arXiv:1707.09854v1 [math.KT] 27 Jul 2017

The IA-congruence kernel of high rank free
Metabelian groups
David El-Chai Ben-Ezra
August 1, 2017
Abstract
The congruence subgroup problem for a finitely generated group Γ
and G ≤ Aut(Γ) asks whether the map Ĝ → Aut(Γ̂) is injective, or more
generally, what is its kernel C (G, Γ)? Here X̂ denotes the profinite completion of X. In this paper we investigate C (IA(Φn ), Φn ), where Φn is a
free metabelian group on n ≥ 4 generators, and IA(Φn ) = ker(Aut(Φn ) →
GLn (Z)).
We introduce surjective representations of IA(Φn ) onto the group
x7→1
ker(GLn−1 (Z[x±1 ]) −→ GLn−1 (Z)) which come via the classical Magnus representation of IA(Φn ). Using this representations combined with
some methods and results from Algebraic K-theory, we prove that for every n ≥ 4, C (IA(Φn ), Φn ) contains a product of n copies of the congruence
\
\
±1 ])) which is central in IA(Φ
\n ).
kernel ker(SLn−1
(Z[x±1 ]) → SLn−1 (Z[x
It enables us to show that contrary to free nilpotent cases, C (IA(Φn ), Φn )
is not trivial and not even finitely generated.
We note that using some results of this paper we show in an upcoming
paper that actually, all the elements of C (IA(Φn ), Φn ) lie in the center
\n ).
of IA(Φ

Mathematics Subject Classification (2010): Primary: 19B37, 20H05, Secondary: 20E36, 20E18.
Key words and phrases: congruence subgroup problem, automorphism groups,
proﬁnite groups, free metabelian groups.

Contents
1 Introduction

2

2 Some background in algebraic K-theory

6

3 IA (Φn ) and its subgroups

8

4 The subgroups Ci

13
1

5 The centrality of Ci

20
m

6 Some elementary elements of hIA (Φn ) i
7 A main lemma
7.1 Decomposing the proof . .
7.2 Some needed computations
7.3 Elements of form 2 . . . . .
7.4 Elements of form 3 . . . . .
7.5 Elements of form 4 . . . . .

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25
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47

Introduction

The classical congruence subgroup problem (CSP) asks for, say, G = SLn (Z) or
G = GLn (Z), whether every ﬁnite index subgroup of G contains a principal congruence subgroup, i.e. a subgroup of the form G (m) = ker (G → GLn (Z/mZ))
for some 0 6= m ∈ Z. It is a classical 19th century result that the answer is
negative for n = 2. On the other hand, quite surprisingly, it was proved in the
sixties by Mennicke [Men] and by Bass-Lazard-Serre [BaLS] that for n ≥ 3 the
answer to the CSP is aﬃrmative. A rich theory of the CSP for more general
arithmetic groups has been developed since then.
By the observation GLn (Z) ∼
= Aut (Zn ), the CSP can be generalized to
automorphism groups as follows: Let Γ be a group and G ≤ Aut (Γ). For a
ﬁnite index characteristic subgroup M ≤ Γ denote:
G (M ) = ker (G → Aut (Γ/M )) .
Such a G (M ) will be called a “principal congruence subgroup” of G and a
ﬁnite index subgroup of G which contains G (M ) for some M will be called a
“congruence subgroup”. The CSP for the pair (G, Γ) asks whether every ﬁnite
index subgroup of G is a congruence subgroup.
One can easily see that the CSP is equivalent to the question: Is the congruence map Ĝ = limG/U → limG/G (M ) injective? Here, U ranges over all
←−
←−
ﬁnite index normal subgroups of G, and M ranges over all ﬁnite index characteristic subgroups of Γ. When Γ is ﬁnitely generated, it has only ﬁnitely many
subgroups of given index m, and thus, the charateristic subgroups: Mm =
∩ {∆ ≤ Γ | [Γ : ∆] = m} are of ﬁnite index in Γ. Hence, one can write Γ̂ =
1
lim
←−m∈N Γ/Mm and have :
limG/G (M ) =
←−
≤

lim
G/G (Mm ) ≤ limm∈N Aut(Γ/Mm )
←−m∈N
←−
Aut(lim
(Γ/M
))
= Aut(Γ̂).
m∈N
m
←−

1 By the celebrated theorem of Nikolov and Segal which asserts that every finite index
subgroup of a finitely generated profinite group is open [NS], the second inequality is actually
an equality. However, we do not need it.

2

Therefore, when Γ is ﬁnitely generated, the CSP is equivalent to the question:
Is the congruence map: Ĝ → Aut(Γ̂) injective? More generally, the CSP asks
what is the kernel C (G, Γ) of this map. For G = Aut (Γ) we will also use the
simpler notation C (Γ) = C (G, Γ). The classical congruence subgroup result
n
mentioned above can therefore
 be reformulated as C (Z ) = {e} for n ≥ 3, and
2
it is also known that C Z = F̂ω , where F̂ω is the free proﬁnite group on a
countable number of generators (cf. [Mel], [L]).
Very few results are known when Γ is non-abelian. Most of the results are
related to Γ = π(Sg,n ), the fundamental group of the closed surface of genus g
with n punctures (see [DDH], [Mc], [A], [Bo1], [Bo2]). As observed in [BER], the
result of Asada in [A] actually gives an aﬃrmative solution to the case Γ = F2 ,
G = Aut (F2 ) (see also [BL]). Note that for every n > 0, π(Sg,n ) ∼
= F2g+n−1 =
the free group on 2g + n − 1 generators. Hence, the above mentioned results
relate to various subgroups of the automorphism group of ﬁnitely generated free
groups. However, the CSP for the full Aut (Fd ) when d ≥ 3 is still unsettled.
Denote now the free metabelian group on n generators by Φn = Fn /Fn′′ .
Considering the metabelian case, it was shown in [BL] (see also [Be1]) that
C (Φ2 ) = F̂ω . In addition, it was proven there that C (Φ3 ) ⊇ F̂ω . The basic
motivation which led to this paper was to complete the picture in the free
metabelian case and investigate C (Φn ) for n ≥ 4. Now, denote IA(Φn ) =
ker(Aut(Φn ) → GLn (Z)). Then, the commutative exact diagram:
1 → IA (Φn ) → Aut (Φn )
ց
↓
Aut(Φ̂n )

→ GLn (Z) → 1
↓
→ GLn (Ẑ)

gives rise to the commutative exact diagram (see Lemma 2.1 in [BER]):
\
\
\
IA
(Φn ) → Aut
(Φn ) → GL
n (Z)
ց
↓
↓
Aut(Φ̂n ) → GLn (Ẑ)

→ 1
.

\
Hence, by using the fact that GL
n (Z) → GLn (Ẑ) is injective for n ≥ 3, one
can obtain that C (Φn ) is an image of C (IA (Φn ) , Φn ). Thus, for investigating
C (Φn ) it seems to be worthwhile to investigate C (IA (Φn ) , Φn ). The main
result of the paper is the following:
Theorem 1.1. For every n ≥ 4, C (IA (Φn ) , Φn ) contains a subgroup C which
satisfies the following properties:
Qn
1. C is isomorphic to a product C = i=1 Ci of n copies of:
\
\
±1 ])).
(Z[x±1 ]) → SLn−1 (Z[x
Ci ∼
= ker(SLn−1

\
2. C is contained in the center of IA
(Φn ).
3. C is a direct factor of C (IA (Φn ) , Φn ), i.e. there is a normal subgroup
N ⊳ C (IA (Φn ) , Φn ) such that C (IA (Φn ) , Φn ) = N × C.
3

