Download 12 Recognizing invertible elements and full ideals using finite

12 Recognizing invertible elements and full ideals using finite quotients
12.1 Definitions. Let G be a group.
(i) An element og Z[G] is residually invertible if its image in Mn (Z[G/H]) is invertible for every normal
subgroup H ⊂ G of finite index. The ring Z[G] has finitely detectable invertibles if every residually invertible
element is invertible. Clearly, this implies that G is residually finite.
(ii) More generally, let n be a positive integer. An element of Mn (Z[G]) is residually invertible if its
image in Mn (Z[G/H]) is invertible for every normal subgroup H ⊂ G of finite index. The ring Mn (Z[G])
has finitely detectable invertibles if every residually invertible element is invertible.
(iii) Let ξ be an element of N (G) = Hom(G, R) \ {0}. For H < G a normal subgroup, we denote ξ the
induced morphism G/(H ∩ ξ) → R. If n ∈ N∗ , an element of Mn (Z[G]ξ ) is residually invertible if its image
in Mn (Z[G/(H ∩ ker ξ)]ξ ) is invertible for every normal subgroup H ⊂ G of finite index. The ring Mn (Z[G]ξ )
has finitely detectable invertibles if every residually invertible element is invertible.
Remarks. 1) The interest of of the last property is that, from the results of Section 11, we obtain the
Proposition. Let M be a closed 3-dimensional manifold and ξ ∈ H 1 (M, R) \ {0}, such that every twisted
Alexander polynomial ∆H
M,u is unitary. Assume that Mn (Z[π1 (M )]ξ ) has finitely detectable invertibles for
all n∗ ∈ N. Then ξ is represented by a nonsingular closed one-form.
Proposition. The finite detectability of invertibles is hereditary for subgroups of finite index: if G has it in
some form, every subgroup H < G of finite index has it in the same form.
Proof. It suffices to treat the form (i), the other cases being similar. Assume that λ ∈ Z[H] is residually
invertible. If K is a normal subgroup of finite index of G, K ∩ H is a normal subgroup of finite index of H,
thus λ ∈ Z[H/K ∩ H] is invertible. The image of the inverse in Z[G/K] is an inverse of λ in Z[G/K], thus λ
is residually invertible in Z[G]. By hypothesis, λ is invertible in Z[G]. The projection of its inverse to Z[H]
is an inverse to λ in Z[H], qed.
2) Let G be a group which is finitely generated, residually finite and right-orderable. It seems likely that
Z[G] and Z[G]ξ have finitely detectable invertibles. For matrix rings, this is perhaps somewhat optimistic.
Note that such groups include the case G = π1 (M 3 ), M closed with b1 (M ) > 0 by [Boyer-Rolfsen-Wiest]
(and Perelman for the residual finiteness).
3) T. Delzant [De 2008b] has given an example of a finitely generated and residually finite group G such
that there exists λ ∈ Z[G] residually invertible invertible but not invertible: take G to be the wreath product
H o Z where H = SL(n, Z) or a finite simple group. Quel est l’exemple ?
4) Recall that for the rings Mn (Z[G]) and Mn (Z[G]ξ ), every element which is left or right invertible is
invertible.
12.2 Z[Z] has residually detectable invertibles
Proof. Let P ∈ Z[t, t−1 ] be residually invertible, ie its image is invertible in Z[t, t−1 ]/htk − 1i for every k > 0.
We want to show that P = ±tm . Multiplying by ±tm , we can assume that λ = P (t) = a0 + · · · + ad td ∈ Z[t],
with a0 6= 0 and ad > 0. We want to show that ad = 1 and d = 0.
Fix k ∈ N∗ . The invertibility of P (t) in Z[t, t−1 ]/htk − 1i is equivalent to that of P (t) in Z[t]/htk − 1i,
thus to that of P (ω) in Z[ω] for every k-eth root of unity ω.
Y Thus P (ω) is an algebraic unit (an algebraic integer whose inverse is an algebraic integer). Thus
P (ω) is an algebraic unit belonging to Q, ie ±1. On the other hand, the theory of the resultant gives
ω k =1
Y
P (ω) = ±Res(P (t), tk − 1) = ±akd
ω k =1
Y
P (α)=0
thus
Y
|αk − 1| = a−k
d .
