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As a consequence, similar theorems for Lie p-algebras with one ... The periodicity of cohomologies of cyclic groups and cyclic Lie p-algebras is well known (see ...
Journal of Mathematical Sciences, Vol. 160, No. 6, 2009

COMBINATORIAL ASPECTS OF FREE ASSOCIATIVE ALGEBRAS AND COHOMOLOGIES OF LIE p-ALGEBRAS WITH ONE DEFINING RELATION G. Rakviashvili

UDC 512.554

Abstract. Let R and Q be elements of a free, associative, finitely generated algebra AX over a field k. Assume that a leading homogeneous part Qv does not have two-sided divisors and RQ = QR and Rv = Qtv . In this paper, the solutions of the equation Σi xi Ryi = z in AX are found and with their help, the identity theorem and Freiheitssatz for a finitely generated associative algebra kX; R = 0 with one defining relation are proved. As a consequence, similar theorems for Lie p-algebras with one defining relation are proved; these results are applied to the proof of the periodicity of cohomologies of Lie p-algebras with one defining relation.

CONTENTS 1. 2. 3. 4. 5.

Introduction . . . . . . . . . . . . . . . . . . . . . . Resolution in the General Case . . . . . . . . . . . Free Differential Calculus in Associative Algebras . Construction of Differentials . . . . . . . . . . . . . Zero Divisors in u(L) . . . . . . . . . . . . . . . . . xi Ryi = z, Identity Theorem, and The Equation i

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Freiheitssatz

. . . . . .

. . . . . .

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822 823 823 825 827 828

6. Applications for Lie p-Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 830 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 832

Introduction The periodicity of cohomologies of cyclic groups and cyclic Lie p-algebras is well known (see [1, 8]). In 1950, R. Lyndon proved [5] that cohomologies of groups with one defining relation are periodic with period 2. To obtain this result, he proved the identity theorem applying the Magnus Freiheitssatz [6]. In the present work, similar results for cohomologies of a finitely generated Lie p-algebra L over a field k, char k = p > 0 (see Sec. 6), with one defining relation are obtained; in particular, the free periodic resolution for a trivial L-module k is constructed (Theorem 6.2). Necessary results of the combinatorial algebra theory are proved for an object the special case of which is a restricted universal enveloping algebra u(L). In Sec. 1, we show how to construct a free resolution for an augmented, finitely generated associative algebra. In Sec. 2, necessary statements of a free differential calculus for associative algebras are described. The general method advanced in Secs. 1 and 3 with the help of a free differential calculus is applied to a restricted universal enveloping algebra u(L). In Sec. 4, zero divisors in u(L) are investigated for the purpose of further extending of the resolution from Sec. 3. In Sec. 5, the basic result xi Ryi = z, where xi , yi , z, and R are elements of of Gerasimov [3] about solutions of the equation i

a free associative algebra, is generalized, which enables generalizing his other theorems about identity of words and Freiheitssatz (these theorems for Lie algebras were considered by A. I .Shirshov in [9, Translated from Sovremennaya Matematika i Ee Prilozheniya (Contemporary Mathematics and Its Applications), Vol. 59, Algebra and Geometry, 2008.

822

c 2009 Springer Science+Business Media, Inc. 1072–3374/09/1606–0822 

10]). The author’s results from this section have intersections with [4, 7]. In Sec. 6, all results are combined for the construction a free periodic resolution of the trivial u(L)-module k. 1.

Resolution in the General Case

Let A be an associative algebra over a field k generated by a finite set X = {x1 , . . . , xn1 }, with an augmentation ε : A → k. Let us consider the sequence n1 

where

∂0 =M 1

ε

0 A− → k → 0, A −−−−→

⎞ x1 − ε(x1 ) ⎟ ⎜ .. M01 = ⎝ ⎠, . xn1 − ε(xn1 )

(1)



∂0 (a1 , · · · , an1 ) = (a1 , · · · , an1 )M01 .

