Torsion in free centre-by-nilpotent-by-abelian Lie rings of rank 2

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by  Ralph Stöhr

TORSION IN FREE CENTRE-BY-NILPOTENT-BY-ABELIAN LIE
RINGS OF RANK 2
RALPHSTO¨HR
Abstract. For c ≥ 2, the free centre-by-(nilpotent-of-class-c-1)-by abelian
7
LieringonasetX isthequotientL/[(L′)c,L]whereListhefreeLieringon
1
0 X, and (L′)c denotes the cth term of the lower central series of the derived
2 idealL′=L2 ofL. Inthispaperwegiveacompletedescriptionofthetorsion
n subgroup of its additive group in the case where |X| = 2 and c is a prime
a number.
J

0
1
1. Introduction
]
A
For an integer c ≥ 2, the free centre-by-(nilpotent-of-class c−1)-by-abelian Lie
R
ring on a set X is the quotient
.
h
t (1.1) L/[(L′)c,L]
a
m where Lis an(absolutely)free Lie ringonX and(L′)c is the cthterm ofthe lower
[ central series of the derived ideal L′ = L2 of L. In this note we determine the

1 torsionsubgroupoftheadditivegroupof (1.1)inthecasewhereLhasrank2,that
v
is X is a set of two elements, and c is a prime number. It is easily observed that
4
9 for any c ≥ 2 and any X with |X| ≥ 2 the torsion subgroup of (1.1) is contained
5 in the central ideal (L′)c/[(L′)c,L]. If c = p, where p is a prime, then the ideal
2
0 (L′)p/[(L′)p,L] is a direct sum of a free abelian group and an infinite elementary
1. abelianp-group. WeexhibitanexplicitZp-basisofthetorsionpartif|X|=2. This
0 is the main result of this paper, see Theorem 5.1 in Section 5. In the earlier paper
7
[1]acompletedescriptionofthetorsionsubgroupoftheadditivegroupofthemore
1
: reduced quotient
v
Xi (1.2) L/([(L′)c,L]+L′′′)
r
a was obtained, again in the case of rank 2, but for arbitrary c≥2. Note that (1.2)
coincides with (1.1) for c =2 and c =3, but not for c ≥4. We make essential use
of the results from [1].
The interest in torsion in relatively free Lie rings of the form (1.1) as well as
their group-theoretic counterparts, the relatively free groups F/[γ (F′),F],where
c
F is a free group and γ (F′) is the cth term of the lower central series of the
c
derivedsubgroupF′,hasalonghistory. Ifc=2,theseturnintothefreecentre-by-
metabelian Lie rings and the free centre-by-metabeliangroups,respectively, and it
was the latter in which Kanta Gupta [8] first discovered torsion elements, a major
Date:January11,2017.
2010 Mathematics Subject Classification. Primary17B01, 17B55.
1
2 RALPHSTO¨HR
surprise at the time. Later torsion was detected in F/[γ (F′),F] if c is a prime
c
[15] and if c = 4 [16]. Quite surprising, though, it turned out that F/[γ (F′),F]
c
is torsion free if c is divisible two distint primes [10]. As to Lie rings, early work
in [12] on the free centre-by-metabelian Lie rings L/[L′′,L], and in particular on
torsionintheadditivegroup,turnedoutinneedofsomerectification,andthiswas
eventually accomplishedin [14] and [11]. For larger values of c, Drensky [7] proved
that for any prime p the free Lie ring L/[(L′)p,L] contains non-trivial multilinear
elements of degree 2p+1 which have order p.
The paper is organized as follows. In Section 2 we introduce notation and as-
sembleanumberofpreliminaryresultsonthehomogeneouscomponentsoffreeLie
rings. These are further examined in Section 3 where we derive our main result on
Lie powers of free modules in prime degree, postponing, however, one key ingredi-
ent on Lie powers of prime degree p over fields of characteristic p to Section 4. In
thefinalSection5weexploitthe resultsoftheprevioussectionstoderiveourmain
result.
