denote by S,(a,, a,, . . . ,a,) the sum Sn(a,,a2,...,a,J=
C
(il.iz..
.. . i d
123
v(ai,, at2
.-
ai),
=24
6.
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
where u ( x , , x2, . . . ,x,) is some fixed nonassociative word of length n. By Lemma 1.4 we have S,(a,,
. . . ,a,) = u(a1 +
+ a,, . . . ,a, + . . . + a,)
* * *
i= 1
+ 1 + +
U(Ul+. l_
a^i
+a^,+.
* * *
+ a^, + * + i i j + + a,, . . . ,a, +. . . . + a,) * *
* *
* * +
'
n
In view of the fact that A is a power-associative algebra, the right-hand side of this equality does not depend on how the parentheses are arranged in the monomial u. Consequently, the sum S,(a,, . . . ,a,) likewise does not depend on how the parentheses are arranged in the monomial u. For example, one can assume the arrangement of the parentheses to be the standard one from the right. Furthermore, it follows from (1) that S,(a,, . . . ,a,) E I,(A) for any elements a,, . . . ,a, E A . Now let a, b be arbitrary elements of the algebra A. Then we have S,(ab, a,.
n- 1
. . , a) = (n - l ) ! 1 ai(ab)d""-i i=O
n- 1
=~
hence
( -n l)!
1 d b ~ " - ' - '= aS,(b, a, . . .,a),
i=O
aS,(b, a, . . . ,a) E Z,(A).
(2) Linearizing this inclusion in a, for any a,, . . . ,a,, b E A we obtain the inclusion
( n - I)!
n
i= 1
aiS,(b, a,, . . . , G i , . . . . a,)
E
Z,(A).
(3)
Setting a, = a, a2 = a3 = . . . = a, = b in (3), we obtain
(n - l)!aS,(b, b, . . . ,b)
+ (n - l)(n - l)!bS,(a, b, . . . ,b)
whence by (2) it follows
(n!)%b" E Z,(A). It is proved analogously that for any a, b E A This proves the lemma.
(n!)2b"aE Z,(A).
E
Z,(A),
6.1.
set
125
NACATA-HIGMAN THEOREM
COROLLARY. Let A be an alternative algebra. We denote by .!,,(A) the J,(A) = { a E ~ ( ( n ! )E~I,,(A) a for some k}.
Then J,(A) is an ideal of the algebra A , and the quotient algebra A/J,(A) does not contain elements of additive order I n. PROOF. It is clear that J,(A) is a Q-submodule of the @-module A . Now let a E J,(A), b E A, and let (n!)kaE In(A).By Lemma 1 (n!)2b[(n!)ka], ( r ~ ! ) ~ [ ( n ! )E~I nu(]Ab) ,whence ( r ~ ! ) ~ +(n!)k+2ab ~ b a , E I,,@), and ba, ab E J,(A). By the same token we have proved that J,(A) is an ideal of the algebra A . We now note that the ideal J,(A) has the following property: if(n!)kaE J,(A), then a E J,(A). Hence it follows that the quotient algebra A/J,,(A)does not contain elements of additive order ~ n This . proves the corollary.
We can now prove the following assertion: THEOREM 1 (Nagata, Higman). Let A be an arbitrary assocative algebra. Then for any natural number n A2"- 1
s J,(A).
PROOF (Higgins). We shall prove the theorem by induction on n. For n = 1 the assertion is obvious, so assume it is true for n - 1. In view of the fact that &(b, a, . . . , a) E Z,,(A),for any a, b E A we have n-
1
(lc:
i=O
Now let a, b, c be arbitrary elements of the algebra A . We consider the sum
1
n- 1
,n-1-icbjaip-1-jj= 1
i,j=l
(cbj)ai) bn - 1 - j
j= 1
where k E J,(A) by (4). On the other hand, we have
1
n- 1 an-l-icbjaip-l-j=
i,j = 1
n- 1 a n - l - i c ( n ~ l ~ a i b n - l - j
j= I
i=l
-
1
n- 1
i=l
(an-
1 -icai)bn-
1 +
k
- an- I c y -
1 -
where again k , , k , E J,(A). As a result we obtain - 1 cbn - 1 E Jn(A).
1
+ k,,
6.
126
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
Because the elements Q, b, c are arbitrary, it follows that nl, - 1(AMIn - ,(A)
Jn(A),
and furthermore Jn-
1(A)AJn-,(A) C Jn(A)*
By the induction assumption A2"-'-l E J,- l(A), and consequently AP-1
= AZn-'-lAAZn-'-l
C J,,( A).
This proves the theorem. COROLLARY 1. Let A be an associative nil-algebra of index n without elements of additive order I n . Then A is nilpotent of index I2" - 1. For the proof it suffices to note that under the conditions of the corollary J,(A) = 0. COROLLARY 2. Let A be an arbitrary associative algebra. Then for any natural number n there exists a natural number k such that for any a € AZn-' where aiE @ and ai E A.
a&,
(n!)ka= i
For the proof we consider the free associative @-algebraAss[X] from the set of free generators X = {xi, x2, . . .}. By Theorem 1 we have xlxz
* * *
x2n- 1 E
J,(Ass[X]),
and consequently, for some k (n!)kxlx2* * xZnMl=
a&, i
where aiE @ and ui E Ass[X].
As is easy to see, the given number k is the desired one. We note that the bound 2" - 1 obtained in the Nagata-Higman theorem is not exact, since in the case n = 3 it is known A6 E J3(A)while 23 - 1 = 7. The question of a corresponding exact bound f(n) for the case of an arbitrary n remains open. Razmyslov recently showed that f(n) In2. On the other hand, Kuz'min has proved that f(n) < n(n + 1)/2. Exercises
1. Let A be a Jordan algebra. As in the case of alternative algebras, the subsets Z,(A) and Jn(A)can be considered in A. Prove that J,(A) is an ideal of the algebra A for any n.