The equality xi1 xi2 . . . xim =

m

ε(xi1 ) . . . ε(xij−1 )(xij − ε(xij ))xij+1 . . . xim + ε(xi1 )ε(xi2 ) . . . ε(xim )

j=1

implies that Ker(ε) is generated as a right A-module by elements of the form xij − ε(xij ); a similar statement takes place if we consider Ker(ε) as a left A-module. Therefore, if ε(a) = 0, then a = n1  ai (xi − ε(xi )), ai ∈ A. This implies that a = (a1 , . . . , an1 )M01 , i.e., the sequence (1) is exact. Let i=1 2 us denote by M 1 the set of all (a1 , . . . , an1 ) ∈ n1 A such that (a1 , . . . , an1 )M01 = 0. Assume that 2 ⊆ M 21 is a subset with n2 (n2 may be infinite) elements which generates M 21 as a left A-module, M 1 and let M12 be the corresponding matrix with n2 rows and n1 columns. Let us consider the sequence n2 

∂1 =M12

A −−−−→

n1 

∂0 =M 1

ε

0 A− → k → 0. A −−−−→

It is exact since, first, from M12 M01 = 0 it follows that ∂1 ∂0 = 0 and, second, if a ∈ Ker ∂0 , then aM01 = 0, and, hence, a is a linear combination of lines of the matrix M12 ; therefore, choosing appropriately (a1 , . . . , an2 ), we obtain that (a1 , . . . , an2 )M12 = 0. If we continue this construction, we obtain a free resolution of the left A-module k. 2.

Free Differential Calculus in Associative Algebras

Definition 2.1. Let H be a bialgebra with an augmentation ε : H → k and M be a left H-module. Then a linear map d : H → M is called a derivation if d(h1 h2 ) = d(h1 )ε(h2 ) + h1 d(h2 ). Let X = {x1 , . . . , xn } be a finite set. Let AX, LX, and Lp X be a free associative algebra, a free Lie algebra, and a free Lie p-algebra generated by a set X over a field k, respectively. It is known [9] that AX  U (LX)  u(Lp X), where U (−) and u(−) denote the universal enveloping and the restricted universal enveloping algebra, respectively. Derivations in these algebras are defined provided that ε(x) = 0 if x ∈ X. Assume that R1 , R2 , . . . , Rm ∈ Lp X. Let us define L = Lp X/R, where R is the ideal generated by the elements R1 , . . . , Rm ; then L is a Lie p-algebra with m defining relations. Clearly, u(L)  AX/R, where R denotes here the ideal generated by elements R1 , . . . , Rm already in AX (Ri are considered as elements in AX with the help of an embedding Lp X → AX). Let us denote by dAX the free AX-module generated by free generators {dx1 , . . . , dxn }. Let us

i · · · xi ) = xi · · · xi dxi . It is clear that it is a derivation and it define d : AX → dAX as d(x 1 1 j j−1 j 823

can naturally be called universal. Let us define a partial derivative ∂u/∂xi as the coefficient of dxi in

d(u): n

∂u ∂u

dxi , ∈ AX. d(u) = ∂xi ∂xi i=1

Partial derivatives of a product are calculated as derivations of a product. It is clear that for any AX-module M and a derivation d : AX → M , the equality d(u) =

n

∂u d(xi ) ∂xi i=1

holds, i.e., if two derivations coincide on generators, then they coincide everywhere. Let us denote AX+ = ker(ε), where ε : AX → k is the augmentation. Any u ∈ AX can be presented uniquely as the sum u = α + u+ , α ∈ k, u+ ∈ AX+ . Let us define a k-linear map D : AX → AX as D(u) = u+ = u − ε(u). Let us prove that D is a derivation: + + + + + D(u1 u2 ) = D(α1 + u+ 1 , α2 + u2 ) = α2 u1 + α1 u2 + u1 u2 = D(u1 )ε(u2 ) + u1 D(u2 ).