2. Notation and some preliminaries
We write maps on the right and use left-normed notation for Lie products. Let
Abeafreeabeliangroup. ByL(A)wedenotethefreeLieringonA. Forapositive
integerc,weletLc(A)denotethedegreechomogeneouscomponentofL(A),thatis
the spanof all left normed simple Lie products [a ,a ,...,a ] with a ∈A. In par-
1 2 c i
ticular,L1(A)=A. TheuniversalenvelopeofL(A)canbeidentifiedwiththetensor
ring T(A) = Ti(A) where T0(A) = Z and, for c > 0, Tc(A) = A⊗···⊗A,
c≥0
c
the cth tensorLpower of A. By Ac we denote the cth symmetric power of A. The
| {z }
free metabelian Lie ring on A is the quotient of L(A) by its second derived ideal:
M(A) = L(A)/L(A)′′. This too is a graded Lie ring and we let Mc(A) denote its
cth homogeneous component, that is Mc(A) = Lc(A)/(Lc(A)∩L(A)′′). We call
Lc(A) and Mc(A) the cth free Lie power and the cth free metabelian Lie power of
A, respectively. It is well-known that if A is an ordered Z-basis of A, then the left
normedsimple Lie products [b ,b ,...,b ] with b ∈A and b >b ≤···≤b form
1 2 c i 1 2 c
a Z-basis of Mc(A) (see [2, Section 4.2.2]).
The canonical embedding of L(A) into its universal envelope T(A) induces in
each degree c an embedding ν : Lc(A) → Tc(A). By a well-known theorem of
c
Wever (see [13, Chapter 5, Theorem 5.16]), the composite of this embedding with
the natural projection ρ : Tc(A) → Lc(A) defined by a ⊗···⊗a 7→ [a ,...,a ]
c 1 c 1 c
amounts to multiplication by c on Lc(A):
(2.1) Lc(A)−ν→c Tc(A)−ρ→c Lc(A), ν ρ =c.
c c
The definition of Mc(A) gives rise to a short exact sequence
(2.2) 0→Bc(A)→Lc(A)−η→c Mc(A)7→0
FREE CENTRE-BY-NILPOTENT-BY-ABELIAN LIE RINGS 3
where Bc(A)=Lc(A)∩L(A)′′ andη is the naturalprojectionmap. Moreover,for
c
c>2 the metabelian Lie power Mc(A) fits into a short exact sequence
(2.3) 0→Mc(A)−µ→c A⊗Ac−1 −κ→c Ac →0
where the maps µ and κ are given by
c c
[a ,a ,...,a ]7→a ⊗(a ◦···◦a )−a ⊗(a ◦···◦a )
1 2 c 1 2 c 2 1 c
and a ⊗(a ◦···◦a ) 7→ a ◦a ◦···◦a , respectively (see [9, Corollary 3.2]).
1 2 c 1 2 c
Moreover,there is a map λ :A⊗Ac−1 →Mc(A), given by
c
a ⊗(a ◦···◦a )7→[a ,a ,a ,...,a ]+[a ,a ,a ,...,a ]+···+[a ,a ,a ,...,a ],
1 2 c 1 2 3 c 1 3 2 c 1 c 2 c−1
such that the composite of µ and λ amounts to multiplication by c on Mc(A):
c c
(2.4) Mc(A)−µ→c A⊗Ac−1 −λ→c Mc(A), µ λ =c
c c
see [1, Section 3]. Finally, for c≥2 there is a map θ :Mc(A)→Lc(A) given by
c
1
(2.5) [a ,...,a ]7→ a ,a ,...,a − a ,a ,...,a ,
1 c c 1 π(2) π(c) 2 τ(1) τ(c)
!
σ τ
X(cid:2) (cid:3) X(cid:2) (cid:3)
where the sums run over all permutations σ of {2,3,...,n} and all permutations τ
of{1,3,...,n},respectively. AlthoughweworkoverZ,thefactor1/cin(2.5)makes
sense as the expression on the right hand side can be written as a Lie polynomial
withintegercoefficients(see[4,Section2]). ThisLiepolynomialhasbeencalculated
in [5, Proposition7.3],but since it is rather involved,we prefer to use the compact
form(2.5) inwhatfollows. The compositeof θ andthe naturalprojectionη as in
c c
(2.2) amounts to multiplication by (c−2)! on Mc(A):
(2.6) Mc(A)−θ→c Lc(A)−η→c Mc(A), θ η =(c−2)!
c c
(see [4, Section 2]).