6.2.
“7
DOROFEEV’S EXAMPLE
2. Prove that for any prime number p there exists a nonnilpotent associative nil-algebra of index p over a field of characteristic p . 3. (Higman) Prove that the containment A6 E J,(A) is valid for every associative algebra A .
2. DOROFEEV’S EXAMPLE
We return now to alternative algebras. It turns out that, in contrast to the associative case, alternative nil-algebras of bounded index can be nonnilpotent, that is, the Nagata-Higman theorem does not carry over to alternative algebras. This was proved by G . V. Dorofeev, who constructed an example of a nonnilpotent solvable alternative algebra over an arbitrary ring of operators. We present his example in somewhat altered form. Let us consider two sets of symbols: E = {ek}, k = 1, 2,. . . , n , . . . ; V = (x, L j , R,}, i, j = 1,2, . . . , n, . . . . If u is an arbitrary associative word from elements of V , then we shall denote by dR(u)the number of symbols Ri which appear in the composition of u. We call the word u regular if u has one of the following forms: (1) dR(u)= 0, u = x or u = xL,; (2) dR(u)= 1, u = xRJ or u = xLiRj; . Rin, where i, < i, < . . < in and the * ( 3 ) dR(u)2 2, u = xLi,Ri2Ri3 symbol L i , may be absent.
We denote by T( V ) the set of all regular words. Now let u be an arbitrary word from V of the form
u = xLi,Ri2RiJ. . . Rim, A
where dR(u)2 2 and the symbol L i , may be absent. We denote by ii a regular word of the form ii = X L . ~ , R ~ ,. R . * ~Rjm, ,
where { j , , j,, . . . ,in}= {i,, i,, . . . , in} and the symbol L j , appears in the composition of if and only if the symbol L i t appears in u. We denote by t(u) the number of inversions in the permutatlon (i,, i,, . . . , in). If any two of the symbols i,, i 2 , . . . , in are identical, then we set ii = 0. Let us consider the free @-moduleA for which a basis is the set E u T(V ) . We convert A into an algebra by defining multiplication on the basis according to the following rules: 1. x . y = 0, ifx,yE E or x , y ~T ( V ) .
6.
I 28
2. (a) (b) 3. (a) (b)
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
x . ei = x R i , x L j . ei = x L j R i ; if u E T(V)and dR(u)2 1, then u * ei = ( - l ) ‘ ( U R u ) a . e, * x = x L i , ei* x L j = x R j . e , ; if u E T( V ) and dR(u) 2 1, then u = u‘R, and
ej . u = e j . (u’Ri)= (ej . u’
+ u’ . e j ). e i .
As is easy to see, rules (1)-(3) define the products of any basis elements. We now consider the @-module Z = @(T(V ) )generated by the set T(V ) .It is clear that I is an ideal of the algebra A , and in addition I’ = (0) and A’ E I . Consequently (A’)’ = (0), so A is solvable and therefore a nil-algebra of bounded index. Furthermore, the algebra A is nonnilpotent, since for any n the product * ((x . e l ) .e z ) .. - ) . en = x R , R , * * - R , is different from zero. It remains to prove that the algebra A is alternative. We shall divide the proof of this fact into several lemmas. We first note the following obvious equality: ( a
( u * e,) . ej + ( u * ej) e, = 0,
(5)
where u E T(V ) and e i ,e j E E . LEMMA 2. For any u E T( V ) and e,, ek E E there is the equality e i . ( u . ek)= (ei u
+ u . e i ) .ek.
(6)
PROOF. We shall prove this equality by induction on the number dR(u), that is, on the number of symbols Ri appearing in u. If dR(u)= 0, then either u = x or u = x L j , and ( 6 ) follows at once from the rules of multiplication. Now let dR(u)= n 2 1 , and assume equality ( 6 ) is valid for all words from T ( V ) containing less than n symbols R , . Let u = u‘R,. If k 2 n, then (6) follows from rule 3(b). Consequently, one can assume k c n. By the multiplication rules 2(b) and 3(b), the induction assumption, and formula (5), we have e, . ( u . ek)= ei * (u’R, * ek)= ( - l)r(uRk)ei . (V’RkRn) = (- f)r(uRk)(ei . v ’ R k + 2, . ei)* en = (- l)r(URk). ( - l)f(u’Rk)(ei * (u‘ * ek) (0’ * ek) ei) . en
+
+ u’ . e,) . ek] . en + [(u’ e i ) e,] en = [(e,- u’ + u’ - e,) en] - f?k - [(u’ . ei). en] . e, = [e, (u’ - en)] ek + [(u’ e n ) .e,] . e k = (e, . u‘R, + u’Rn. ei) ek = (e, . u + u . ei). e k . = - [(e,. u‘
*
*
*
*
This proves the lemma.
6.2.
129
DOROFEEV’S EXAMPLE
LEMMA 3. For any u E T(V) and e,, ej, ek E E there are the equalities [(e,. u) * ej
+ (ej. u) . ei] . ek = 0,
(7)
ei * (ej . u) = ( 0 . ej) . ei.
(8)
PROOF. We shall first prove equality (7). If u = x or u = xL,, then (7) is valid. Now let u = u’R,, and assume equality (7) is true for words having a smaller number of the symbols R , . Then by (6), (5), and the induction assumption, (ei * U’R,). ej = [(e, . u‘ + u’ * e,) . en] . ej = - [(en . u‘ + u’ . e n ) .ei] . ej = [(en . u‘ + u’ en). ej] . e, = - [(ej * u‘ + u‘ * e j ) .en] . e, = - [ej . (u‘ . en)] . e, = - (ej . u) . ei.