We note that as for every n ≥ 4, IA (Φn ) is ﬁnitely generated [BM], the
\
action of Aut (Φn ) on IA (Φn ) by conjugation induce an action of Aut
(Φn )
\
\
on IA
(Φn ), such that the closer IA (Φn ) of IA (Φn ) in Aut
(Φn ) acts trivially
\
\
\
on Z(IA
(Φn )), the center of IA
(Φn ). Thus, as we have Aut
(Φn )/IA (Φn ) =
\
\
\
GLn (Z) we obtain a natural action of GLn (Z) on Z(IA (Φn )). It will be clear
from the description in the paper that the permutation matrices permute the
copies Ci through this natural action.
As by using the techniques of Kassabov and Nikolov in [KN] one can show
that Ci are not ﬁnitely generated, this gives the imediate following theorem as
a corollary:
Theorem 1.2. For every n ≥ 4, C (IA (Φn ) , Φn ) is not finitely generated.
It will be shown in an upcoming paper that when Γ is a ﬁnitely generated
free nilpotent group of class c, then C (IA (Γ) , Γ) = {e} is always trivial. So
the free metabelian cases behave completely diﬀerent from free nilpotent cases.
The following problem is still open:
Qn
Problem 1.3. Is C (IA (Φn ) , Φn ) = i=1 Ci or it contains more elements?
We get close to solve Problem 1.3 in an upcoming result, which asserts [Be2]:

\
Theorem 1.4. For every n ≥ 4, C (IA (Φn ) , Φn ) is central in IA
(Φn ).
We note that considering our basic motivation, as C (Φn ) is an image of
C (IA (Φn ) , Φn ) we actually obtain that when n ≥ 4, the situation is dramatically diﬀerent from the cases of n = 2, 3 described above, and:
Theorem 1.5. For every n ≥ 4, C (Φn ) is abelian.
We remark that despite the result of the latter theorem, we do not know
whether C (Φn ) is also not ﬁnitely generated. In fact we can not even prove at
this point that it is not trivial.
The main line of the proof of Theorem 1.1, is organized along the paper as
follows. For a ring R, ideal J ⊳ R and d ∈ N denote:
GLd (R, J) = ker(GLd (R) → GLd (R/J)).
±1
n
For n ∈ N denote also the ring Rn = Z[x±1
1 , . . . , xn ] = Z[Z ]. Using the
Magnus embedding of IA (Φn ), in which IA (Φn ) can be viewed as:


 

x1 − 1 
x1 − 1




 

..
..
IA (Φn ) = A ∈ GLn (Rn ) | A 
=



.
.




xn − 1
xn − 1
u
we obtain in 3, for every 1 ≤ i ≤ n, a natural embedding of:

GLn−1 (Rn , (xi − 1)Rn ) ֒→ IA (Φn )
4

and a surjective natural homomorphism:
ρi

IA (Φn ) ։ GLn−1 (Z[xi±1 ], (xi − 1)Z[x±1
i ])
in which the obvious copy of the subgroup GLn−1 (Z[xi±1 ], (xi − 1)Z[x±1
i ]) in
GLn−1 (Rn , (xi − 1)Rn ) is mapped onto itself via the composition map. This
description,
u after quoting some
u classical notions and results from Algebraic Ktheory in 2, enables us in 4 to show that for every n ≥ 4 and 1 ≤ i ≤ n,
C (IA (Φn ) , Φn ) contains a copy of:
Ci

∼
=

±1
\
ker(GLn−1 (Z[xi±1\
], (xi − 1)Z[x±1
i ]) → GLn−1 (Z[xi ]))

∼
=

±1
\
\
ker(SLn−1
(Z[x±1
i ]) → SLn−1 (Z[xi ]))

such that C (IA (Φn ) , Φn ) is maped onto Ci through the map ρ̂i which induced
by ρi . The second isomorphism in the obove equation is obtained by using a
main lemma which is left to the end of the paper and some classical results from
Algebraic K-theory. In particular, we get that for every 1 ≤ i ≤ n:
u

C (IA (Φn ) , Φn ) = (C (IA (Φn ) , Φn ) ∩ ker ρ̂i ) ⋊ Ci .

In 4 we aslo show that the copies Ci intersect each other trivially. Then,
following the techniques of Kassabov and Nikolov in [KN] we show that Ci is
not ﬁnitely generated, and therefore
deduce that C (IA (Φn ) , Φn ) is not ﬁnitely
u
generated either. Then, in 5 we show that the copies Ci lie in the center of
\
IA
(Φn ), using classical results from Algebraic K-theory and the main reminded
lemma. In particular we obtain that:
C (IA (Φn ) , Φn ) = (C (IA (Φn ) , Φn ) ∩ni=1 ker ρ̂i ) ×

n
Y

Ci

i=1

This complete
the proof of Theorem 1.1.
u
m
In 6 we compute some elements
u in hIA (Φn ) i which help us later to prove
the main reminded lemma. In 7, using classical results from algebraic Ktheory, we end the paper by proving a main lemma which asserts that for every
1 ≤ i ≤ n, we have:

m
GLn−1 (Rn , (xi − 1)Rn ) ∩ E n−1 Rn , Hn,m2 ⊆ hIA (Φn ) i

when:

• GLn−1 (Rn , (xi − 1)Rn ) denotes its appropriate copy in IA (Φn ) described
above.

• E n−1 Rn , Hn,m2 is the subgroup of En−1 (Rn ) = hIn−1 + rEi,j | r ∈ Rn i
which is generated as a normal subgroup by the elementary
 matrices of the
form In−1 + hEi,j for h ∈ Hn,m2 = ker Rn → Zm2 [Znm2 ] , 1 ≤ i 6= j ≤ n.
Here, In−1 is the (n − 1) × (n − 1) unit matrix and Ei,j is the matrix which
has 1 in the (i, j)-th entry and 0 elsewhere.
5

• The intersection is obtained by viewing the copy of GLn−1 (Rn , (xi −1)Rn )
in IA(Φn ) as a subgroup of GLn−1 (Rn ).
Acknowledgements: I wish to oﬀer my deepest thanks to my great supervisor
Prof. Alexander Lubotzky for his sensitive and devoted guidance, and to the
Rudin foundation trustees for their generous support during the period of the
research.

2

Some background in algebraic K-theory

In this section we ﬁx some notations and recall some deﬁnitions and background
in algebraic K-theory which will be used through the paper. One can ﬁnd more
general information in the references ([Ros], [Mi], [Bas]). In this section R will
always denote a commutative ring with identity. We start with recalling the
following notations: Let R be a commutative ring, H ⊳ R an ideal, and d ∈ N.
Then:
• GLd (R) = {A ∈ Mn (R) | det (A) ∈ R∗ }.
• SLd (R) = {A ∈ GLd (R) | det (A) = 1}.
• Ed (R) = hId + rEi,j | r ∈ R, 1 ≤ i 6= j ≤ di.
• GLd (R, H) = ker (GLd (R) → GLd (R/H)).
• SLd (R, H) = ker (SLd (R) → SLd (R/H)).
• Ed (R, H) = the normal subgroup of Ed (R), which is generated as a normal subgroup by the elementary matrices of the form Id + hEi,j for h ∈ H.
As for every d ≥ 3, Ed (R, H) is normal in GLd (R) (see Corollary 1.4 in [Su]),
one can have the following deﬁnitions:
K1 (R; d) = GLd (R) /Ed (R)
SK1 (R; d) = SLd (R) /Ed (R)

K1 (R, H; d) = GLd (R, H) /Ed (R, H)
SK1 (R, H; d) = SLd (R, H) /Ed (R, H) .

We now go ahead with the following deﬁnition:
Definition 2.1. Let R be a commutative ring, and 3 ≤ d ∈ N. We deﬁne
the “Steinberg group” Std (R) to be the group which generated by the elements
xi,j (r) for r ∈ R and 1 ≤ i 6= j ≤ d, under the relations:
• xi,j (r1 ) · xi,j (r2 ) = xi,j (r1 + r2 ).
• [xi,j (r1 ) , xj,k (r2 )] = xi,k (r1 · r2 ).
• [xi,j (r1 ) , xk,l (r2 )] = 1.

6

for every diﬀerent 1 ≤ i, j, k, l ≤ d and every r1 , r2 ∈ R.
As the elementary matrices Id + rEi,j satisfy the relations which deﬁne
Std (R), the map xi,j (r) 7→ Id + rEi,j deﬁnes natural homomorphism φd :
Std (R) → E d (R). The kernel of this map is denoted by K2 (R; d) = ker (φd ).
Now, for two invertible elements u, v ∈ R∗ and 1 ≤ i 6= j ≤ d deﬁne the
“Steinberg symbol” by:
{u, v}i,j = hi,j (uv) hi,j (u)−1 hi,j (v)−1 ∈ Std (R)
when hi,j (u) = wi,j (u) wi,j (−1)
and


wi,j (u) = xi,j (u) xj,i −u−1 xi,j (u) .