Now suppose that A carries the structure of a module for the polynomial ring
U =Z[X]whereX isafinitesetofvariables. Thenalltheobjectsintroducedinthis
section such as Lie powers, symmetric powers etc. will be regarded as U-modules
under the derivation action. For example, for x∈X, a ∈A,
i
c
[a ,a ,...,a ]x= [a ,...,a x,...,a ],
1 2 c 1 i c
i=1
X
Notethatallthemapsintroducedinthissectionarecompatiblewiththederivation
action,thatis, allthese maps are,in fact, Z[X]-module homomorphisms. This will
be used in what follows without further reference being given. The ring of integers
Z will be regarded as a trivial U-module.
3. Lie powers of free modules
InthissectionweretainthenotationintroducedinSection2,butnowweassume
throughout that A is a free U-module for the polynomial ring U = Z(X). All
homologygroupsinthissectionwillbeoverthegroundringU. Forbrevity,ifW is
a U-module, the homology groups H (U,W)=TorU(W,Z), k ≥0, will be written
k k
as H (W).
k
4 RALPHSTO¨HR
Itis wellknown(see, forexample,[14, Lemma 5.2])that ifA is afree U-module
then both Tc(A) and A ⊗ Ac−1 with c ≥ 2 are also free U-modules under the
derivation action.
Lemma 3.1. Let A be a free U-module, c≥2. Then
(i) the tensor products L (A)⊗ Z and M (A)⊗ Z are direct sums of a free
c U c U
abelian group and a torsion group of exponent dividing c,
(ii) for k ≥1 the homology groups H (L (A)) and H (M (A)) are torsion groups
k c k c
of exponent dividing c.
Proof. Byapplyingthehomologyfunctortothemapsin(2.1)wegetthatH (ν ρ )
k c c
is multiplication by c on H (L (A)) for all k ≥ 0. Since A is a free U-module,
k c
H (T (A))isfreeabelianandH (T (A))=0fork ≥1. Then,foranyu∈H (LcA)
0 c k c k
with k >0,
cu=uH (ν ρ )=(uH (ν ))H (ρ )=0H (ρ )=0.
k c c k c k c k c
The same holds if k = 0 and u ∈ Ker(H (ν )). Hence multiplication by c
0 c
annihilates both the homology groups H (L (A)) for k ≥ 1 and the kernel of the
k c
homomorphism H (ν ). The image of this homomorphism is contained in the free
0 c
abelian group H (T (A)), and hence itself free abelian. The results (i) and (ii)
0 c
for L (A) follow. The proof of (i) and (ii) for M (A) are obtained by a similar
c c
argument using the maps in (2.4) instead of those in (2.1). (cid:3)
Nowwe considerLie powersofprime degree. Letp be a prime. By applyingthe
homology functor to the short exact sequence (2.2) we obtain the exact sequence
(3.1) ···→H (Mp(A))→Bp(A)⊗ Z→Lp(A)⊗ Z−η−p−⊗→1 Mp(A)⊗ Z7→0.
1 U U U
ByLemma3.1.,Lp(A)⊗ Zisadirectsumofafreeabeliangroupandanelementary
U
abelian p-group,and H (Mp(A)) is an elementary abelian p-group. Note that this
1
does not exclude the possibility that these torsion subgroups are trivial. Now the
exactsequence(3.1) yields thatBp(A)⊗ Z is a directsum ofa free abeliangroup
U
and a (possibly trivial) p-groupof exponent dividing p2. In fact, we will show that
thisgrouphasactuallynotorsion,inotherwords,wewillprovethefollowingresult.
Lemma 3.2. Let A be a free U-module and p a prime. Then the tensor product
Bp(A)⊗ Z is a free abelian group.