-
Thus ifd,(u) > 0, then an even stronger equality than (7) is valid: (e, . u) . ej + (ej. u) . e, = 0. We shall now prove (8). If dR(u)= 0, then (8) is valid. Now let u = u’R,. By (3,(6), (7), and the induction assumption, we have
e i . (ej . U’R,)= e , . [(ej * u’ + u’ * ej). en] = [ei . (ej . u’ + u’ . ej) + ( e j . u’ + u’ * ej) * ei] . en = [(u’ . ej). e, (e, . u‘ u’ . e,). e j (ej. u’) . e, (u’ . ej). e,] . en = [(ei * u’) . ej + (ej. u’) . ei] . en + [(u’ . en). ej] . ei = ( u * ej) * ei.
+
+
+
+
This proves the lemma. LEMMA 4. The algebra A is alternative. PROOF. From the multiplication table it is clear that only associators which contain one element from T(V )and two elements from E are nonzero. By equality ( 5 ) we have
(u, e,, ej)
+ (u, e j , ei) = 0.
Furthermore, from (8) and ( 5 ) we also have (e,, ej, u)
+ (ej, e,, u) = 0.
We now consider the expression (e,, u, ej) + (ei, ej. u) = (e, . u) . ej - e , . ( u . ej) - e, (ej. u) = (e,. 0 ) . ej - (ei. u ) . ej - ( u . ei). ej - ( u * e.). J e,
=O
130
6.
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
by (6),(8), and (5). Moreover, again by (6) we obtain (u, e i , ej)
+ (ei,u, ej) = (u
ei) * ej = ( u * ei) * ej = 0. *
+ (ei u) + (ei - u) *
e j - ei * (u - ej) * ej - (ei * u) * e j - (u ei) * ej
*
Finally, it is easy to see that (0,
ei, ei) = (ei,u, ei) = (ei, ei, u) = 0.
This proves the lemma. Exercises 1. Let A be the solvable nonnilpotent algebra constructed in Section 2. Prove that its annihilator, Ann A = {a E A1 aA = Aa = (0)},is generated as a @-moduleby elements of the form xLiRi, xLiRj + xLjRi. Also prove that the quotient algebra A = A/Ann A is a nonnilpotent, solvable, nil-algebra of index 3 and that Ann A = (8). 2. Prove that the algebra A defined in the previous exercise has zero associative center. 3. Let Tkbe the ideal of identities satisfied by the free algebra with k generators in the variety of alternative algebras of solvable index 2. Prove that the chain of T-ideals
does not stabilize at any finite step. 3. WEVLAKOV'STHEOREM
In the previous section we established that the Nagata-Higman theorem does not carry over verbatim to alternative algebras. We proved an even stronger assertion, namely, the nilpotency of an alternative algebra A does not, in general, follow from its solvability. (We recall that the analogous fact also holds in the theory of Jordan algebras.) In connection with all of this there arises a natural question: Is every alternative nil-algebra of bounded index solvable? As proved by K. A. Zhevlakov, over a sufficiently nice ring of operators the answer to this question turns out to be yes. For the proof of Zhevlakov's theorem we need 3 series of lemmas. Henceforth A is an arbitrary alternative algebra. LEMMA 5. In the algebra A there is the identity (zx xz, y, z ) = (xzx, y, 22). 0
(9)
6.3.
ZHEVLAKOV'S THEOREM
'3'
PROOF. Applying the Moufang identities and their linearizations, we obtain ( z x xz, y, z ) = ( z x ) (xz, y , z ) 0
0
= (4O ( x , Y Z , 4
+ (xz)
0
+ (xz)
( z x ,y , z )
( x ,Z Y , z ) = ( z x x , yz, z ) - x ( z x , yz, z ) + ( x z x , zy, z) - x (xz, zy, z ) = ( Z X Z , yz, z) + ( X Z Z , zy, z ) + ( x z x , y z, z ) - 2x ( x , zyz, z ) = 2(x2, zyz, z) + (xzx,y, Z Z ) - 2(x2, zyz, z ) = ( x z x ,y , ZZ). O
0
0
0
0
0
0
This proves the lemma. LEMMA 6. In the quotient algebra A = A/J,(A) there is the identity z ( Y - 1 , y , z)z = 0.
PROOF. We first note that in 2 the identity n- 1
o = s,,(z, x , . . . ,x ) = (n - I)! C
X'ZY-
i=O
' -'
is valid, whence from the properties of the ideal Jn(A)we obtain i=O
Now in view of (10) we have
c (zx')
n-1
0
(x"-'-'z) = nzx"-'z+
n- 1
1 Xn-l-iZZXi
= nzx"-'z,
i=O
i=O
whence, from linearization in x of identity (9) and the Moufang identities, we have nz(x"-',y,z)z = (nzxn-lz, y,z) = z), Y , z
This proves the lemma. LEMMA 7. In the algebra
a there is the identity
(Y-l,yn-l,zZ)
= 0,
n 2 2.
PROOF. We note first of all that in every alternative algebra the identity n- 1
(9,y , z ) =
1 xyx, y , z)x" -
i=O
-i
6.
132
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
is valid. In fact, by Artin's theorem (x", y,x) = 0, whence
and then by the Moufang identities, we obtain (11). By Lemma 6 for any k, I 2 1 we have zk(x"-
1,
y, z)z' = 0.
Hence by identity (11) it follows that for any m 2 1 (x"-',y,f) = (x"-l,y.z)
zm-l.
0
We now have 0 = (x"-
1,
y, z") = [(x"-
1,
y, z) 0 zn- '3 = { . . . [(x"- 1.y,z)
>Z]
0
z.-}
0
z.
Linearization of this identity in z gives 0=
c.. .
(il.iz..