P (α)=0
1
(αk − 1),
Denote P− (k), P0 (k) and P+ (k) the products of |αk − 1| over the zeros of P satisfying |α| < 1, |α| = 1 and
|α| > 1. We thus have P− (k)P+ (k)P0 (k) = a−k
d . Moreover, as k → +∞, lim P− (k) = 1 and (lim P+ (k) = +∞
if the product is nonempty, 1 otherwise). Thus
(
1 if P has no zero of modulus > 1 and ad = 1
lim P0 (k) =
k→∞
0 otherwise.
Let us show by contradiction that P has no zero of modulus 1. Let g = (α1 , · · · , αr ) ∈ (S 1 )r , where r > 0
and α1 , · · · , αr are all such such zeros. By compactness of the group (S 1 )r , the ω-limit set of (g k ) contains
1 and g, thus lim inf max |αik − 1| = 0 = lim inf max |αik − αi |. Thus we have
k→+∞ 1≤i≤r
k→+∞ 1≤i≤r
lim inf P0 (k) = 0 , lim sup P0 (k) ≥
k→+∞
k→+∞
r
Y
|αi − 1| > 0,
i=1
contradicting the existence of the limit.
Thus there is no zero of modulus 1, and the limit is 1. Thus we are in the first case: ad = 1 and all
zeros of P have modulus ≤ 1, in fact < 1. Since their product is ±a0 , there is no zero at all, ie d = 0, qed.
Remark. It suffices that P (t) be invertible in Z[ω] for ω a k-eth root of unity with k large enough. Indeed, replace tk − 1 by the cyclotomic polynomial Φk (t) in the argument above. The same proof gives
k
P− (k)P0 (k)P+ (k) = a−d
, dk = deg(Φk ), and now P− (k), P0 (k) and P+ (k) are products of |Φk (α)|. And
d
also
(
1 if P has no zero of modulus > 1 and ad = 1
lim P0 (k) =
k→∞
0 otherwise.
If the limit is zero, P has some zero of modulus 1. Let g = (α1 , · · · , αr ) ∈ (S 1 )r , where r > 0 and α1 , · · · , αr
are all such such zeros. We have
r
Y
lim
|Φk (αi )| = 0.
k→+∞
Using tk − 1 =
Y
Φl (t), we have lim
`|k
r
Y
i=1
|αik − 1| = 0. Since there exists k = kj → +∞ such that αik → αi
i=1
Q
|αi − 1| = 0 ie some αi = 1.
Y
Since |Φk (1)| ≥ 1, we have lim
|Φk (αi )| = 0. By the argument just seen, this product is empty ie
for all i,
αi 6=1
Q
every αi = 1. But then |Φp (αi )| = pr for p prime, giving a contradiction.
Thus the limit is 1, so P has no zero of modulus > 1 and ad = 1. Symmetrically, P has no zero of
modulus < 1. Thus P = (t − α1 ) · · · (t − αr ) with r ≥ 0 and |αi | = 1. Thus, for k large, we have
P0 (k) =
r
Y
|Φk (αi )| = 1.
i=1
r
Y
|αik − 1| is constant for k large. Since lim inf max |αik − 1| = 0 = lim inf max |αik − αi |,
k→+∞ 1≤i≤r
k→+∞ 1≤i≤r
i=1Q
this constant is 0 = |αi − 1| thus some αi = 1.
Write P (t) = (t − 1)s Q(t) with Q(1) 6= 0. Applying this argument to Q, Q is a constant thus Q(t) = 1.
Then P0 (p) = |Φp (1)|s = ps , thus s = 0, P = 1, qed.
This implies that
Question. Does Z/pZ[Z] have finitely detectable invertibles?
12.3 The case of virtually Abelian groups
Proposition 1. For every r ∈ N and n ∈ N∗ , Mn (Z[Zr ]) and Mn (Z[Zr ]ξ ) have finitely detectable invertibles.
2
Proof. Using the determinant, it suffices to treat the case n = 1. For r = 0, the proposition is obvious. For
r = 1, the case of Z[Z] has been proved in the previous section. The case of Z[Z]ξ is obvious since ker ξ = 0.
Assume that r > 1 and the proposition has been proved for Zr−1 .
Let λ ∈ Z[Zr ] be residually invertible. Since supp(λ) is finite, there exists u ∈ Hom(Zr , Z) which is
primitive and injective on supp(λ). Then u(λ) ∈ Z[Z] = Z[Zr / ker u] is residually invertible, thus u(λ) = ±tm ,
thus λ = ±g is invertible.