It follows from the definition of D that n

∂u+ + xi , u = ∂xi

+ ) = 0 ⇐⇒ u+ = 0. d(u

(2)

i=1

Let du(L) be the free u(L)-module generated by the free generators {dx1 , · · · , dxn }. Let us define d = π d : AX → du(L), where π is the natural projection AX → u(L). For simplicity, we shall write further a ≡ b instead of π(a) = π(b), a, b ∈ AX. Let d : AX → du(L)  d(AX/R) be any derivation. If r ∈ R, then ru(L) = 0 and ε(r) = 0; therefore, we have d(arb) = d(ar)ε(b) + ard(b) ≡ aε(b)d(r). Lemma 2.1. Let d : AX → du(L)  d(AX/R) be any derivation. Then d(u) ≡ 0 ⇐⇒ u = α +

n

α ∈ k,

ri xi ,

ri ∈ R.

i=1

Proof. Assume that u=α+

n

ri xi ,

α ∈ k,

ri ∈ R;

i=1

then d(u) ≡ 0. Conversely, if u=α+

n

ri xi

i=1

and d(u) ≡ 0, we have that ri ∈ R since d(u) = d(α) +

n

d(ri )ε(xi ) +

i=1

n

i=1

ri d(xi ) ≡

n

i=1

and dxi are free generators. Lemma 2.2. Let u1 , . . . , un ∈ AX. Then n

ui xi ≡ 0

i=1

if and only if there exists r ∈ R such that ∂r/∂xi ≡ ui , i = 1, . . . , n. 824

ri d(xi )

Proof. Let r=

n

ui xi ∈ AX.

i=1 ∂r ≡ ui , i = 1, . . . , n. Assume that r ≡ 0; then r ∈ R and ∂x i Conversely, assume that r ∈ R and ∂r/∂xi ≡ ui , i = 1, . . . , n; then r ∈ AX+ . Therefore, it follows from (2) that n

∂r xi = r ≡ 0. ∂xi i=1

3.

Construction of Differentials

Let L = (x1 , . . . , xn )/(R1 , . . . , Rm ) be a Lie p-algebra with n generators and with m defining relations. Our purpose is to continue the exact sequence (1) when A = u(L). Proposition 3.1. Assume that L = (x1 , . . . , xn )/(R1 , . . . , Rm ) and A = u(L). Then it is possible to assume that ⎛ ⎞ x1   ∂Ri ⎜ .. ⎟ 1 2 . M0 = ⎝ . ⎠ , M1 = ∂xj i,j xn1 Proof. It is already proved (see (1)) that M01 is of such a type as in the proposition. Assume in (1) that n

∂0 (u1 , . . . , un ) = (u1 , . . . , un )M01 = ui xi ≡ 0. i=1

Assume that r=

n

ui xi ,

r = 0,

i=1

since here r is considered as an element from AX). Then r ∈ R and ∂r/∂xi = ui , i = 1, . . . , n. Therefore, it suffices to prove that the linear combination of the rows of the matrix M12 = (∂Ri /∂xj )i,j with coefficients from AX and the set {(∂r/∂x1 , . . . , ∂r/∂xn )}, r ∈ R, coincide. Indeed, if r ∈ R, then we have n

∂r xi = r ≡ 0 ∂xi i=1

by Lemma 2.2 and this means that (∂r/∂x1 , . . . , ∂r/∂xn )M01 ≡ 0. Since r ∈ R, we should have

r= ai Rti bi , i

and consequently d(r) =



ai ε(bi )dRti ,

i

i.e., (∂r/∂x1 , . . . , ∂r/∂xn ) is a linear combination of rows of the matrix (∂Ri /∂xj )i,j with coefficients from AX. Lemma 3.2. Let M01 and M12 be matrices from Proposition 3.1. For T =

i

ai Rti bi ∈

n

(R)xi ,

i=1

825

assume that ut =



ai ε(bi ),

t = 1, . . . , m.

ti =t 3

3

Then u = (u1 , . . . , um ) ∈ M 2 . Conversely, if (u1 , . . . , um ) is a row from M 2 , then m

ui Ri ∈

n

i=1

(R)xi .

i=1

Proof. We have by Lemma 2.1 dT =



ai ε(bi )dRti =

m



ai ε(bi )dRt =

m

t=1 ti =t

i

ut dRt

t=1

=

m

ut

t=1

Since dxi are free generators,

m

ut

t=1

n

∂Rt i=1

∂xi

dxi =

m n



i=1

t=1

∂Rt ≡ 0, ∂xi

that is, (u1 , . . . , um )M12 = 0. 3 Conversely, assume that u = (u1 , . . . , um ) ∈ M 2 . Assume that T =

m

ui Ri .