{. . . [(X"-l,y,Zi,)
in)
z i s . . .}
0
Zi2]
0
0
zin.
(12)
We set here z1 = z z = z, z3 = z4 = . . * = z, = y. Since for any k 2 0
{ [. . . ( (x" - 1, y, z) y) y * . * ] y } 0
0
0
0
2
k
= ([.
. . (x"-
- ([.
* *
1
0
y) 0 y
* *
a ]
0
y, y, z) 0 z
k
(Xn-1 0 y) 0 y
. . .] y, y, Z Z ) 0
V
k
= { . ' . [(x"-
1, y, 22)
0
y]
0
y
* *
a }
0
y
V
k
= (x"-
y, ZZ) 0 y",
l,
then, after we carry out the permutations, identity (12) takes the form
o = 2(n - 1)!(xn-',y,z2)O
f - '= 2(n - 1)!(xn-1,f-1,z2).
This proves the lemma. LEMMA 8. A 4 s J,(A). PROOF. For any a, b E A we have a', a 0 b E J,(A) and 2aba = (a 0 b) 0 a - az 0 b E J,(A), whence also aba E J,(A). Now, as in the proof of Proposi-
tion 5.3,we obtain that the quotient algebra A = A / J , ( A ) is anticommutative
6.3.
=33
ZHEVI.AKOV'S THEOREM
and antiassociative. In the algebra 2 we consider the associator
+
(ab,C,d ) = [(ab)c]d- (ab)(cd)= [a(bc)]d a [ b ( c d ) ] = a[(bc)d] - a[(bc)d] = 0.
On the other hand, we have ( a b , ~ , d=) [(ab)c]d- (ab)(cd) = [(ab)c]d [(ab)c]d= 2[(ab)c]d.
+
The equality obtained proves that
A4 = (0). This proves the lemma.
Now we can prove THEOREM 2 (Zheolakou). Let A be an arbitrary alternative algebra. Then for any natural number n A("("+ I )/2) G J , ( A ) .
PROOF. For n = 1 the assertion of the theorem is obvious. For n = 2 by Lemma 8 we have G A2 . A' G A4 G J,(A), that is, the assertion is l(A), - n > 2. We consider the quotient also true. Now let A("("-')I2) c .I, algebra 2 = A/J,(A). By the induction assumption we have
A("("-l ) / 2 ) E J n - l(A). Furthermore, in view of the containment J,-,(A) E J,(A), from Lemma 7 it follows that J , - , ( A )is an associative nil-algebra of index n. By the NagataHigman theorem the algebra J , - ,(A) is nilpotent of index 2" - 1, whence then finally we obtain (Jn-,(A))(") = (0). Since (A(k))(") = , p n + 1)/2) =
( -A( n ( n -
1)/2) ( n )
)
E
( J n - I(A))'")
= (01,
whence A("("+11/21 c - J,(A).
This proves the theorem. COROLLARY 1. Let A be an alternative nil-algebra of index n without elements of additive order In.Then the algebra A is solvable of index ~ n ( n 1)/2.
+
COROLLARY 2. Let A be an arbitrary alternative algebra. Then for any natural number n there exists a natural number k such that for any LI
A("("
f
I
), 2 )
qa;,
( n ! ) k a= i
where u iE 0 and a, E A .
6.
'34
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
Thus every alternative nil-algebra of bounded index is locally nilpotent, over a sufficiently nice ring of operators even solvable, but over an arbitrary ring of operators it is possible it may not be nilpotent. In addition, as for the Nagata-Higman theorem, it is rather easy to see that the restriction on the characteristic is essential for solvability. We note for comparison that, in the case of Jordan algebras, the situation is far from being so clear. We already remarked in Chapter 5 that the answer to the question on local nilpotency of Jordan nil-algebras of bounded index is unknown for Jordan algebras that are not special. The question on solvability of Jordan nil-algebras of bounded index is not solved even for special Jordan nil-algebras of index 3. Exercises
4
Below A is an alternative @-algebrawhere E 0. 1. Prove that for any natural number n the set A, = (A(+))(") is a subalgebra of the algebra A. Hint.Prove that A, = Z2(A,- 1), then apply Lemma 1 and induction on n. 2. Prove that A'in) E (A(+))("). In particular, the algebra A is solvable if and only if the Jordan algebra A ( + )is solvable. Hint.Apply Lemma 8 and induction on n. 3. Let I be an ideal of the algebra A, M be some submodule of the 0module A, and also I 0 A ( + )5 M. Prove that the containment u(Z, A, A, A) G M is valid for any nonassociative word u(x 1, x 2 ,x3,xq) of length 4. Hint.Consider the free alternative @-algebraand apply Lemma 8. 4. Prove that A3"+' G (A(+))"+'. In particular, the algebra A is nilpotent if and only if the Jordan algebra A ( + )is nilpotent. Hint.Apply induction on n and use Exercise 3.
LITERATURE Dorofeev [55], Zhevlakov [66], Kuz'min [103], Nagata [157], Razmyslov [175], Higman r2531. - Results on connections between solvabilityand nilpotency in some other varieties ofalgebras: Anderson [13,141,Dorofeev [59],Zhevlakov [69],Zhevlakov and Shestakov [go], Markovichev [144], Nikitin [ l a ] , Pchelintsev [172, 1741, Roomel'di [184], Shestakov [267].
6
Solvability and Nilpotency of Alternative Algebras
1.
THE NAGATA-HIGMAN THEOREM
In connection with Shirshov’s theorem on the local nilpotency of an alternative nil-algebra of bounded index, there arises a natural question: To what extent is the localness condition in that theorem essential, that is, is not every alternative nil-algebra of bounded index nilpotent? In this section we show that in the case of associative algebras, with a sufficiently nice ring of operators, the answer to this question turns out to be yes. We shall need the following
LEMMA 1. Let A be an alternative algebra. We denote by Z,(A) the set z,,(A)=
1aial)aiE 0,ai E A
( i
Then (n!)2Z,(A)A+ (n!)’AZ,(A)E Z,,(A).