One can show that {u, v}i,j ∈ K2 (R; d) and lie in the center of Std (R). In
addition, for every 3 ≤ d ∈ N, the Steinberg symbols {u, v}i,j do not depend
on the indices i, j, so they can be denoted simply by {u, v} (see [DS]). The
Steinberg symbols satisfy many identities. For example:
{uv, w} = {u, w} {v, w}

, {u, vw} = {u, v} {u, w} .

(2.1)

In the semi-local case we have the following ([SD], Theorem 2.7):
Theorem 2.2. Let R be a semi-local commutative ring and d ≥ 3. Then,
K2 (R; d) is generated by the Steinberg symbols {u, v} for u, v ∈ R∗ . In particular, K2 (R; d) is central in Std (R).
Let now R be a commutative ring, H ⊳ R ideal
 and d ≥ 3. Denote R̄ = R/H.
Clearly, there is a natural map E d (R) → E d R̄ . It is easy to see that Ed (R, H)
is in the kernel of the latter map, so we have a map:

πd : Ed (R) /Ed (R, H) → Ed R̄ .

In addition, it is easy to see that we have a surjective map:

ψd : Std R̄ ։ Ed (R) /Ed (R, H)


deﬁned by: xi,j (r̄) → Id + rEi,j , such that φd : Std R̄ → Ed R̄ satisﬁes:
φd = πd ◦ ψd . Therefore, we obtain the surjective map:

ψd
K2 R̄; d = ker (φd ) ։ ker (πd ) =
≤

(Ed (R) ∩ SLd (R, H)) /Ed (R, H)
SK1 (R, H; d) .

In particular, it implies that if Ed (R) = SLd (R), then we have a natural
surjective map:
K2 (R/H; d) ։ SK1 (R, H; d) .
From here one can easily deduce the following corollary, which will be needed
later in the paper.
7

Corollary 2.3. Let R be a commutative ring, H ⊳ R ideal of finite index and
d ≥ 3. Assume also that Ed (R) = SLd (R). Then:
1. SK1 (R, H; d) is a finite group.
2. SK1 (R, H; d) is central in GLd (R) /Ed (R, H).
3. Every element
 of SK1 (R, H; d) has a representative in SLd (R, H) of the
A
0
such that A ∈ SL2 (R, H).
form
0 Id−2
Proof. The ring R̄ = R/H is ﬁnite. In particular,
R̄ is Artinian and hence semi
local. Thus, by Theorem 2.2, K2 R̄; d is an abelian group which generated
by the Steinberg symbols {u, v} for u, v ∈ R̄∗ . As R̄ is ﬁnite, so is the number
of the Steinberg symbols. By equation 2.1 we obtain that the order of any
abelian group
Steinberg symbol is ﬁnite. So K2 R̄; d is a ﬁnitely generated

d
is
ﬁnite.
Moreover,
as
which its generators are of ﬁnite order. Thus, K2 R̄;


R̄ is semi-local, Theorem 2.2 implies that K2 R̄; d is central Std R̄ . Now,
as we assume Ed (R) = SLd (R), SK1 (R, H; d) is the image of K2 R̄; d under
the surjective map

Std R̄ ։ Ed (R) /Ed (R, H) = SLd (R) /Ed (R, H) .

This implies part 1 and that SK1 (R, H; d) is central in SLd (R) /Ed (R, H).
Moreover, as d ≥ 3, we have {u, v} = {u, v}1,2 for every u, v ∈ R̄∗ . Now,
it is easy to check from the deﬁnition that the image of {u, v}1,2 under the



A
0
· Ed (R, H) for
map Std R̄ ։ SLd (R) /Ed (R, H) is of the form
0 Id−2
some A ∈ SL2 (R, H). So as SK1 (R, H; d) is generated by the images of the
Steinberg symbols, the same holds for every element in SK1 (R, H; d). So we
obtain part 3. Now, as d ≥ 3 we can write:
GLd (R) = SLd (R) · {Id + (r − 1) E3,3 | r ∈ R∗ }
∗
and as all the elements of theform Id + (r
 − 1) E3,3 for r ∈ R commute with
A
0
for A ∈ SL2 (R, H), the centrality
all the elements of the form
0 Id−2
of SK1 (R, H; d) in SLd (R) /Ed (R, H) shows that actually SK1 (R, H; d) is
central in GLd (R) /Ed (R, H), as required in part 2.

3

IA (Φn ) and its subgroups

We start to deal with the IA-automorphism group of the free metabelian group,
G = IA (Φn ) = ker (Aut (Φn ) → Aut (Φn /Φ′n ) = GLn (Z)), by presenting some
of its properties and subgroups. We start with the following notations:
• Φn = Fn /Fn′′ = the free metabelian group on n elements. Here Fn′′ denotes
the second derivative of Fn , the free group on n elements..
′

m

′ m
• Φn,m = Φn /Mn,m , where Mn,m = (Φ′n Φm
n ) (Φn Φn ) .

8

• IGn,m = G(Mn,m ) = ker (IA (Φn ) → Aut (Φn,m )) .
• IAn,m = ∩ {N ⊳ IA (Φn ) | [IA (Φn ) : N ] | m}
±1
n
• Rn = Z[Zn ] = Z[x±1
1 , . . . , xn ] where x1 , . . . , xn are the generators of Z .

• Zm = Z/mZ.
• σi = xi − 1 for 1 ≤ i ≤ n. We also denote by ~σ the column vector which
has σi in its i-th entry.
Pn
• An = i=1 σi Rn = the augmentation ideal of Rn .
Pn
• Hn,m = ker (Rn → Zm [Znm ]) = i=1 (xm
i − 1) Rn + mRn .

By the well known Magnus embedding (see [Bi], [RS], [Ma]), one can identify
Φn with the matrix group:
)
(

n
X
g a 1 t1 + . . . + a n tn
ai (xi − 1)
| g ∈ Zn , ai ∈ Rn , g − 1 =
Φn =
0
1
i=1

where ti is a free basis for Rn -module, under the identiﬁcation of the generators
of Φn with the matrices


xi ti
1 ≤ i ≤ n.
0 1
Moreover, for every α ∈ IA (Φn ), one can describe α by its action on the
generators of Φn , by:




xi ti
xi ai,1 t1 + . . . + ai,n tn
α:
7→
0 1
0
1
and this description gives an injective homomorphism (see [Bac], [Bi]):
IA (Φn ) ֒→
deﬁned by α

7→

GLn (Rn )

a1,1 · · ·
 ..
 .
an,1


a1,n
.. 
. 
· · · an,n

which gives an identiﬁcation of IA (Φn ) with the group:
IA (Φn ) =
=

{A ∈ GLn (Rn ) | A~σ = ~σ }
n
o
In + A ∈ GLn (Rn ) | A~σ = ~0 .

Proposition 3.1. Let In + A ∈ IA (Φn ) and denote
Pnthe entries of A by ak,l for
1 ≤ k, l ≤ n. Then, for every 1 ≤ k, l ≤ n, ak,l ∈ l6=i=1 σi Rn ⊆ An .
9

Proof. For a given 1 ≤ k ≤ n, the condition A~σ = ~0 gives the equality:
0 = ak,1 σ1 + ak,2 σ2 + . . . + ak,n σn .
Thus, for a given 1 ≤ l ≤ n, the map Rn → Z[x±1
l ] which deﬁned by xi 7→ 1 for
±1
every i 6= l, maps 0 = ak,1 σ1 + ak,2 σ2 + . . . + ak,n σn 7→ āk,l σP
l ∈ Z[xl ]. Hence,
n
±1
±1
as Z[xl ] is a domain, āk,l = 0 ∈ Z[xl ]. Therefore: ak,l ∈ l6=i=1 σi Rn ⊆ An ,
as required.
Proposition 3.2.
Qn Let In + A ∈ IA (Φn ). Then det (In + A) is of the form:
det (In + A) = r=1 xsrr for some sr ∈ Z.
Qn
Proof. The invertible elements in Rn are the elements of the form ± i=1 xsrr
(see
Qn[CF], chapter 8). Thus, as In + A ∈ GLn (Rn ) we have: det (In + A) =
± i=1 xsrr . However, according to Proposition 3.1, for every entry ak,l of A
we have: ak,l ∈ An . Hence, under the
xi 7→ 1 for every 1 ≤ i ≤ n,
Q projection
sr
one has In + A 7→ In and thus, ± ni=1
x
=
det
(In + A) 7→ det (In ) = 1.
Qn r
Therefore, the option det (In + A) = − i=1 xsrr is impossible, as required.