i=1

Since

m

uj

j=1

we have dT =

m

ui dRi =

i=1

m

ui

i=1

∂Rj ≡ 0, ∂xi

n

∂Ri j=1

and by Lemma 2.1 we have that T ∈

n

m

∂Ri dxj = ( ui )dxj ≡ 0, ∂xj ∂xj j=1 i=1

n

(R)xi

i=1

since ε(T ) = 0. Corollary 3.3. In Lemma 3.2, it suffices to consider elements of the type

ai Rti bi = 0. i

Proof. Assume that T =

i

ai Rti bi ∈

n

(R)xi .

i=1

Then by Lemma 3.2, the row u = (u1 , . . . , um ), where

ai ε(bi ), t = 1, . . . , m, ut = ti =t

826

∂Rt ut dxi ∂xi

 ≡ 0.

corresponds to the element T . But since T ∈

n

(R)xi ,

i=1

there exist T =



ci , di ∈ AX,

ci Rhi di xj ,

i

and



vh =

ci ε(di xj ) = 0,

h = 1, . . . , m;

hi =h

therefore, the difference



ai Rti bi −



i

ci Rhi di xj ,

i 3

i.e., the element 0 from AX, and the element T corresponds to the same element u ∈ M 2 . 4.

Zero Divisors in u(L)

In the present section, we consider Lie p-algebras only with one defining relation R. Let (R) denote q q    i i R and let fpq (x) and fpq (x) denote a p-polynomial αi xp and its derivative αi xp −1 , respectively, i=0

where αi ∈ k, αq = 0.

i=0

Proposition 4.1. Assume that R ∈ Lp X and R = fpq (Q) for some Q ∈ Lp X. Then from U Q ≡ 0   in u(L)  AX/(R) it follows that U ≡ V fpq (Q) for some V and from U fpq (Q) ≡ 0 it follows that U ≡ V Q for some V . q

Proof. The set {Q,Q[p] , . . ., Q[p] −1 } is linearly independent in L = AX/(R). We add this set to a q kxi ⊕ kQ ⊕ · · · ⊕ kQ[p] −1 in such a manner that (we assume that the index k-linear basis L = of Qi is Qi )

i∈I

q −1

I ∩ {Q, Q[p] , . . . , Q[p]

} = ∅. r

s

Let us order I linearly and assume that ∀i ∈ I, i < Q, r < s ⇒ Q[p] < Q[p] . Then the basis in u(L) has the form xji11 . . . xjirr Qj , i1 < · · · < ir , 0 ≤ js < p, 0 ≤ j < pq . Therefore, if U ∈ u(L), then U=

q −1 p

ui Qi ,

i=0

where ui is contained in the subspace of the algebra u(L) generated by the elements xji11 . . . xjirr , i1 < · · · < ir , 0 ≤ js < p. Assume that U Q ≡ 0; since fpq (Q) = 0 in u(L) we have UQ =

q −1 p

i+1

ui Q

i=0

=

q −2 p

i=0

ui Qi+1 −

q−1

1 i upq −1 αi Qp ≡ 0. αq i=0

This implies that ui = 0 if i + 1 = pJ , j = 0, 1, . . . , q − 1, and upJ −1 = Thus,

1 upq −1 αj , j = 0, 1, . . . , q − 1. αq

q−1

1 1 j  U= upq −1 αj Qp −1 + upq −1 Qpq −1 = upq −1 fpq . αq αq j=0

827









Assume that U fpq ≡ 0; then, since Qfpq Q ≡ 0, we have U fpq ≡ u0 fpq ≡ 0, from which it follows that u0 = 0, i.e.,  pq −1

ui Qi−1 Q. U= i=1

 Corollary 4.2. If Q is one of the free generators of a free Lie p-algebra Lp X and i ai Rbi = 0, then there exists V such that i ai ε(bi ) ≡ V Q.   Proof. Since Q is a free generator, dR = dfpq (Q) = fpq (Q)dQ. From i ai Rbi = 0 it follows that  



 ai Rbi ≡ ai ε(bi )dR = ai ε(bi )fpq (Q)dQ; 0≡d therefore, tion (4.1).