The case of Z[Zr ]ξ is a little harder. Let λ ∈ Z[Zr ]ξ be residually invertible. We want to prove that its
ξ-minimal part λ0 ∈ Z[ker ξ] is invertible. We can assume that ker ξ = Zs × {0}, s = r − rk(ξ). If H = kZr ,
H ∩ ker ξ = kZs × {0} and Zr /(H ∩ ker ξ) = (Z/kZ)s ⊕ Zr−s (canonical isomorphism).
There exists f ∈ Hom(Zs , Z) which is injective on supp(λ0 ). Then the invertibility of λ0 is equivalent
to that of f (λ0 ) ∈ Z[Z]. Moreover, one can extend f to a homomorphism f : Zr → Z × Zr−s by sending
all the other generators to themselves. This gives a ring homomorphism f : Z[Zr ]ξ → Z[Z × Zr−s ]ξ . The
hypothesis implies that f (λ) is again residually invertible, thus it suffices to treat the case s = 1.
We can assume that λ0 = P (t) ∈ Z[t]. The image of λ in Z[Z/kZ][Zr−s ]ξ is invertible, and so is its image
in Z[ωk ][Zr−s ]ξ for ωk a primitive k-eth root of unity. For k large enough (it suffices that deg(Φk ) > d),
P (ωk ) 6= 0, thus it is the ξ-minimal part of this image. Thus P (ωk ) is invertible in Z[ωk ] for k large. By the
remark at the end of the previous section, P (t) = ±tr , qed.
Proposition 2. Let Γ < G be a subgroup of finite index such that every Mn (Z[Γ]) (resp. Mn (Z[Γ]ξ ) has
finitely detectable invertibles. Then every Mn (Z[G]) (resp. Mn (Z[G]ξ ) has finitely detectable invertibles.
Proof. By hereditarity to subgroups of finite index, we can assume that H is normal. By hypothesis, we
have a short exact sequence Γ → G → Q where Q is a finite group. Using a section σ : Q → G, we have an
isomorphism of Z[Γ]-modules Z[G] ≈ (Z[Γ])|Q| , |Q| = card(Q). Thus Z[G] can be represented as a subring
of M|Q| (Z[Γ]), and more generally Mn (Z[G]) can be represented as a subring of Mn|Q| (Z[Γ]).
X
Let λ ∈ Mn (Z[G]) be residually invertible. We have λ = (λi,j ) with λi,j =
ai,j,q σ(q), ai,j,q ∈ Z[Γ],.
q∈Q
The associated matrix Mλ ∈ Mn|Q| (Z[Γ]) satisfies
σ(q)
(Mλ )i,q;j,q0 = ai,j,q−1 q0 c(q, q −1 q 0 )
where c : Q × Q → Γ is the 2-cocycle associated to σ and ag = gag −1 . In particular, (MX
λ )i,1;j,q = ai,j,q . This
matrix Mλ is residually invertible, thus by (ii) it has an inverse N = (bi,q;j,q0 ), we have
ai,k,q bk,q;j,1 = δi,j .
k,q
Thus, setting µ = (µi,j ) with µi,j =
X
−1
σ(q)
bi,q;j,1 , we have λµ = Idn , thus λ is invertible.
q∈Q
The proof for Mn (Z[G]ξ ) is entirely similar.
Corollary. If G finitely generated and virtually free Abelian, Mn (Z[G]) and Mn (Z[G]ξ )) have finitely detectable invertibles.
12.4 A partial result for torsion free nilpotent groups
Proposition 2. Let G be a finitely generated torsion free nilpotent group, with center C. Assume that
Mn (Z[G/C])) has finitely detectable invertibles for all n ∈ N∗ . Then Z[G] has finitely detectable invertibles.
Proof. Fix a section σ : G/C → G, and bi-invariant orders on G/C and G such that σ is increasing (this is
possible with an arbitrary choice of σ and the order on G/C).
Let λ ∈ Z[G] be residually invertible, we want to prove that it is invertible. We can write λ =
k
X
λi σ(gi ),
i=0
with λi ∈ Z[Z], λ0 6= 0, g0 < g1 < · · · < gk . It suffices to prove that λ0 is invertible: indeed, the
X same
argument for the reverse order will prove that λk is invertible, thus the image of λ in Z[G/C] is λ =
λi gi
0≤i≤k
3
with λi and λ0 , λk invertible [the important point is that λk 6= 0]. By induction, it must be invertible, thus
a monomial, thus k = 0.