Consider now the map:



Pn
g a 1 t1 + . . . + a n tn
n
Φn =
| g ∈ Z , ai ∈ Rn , g − 1 = i=1 ai (xi − 1)
0
1
↓



Pn
g a 1 t1 + . . . + a n tn
n
n
| g ∈ Zm , ai ∈ Zm [Zm ], g − 1 = i=1 ai (xi − 1)
0
1

which induced by the projections Zn → Znm , Rn = Z[Zn ] → Zm [Znm ]. Using
result of Romanovskiı̆ [Rom], it is shown in [Be1] that this map is surjective and
that Φn,m is canonically isomorphic to its image. Therefore, we can identify the
principal congruence subgroup of IA (Φn ), IGn,m , with:
IGn,m

=
=

{A ∈ ker (GLn (Rn ) → GLn (Zm [Znm ])) | A~σ = ~σ }
n
o
In + A ∈ GLn (Rn , Hn,m ) | A~σ = ~0 .

Let us step forward with the following deﬁnitions:

Definition 3.3. Let A ∈ GLn (Rn ), and for 1 ≤ i ≤ n denote by Ai,i the minor
which obtained from A by erasing its i-th row and i-th column. Now, for every
1 ≤ i ≤ n, deﬁne the subgroup IGLn−1,i ≤ IA (Φn ), by:


The i-th row of A is 0,
IGLn−1,i = In + A ∈ IA (Φn ) |
.
In−1 + Ai,i ∈ GLn−1 (Rn , σi Rn )
Proposition 3.4. For every 1 ≤ i ≤ n we have: IGLn−1,i ∼
= GLn−1 (Rn , σi Rn ).
Proof. The deﬁnition of IGLn−1,i gives us a natural projection from IGLn−1,i →
GLn−1 (Rn , σi Rn ) which maps an element In + A ∈ IGLn−1,i to In−1 + Ai,i ∈
10

GLn−1 (Rn , σi Rn ). Thus, all we need is to explain why this map is injective
and surjective.
Injectivity: Here, It is enough to show that given an element In + A ∈
IA (Φn ), every entry in the i-th column is determined uniquely by the other
entries in its row. Indeed, as A satisﬁes the condition A~σ = ~0,
P for every 1 ≤
− n
i6=l=1 ak,l σl
, i.e.
k ≤ n we have: ak,1 σ1 + ak,2 σ2 + . . . + ak,n σn = 0 ⇒ ak,i =
σi
we have a formula for ak,i by the other entries in its row.
Surjectivity: Without loss of generality we assume i = n. Let In−1 + σn C ∈
GLn−1 (Rn , σn Rn ), and denote by ~cl the column vectors of C. Deﬁne:


Pn−1
In−1 + σn C − l=1 σl~cl
∈ IGLn−1,n
0
1
and this is clearly a preimage of In−1 + σn C.
Under the above identiﬁcation of IGLn−1,i with GLn−1 (Rn , σi Rn ), we will
use through the paper the following notations:
Definition 3.5. Let H ⊳ Rn . We deﬁne:
ISLn−1,i (H) = IGLn−1,i ∩ SLn−1 (Rn , H)
IEn−1,i (H) = IGLn−1,i ∩ E n−1 (Rn , H) ≤ ISLn−1,i (H) .

Observe now that as for every 1 ≤ i ≤ n, GLn−1 Z[xi±1 ], σi Z[x±1
i ] ≤
GLn−1 (Rn , σi Rn ), the latter
isomorphism gives also a natural embedding of

±1
],
σ
Z[x
]
as
a
subgroup of IA (Φn ). Actually:
GLn−1 Z[x±1
i
i
i

Proposition 3.6. For every 1 ≤ i ≤ n, there is a canonical surjective homomorphism:


ρi
±1
±1
±1
GLn−1 Z[x±1
i ], σi Z[xi ] ֒→ IA (Φn ) ։ GLn−1 Z[xi ], σi Z[xi ]

such that the composition map is the identity. Hence:


±1
IA (Φn ) = ker ρi ⋊ GLn−1 Z[x±1
i ], σi Z[xi ] .

Proof. Without loss of generality we assume i = n. First, consider the homomorphism IA (Φn ) → GLn Z[x±1
n ] , which induced by the projection Rn →
Z[x±1
]
which
deﬁned
by
x
→
7
1
for every j 6= n. By Proposition 3.1, given
j
n
Pn−1
In + A ∈ IA (Φn ), all the entries of the n-th column of A are in j=1 σj Rn .

Hence, the above map IA (Φn ) → GLn Z[x±1
n ] is actually a map:
o
n

IA (Φn ) → In + Ā ∈ GLn Z[xn±1 ] | the n-th column of Ā is ~0 .
Observe now, that the right side of the above map is mapped naturally to
GLn−1 Z[x±1
n ] by erasing the n-th column and the n-th row from every element. Hence we obtain a map:

IA (Φn ) → GLn−1 Z[x±1
n ] .
11

Now, by Proposition 3.1, every entry of A such that In + A ∈ IA (Φn), is in
An . Thus, the entries of every Ā such that In−1 + Ā ∈ GLn−1 Z[x±1
n ] is an
image of In + A ∈ IA (Φn ), are all in σn Z[x±1
].
Hence,
we
actually
obtain a
n
homomorphism:

±1
ρn : IA (Φn ) → GLn−1 Z[x±1
n ], σn Z[xn ] .

Observing that the copy of GLn−1 Z[xn±1 ], σn Z[x±1
n ] in IGLn−1,n is mapped
isomorphically to itself by ρn , ﬁnishes the proof.
Let have now the following propositions:
∼
Proposition 3.7.
For 1 ≤ i ≤ n denote R̄ = Z[x±1
i ]. Then, viewing Im(ρi ) =

GLn−1 R̄, σi R̄ , for every m ∈ N one has:

Im(ρi ) ∩ IGn,m = GLn−1 R̄, σi ((xm
i − 1)R̄ + mR̄)

Proof. It is easy to see that the elements of IGL
 n−1,i which correspond to
the elements of GLn−1 R̄, σi ((xm
i − 1)R̄ + mR̄) are clearly in Imρi ∩ IGn,m .
On the other hand, without loss of generality we assume that i = n and let
In + A ∈ Imρn ∩ IGn,m . Then In + A has the form:


Pn−1
In−1 + σn C − l=1 σl~cl
∈ IGLn−1,n
0
1
Pn−1
when the entries of C admit ck,l ∈ R̄ and
j=1 σl ck,l ∈ Hn,m . So ck,l ∈
m
Hn,m ∩ R̄ = (xn − 1)R̄ + mR̄, and the claim follows.
Proposition 3.8. For 1 ≤ i ≤ n denote R̄ = Z[xi±1 ]. Then, for every m ∈ N
one has:

ρi IGn,m2 ⊆ Im(ρi ) ∩ IGn,m ⊆ ρi (IGn,m )

Proof. As every element in Imρi ∩ IGn,m is mapped to itself via ρi we clearly
have Imρi ∩ IGn,m ⊆ ρi (IGn,m). On the other hand, if In + A ∈ IGn,m2 then
viewing Imρi ∼
= GLn−1 R̄, σi R̄ , the entries of ρi (In + A) = In−1 +B belong to
Pm−1
2
2
⊆ (xm
(xm
−1)
R̄+m
σi R̄. Observe now that we have: r=0 xmr
i
i −1)R̄+mR̄,
i
so:
2
xm
i

− 1 = σi

2
m
−1
X

r=1

xri = σi

m−1
X
r=0

xri

m−1
X
r=0


xmr
∈ σi (xm
i − 1)R̄ + mR̄ .
i

So by proposition 3.7 ρi (In + A) ∈ Imρi ∩ IGn,m as required.
Proposition 3.9. For every m ∈ N and 1 ≤ i ≤ n one has:
ρi (IAn,m ) = Im(ρi ) ∩ IAn,m .