i q

i ai ε(bi )fp

5.

i

i

(Q) ≡ 0 since dQ is a free generator, and the statement follows from Proposi-

The Equation

 i

xi Ryi = z, Identity Theorem, and Freiheitssatz

Further, we assume that X = {x1 , . . . , xn } is a finite set and on AX, a positive integral-valued degree function v is fixed such that on AX there exist a weak algorithm (see [2, Sec. 2.2] and [3]) relative to v; in particular, we can assume that v is the standard degree function for which v(xi ) = 1, where xi are free generators [2]. Denote by AXα , α ≥ 0, the subspace generated by monomials of degree α. The following theorem was obtained for fields of any characteristic (Gerasimov [3]). Theorem 5.1. Let w be any homogeneous element of the algebra AX. Then the set {AXα wAXβ }α,β≥0 generates in AX the distributive lattice of subspaces. Denote by uv the leading homogeneous part of an element u ∈ AX relative to the degree function v. If R ∈ AX, then (3) Rv AXv ∩ AXv Rv = (RAX ∩ AXR)v . As a consequence of this theorem, Gerasimov obtained the statement on the kind of solutions of the  equation i xi Ryi = z, and, for associative algebras with one defining relation R generated by element R, the identity theorem and Freiheitssatz. Recall that h ∈ AX is called a proper two-sided divisor of a ∈ AX if a = hb1 = b2 h for some b1 , b2 ∈ AX, where 0 < v(h) < v(a). Proposition 5.2. Assume that for R, Q ∈ AX, an element Qv does not have proper two-sided divisors and Rv = Qtv and RQ = QR. Then the element R satisfies condition (3). Proof. Assume that w = Rx = yR ∈ RAX ∩ AXR. Then wv = Rv xv = yv Rv ∈ Rv AXv ∩ AXv Rv . Conversely, assume that w ∈ Rv AXv ∩ AXv Rv ; then w ∈ Rv AXα ∩ AXα Rv , i.e., w = Qtv x = yQtv , where x, y ∈ AXα . It is well known (see [2, Proposition 6.6.3]) that a multiplicative semigroup of the homogeneous elements of a free associative algebra has a property that if ab = cd, then a = ch, d = hb or c = dh, b = hd. Here two cases are possible: 828

(a) x = hQtv , y = Qtv h, where h is a homogeneous element in AX. We have w = Qtv hQtv . Assume that w1 = hR, w2 = Rh, and w0 = RhR; then w0 = Rw1 = w2 R ∈ RAX ∩ AXR and since hv = h, we have (w0 )v = Qtv hQtv = w, i.e., (a) is proved. (b) Qtv = yh = hx. Since Qv does not have proper two-sided divisors, we have x = y = Qiv for some (t+i) i ≤ t. Then w = Qtv x = yQtv = Qv . Assume that z = RQi ; it is clear that z ∈ RAX ∩ AXR (t+i) and zv = Qv = w, i.e., (b) is proved. Proposition 5.3. Assume for R, Q ∈ AX that Qv does not have proper two-sided divisors and Rv = Qtv and RQ = QR. Then if Rx = yR, we have x = zR + ϕ(Q), y = Rz + ϕ(Q),

(4)

where z and ϕ are any elements from AX and the polynomial ring k[x], respectively. Proof. It is clear that (4) satisfies the equation Rx = yR; let us prove that this equation has no other solutions. Let us apply induction on v(x) = v(y). If v(x) = 0, then the statement is valid. Assume that v(x) > 0 and Rx = yR; then Rv xv = yv Rv . The following two cases are possible: (a) v(Rv ) ≤ v(yv ). Then xv = hRv and yv = Rv h for some homogeneous element h ∈ AX; this implies that v(x − hR) = v(y − Rh) < v(x) = v(y) and R(x − hR) = (y − Rh)R. By induction, y − Rh = Rz1 + ϕ(Q),

x − hR = z1 R + ϕ(Q), and hence x = (z1 + h)R + ϕ(Q),

y = R(z1 + h) + ϕ(Q),

Rx = yR.