I
.
PROOF. We denote by S,(a,, a,, . . . ,a,) the sum Sn(a,,a2,...,a,J=
C
(il.iz..
.. . i d
123
v(ai,, at2
.-
ai),
=24
6.
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
where u ( x , , x2, . . . ,x,) is some fixed nonassociative word of length n. By Lemma 1.4 we have S,(a,,
. . . ,a,) = u(a1 +
+ a,, . . . ,a, + . . . + a,)
* * *
i= 1
+ 1 + +
U(Ul+. l_
a^i
+a^,+.
* * *
+ a^, + * + i i j + + a,, . . . ,a, +. . . . + a,) * *
* *
* * +
'
n
In view of the fact that A is a power-associative algebra, the right-hand side of this equality does not depend on how the parentheses are arranged in the monomial u. Consequently, the sum S,(a,, . . . ,a,) likewise does not depend on how the parentheses are arranged in the monomial u. For example, one can assume the arrangement of the parentheses to be the standard one from the right. Furthermore, it follows from (1) that S,(a,, . . . ,a,) E I,(A) for any elements a,, . . . ,a, E A . Now let a, b be arbitrary elements of the algebra A. Then we have S,(ab, a,.
n- 1
. . , a) = (n - l ) ! 1 ai(ab)d""-i i=O
n- 1
=~
hence
( -n l)!
1 d b ~ " - ' - '= aS,(b, a, . . .,a),
i=O
aS,(b, a, . . . ,a) E Z,(A).
(2) Linearizing this inclusion in a, for any a,, . . . ,a,, b E A we obtain the inclusion
( n - I)!
n
i= 1
aiS,(b, a,, . . . , G i , . . . . a,)
E
Z,(A).
(3)
Setting a, = a, a2 = a3 = . . . = a, = b in (3), we obtain
(n - l)!aS,(b, b, . . . ,b)
+ (n - l)(n - l)!bS,(a, b, . . . ,b)
whence by (2) it follows
(n!)%b" E Z,(A). It is proved analogously that for any a, b E A This proves the lemma.
(n!)2b"aE Z,(A).
E
Z,(A),
6.1.
set
125
NACATA-HIGMAN THEOREM
COROLLARY. Let A be an alternative algebra. We denote by .!,,(A) the J,(A) = { a E ~ ( ( n ! )E~I,,(A) a for some k}.
Then J,(A) is an ideal of the algebra A , and the quotient algebra A/J,(A) does not contain elements of additive order I n. PROOF. It is clear that J,(A) is a Q-submodule of the @-module A . Now let a E J,(A), b E A, and let (n!)kaE In(A).By Lemma 1 (n!)2b[(n!)ka], ( r ~ ! ) ~ [ ( n ! )E~I nu(]Ab) ,whence ( r ~ ! ) ~ +(n!)k+2ab ~ b a , E I,,@), and ba, ab E J,(A). By the same token we have proved that J,(A) is an ideal of the algebra A . We now note that the ideal J,(A) has the following property: if(n!)kaE J,(A), then a E J,(A). Hence it follows that the quotient algebra A/J,,(A)does not contain elements of additive order ~ n This . proves the corollary.
We can now prove the following assertion: THEOREM 1 (Nagata, Higman). Let A be an arbitrary assocative algebra. Then for any natural number n A2"- 1
s J,(A).
PROOF (Higgins). We shall prove the theorem by induction on n. For n = 1 the assertion is obvious, so assume it is true for n - 1. In view of the fact that &(b, a, . . . , a) E Z,,(A),for any a, b E A we have n-
1
(lc:
i=O
Now let a, b, c be arbitrary elements of the algebra A . We consider the sum
1
n- 1
,n-1-icbjaip-1-jj= 1
i,j=l
(cbj)ai) bn - 1 - j
j= 1
where k E J,(A) by (4). On the other hand, we have
1
n- 1 an-l-icbjaip-l-j=
i,j = 1
n- 1 a n - l - i c ( n ~ l ~ a i b n - l - j
j= I
i=l
-
1
n- 1
i=l
(an-
1 -icai)bn-
1 +
k
- an- I c y -
1 -
where again k , , k , E J,(A). As a result we obtain - 1 cbn - 1 E Jn(A).
1
+ k,,
6.
126
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
Because the elements Q, b, c are arbitrary, it follows that nl, - 1(AMIn - ,(A)
Jn(A),
and furthermore Jn-
1(A)AJn-,(A) C Jn(A)*
By the induction assumption A2"-'-l E J,- l(A), and consequently AP-1
= AZn-'-lAAZn-'-l
C J,,( A).
This proves the theorem. COROLLARY 1. Let A be an associative nil-algebra of index n without elements of additive order I n . Then A is nilpotent of index I2" - 1. For the proof it suffices to note that under the conditions of the corollary J,(A) = 0. COROLLARY 2. Let A be an arbitrary associative algebra. Then for any natural number n there exists a natural number k such that for any a € AZn-' where aiE @ and ai E A.
a&,
(n!)ka= i
For the proof we consider the free associative @-algebraAss[X] from the set of free generators X = {xi, x2, . . .}. By Theorem 1 we have xlxz
* * *
x2n- 1 E
J,(Ass[X]),
and consequently, for some k (n!)kxlx2* * xZnMl=
a&, i
where aiE @ and ui E Ass[X].