We now prove that λ0 is invertible. There exists a homomorphism f : C → Z which is injective on
supp(λ0 ). Then we have a short exact sequence Z → (G/ ker f ) → G/C, and f (λ) ∈ Z[G/ ker f ] can be
m
X
written
f (λi )σ(gi ), with f (λ0 ) 6= 0. Then the invertibility of λ0 is equivalent to that of f (λ0 ) ∈ Z[Z].
i=0
The hypothesis implies that f (λ) ∈ Z[G/ ker f ]ξ is again residually invertible, thus it suffices to treat the
case C ≈ Z.
d
X
Let t be a generator of C. Write λ0 =
ai ti = P (t). The image λk of λ in Z[G/htk i]ξ is residually
i=0
invertible, thus invertible since G/htk i contains G/C as a finite index subgroup and Mn (Z[G/C]) has finitely
detectable invertibles. We can send λk to the twisted group algebra Z[ωk ][G/C]ξ , where the product eg .eh =
c(g, h)egh , with c the two-cocycle associated to the central extension C → G → G/C. The image is again
invertible, and its minimal part with respect to the order on G/C is P (ωk ) for k prime > d. Thus P (ωk ) is
invertible in Z[ωk ] for every prime k > d, whence P (t) = ±tr , qed.
12.5 Finite detection of full ideals
Conventions. An ideal in a non commutative ring R is always a left ideal. Similarly, a module over R is
always a left module.
Definitions. Let G be a group. An ideal I ⊂ Z[G] is residually full if its image in every quotient Z[G/H],
H ⊂ G normal subgroup of finite index, is equal to Z[G/H]. Similarly, if ξ is a nonzero morphism from G
to R, a ideal of Z[G]ξ is residually full if its image is full in every quotient Z[G/H ∩ ker ξ]ξ , H ⊂ G normal
subgroup of finite index.
Similarly, a submodule M ⊂ Z[G]n is residually full if its image in every quotient Z[G/H]n , H ⊂ G
normal subgroup of finite index, is equal to Z[G/H]n . An a submodule M ⊂ Z[G]ξ is residually full if its
image is full in every quotient Z[G/H ∩ ker ξ]nξ , H ⊂ G normal subgroup of finite index.
The group ring Z[G] (resp. the Novikov ring Z[G]ξ ) has finitely detectable full ideals if every ideal
I ⊂ Z[G] (resp. I ⊂ Z[G]ξ ) which is residually full is equal to Z[G]. (resp. to Z[G]ξ ).
Since an element of Z[G] or Z[G]ξ is invertible if an only if it is left invertible, the finite detectability of
invertibles is equivalent to the finite detectability of principal ideals.
Also, the finite detectability of full ideals implies the finite detectability of full submodules of Z[G]n and
of the invertibility of matrices:
Proposition. Assume that Z[G] (resp. Z[G]ξ ) has finitely detectable full ideals.
(i) Let M be a residually full submodule of Z[G]n (resp. Z[G]nξ ). Then M = (Z[G])n (resp. Z[G]nξ ).
(ii) The ring Mn (Z[G]) (resp. Mn (Z[G]ξ ) has finitely detectable invertibles for all n.
Proof. It suffices to treat the cas of Z[G], the case of Z[G]ξ being completely analogous.
(i) By induction on n, the result being the hypothesis if n = 1. Assume that n > 1 and the result is true
for n − 1. Consider the set I of λ ∈ Z[G] such that there exists λ1 , · · · , λn−1 ∈ Z[G] with (λ1 , · · · , λn−1 , λ) ∈
M . It is an ideal, and its image is full in every quotient Z([G/H]) with G/H finite. Thus I = Z[G], ie M
contains an element x = (λ1 , · · · , λn−1 , 1).
One has a direct sum decomposition Z[G]n = Z[G]n−1 ⊕ Z[G]x. Substracting µn x from every element
(µ1 , · · · , µn ) ∈ M , ones sees that M = (M ∩ (Z[G]n−1 ) ⊕ Z[G]x. It suffices to prove that M ∩ Z[G]n−1 =
Z[G]n−1 . Clearly, M ∩ Z[G])n−1 is a submodule which has a full image in every quotient (Z[G/H])n−1 ,
H ⊂ G normal subgroup of finite index. By the induction hypothesis, M ∩ Z[G]n−1 = Z[G]n−1 , qed.
(ii) Let A ∈ Mn (Z[G]) be residually invertible. The right multiplication by A is a left-linear map u from
Z[G]n to itself. By hypothesis, im(u) has full image in every quotient (Z[G/H])n , H ⊂ G normal subgroup
of finite index. By (i), im(u) = Z[G]n , ie A is left invertible, thus invertible, qed.