U
Proof. Since we already know that Bp(A)⊗ Z is a direct sum of a free abelian
U
group and a p-group of finite exponent, it is sufficient to show that no non-zero
elementinBp(A)⊗ Zis annihilatedby p. We use reductionmodulo p,that isthe
U
short exact sequence
0→Bp(A)−→p Bp(A)→Bp(A)⊗Z →0
p
which, in its turn, gives rise to the exact sequence
(3.2) ···→H (Bp(A)⊗Z )→Bp(A)⊗ Z−→p Bp(A)⊗ Z→Bp(A)⊗ Z →0.
1 p U U U p
The Lemma will be proved once we show that the homology group on the left is
zero, and this will certainly follow if we can verify that Bp(A)⊗Z , regarded as a
p
FREE CENTRE-BY-NILPOTENT-BY-ABELIAN LIE RINGS 5
module for the polynomial ring Z [X], is projective. The proof of this fact will be
p
given in the next section (see Corollary 4.2). This will then complete the proof of
Lemma 3.2. (cid:3)
Nowwe haveallthe ingredientsin placeto provethe mainresultofthis section.
Recall the homomorphism θ :Mc(A)→Lc(A) defined by (2.5).
c
Proposition 3.3. Let A be a free U-module and p a prime. Then the torsion
subgroups of Lp(A)⊗ Z and Mp(A)⊗ Z are isomorphic, and the homomorphism
U U
θ ⊗1 maps the latter isomorphically onto the former.
p
Proof. By Lemma 3.1(i) both Lp(A)⊗ Z and Mp(A)⊗ Z are direct sums of a
U U
free abeliangroupand an elementary abelianp-group. Consider the maps in (2.6).
By trivializing the U-action we obtain homomorphisms
Mp(A)⊗ Z−θ−p−⊗→1 Lp(A)⊗ Z−η−p−⊗→1 Mp(A)⊗ Z, θ η ⊗1=(p−2)!
U U U k k
So the restriction of the composite θ η ⊗1 to the torsion subgroup of the tensor
k k
product Mp(A)⊗ Z, an elementary abelian p-group,is multiplication by (p−2)!,
U
that is, it is an isomorphism. It follows that the homomorphism θ ⊗ 1 maps
p
the torsion subgroup of Mp(A)⊗ Z isomorphically into the torsion subgroup of
U
Lp(A)⊗ Z. To prove the proposition,we need to verify that this map is also sur-
U
jective,thatis,thehomomorphismθ ⊗1mapsthetorsionsubgroupofMp(A)⊗ Z
k U
isomorphicallyonto thetorsionsubgroupofLp(A)⊗ Z. Butthisistruesinceoth-
U
erwisethe restrictionofη ⊗1to the torsionsubgroupofLp(A)⊗ Zwouldhavea
p U
non-trivial kernel. This, however, is not the case, as follows from the exactness of
(3.1). Since Bp(A)⊗ Z is free abelian by Lemma 3.2, and H (Mp(A)) is torsion
U 1
byLemma3.1(ii),therecannotbeanytorsionelementsinthekernelofη ⊗1. This
p
proves the proposition. (cid:3)
In the next section we fill in the gap left in the proof of Lemma 3.2.
4. The degree p Lie power in characteristic p
In this sectionV denotes a vector space overa field K of prime characteristicp.
Moreover,wewillassumethatV isamoduleforthepolynomialringK[X]whereX
is a finite set of indeterminates. Otherwise we will use all the notation introduced
in Section 2, in particular, Lp(V), Mp(V) and Tp(V) are the pth Lie, metabelian
Lie, and tensor powers of V, respectively, Bp(V) = Lp(V)∩L(V)′′ is the kernel
of the natural projection Lp(V) → Mp(V), and all of these will be regarded as
K[X]-modules under the derivation action. Recall that Lp(V) may be regardedas
a submodule of the tensor power Tp(V), and hence Bp(V) is also a submodule of
Tp(V).
Lemma 4.1. The submodule Bp(V) is a direct summand of the K[X]-module
Tp(V).
This result is essentially proved as Theorem 3.1 in [6], except that there the
module V is assumed to be finite-dimensional. In what follows we reproduce the
6 RALPHSTO¨HR
proof from [6] with some minor amendments necessary to accommodate infinite
dimensional modules.