As is easy to see, the given number k is the desired one. We note that the bound 2" - 1 obtained in the Nagata-Higman theorem is not exact, since in the case n = 3 it is known A6 E J3(A)while 23 - 1 = 7. The question of a corresponding exact bound f(n) for the case of an arbitrary n remains open. Razmyslov recently showed that f(n) In2. On the other hand, Kuz'min has proved that f(n) < n(n + 1)/2. Exercises
1. Let A be a Jordan algebra. As in the case of alternative algebras, the subsets Z,(A) and Jn(A)can be considered in A. Prove that J,(A) is an ideal of the algebra A for any n.
6.2.
“7
DOROFEEV’S EXAMPLE
2. Prove that for any prime number p there exists a nonnilpotent associative nil-algebra of index p over a field of characteristic p . 3. (Higman) Prove that the containment A6 E J,(A) is valid for every associative algebra A .
2. DOROFEEV’S EXAMPLE
We return now to alternative algebras. It turns out that, in contrast to the associative case, alternative nil-algebras of bounded index can be nonnilpotent, that is, the Nagata-Higman theorem does not carry over to alternative algebras. This was proved by G . V. Dorofeev, who constructed an example of a nonnilpotent solvable alternative algebra over an arbitrary ring of operators. We present his example in somewhat altered form. Let us consider two sets of symbols: E = {ek}, k = 1, 2,. . . , n , . . . ; V = (x, L j , R,}, i, j = 1,2, . . . , n, . . . . If u is an arbitrary associative word from elements of V , then we shall denote by dR(u)the number of symbols Ri which appear in the composition of u. We call the word u regular if u has one of the following forms: (1) dR(u)= 0, u = x or u = xL,; (2) dR(u)= 1, u = xRJ or u = xLiRj; . Rin, where i, < i, < . . < in and the * ( 3 ) dR(u)2 2, u = xLi,Ri2Ri3 symbol L i , may be absent.
We denote by T( V ) the set of all regular words. Now let u be an arbitrary word from V of the form
u = xLi,Ri2RiJ. . . Rim, A
where dR(u)2 2 and the symbol L i , may be absent. We denote by ii a regular word of the form ii = X L . ~ , R ~ ,. R . * ~Rjm, ,
where { j , , j,, . . . ,in}= {i,, i,, . . . , in} and the symbol L j , appears in the composition of if and only if the symbol L i t appears in u. We denote by t(u) the number of inversions in the permutatlon (i,, i,, . . . , in). If any two of the symbols i,, i 2 , . . . , in are identical, then we set ii = 0. Let us consider the free @-moduleA for which a basis is the set E u T(V ) . We convert A into an algebra by defining multiplication on the basis according to the following rules: 1. x . y = 0, ifx,yE E or x , y ~T ( V ) .
6.
I 28
2. (a) (b) 3. (a) (b)
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
x . ei = x R i , x L j . ei = x L j R i ; if u E T(V)and dR(u)2 1, then u * ei = ( - l ) ‘ ( U R u ) a . e, * x = x L i , ei* x L j = x R j . e , ; if u E T( V ) and dR(u) 2 1, then u = u‘R, and
ej . u = e j . (u’Ri)= (ej . u’
+ u’ . e j ). e i .
As is easy to see, rules (1)-(3) define the products of any basis elements. We now consider the @-module Z = @(T(V ) )generated by the set T(V ) .It is clear that I is an ideal of the algebra A , and in addition I’ = (0) and A’ E I . Consequently (A’)’ = (0), so A is solvable and therefore a nil-algebra of bounded index. Furthermore, the algebra A is nonnilpotent, since for any n the product * ((x . e l ) .e z ) .. - ) . en = x R , R , * * - R , is different from zero. It remains to prove that the algebra A is alternative. We shall divide the proof of this fact into several lemmas. We first note the following obvious equality: ( a
( u * e,) . ej + ( u * ej) e, = 0,
(5)
where u E T(V ) and e i ,e j E E . LEMMA 2. For any u E T( V ) and e,, ek E E there is the equality e i . ( u . ek)= (ei u
+ u . e i ) .ek.
(6)
PROOF. We shall prove this equality by induction on the number dR(u), that is, on the number of symbols Ri appearing in u. If dR(u)= 0, then either u = x or u = x L j , and ( 6 ) follows at once from the rules of multiplication. Now let dR(u)= n 2 1 , and assume equality ( 6 ) is valid for all words from T ( V ) containing less than n symbols R , . Let u = u‘R,. If k 2 n, then (6) follows from rule 3(b). Consequently, one can assume k c n. By the multiplication rules 2(b) and 3(b), the induction assumption, and formula (5), we have e, . ( u . ek)= ei * (u’R, * ek)= ( - l)r(uRk)ei . (V’RkRn) = (- f)r(uRk)(ei . v ’ R k + 2, . ei)* en = (- l)r(URk). ( - l)f(u’Rk)(ei * (u‘ * ek) (0’ * ek) ei) . en
+
+ u’ . e,) . ek] . en + [(u’ e i ) e,] en = [(e,- u’ + u’ - e,) en] - f?k - [(u’ . ei). en] . e, = [e, (u’ - en)] ek + [(u’ e n ) .e,] . e k = (e, . u‘R, + u’Rn. ei) ek = (e, . u + u . ei). e k . = - [(e,. u‘
*
*
*
*
This proves the lemma.
6.2.
129
DOROFEEV’S EXAMPLE
LEMMA 3. For any u E T(V) and e,, ej, ek E E there are the equalities [(e,. u) * ej
+ (ej. u) . ei] . ek = 0,
(7)
ei * (ej . u) = ( 0 . ej) . ei.