12.6 A useful lemma
4
Lemma. Let G be any group, and that I ⊂ Z[G] is a residually full ideal. Then every finite quotient of
Z[G]/I is zero. In particular, if Z[G]/I is finite it is zero ie I = Z[G].
Proof. Such a quotient is of the form Z[G]/J where J is an ideal containing I. Then the subgroup H :=
{g ∈ G | g = 1 in Z[G]/J} has finite index in G. Since J contains I, it is residually full, thus its image in
Z[G/H] is full. By definition of H, we have a factorization
Z[G]



y
−
→
Z[G]/J
%
Z[G/H].
Thus J surjects onto Z[G]/J, which means that Z[G]/J = 0.
12.7 Finite detectability of full ideals of Z[Z]
Lemma. Let I ⊂ Z[Z] = Z[t, t−1 ] be an ideal which is not contained in a proper principal ideal. Then
(i) I contains a nonzero integer a.
(ii) I contains a bi-unitary polynomial P = tn + · · · + 1 ∈ Z[t].
Proof. (i) Since Q[t, t−1 ] is a principal ideal domain, there exists A ∈ Q[t, t−1 ] such that QI = AQ[t, t−1 ].
We can assume that A is in Z[t, t−1 ] and is primitive ie not divisible by an integer > 1.
For every P ∈ I there exists n ∈ N∗ such that nP ∈ AZ[t, t−1 ]. Since A is primitive and Z[t, t−1 ] is
factorial, P ∈ AZ[t, t−1 ], thus I ⊂ AZ[t, t−1 ]. By the hypothesis, A is a unit of Z[t, t−1 ], ie A = ±tn . Since
A ∈ QI, this implies that I contains an nonzero integer.
(ii) We know that Z[[t]][t−1 ] is a principal ideal domain, and that the ideal Z[[t]][t−1 ]I is generated over
Z[[t]][t−1 ] by an element P + ∈ I. Thus aZ[[t]][t−1 ] ⊂ P + Z[[t]][t−1 ], which implies that P + is associated in
Z[[t]][t−1 ] to b for some b ∈ N∗ dividing a. Thus
I ⊂ Z[[t]][t−1 ]I = P + Z[[t]][t−1 ] = bZ[[t]][t−1 ].
The hypothesis implies that b = 1, thus up to multiplication by ±tn we have P + = 1 + a1 t + . . . + an tn .
Similarly, I contains P − = 1+b1 t−1 +· · ·+bm t−m . Thus if N = max(n, m)+1, I contains P + +tN P − =
N
t + · · · + 1, qed.
Proposition. The group ring Z[Z] = Z[t, t−1 ] has finitely detectable full ideals.
Proof. Let I ⊂ Z[t, t−1 ] be an ideal whose image in Z[t]/htk − 1i is full for every k ∈ N∗ . We want to prove
that I = Z[t, t−1 ].
If I is principal, this follows from the finite detectability of invertibles. Assume that I is not principal.
Again by the finite detectability of invertibles, I is not contained in a proper principal ideal. By Lemma 1,
I contains an integer a > 0 and a bi-unitary polynomial P . We can assume that a > 1 otherwise the result
is already proved.
Since P is bi-unitary, Z[t, t−1 ]/hP i is finitely generated as a Z-module, thus Z[t, t−1 ]/ha, P i is finite and
a fortiori Z[t, t−1 ]/I is finite. By the useful lemma, I = Z[t, t−1 ].
12.8 Finite detectability of full ideals of Z[Zn ]
Proposition. The group G = Zn has finitely detectable full ideals.
Proof. By induction on n. The case n = 1 is already known, let us assume that n > 1 and that the result
holds for n − 1. Let I ⊂ Z[Zn ] which is residually full. The image of I in Z[{0} × Zn−1 ] is residually full,
thus by the induction hypothesis is is equal to Z[{0} × Zn−1 ].
Let k be a positive integer. The image I k of I in Z[Z/kZ × Zn−1 ] is residually full, and the quotient
Z[Z/kZ×Zn−1 ]/I k is an image of Z[Z/kZ×Zn−1 ]/Z[{0}×Zn−1 ], thus is finitely generated as a Z-module. By
±1
the previous proposition, I k = Z[Z/kZ × Zn−1 ]. Writing multiplicatively Z[G] = Z[t±1
1 , · · · , tn ], this means
5
k
k
±1
±1
±1
that t1 − 1 is invertible in Z[t±1
1 , · · · , tn ]/I for every k > 0. Similarly, t1 − 1 is invertible in Z[t1 , · · · , tn ]/I
for every i = 2, · · · , n and every k > 0.