12

Proof. As every element in the intersection Imρi ∩ IAn,m is mapped to itself via
ρi we clearly have Imρi ∩IAn,m ⊆ ρi (IAn,m ). For the opposite, assume that α ∈
IAn,m , and denote ρi (α) = β ∈ Imρi . We want to show that β ∈ IAn,m . So let
N ⊳ IA(Φn ) such that [IA(Φn ) : N ]|m. Then obviously [Imρi : (N ∩ Imρi )]|m.
−1
Thus, as ρi is surjective [IA(Φn ) : ρ−1
i (N ∩ Imρi )]|m so α ∈ ρi (N ∩ Imρi ) and
hence β = ρi (α) ∈ N ∩ Imρi ≤ N . As this is valid for every such N , we have:
β ∈ IAn,m as required.
We close this section with the following deﬁnition:
Definition 3.10. For every 1 ≤ i ≤ n, denote:
IGL′ n−1,i = {In + A ∈ IA (Φn ) | The i-th row of A is 0} .
Obviously, IGLn−1,i ≤ IGL′ n−1,i , and by the same injectivity argument as
in the proof of Proposition 3.4, one can deduce that:
Proposition 3.11. The subgroup IGL′ n−1,i ≤ IA (Φn ) is canonically embedded
in GLn−1 (R), by the map: In + A 7→ In−1 + Ai,i .
Remark 3.12. Note that in general IGLn−1,i  IGL′ n−1,i . For example, I4 +
σ3 E1,2 − σ2 E1,3 ∈ IGL′ 3,4 \ IGL3,4 .

4

The subgroups Ci

In this section we deﬁne the subgroups Ci ≤ C (IA (Φn ) , Φn ), and we show that
C (IA (Φn ) , Φn ) can be viewed as a semi-direct product of each one of them.
We also show that when n ≥ 4:
±1 ]))
\
\
(Z[x±1 ]) → SLn−1 (Z[x
Ci ∼
= ker(SLn−1

and use it to show that C (IA (Φn ) , Φn ) is not ﬁnitely generated. Along the
section we will simplify the notations and write IAm = IAn,m and IGm =
IGn,m .
It is proven in [Be1] that Φ̂n = lim
←−Φn,m . So, as for every m ∈ N ker(Φn →
Φn,m ) is characteristic in Φn , we can write explicitly:
\
= ker(IA
(Φn ) → Aut(Φ̂n ))
\
= ker(IA
(Φn ) → lim
←−Aut (Φn,m ))
\
= ker(IA
(Φn ) → lim
←− (IA (Φn ) /IGm ))
\
= ker(IA
(Φn ) → lim (IA (Φn ) /IGm2 )).
←−
Now, as for every n ≥ 4 we know that IA (Φn ) is ﬁnitely generated (see [BM]),
\
we have IA
(Φn ) = lim
←− (IA (Φn ) /IAm ). Hence:
C(IA (Φn ) , Φn ) = ker(lim
←− (IA (Φn ) /IAm ) → lim
←− (IA (Φn ) /IGm ))
C (IA (Φn ) , Φn )

= lim (IAm · IGm /IAm )
←−
2
= lim
←− (IAm · IGm /IAm ) .
13

Observe now that for every 1 ≤ i ≤ n the composition map:


ρi
±1
±1
±1
GLn−1 Z[x±1
i ], σi Z[xi ] ֒→ IA (Φn ) ։ GLn−1 Z[xi ], σi Z[xi ]

induce a similar composition map for the proﬁnite completions:



±1
±1
\
±1
±1
\
\ ρ̂i
GLn−1 Z[x
i ], σi Z[xi ] ֒→ IA (Φn ) ։ GLn−1 Z[xi ], σi Z[xi ] .

\
This enables us to write: IA (Φn ) = ker ρi ⋊ Imρi and IA
(Φn ) = ker ρ̂i ⋊ Imρ̂i .
Deﬁne now:
Definition 4.1. Ci = C (IA (Φn ) , Φn ) ∩ Imρ̂i = ker(Imρ̂i → Aut(Φ̂n )).
Proposition 4.2. If 1 ≤ i 6= j ≤ n, then Ci ∩ Cj = {e}.
Proof. By the above description for C(IA (Φn ) , Φn ), one can write:
Ci

=

ker(Imρ̂i → Aut(Φ̂n ))

=

2
ker(lim
←− (IAm · Imρi /IAm ) → lim
←− (IA (Φn ) /IGm ))
2
lim
←
− ((IAm · Imρi ) ∩ (IAm · IGm )) /IAm .

=

We claim now that:
(IAm · Imρi ) ∩ (IAm · IGm2 ) ⊆ IAm · (Imρi ∩ IGm ) .
So we have to show that if ar = bs such that a, b ∈ IAm , r ∈ Imρi and s ∈ IGm2 ,
then there exist c ∈ IAm and t ∈ Imρi ∩ IGm such that
 ar = bs = ct. Indeed,
write: Imρi ∋ r = a−1 bs. Then: r = ρi (r) = ρi a−1 b ρi (s), and:

ρi a−1 b ∈ ρi (IAm ) = Imρi ∩ IAm

ρi (s) ∈ ρi (IGm2 ) ⊆ Imρi ∩ IGm .

Therefore, by deﬁning c = a · ρi a−1 b , and t = ρi (s) we get the required
inclusion. Thus, we have:
Ci ∩ Cj ⊆ lim
←− ((IAm · (Imρi ∩ IGm ) ∩ IAm · (Imρj ∩ IGm )) /IAm ) .
We claim now that:
IAm · (Imρi ∩ IGm ) ∩ IAm · (Imρj ∩ IGm ) = IAm .
Indeed, if ar = bs such that a, b ∈ IAm , r ∈ Imρi ∩ IGm and s ∈ Imρj ∩ IGm ,
then:

ar = aρi (r) = aρi a−1 b ρi (s)

and we have aρi a−1 b ∈ IAm . I addition, it is not diﬃcult to observe that:
ρi (s)

∈ ρi (Imρj ∩ IGm )
⊆ hIn + m(σi Ek,j − σj Ek,i ) | k 6= i, ji
m

= hIn + σi Ek,j − σj Ek,i | k 6= i, ji
Hence, ar ∈ IAm , and Ci ∩ Cj = {e} as required.
14

⊆ IAm .

The proof of the above proposition shows that for every 1 ≤ i ≤ n and for
every m ∈ N we have:
(IAm · Imρi )∩(IAm · IGm2 ) ⊆ IAm ·(Imρi ∩ IGm ) ⊆ (IAm · Imρi )∩(IAm · IGm )
and by proposition 3.8 we have also ρi (IGm2 ) ⊆ Im(ρi ) ∩ IGn,m ⊆ ρi (IGm ). It
ρ̂i

follows that Ci ֒→ C(IA (Φn ) , Φn ) ։ Ci such that the composition map is the
identity. Indeed:
Ci

=
=
=
֒→
ρ̂i

։
=
=

lim
←
− ((IAm · Imρi ) ∩ (IAm · IGm )) /IAm
lim
←
−IAm · (Imρi ∩ IGm ) /IAm
limIA · (Imρi ∩ IGm2 ) /IAm
←− m
2
lim
←−IAm · IGm /IAm = C(IA (Φn ) , Φn )
2
lim
←−ρi (IAm ) · ρi (IGm ) /ρi (IAm )
lim
←
− (Imρi ∩ IAm ) · (Imρi ∩ IGm ) / (Imρi ∩ IAm )

limIA · (Imρi ∩ IGm ) /IAm = Ci .
←− m

Hence, one gets the following corollary:
Corollary 4.3. For every 1 ≤ i ≤ n we have:
C (IA (Φn ) , Φn ) = (ker ρ̂i ∩ C (IA (Φn ) , Φn )) ⋊ Ci .
We would like now to compute Ci and to show that it is canonically isomorphic to:
\
\
±1 ])).
ker(SLn−1
(Z[x±1 ]) → SLn−1 (Z[x
So set n ≥ 4, 1 ≤ i0 ≤ n, denote: x = xi0 , σ = σi0 = xi0 − 1, R = Z[x±1 ] =
m
Z[x±1
i0 ] and for m ∈ N denote: Hm = (x − 1) R + mR. Denote also:
ρ