(b) v(Rv ) > v(yv ). Then, as in part (b) of the proof of Proposition 5.2, we have that xv = yv = Qiv . Assume that x1 = x − Qi and y1 = y − Qi . It is obvious that v(x1 ) = v(y1 ) < v(x) = v(y) and Rx1 = y1 R. Therefore, by induction, x1 = z0 R +ϕ(Q) and y1 = Rz0 +ϕ(Q). Assume that x = x1 +Qi and y = y1 + Qi ; then it is obvious that x and y satisfy the conditions of the proposition. Note that Propositions 5.2 and 5.3 for the special case t = 1 were proved in [3]. Proposition 5.4. Assume that R, Q ∈ AX and Qv does not have a proper two-sided divisor, Rv = Qtv , and RQ = QR. Then for any element S ∈ AX, the following conditions are equivalent:  (a) If the operator T (z) = Xi zYi satisfies the condition T (R) = S, then

Tj (zxj − yj z), T (z) = T0 (z) + j

where Tj are some operators, xj , yj ∈ AX, Rxj = yj R, and the operator T0 (z) = satisfies the condition  v(ai ) +j v(bi ) ≤ v(S) − v(R). (b) Denote AXi = j≤i AX , i ≥ 0. Then for any i ≥ 0, we have

AXi ∩ AXRAX = AXα RAXβ .



ai zbi

α+β+v(R)≤i

Proof. The validity of this statement immediately follows from Proposition 5.2 and [3, Propositions 1 and 2]. Theorem 5.5 (Freiheitssatz). Let Q, R ∈ AX, y ∈ X, Q ∈ / AX\{y}, and let Qy denote the leading part of Q with respect to the variable x. Assume that Rw = Qtw for any degree function w, RQ = QR, and (Qy )v does not have proper two-sided divisors. Then the associative algebra with one defining relation kX; R = 0 satisfies the Freiheitssatz, i.e., the set X\{y} generates in kX; R = 0 a free associative subalgebra. 829

Proof. We shall follow argumentations from the proof of [3, Proposition 3]. Assume that f ∈ AX\{y} ∩ AXRAX. Introduce the new degree function v1 (x) = {v(x), x ∈ X\{y}; top max{v(Q), v(f )} + 1,

x = y.

(5)

Then we have (Qy )v = Qv1 . Indeed, let N be the exponent of the element Q with respect to the variable y and let a monomial M enter in Q with a nonzero coefficient. If the exponent of M on y is equal to N , then v1 (M ) ≥ N v1 (y) > (N − 1)v1 (y) + v(Q). If the exponent of M on y is less than N , then v1 (M ) ≤ (N − 1)v1 (y) + v(Q). Thus, (Qy )v = Qv1 , and by Proposition 5.4(b) we have

AXv1 (f ) ∩ AXRAX = AXα RAXβ α+β+v1 (R)≤v1 (f )

(here the homogeneity is understood relative to the function v1 ). However, since v1 (f ) = v(f ) < v1 (y) ≤ v1 (R), there are no α, β ≥ 0 such that the inequality α + β + v1 (R) ≤ v1 (f ) holds. This implies that f ∈ AXv1 (f ) ∩ AXRAX = {0}, i.e., f = 0. The theorem is proved. Proposition 5.6. Assume that R, Q ∈ AX, Qv does not have proper two-sided divisors, and Rv = Qtv and RQ = QR. Then in the algebra with one defining relation kX; R = 0, the identity problem is solvable. Proof. This follows from Proposition 5.2 and [3, Proposition 2]. 6.