Proof. For each r ≥ 1 choose a basis B(r) of Lr(V) and let B = B(r). Thus B
r
is a basis of L(V). For b∈B, let deg(b) denote the degree of b, that is, deg(b)=r
S
for b ∈ B(r). Order B in any way subject to b < b′ whenever deg(b) < deg(b′).
By the Poincar´e–Birkhoff–Witt Theorem, Tp(V) has a basis C consisting of all
products of the form b ⊗b ⊗···⊗b with b ,...,b ∈B, b ≤b ≤···≤b and
1 2 k 1 k 1 2 k
deg(b )+···+deg(b )=p. Morespecifically,anybasiselementc∈C hasthe form
1 k
(1) (1) (2) (2) (p) (p)
(4.1) c=b ⊗···⊗b ⊗b ⊗···⊗b ⊗···⊗b ⊗···⊗b ,
1 k1 1 k2 1 kp
where k ,...,k are non-negative integers such that k +2k +···+pk = p and
1 p 1 2 p
where,fori=1,...,p,wehaveb(i),...,b(i) ∈B(i) andb(i) ≤···≤b(i). Wecallthe
1 ki 1 ki
p-tuple (k ,...,k ) the type of c and denote it by type(c). Let Ω denote the set of
1 p
all such types. We order Ω (lexicographically) by (k ,...,k ) > (k′,...,k′) if for
1 p 1 p
some j ∈{1,...,p} we have k =k′ for all i<j but k >k′. Using this ordering,
i i j j
write Ω = {ω ,ω ,...,ω } where ω > ω > ··· > ω . Thus ω = (p,0,...,0)
1 2 m 1 2 m 1
and ω =(0,...,0,1).
m
For i = 1,...,m, define C = {c ∈ C : type(c) = ω } and let X denote the
i i i
subspace of Tp(V) spanned by C ∪C ∪···∪C . Also, write X =0. Thus
i i+1 m m+1
Tp(V)=X >X >···>X >X =0.
1 2 m m+1
Note that X /X has basis C modulo X . Furthermore, C consists of all
i i+1 i i+1 1
products b(1)⊗b(1)⊗···⊗b(1) with b(1),...,b(1) ∈B(1) and b(1) ≤···≤b(1). Also,
1 2 p 1 p 1 p
C =B(p) and X =Lp(V).
m m
Let i ∈ {1,...,m} where ω = (k ,...,k ). For c ∈ C written as in (4.1) it is
i 1 p i
wellknownandeasytoverifythatthevalueofcmoduloX isunchangedbyany
i+1
permutation of the factors b(1),...,b(p). In particular, for all π ∈ Sym(k ), ...,
1 kp 1 1
π ∈Sym(k ), we have
p p
(4.2) b(1) ⊗···⊗b(1) ⊗···⊗b(p) ⊗···⊗b(p) +X =c+X .
π1(1) π1(k1) πp(1) πp(kp) i+1 i+1
Itfollows easilythat X is aK[X]-submoduleof Tp(V). Forc writtenas beforelet
i
c∈Sk1(L1(V))⊗···⊗Skp(Lp(V)) be defined by
c=(b(1)◦···◦b(1))⊗(b(2)◦···◦b(2))⊗···⊗(b(p)◦···◦b(p)).
1 k1 1 k2 1 kp
Clearly {c : c ∈ Ci} is a basis of Sk1(L1(V))⊗···⊗Skp(Lp(V)). Furthermore, it
followseasilyfrom(4.2)thatthelinearmapgivenbyc+X 7→cisaK[X]-module
i+1
isomorphism from Xi/Xi+1 to Sk1(L1(V))⊗···⊗Skp(Lp(V)). Thus
(4.3) Xi/Xi+1 ∼=Sk1(L1(V))⊗···⊗Skp(Lp(V)).