=

ρ̂

=

ρi0 : IA (Φn ) ։ GLn−1 (R, σR)
\
\
(Φn ) ։ GLn−1
(R, σR).
ρ̂i : IA
0

Now, write (the last equality is by proposition 3.7):
Ci0

=
=

ker(Imρ̂ → Aut(Φ̂n ))
\
ker(GLn−1
(R, σR) → Aut(Φ̂n ))

=

\
ker(GLn−1
(R, σR) → lim
←− (IA (Φn ) /IGm ))

=

\
ker(GLn−1
(R, σR) → lim
←− (GLn−1 (R, σR) · IGm /IGm ))

=

\
ker(GLn−1
(R, σR) → limGLn−1 (R, σR) / (GLn−1 (R, σR) ∩ IGm ))
←−
\
ker(GLn−1 (R, σR) → lim
←−GLn−1 (R, σR) /GLn−1 (R, σHm )).

=

15

Now, by the same computation as in Proposition 3.8 one can show that for every
m ∈ N one has (Hm2 ∩ σR) ⊆ σHm ⊆ (Hm ∩ σR), so the latter is equals to:
\
ker(GLn−1
(R, σR) → lim
←−GLn−1 (R, σR) / (GLn−1 (R, σR) ∩ GLn−1 (R, Hm )))
\
= ker(GLn−1
(R, σR) → lim
←−GLn−1 (R) /GLn−1 (R, Hm )
\
= ker(GLn−1
(R, σR) → limGLn−1 (R/Hm )).
←−
Now, if R̄ is a ﬁnite quotient of R, then as x is invertible in R, its image x̄ ∈ R̄
is invertible in R̄. Thus, there exists r ∈ N such that x̄r = 1R̄ . In addition, there
exists s ∈ N such that 1R̄ + . . . + 1R̄ = 0R̄ . Therefore, for m = r · s the map
|
{z
}
s

R → R̄ factorizes through Zm [Zm ] ∼
lim (R/Hm ),
= R/Hm . Thus, we have R̂ = ←
−
which implies that: GLn−1 (R̂) = ←
lim
GL
(R/H
).
Therefore:
n−1
m
−
\
(R, σR) → GLn−1 (R̂)).
Ci0 = ker(GLn−1

Now, the short exact sequence:
1 → GLn−1 (R, σR) → GLn−1 (R) → GLn−1 (Z) → 1
gives rise to the exact sequence (see [BER], Lemma 2.1):
\
\
GLn−1
(R, σR) → GL\
n−1 (R) → GLn−1 (Z) → 1
which gives rise to the commutative diagram:
\
\
GLn−1
(R, σR) → GL\
n−1 (R) → GLn−1 (Z)
ց
↓
↓
GLn−1 (R̂) → GLn−1 (Ẑ)

→ 1
→ 1

.

Assuming n ≥ 4 and using the aﬃrmative answer to the classical congruence
subgroup problem ([Men], [BaLS]), the map: GL\
n−1 (Z) → GLn−1 (Ẑ) is injec\
tive. Thus, by diagram chasing we obtain that the kernel ker(GLn−1
(R, σR) →
\
GLn−1 (R̂)) is mapped onto ker(GLn−1 (R) → GLn−1 (R̂)). Let have now the
following lemma:


Lemma 4.4. Let d ≥ 3 and denote: Dm = Id + xk·m − 1 E1,1 | k ∈ Z for
m ∈ N. Then:
\
GL
d (R) =

lim (GLd (R) / (Dm Ed (R, Hm )))
←−

\
SL
d (R) =

lim
←
− (SLd (R) /Ed (R, Hm )) .

Proof. We will prove the ﬁrst part and the second is similar but easier. We
ﬁrst claim that Dm Ed (R, Hm ) is a ﬁnite index normal subgroup of GLd (R):
16

By well-known result of Suslin [Su], SLd (R) = Ed (R). Thus, by Corollary
2.3, SK1 (R, Hm ; d) = SLd (R, Hm ) /Ed (R, Hm ) is ﬁnite. As the subgroup
SLd (R, Hm ) is of ﬁnite index in SLd (R), so is Ed (R, Hm ). Now, it is not
diﬃcult
elements of R is equals to R∗ =
 k to see that the group of invertible
 k·m
±x | k ∈ Z (see [CF], chapter 8). So as x
| k ∈ Z is of ﬁnite index in R∗ ,
Dm SLd (R) is of ﬁnite index in GLd (R). We deduce that also Dm Ed (R, Hm ) is
of ﬁnite index in GLd (R). It remains to showuthat Dm Ed (R, Hm ) is normal in
GLd (R). We already stated previously (see 2) that Ed (R, Hm ) is normal in
GLd (R). Thus, it is easy to see that it is enough to show that the commutators
of the elements of Dm with any set of generators of GLd (R), are in Ed (R, Hm ).
By the above mentioned result of Suslin and as R∗ = {±xr | r ∈ Z}, GLd (R) is
generated by the elements of the forms:
1. Id + (±x − 1) E1,1
2.
3.

Id + rEi,j
Id + rE1,j

r ∈ R, 2 ≤ i 6= j ≤ d
r ∈ R, 2 ≤ j ≤ d

4.

Id + rEi,1

r ∈ R, 2 ≤ i ≤ d.

Now, obviously, the elements of Dm commute with the elements of the forms 1
and 2. In addition, for the elements of the forms 3 and 4, one can easily compute
that:




= Id + r xk·m − 1 E1,j ∈ Ed (R, Hm )
Id + xk·m − 1 E1,1 , Id + rE1,j




Id + xk·m − 1 E1,1 , Id + rEi,1 = Id + r x−k·m − 1 Ei,1 ∈ Ed (R, Hm )

for every 2 ≤ i, j ≤ d, as required.
Now, every ﬁnite index normal subgroup of GLd (R) contains Dm for some
m ∈ N. In addition, It is not hard to show that when d ≥ 3, every ﬁnite index
normal subgroup N ⊳ GLd (R) contains Ed (R, H) for some ﬁnite index ideal
H ⊳ R (see [KN], Section 1). Thus, as every ﬁnite index ideal H ⊳ Rn contains
\
Hm for some m, GL
d (R) = lim (GLd (R) / (Dm Ed (R, Hm ))), as required.
←−
Let have now the following proposition:
\
Proposition 4.5. Let n ≥ 4. Then, the map GLn−1
(R, σR) → GL\
n−1 (R) is
injective. Hence, the surjective map:
\
(R, σR) → GLn−1 (R̂)) ։ ker(GL\
Ci0 = ker(GLn−1
n−1 (R) → GLn−1 (R̂))
is an isomorphism.
Proof. We showed in the previous lemma that
GL\
n−1 (R) = lim
←−GLn−1 (R) / (Dm En−1 (R, Hm ))


where: Dm = In−1 + xk·m − 1 E1,1 | k ∈ Z and Hm = (xm − 1) R + mR.
\
Hence, the image of GLn−1
(R, σR) in GL\
n−1 (R) is:
lim
←−GLn−1 (R, σR) / (GLn−1 (R, σR) ∩ Dm En−1 (R, Hm ))
17

which by using that Dm ⊆ GLn−1 (R, σR), one can see that it is equals to:
lim
←−GLn−1 (R, σR) / (Dm (GLn−1 (R, σR) ∩ En−1 (R, Hm ))) .
Now, by the surjective map ρ : IA (Φn ) ։ GLn−1 (R, σR) we have:
m

hIA (Φn ) i
IEn−1,i0 (Hn,m )

ρ

m

։ hGLn−1 (R, σR) i
ρ

։ GLn−1 (R, σR) ∩ En−1 (R, Hm ) .