(8)
PROOF. We shall first prove equality (7). If u = x or u = xL,, then (7) is valid. Now let u = u’R,, and assume equality (7) is true for words having a smaller number of the symbols R , . Then by (6), (5), and the induction assumption, (ei * U’R,). ej = [(e, . u‘ + u’ * e,) . en] . ej = - [(en . u‘ + u’ . e n ) .ei] . ej = [(en . u‘ + u’ en). ej] . e, = - [(ej * u‘ + u‘ * e j ) .en] . e, = - [ej . (u‘ . en)] . e, = - (ej . u) . ei.
-
Thus ifd,(u) > 0, then an even stronger equality than (7) is valid: (e, . u) . ej + (ej. u) . e, = 0. We shall now prove (8). If dR(u)= 0, then (8) is valid. Now let u = u’R,. By (3,(6), (7), and the induction assumption, we have
e i . (ej . U’R,)= e , . [(ej * u’ + u’ * ej). en] = [ei . (ej . u’ + u’ . ej) + ( e j . u’ + u’ * ej) * ei] . en = [(u’ . ej). e, (e, . u‘ u’ . e,). e j (ej. u’) . e, (u’ . ej). e,] . en = [(ei * u’) . ej + (ej. u’) . ei] . en + [(u’ . en). ej] . ei = ( u * ej) * ei.
+
+
+
+
This proves the lemma. LEMMA 4. The algebra A is alternative. PROOF. From the multiplication table it is clear that only associators which contain one element from T(V )and two elements from E are nonzero. By equality ( 5 ) we have
(u, e,, ej)
+ (u, e j , ei) = 0.
Furthermore, from (8) and ( 5 ) we also have (e,, ej, u)
+ (ej, e,, u) = 0.
We now consider the expression (e,, u, ej) + (ei, ej. u) = (e, . u) . ej - e , . ( u . ej) - e, (ej. u) = (e,. 0 ) . ej - (ei. u ) . ej - ( u . ei). ej - ( u * e.). J e,
=O
130
6.
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
by (6),(8), and (5). Moreover, again by (6) we obtain (u, e i , ej)
+ (ei,u, ej) = (u
ei) * ej = ( u * ei) * ej = 0. *
+ (ei u) + (ei - u) *
e j - ei * (u - ej) * ej - (ei * u) * e j - (u ei) * ej
*
Finally, it is easy to see that (0,
ei, ei) = (ei,u, ei) = (ei, ei, u) = 0.
This proves the lemma. Exercises 1. Let A be the solvable nonnilpotent algebra constructed in Section 2. Prove that its annihilator, Ann A = {a E A1 aA = Aa = (0)},is generated as a @-moduleby elements of the form xLiRi, xLiRj + xLjRi. Also prove that the quotient algebra A = A/Ann A is a nonnilpotent, solvable, nil-algebra of index 3 and that Ann A = (8). 2. Prove that the algebra A defined in the previous exercise has zero associative center. 3. Let Tkbe the ideal of identities satisfied by the free algebra with k generators in the variety of alternative algebras of solvable index 2. Prove that the chain of T-ideals
does not stabilize at any finite step. 3. WEVLAKOV'STHEOREM
In the previous section we established that the Nagata-Higman theorem does not carry over verbatim to alternative algebras. We proved an even stronger assertion, namely, the nilpotency of an alternative algebra A does not, in general, follow from its solvability. (We recall that the analogous fact also holds in the theory of Jordan algebras.) In connection with all of this there arises a natural question: Is every alternative nil-algebra of bounded index solvable? As proved by K. A. Zhevlakov, over a sufficiently nice ring of operators the answer to this question turns out to be yes. For the proof of Zhevlakov's theorem we need 3 series of lemmas. Henceforth A is an arbitrary alternative algebra. LEMMA 5. In the algebra A there is the identity (zx xz, y, z ) = (xzx, y, 22). 0
(9)
6.3.
ZHEVLAKOV'S THEOREM
'3'
PROOF. Applying the Moufang identities and their linearizations, we obtain ( z x xz, y, z ) = ( z x ) (xz, y , z ) 0
0
= (4O ( x , Y Z , 4
+ (xz)
0
+ (xz)
( z x ,y , z )
( x ,Z Y , z ) = ( z x x , yz, z ) - x ( z x , yz, z ) + ( x z x , zy, z) - x (xz, zy, z ) = ( Z X Z , yz, z) + ( X Z Z , zy, z ) + ( x z x , y z, z ) - 2x ( x , zyz, z ) = 2(x2, zyz, z) + (xzx,y, Z Z ) - 2(x2, zyz, z ) = ( x z x ,y , ZZ). O
0
0
0
0
0
0
This proves the lemma. LEMMA 6. In the quotient algebra A = A/J,(A) there is the identity z ( Y - 1 , y , z)z = 0.
PROOF. We first note that in 2 the identity n- 1
o = s,,(z, x , . . . ,x ) = (n - I)! C
X'ZY-
i=O
' -'
is valid, whence from the properties of the ideal Jn(A)we obtain i=O
Now in view of (10) we have
c (zx')
n-1
0
(x"-'-'z) = nzx"-'z+
n- 1
1 Xn-l-iZZXi
= nzx"-'z,
i=O
i=O
whence, from linearization in x of identity (9) and the Moufang identities, we have nz(x"-',y,z)z = (nzxn-lz, y,z) = z), Y , z
This proves the lemma. LEMMA 7. In the algebra
a there is the identity
(Y-l,yn-l,zZ)
= 0,
n 2 2.
PROOF. We note first of all that in every alternative algebra the identity n- 1
(9,y , z ) =
1 xyx, y , z)x" -
i=O
-i
6.
132
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
is valid. In fact, by Artin's theorem (x", y,x) = 0, whence
and then by the Moufang identities, we obtain (11). By Lemma 6 for any k, I 2 1 we have zk(x"-
1,
y, z)z' = 0.