Applications for Lie p-Algebras q

Lemma 6.1. Let R, Q ∈ AX and R = fpq (Q). Then Rv = Qpv and RQ = QR. Proof. The proof is obvious. Theorem 6.2. Let L = Lp X/(R) be a Lie p-algebra with one defining relation R = fpq (Q), where X = {x1 , . . . , xn }, and Qv does not have proper two-sided divisors. Then the following sequence is the free resolution of the trivial u(L)-module k, periodic with period 2 starting from second dimension: ⎛x ⎞ 1

.. ⎠ q n ∂R ∂R .  ( ,..., ) f (Q) ∂x ∂xn Q Q p x ε → u(L) −−−−→ u(L) − → u(L) −−−1−−−−−→ → k → 0, u(L) −−−n−→ u(L) − ··· − ⎝

u

→ denotes multiplication by u from the right. where − Proof. The exactness in dimension i = 0 follows from (1), in dimension i = 1 from Proposition 3.1, and in dimensions i = 3, 4, . . . from Proposition 4.1. We need to prove the exactness only in dimension 2. Since ∂Q ∂R  = fpq (Q) , ∂xi ∂xi we have     ∂R ∂Q ∂Q ∂R  ,..., ,..., = fpq (Q) . ∂x1 ∂xn ∂x1 ∂xn 

Therefore, since Qfpq (Q) = R ≡ 0, we have   ∂R ∂R ≡ 0. ,..., Q ∂x1 ∂xn 3

Then by Lemma 3.2 and Corollary 3.3 we have that  M 2 is exhausted by u ∈ AX such that u =  ai εi (bi ), where ai and bi satisfy the condition ai Rbi = 0. By Proposition 5.4, all solutions of i i   ai Rbi = 0 have the form Ti (zxi − yi z), where Rxi = yi R, and Ti is any operator the equation i

830

i

Ti (z) =

 j

aij zbij . But then by Proposition 5.3, assuming that xi = gi R + fi (Q) and yi = Rgi + fi (Q),

where gi ∈ AX and fi ∈ k[x], we have



Ti (zxi − yi z) = aij z(gi R + fi (Q))bij − aij (Rgi + fi (Q))zbij , i

i,j

from whence it follows that u=



aij ε((gi R + fi (Q))bij ) − aij (Rgi + fi (Q))e(bij ).

i,j

Since ε(u(L)+ ) = 0, it is possible to suppose that bij = 1; then



(aij ε(fi (Q)) − aij (Rgi + fi (Q))) ≡ aij (ε(fi (Q)) − fi (Q)). u= i,j

i,j

It is clear that ε(fi (Q)) − fi (Q) = wi Q for some wi ∈ AX, i.e., the theorem is proved (see §1). Corollary 6.3. Let L = Lp X/(R) be a Lie p-algebra with one defining relation R = fpq (Q), where X = {x1 , . . . , xn }. Assume that Qv does not have proper two-sided divisors and M is some L-module. Then  n 

∂Q  M , H 2 (L, M )  Q M/fpq (Q) ∂xi i=1

H 2i+1 (L, M )  f  q (Q) M/QM, i ≥ 1, p



H 2i (L, M )  Q M/fpq (Q)M, i ≥ 2. Proof. This follows from Lemma 6.1 and Theorem 6.2. Corollary 6.4. Let L = Lp X/(R) be a Lie p-algebra with one defining relation R = fpq (Q), where X = {x1 , . . . , xn }. Assume that Qv does not have proper two-sided divisors and M is some L-module. Let us denote by C the (cyclical ) p-subalgebra of L generated by an element Q. Then H i (L, M )  H i (C, M ),

i ≥ 3.

Proof. This follows from Corollary 6.3 and from [P68]. Theorem 6.5 (Freiheitssatz). Let L = Lp X/(R) be a Lie p-algebra with one defining relation R = fpq (Q), where X = {x1 , . . . , xn , y}. If Q contains y explicitly and (Qy )v does not have proper two-sided divisors, where Qy denotes the leading part of Q with respect to y, then L satisfies the Freiheitssatz, i.e., the set {x1 , . . . , xn } generates in L = Lp X/(R) a free Lie p-subalgebra. Proof. Let us consider the universal enveloping algebra u(L); then the theorem follows from Lemma 6.1 and Theorem 5.5. Theorem 6.6. Let L = Lp X/(R) be a Lie p-algebra with one defining relation R = fpq (Q), where X = {x1 , . . . , xn }. Assume that Qv does not have proper two-sided divisors. Then in the Lie p−algebra L, the identity problem is solvable. Proof. Let us consider the universal enveloping algebra u(L); then the theorem follows from Lemma 6.1 and Proposition 5.6. 831

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