m
So as by the main Lemma (Lemma 7.1) we have IEn−1,i0 Hn,m2 ⊆ hIA (Φn ) i,
we have also:
m

GLn−1 (R, σR) ∩ En−1 (R, Hm2 ) ⊆ hGLn−1 (R, σR) i .
m

As obviously Dm2 ⊆ hGLn−1 (R, σR) i, we deduce the following natural surjective maps:
2
2
lim
←−GLn−1 (R, σR) / (Dm (GLn−1 (R, σR) ∩ En−1 (R, Hm )))
m
։ lim
←−GLn−1 (R, σR) / hGLn−1 (R, σR) i

\
։ GLn−1
(R, σR)
։ limGLn−1 (R, σR) / (Dm (GLn−1 (R, σR) ∩ En−1 (R, Hm )))
←−
2
2
= lim
←−GLn−1 (R, σR) / (Dm (GLn−1 (R, σR) ∩ En−1 (R, Hm )))
such that the composition gives the identity map. Hence, these maps are also
injective, and in particular, the map:
\
GLn−1
(R, σR) → lim
←−GLn−1 (R, σR) / (Dm (GLn−1 (R, σR) ∩ En−1 (R, Hm )))
is injective, as required.
Proposition 4.6. Let d ≥ 3. Then, the natural embedding SLd (R) ≤ GLd (R)
induce a natural isomorphism:
\
\
∼
ker(SL
d (R) → SLd (R̂)) = ker(GLd (R) → GLd (R̂)).
Proof. By Lemma 4.4 we have:
\
ker(GL
d (R) → GLd (R̂) = lim
←−GLd (R/Hm ))
= ker(limGLd (R) /Dm Ed (R, Hm ) → limGLd (R) /GLd (R, Hm ))
←−
←−
=←
lim
GL
(R,
H
)
/D
E
(R,
H
)
.
d
m
m
d
m
−
Notice now that under the map R → Zm [Zm ], every A ∈ GLd (R, Hm ) is
mapped to Id and hence det (A) 7→ 1, which implies that det (A) = xk·m for some
k ∈ Z (provided m > 2. When m = 2, det (A) = ±x2k , but this exception does

18

not inﬂuence the inverse limit below). Thus: GLd (R, Hm ) = Dm SLd (R, Hm )
(for every m > 2). Therefore, since Dm ∩ SLd (R, Hm ) = {Id }, we deduce that:
\
ker(GL
d (R) → GLd (R̂))

=
=

lim
←
−Dm SLd (R, Hm ) /Dm Ed (R, Hm )
lim
←−SLd (R, Hm ) /Ed (R, Hm )

=

\
lim
←− ker(SLd (R) → SLd (R̂)).

The immediate corollary from Propositions 4.5 and 4.6 is:
Corollary 4.7. For every n ≥ 4, Ci0 ∼
= ker(SL\
n−1 (R) → SLn−1 (R̂)).
We close the section by showing that ker(SL\
n−1 (R) → SLn−1 (R̂)) is not
ﬁnitely generated, using the techniques in [KN]. It is known that the group
ring Z[x±1 ] = Z [Z] is Noetherian (see [I], [BrLS]). In addition, it is known that
dim (Z) = 1 and thus dim (Z [Z]) = 2 (see [Sm]). Therefore, by Proposition 1.6
in [Su], as n − 1 ≥ 3, for every H ⊳ R, the canonical map:
SK1 (R, H; n − 1) → SK1 (R, H) := lim
−→ SK1 (R, H; d)
d∈N

is surjective. Hence, the canonical map (when H ⊳ R ranges over all ﬁnite
index ideals of R):
ker(SL\
n−1 (R) → SLn−1 (R̂))

=
=

lim
←− (SLn−1 (R, H) /En−1 (R, H))
lim
←−SK1 (R, H; n − 1)

→ lim
←−SK1 (R, H)
is surjective, so it is enough to show that lim
←−SK1 (R, H) is not ﬁnitely generated.
By a result of Bass (see [Bas], chapter 5, Corollary 9.3), for every H ⊳ K ⊳ R
of ﬁnite index in R, the map: SK1 (R, H) → SK1 (R, K) is surjective. Hence,
it is enough to show that for every l ∈ N there exists a ﬁnite index ideal H ⊳ R
such that SK1 (R, H) is generated by at least l elements. Now, as SK1 (R) = 1
[Su], we obtain the following exact sequence for every H ⊳ R (see Theorem 6.2
in [Mi]):
K2 (R) → K2 (R/H) → SK1 (R, H) → SK1 (R) = 1.
In addition, by classical result of Quillen ([Q], [Ros] Theorem 5.3.30), we have:

K2 (R) = K2 Z[x±1 ] = K2 (Z) ⊕ K1 (Z)

so by the classical facts K2 (Z) = K1 (Z) = {±1} (see [Mi] chapters 3 and 10)
we deduce that K2 (R) is of order 4. Hence, it is enough to prove that for every
l ∈ N there exists a ﬁnite index ideal H ⊳ R such that K2 (R/H) is generated
by at least l elements. Following [KN], we state the following proposition (by
the proof of Theorem 2.8 in [SD]):
19

Proposition 4.8. Let p be a prime, l ∈ N and denote by P ⊳ Z [y] the ideal

l
which generated by p2 and y p . Then, for R̄ = Z [y] /P , K2 R̄ is an elementary
abelian p-group of rank ≥ l.
Observe now that for every l ≥ 0:
pl+1

(y + 1)

l

= (y p + 1 + p · a (y))p = 1 mod P

so y + 1 is invertible in R̄. Therefore we have a well deﬁned sujective homomorphism R → R̄ which deﬁned by sending x 7→ y + 1. In particular,
H = ker R → R̄ is a ﬁnite index ideal of R which satisﬁes the above requirements. This shows that indeed ker(SL\
n−1 (R) → SLn−1 (R̂)) is not ﬁnitely
generated, and C (IA (Φn ) , Φn ) is not ﬁnitely generated either.

5

The centrality of Ci

In this section we will prove that for every n ≥ 4, the copies Ci lie in the
\
center of IA
(Φn ). Along the section we will assume that n ≥ 4 is constant,
and will show it for i = n. Symmetrically, it will be valid for every i. So for
±1
simplifying
we will denote: R = Rn = Z[x±1
1 , . . . , xn ] and Hm =
Pnnotations
m
Hn,m = i=1 (xi − 1) Rn + mRn (note that in the previous section we used
these notations for diﬀerent objects, i.e. for Z[x±1 ] and (xm −1)Z[x±1 ]+mZ[x±1 ]
respectively). Also here we will simplify the notations and write IAm = IAn,m
and IGm = IGn,m . We saw in Section 4 that we can write:
C(IA (Φn ) , Φn )

= ker(lim (IA (Φn ) /IAm ) → lim (IA (Φn ) /IGm2 ))
←−
←−
2 /IAm ) .
= lim
(IA
·
IG
m
m
←−

Similarly, we can write:
Cn

=
≤
≤

2
lim
←
− (IAm · (Imρn ∩ IGm ) /IAm )
2
lim
←− (IAm · IGm /IAm )

\
lim
←− (IA (Φn ) /IAm ) = IA (Φn ).

\
Hence, if we want to show that Cn lies in the center of IA
(Φn ), it suﬃces to
show that for every m ∈ N, the group IAm · (Imρn ∩ IGm2 ) /IAm lies in the
center of IA (Φn ) /IAm .

Denote R̄ = Z[x±1
n ], and recall Imρn ∩ IGn,m2 ≃ GLn−1 R̄, σn Hm2 ∩ R̄)
(see Proposition 3.7). We claim that by this isomorphism one has:


IAm · Imρn ∩ IGn,m2 /IAm = IAm · SLn−1 R̄, σn Hm2 ∩ R̄) /IAm .