Hence by identity (11) it follows that for any m 2 1 (x"-',y,f) = (x"-l,y.z)
zm-l.
0
We now have 0 = (x"-
1,
y, z") = [(x"-
1,
y, z) 0 zn- '3 = { . . . [(x"- 1.y,z)
>Z]
0
z.-}
0
z.
Linearization of this identity in z gives 0=
c.. .
(il.iz..
{. . . [(X"-l,y,Zi,)
in)
z i s . . .}
0
Zi2]
0
0
zin.
(12)
We set here z1 = z z = z, z3 = z4 = . . * = z, = y. Since for any k 2 0
{ [. . . ( (x" - 1, y, z) y) y * . * ] y } 0
0
0
0
2
k
= ([.
. . (x"-
- ([.
* *
1
0
y) 0 y
* *
a ]
0
y, y, z) 0 z
k
(Xn-1 0 y) 0 y
. . .] y, y, Z Z ) 0
V
k
= { . ' . [(x"-
1, y, 22)
0
y]
0
y
* *
a }
0
y
V
k
= (x"-
y, ZZ) 0 y",
l,
then, after we carry out the permutations, identity (12) takes the form
o = 2(n - 1)!(xn-',y,z2)O
f - '= 2(n - 1)!(xn-1,f-1,z2).
This proves the lemma. LEMMA 8. A 4 s J,(A). PROOF. For any a, b E A we have a', a 0 b E J,(A) and 2aba = (a 0 b) 0 a - az 0 b E J,(A), whence also aba E J,(A). Now, as in the proof of Proposi-
tion 5.3,we obtain that the quotient algebra A = A / J , ( A ) is anticommutative
6.3.
=33
ZHEVI.AKOV'S THEOREM
and antiassociative. In the algebra 2 we consider the associator
+
(ab,C,d ) = [(ab)c]d- (ab)(cd)= [a(bc)]d a [ b ( c d ) ] = a[(bc)d] - a[(bc)d] = 0.
On the other hand, we have ( a b , ~ , d=) [(ab)c]d- (ab)(cd) = [(ab)c]d [(ab)c]d= 2[(ab)c]d.
+
The equality obtained proves that
A4 = (0). This proves the lemma.
Now we can prove THEOREM 2 (Zheolakou). Let A be an arbitrary alternative algebra. Then for any natural number n A("("+ I )/2) G J , ( A ) .
PROOF. For n = 1 the assertion of the theorem is obvious. For n = 2 by Lemma 8 we have G A2 . A' G A4 G J,(A), that is, the assertion is l(A), - n > 2. We consider the quotient also true. Now let A("("-')I2) c .I, algebra 2 = A/J,(A). By the induction assumption we have
A("("-l ) / 2 ) E J n - l(A). Furthermore, in view of the containment J,-,(A) E J,(A), from Lemma 7 it follows that J , - , ( A )is an associative nil-algebra of index n. By the NagataHigman theorem the algebra J , - ,(A) is nilpotent of index 2" - 1, whence then finally we obtain (Jn-,(A))(") = (0). Since (A(k))(") = , p n + 1)/2) =
( -A( n ( n -
1)/2) ( n )
)
E
( J n - I(A))'")
= (01,
whence A("("+11/21 c - J,(A).
This proves the theorem. COROLLARY 1. Let A be an alternative nil-algebra of index n without elements of additive order In.Then the algebra A is solvable of index ~ n ( n 1)/2.
+
COROLLARY 2. Let A be an arbitrary alternative algebra. Then for any natural number n there exists a natural number k such that for any LI
A("("
f
I
), 2 )
qa;,
( n ! ) k a= i
where u iE 0 and a, E A .
6.
'34
SOLVABILITY, NILPOTENCY OF ALTERNATIVE ALGEBRAS
Thus every alternative nil-algebra of bounded index is locally nilpotent, over a sufficiently nice ring of operators even solvable, but over an arbitrary ring of operators it is possible it may not be nilpotent. In addition, as for the Nagata-Higman theorem, it is rather easy to see that the restriction on the characteristic is essential for solvability. We note for comparison that, in the case of Jordan algebras, the situation is far from being so clear. We already remarked in Chapter 5 that the answer to the question on local nilpotency of Jordan nil-algebras of bounded index is unknown for Jordan algebras that are not special. The question on solvability of Jordan nil-algebras of bounded index is not solved even for special Jordan nil-algebras of index 3. Exercises
4
Below A is an alternative @-algebrawhere E 0. 1. Prove that for any natural number n the set A, = (A(+))(") is a subalgebra of the algebra A. Hint.Prove that A, = Z2(A,- 1), then apply Lemma 1 and induction on n. 2. Prove that A'in) E (A(+))("). In particular, the algebra A is solvable if and only if the Jordan algebra A ( + )is solvable. Hint.Apply Lemma 8 and induction on n. 3. Let I be an ideal of the algebra A, M be some submodule of the 0module A, and also I 0 A ( + )5 M. Prove that the containment u(Z, A, A, A) G M is valid for any nonassociative word u(x 1, x 2 ,x3,xq) of length 4. Hint.Consider the free alternative @-algebraand apply Lemma 8. 4. Prove that A3"+' G (A(+))"+'. In particular, the algebra A is nilpotent if and only if the Jordan algebra A ( + )is nilpotent. Hint.Apply induction on n and use Exercise 3.
LITERATURE Dorofeev [55], Zhevlakov [66], Kuz'min [103], Nagata [157], Razmyslov [175], Higman r2531. - Results on connections between solvabilityand nilpotency in some other varieties ofalgebras: Anderson [13,141,Dorofeev [59],Zhevlakov [69],Zhevlakov and Shestakov [go], Markovichev [144], Nikitin [ l a ] , Pchelintsev [172, 1741, Roomel'di [184], Shestakov [267].