Suppose that i ∈ {2,...,m−1} where ω = (k ,...,k ). Thus k ,...,k < p
i 1 p 1 p
(and, in fact, kp = 0). Let σi : Sk1(L1(V))⊗···⊗Skp(Lp(V)) → Tp(V) be the
linear map defined on the basis {c:c∈C } by
i
1
(1) (1) (p) (p)
σ (c)= b ⊗···⊗b ⊗···⊗b ⊗···⊗b ,
i k1!···kp! π1(1) π1(k1) πp(1) πp(kp)
π1π∈pS∈yXSmy(mk(1k)p,.)..,
FREE CENTRE-BY-NILPOTENT-BY-ABELIAN LIE RINGS 7
where c is written as in (4.1) It is straightforward to verify that σ is a K[X]-
i
module homomorphism. It follows easily from (4.2) that σ (c) ∈ X and, indeed,
i i
σ (c)+X = c+X for all c ∈ C . Thus the map σ is injective and we have
i i+1 i+1 i i
X =Im(σ )⊕X . Since X =Lp(V) we therefore have
i i i+1 m
X =Im(σ )⊕···⊕Im(σ )⊕Lp(V).
2 2 m−1
Let W be the submodule of X given by
2
(4.4) W =Im(σ )⊕···⊕Im(σ )⊕(L(V)′′∩Lp(V)).
2 m−1
Consider the maps
α:Tp(V)→V ⊗Vp−1, β :V ⊗Vp−1 →Tp(V)
defined by
a ⊗a ⊗···⊗a 7→a ⊗(a ◦···◦a )
1 2 p 1 2 p
and
1
a ⊗(a ◦···◦a )7→ a ⊗a ⊗···⊗a ,
1 2 p (p−1)! 1 π(2) π(p)
π
X
wherea ∈V andπ runsoverallpermutationsof{2,3,...,c}. Thenαissurjective
i
and the composite βα is the identity map on V ⊗Vp−1. Hence we have a direct
decomposition
Tp(V)=Ker(α)⊕Im(β)∼=Ker(α)⊕ (V ⊗Vp−1).
We claim that Ker(α) = W. It is easily seen that W is contained in Ker(α). To
verify that we have actually equality, note that the elements
(4.5) b(1)⊗b(1)⊗···⊗b(1) with b(1),...,b(1) ∈B(1) and b(1) ≤···≤b(1)
1 2 p 1 p 1 p
form a basis of Tp(V) modulo X . Furthermore, the Lie products
2
(4.6) [b(1),b(1),...,b(1)] with b(1),...,b(1) ∈B(1) and b(1) >b(1) ≤···≤b(1)
1 2 p 1 p 1 2 p
form a basis of Lp(V) modulo Lp(V)∩L(V)′′ (see Section 2). It follows that the
elements (4.5) together with the elements (4.6) form a basis of Tc(V) modulo W.
Moreover,the imagesofthese elementsunder the mapαforma basisofV ⊗Vp−1.
Indeed, we have
(4.7) (b(1)⊗b(1)⊗···⊗b(1))α=b(1)⊗(b(1)◦···◦b(1))
1 2 p 1 2 p
and
(4.8) ([b(1),b(1),...,b(1)])α=b(1)⊗(b(1)◦···◦b(1))− b(1)⊗(b(1)◦···◦b(1)).
1 2 p 1 2 p 2 1 p
This can easily be seen from the short exact sequence (2.3) (with V instead of
A). Indeed, the elements (4.7) are mapped by κ one-to-one onto the canonical
p
basis of Vp, and the elements (4.8) are precisely the images of the canonical basis
elements of Mp(V) under the map µ . Consequently, we have the desired equality
p
Ker(α) = W, and so Tp(V) = W ⊕Im(β). By (4.4), Lp(V)∩L(V)′′ is a direct
summand of W. Thus Lp(V)∩L(V)′′ is a direct summand of Tp(V) and we have
Lemma 4.1. (cid:3)
Now the result we need to complete the proof of Proposition 3.3 follows easily.
8 RALPHSTO¨HR
Corollary 4.2. If V is a free K[X]-module, then Bp(V) is a projective K[X]-
module.
Proof. IfV isafreeK[X]-module,thenthetensorpowerTp(V)isalsoafreeK[X]-
module, see [14, Lemma 5.2]). Since Bp(V) is a direct summand of Tp(V), it is
projective. (cid:3)
5. The main result
InthisSectionLdenotesafreeLieringoffiniterankwithfreegeneratingsetX.