Fix a prime number p, and let Ip be the image of I in Z/pZ[Zn ]. Let Fp be the algebraic closure of
Z/pZ = Fp . Consider the affine toric manifold
Vp = {(z1 , · · · , zn ) ∈ Fp
∗n
| (∀P ∈ Ip ) P (z1 , · · · , zn ) = 0}.
k
±1
Since ti − 1 is invertible in Z[t±1
1 , · · · , tn ]/I for every i = 1, · · · , n and every k > 0, the i-eth coordinate
∗
function zi : Vp → Fp never takes a value which is a root of unity, thus it is constant on every connected
component of Vp , and takes a finite number of values on Vp . Thus there exists a nonzero polynomial
Pi ∈ Fp [ti ] such that Pi (zi ) = 0 on Vp for every i = 1, · · · , n. Multiplying by its Galois conjugates and by
a suitable integer, we can assume that Pi ∈ Z/pZ[ti ]. By the (toric) Nullstellensatz, up to replacing Pi by
some power and multiplying a suitable integer prime with p, we have Pi ∈ Ip .
Thus Ip contains the ideal hP1 (t1 ), · · · , Pn (tn )i, whence (Z/pZ[Zn ])/Ip is a quotient of
±1
Z/pZ[t±1
1 , · · · , tn ]/hP1 (t1 ), · · · , Pn (tn )i =
n
O
(Z/pZ[t±1
i ])/hPi (ti )i.
i=1
Since the Pi are not zero, this ring is finite, and so is (Z/pZ[Zn ])/Ip . By the useful lemma, it is zero. Thus
if A = Z[Zn ]/I, A/pA = 0 for every prime p.
Since A is finitely generated as an algebra over Z, A = 0. Indeed, the property A = pA for every prime
p means that A is a Q-vector space. Take a basis (ei ), and let (aj ) be a finite generating set for A. The
multiplication table in the basis involves only a finite number of rationals, which are all contained in some
Z[1/m]. Thus A ⊂ Z[1/m]hei i, contradiction.
This finishes the proof that Z[Zn ]/I has finitely detectable full ideals.
12.9 From a finite index subgroup to the group
Proposition. If Γ < G has finite index and Z[Γ] has finitely detectable full ideals, so has Z[G].
We can assume that Γ is normal. By hypothesis, we have a short exact sequence Γ → G → Q where Q
is finite. Let I ⊂ Z[G] be a residually full ideal. As in 12.3, Z[G] can be identified to (Z[Γ])|Q| , so that Z[G]
embeds in M|Q| (Z[Γ]). Then Then I is identified with a submodule M ⊂ (Z[Γ])|Q| .
Let H < G be a normal subgroup of finite index. Then H ∩ Γ < Γ is a normal subgroup of finite index.
The hypothesis implies that the image of M in (Z[Γ/Γ ∩ H])|Q| is full. Since all normal subgroups of finite
index of Γ are of the form Γ ∩ H, this implies that M = (Z[Γ])|Q| , ie I = Z[G].
12.10 A result for nilpotent groups
Proposition. Let G be a nilpotent group, with center C. Assume that Z[G/C] has finitely detectable full
ideals. Then Z[G] has finitely detectable full ideals.
Proof. We make an induction on the class of nilpotency. Let c be the nilpotency class of G, so that
G(c+1) = {1} ie every c + 1-commutator [x0 , [x1 , · · · [xc−1 , xc ] · · ·]] = 1. Thus G(c) ⊂ C, and an easy
(c)
commutator calculus gives Gk ⊂ C k , where Gk is the subgroup of G generated by the k-eth powers. Thus
k
Gk /C has a nilpotency class ≤ c − 1, and by the induction hypothesis Z[Gk /C k ] has finitely detectable full
ideals. Moreover, Gk has finite index in G by induction on c, thus Gk /C k has finite index in G/C k . By the
proposition of the previous section, Z[G/C k ] has finitely detectable ideals for every k > 0.
Let I ⊂ Z[G] be a residually full ideal. Set A = Z[G]/I, which is an associative algebra over Z,
finitely generated as an algebra. The image C of C in A is a subgroup of A∗ , central in A such that
hxk − 1 | x ∈ Ci = A for every k > 0. Moreover every finite quotient of A is zero.
Assume first that C is isomorphic to Z. Let t be a generator, t = x its image in A. Then x is a central
unit of A and xk − 1 is invertible for every k > 0.
6