Our aim is to determine the torsionsubgroupof the additive groupof the quotient
(1.1). In view of the short exact sequence
0→(L′)c/[(L′)c,L]→L/[(L′)c,L]→L/(L′)c →0,
this is a free central extension of the free (nilpotent-of-class-c-1)-by-abelian Lie
ring L/(L′)c. The additive structure of the latter is well-understood. Its under-
lying abelian group is free abelian [3]. Consequently, any torsion elements must
be contained in the central quotient (L′)c/[(L′)c,L], and it is this quotient we will
focus onfromnowon. By the Shirshov-WittTheorem,the derivedidealL′ isitself
afreeLiering,namely,thefreeLieringonL′/L′′: L′ =L(L′/L′′). Thisisagraded
Lie ring and its degree c homogeneous component Lc(L′/L′′) is isomorphic to the
lower central quotient
(5.1) Lc(L′/L′′)∼=(L′)c/(L′)c+1.
The adjoint representation induces on these lower central quotients the structure
ofanL/L′-module,andhence ofa module for its universalenvelopeU =U(L/L′).
ThelattermaybeidentifiedwiththepolynomialringonX: U =Z[X]. Thus(5.1)
is actually a U-module isomorphism. In view of the canonical isomorphism
((L′)c/(L′)c+1⊗U Z∼=(L′)c/[(L′)c,L],
trivializing the U-action on both sides of (4.1) gives an isomorphism
(5.2) (L′)c/[(L′)c,L]∼=Lc(L′/L′′)⊗U Z.
Wewillexploitthisisomorphismtoinvestigatetheadditivestructureofthequotient
on the left hand side by examining the tensor product on the right hand side.
Suppose that L has rank 2 and, say, X = {x,y}. Then L′/L′′ is a free cyclic
moduleoverthepolynomialringU =Z[x,y]withfreegenerator[y,x],see[1,Proof
of Theorem 6.1]. If c is a prime, say c = p, then Proposition 3.3 applies to the
tensor product on the right hand side of (5.2). Hence this tensor product is a
direct sum of a free abelian group and an elementary abelian p-group. Moreover,
the torsion subgroup is the image in Lp(L′/L′′)⊗ Z of the torsion subgroup of
U
Mp(L′/L′′)⊗ Z under the map θ ⊗1. A complete description of the latter is
U p
given in [1, Corollary 6.2]. The elements
[[u,y],[u,x],u,...,u]⊗1
p−2
| {z }
FREE CENTRE-BY-NILPOTENT-BY-ABELIAN LIE RINGS 9
whereu=[y,x,x,...,x,y,...,y]withs,t>0formabasisofthistorsionsubgroup
s t
as a Z -module. Applying θ ⊗1 to such a basis element gives
p p
| {z } | {z }
p−1
(p−2)!
([[u,y],u,...,u,[u,x],u,...,u]−[[u,x],u,...,u,,[u,y],u,...,u])⊗1.
p
i=0
X i p−2−i i p−2−i
Sincethisisaneleme|nt{ozfo}rderpw|ec{aznd}ropthefac|tor{ozf(p}−2)!in|th{ezsta}tement
of our main result, which summarizes the discussion in this final section.
Theorem 5.1. Let L be a free Lie ring of rank 2 with free generators x and y, and
let p be a prime. Then the quotient (L′)p/[(L′)p,L] is a direct sum of a free abelian
group and an elementary abelian p-group. Modulo [(L′)p,L] the elements
p−1
1
([[u,y],u,...,u,[u,x],u,...,u]−[[u,x],u,...,u,,[u,y],u,...,u])
p
i=0
X i p−2−i i p−2−i
where u=[y,x,x,.|..,{xz,y,}...,y] w|ith{zs,}t>0 form|a ba{zsis}of this t|ors{izon}subgroup
s t
as a Z -module. (cid:3)
p
| {z } | {z }
The legality of the factor 1/p in the statement of the theorem is explained in
Section 2.
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SchoolofMathematics,UniversityofManchester,AlanTuringBuilding,Manchester,
M139PL, United Kingdom
E-mail address: [email protected]