Indeed, if α ∈ Imρn ∩ IGm2 then det(α) ∈ 1 + σn Hm2 ∩ R̄. In particular, it is
2
s
in 1 + Hm2 ∩ R̄ so it must be of the form xm
for some s ∈ Z (see proposition
n
3.1). Consider now the element

2
(In + σn E1,1 − σ1 E1,n )m s ∈ IAm ∩ GLn−1 R̄, σn Hm2 ∩ R̄) .
20

2

Then by replacing α with (In + σn E1,1 − σ1 E1,n )−m s · α we deduce that Imρn ∩
IGn,m2 ⊆ IAm · SLn−1 R̄, σn Hm2 ∩ R̄) and we get the above equality. It
\
follows that if we want to show that Cn lies in the center of IA
(Φn ) it suﬃces

\
(Φn ).
to show that IAm · SLn−1 R̄, σn Hm2 ∩ R̄) /IAm lies in the center of IA
However, we are going to show even more. We will show that
IAm · ISLn−1,n (σn Hm2 ) /IAm
lies in the center of IA (Φn ) /IAm .
Let Fn be the free group on f1 , . . . , fn . It is a classical result by Magnus
([MKS], Chapter 3, Theorem N4) that IA (Fn ) is generated by the automorphisms of the form:
(
fr 7→ [ft , fs ] fr
αr,s,t =
fu 7→ fu
u 6= r
when 1 ≤ r, s 6= t ≤ n (notice that we may have r = s). In their paper [BM],
Bachmuth and Mochizuky show that for every n ≥ 4, IA (Φn ) is generated by
the images of these generators under the natural map Aut (Fn ) → Aut (Φn ).
I.e. IA (Φn ) is generated by the elements of the form:
Er,s,t = In + σt Er,s − σs Er,t 1 ≤ r, s 6= t ≤ n.
Therefore, for showing the centrality of Cn , it is enough to show that given:
• an element: λ̄ ∈ IAm · ISLn−1,n (σn Hm2 ) /IAm ,
• and one of generators: Er,s,t = In + σt Er,s − σs Er,t for 1 ≤ r, s 6= t ≤ n,
there exists a representative of λ̄, λ ∈ ISLn−1,n (σn Hm2 ), such that [Er,s,t , λ] ∈
IAm . So assume that we have an element: λ̄ ∈ IAm · ISLn−1,n (σn Hm2 ) /IAm .
Then, a representative for λ̄, λ ∈ ISLn−1,n (σn Hm2 ), has the form:


Pn−1
In−1 + σn B − i=1 σi~bi
λ=
0
1
for some (n − 1) × (n − 1) matrix B which its entries bi,j admit bi,j ∈ Hm2 and
its column vectors denoted by ~bi . Let have now the following proposition:
Proposition 5.1. Let λ̄ ∈ IAm · ISLn−1,n (σn Hm2 ) /IAm . Then, for every
1 ≤ l < k ≤ n − 1, λ̄ has a representative in ISLn−1,n (σn Hm2 ), of the following
form (the following notation means that the matrix is similar to the identity
matrix, except the entries in the l-th and k-th rows):


Il−1
0
0
0
0
0
 0
1 + σn a
0
σn b
0
−σl a − σk b 

 ← l-th row
 0

0
I
0
0
0
k−l−1


 0
σn c
0
1 + σn d
0
−σl c − σk d 

 ← k-th row
 0

0
0
0
In−k−1
0
0
0
0
0
0
1
(5.1)
21

for some a, b, c, d ∈ Hm2 .
Proof. We will demonstrate the proof in the case l = 1, k = 2, and symmetrically, the arguments hold for arbitrary 1 ≤ l < k ≤ n − 1. Consider an arbitrary
representative of λ̄:

Pn−1 ~ 
In−1 + σn B − i=1
σi bi
λ=
∈ ISLn−1,n (σn Hm2 ) .
0
1
Then In−1 + σn B ∈ SLn−1 (R, σn Hm2 ). Consider now the ideal:
′
R ⊲ Hm
2 =

n−1
X

2

2

m
2
(xm
r − 1)R + σn (xn − 1)R + m R.

r=1

′
′
Observe that σn Hm2 ⊳ Hm
2 ⊳ H m2 ⊳ R and that Hm2 ∩ σn R = σn Hm2 .
In addition, by similar computations as in the proof of proposition 3.8, for
4
2
every x ∈ R we have xm − 1 ∈ (x − 1) (xm − 1)R + (x − 1) m2 R, and thus
′
′
H m 4 ⊆ Hm
2 , so Hm2 is of ﬁnite index in R.

′
Now, In−1 + σn B ∈ SLn−1 (R, σn Hm2 ) ⊆ SLn−1 R, Hm
2 . Thus, by the
′
third part of Corollary 2.3, as R ⊲ Hm
2 is an ideal of ﬁnite index, n − 1 ≥ 3
and En−1 (R) = SLn−1 (R) [Su], one can write the matrix In−1 + σn B as:

 ′
A
0
In−1 + σn B = AD when A =
0 In−3


′
′
Now, consider the
and D ∈ E n−1 R, Hm
for some A′ ∈ SL2 R, Hm
2 .
2
images of D and A under the projection σn → 0, which we
denote
by D̄ and Ā,

′
respectively. Observe that obviously, D̄ ∈ E n−1 R, Hm
2 . In addition, observe
that:
AD ∈ GLn−1 (R, σn R) ⇒ ĀD̄ = In−1 .

Thus, we have In−1 + σn B = AĀ−1 D̄−1 D. Therefore, by replacing D by D̄−1 D
and A by AĀ−1 we can assume that:

 ′
A
0
In−1 + σn B = AD for A =
0 In−3

′
where: A′ ∈ SL2 R, Hm
∩ GL2 (R, σn R) = SL2 (R, σn Hm2 ), and:
2
D

∈
⊆

′
E n−1 (R, Hm
2 ) ∩ GLn−1 (R, σn R)
E n−1 (R, H m2 ) ∩ GLn−1 (R, σn R) = IEn−1,n (Hm2 ) .

Now, as we prove in the main lemma (Lemma 7.1) that IEn−1,n (Hm2 ) ⊆
m
hIA (Φn ) i ⊆ IAm , this argument shows that λ can be replaced by a representative of the form 5.1.
We now return to our initial mission. Let λ̄ ∈ IAm ·ISLn−1,n (σn Hm2 ) /IAm ,
and let Er,s,t = In + σt Er,s − σs Er,t for 1 ≤ r, s 6= t ≤ n be one of the above
22

generators for IA (Φn ). We want to show that there exists a representative of λ̄,
λ ∈ ISLn−1,n (σn Hm2 ), such that [Er,s,t , λ] ∈ IAm . We separate the treatment
to two cases:
The first case is: 1 ≤ r ≤ n − 1. In this case one can take an arbitrary representative λ ∈ ISLn−1,n (σn Hm2 ) ∼
= SLn−1 (R, σn Hm2 ). Considering the embedding of IGL′n−1,n in GLn−1 (R), we have: Er,s,t ∈ IGL′n−1,n ⊆ GLn−1 (R).
Thus, as by Corollary 2.3
SK1 (R, Hm2 ; n − 1) = SLn−1 (R, Hm2 ) /En−1 (R, Hm2 )
is central in GLn−1 (R) /En−1 (R, Hm2 ), we have:
[Er,s,t , λ] ∈ [GLn−1 (R) , SLn−1 (R, σn Hm2 )] ⊆ En−1 (R, Hm2 ) .
In addition, as SLn−1 (R, σn Hm2 ) ≤ GLn−1 (R, σn R) and GLn−1 (R, σn R) is
normal in GLn−1 (R), we have:
[Er,s,t , λ] ∈ [GLn−1 (R) , GLn−1 (R, σn R)] ⊆ GLn−1 (R, σn R) .
Thus, we obtain from Lemma 7.1 that:
[Er,s,t , λ]

∈ En−1 (R, Hm2 ) ∩ GLn−1 (R, σn R)
= IEn−1,n (Hm2 ) ⊆ hIA (Φn )m i ⊆ IAm .

The second case is: r = n. This case is a bit more complicated than
the previous one, as Er,s,t is not in IGL′n−1,n . Here, by Proposition 5.1 one
can choose λ ∈ ISLn−1,n (σn Hm2 ) which its t-th row is equals to the standard
vector ~et . As t 6= r = n, we obtain thus that both λ and Er,s,t are in the
canonical embedding of IGL′n−1,t in GLn−1 (R). Moreover, considering the
above mentioned embedding of IGL′n−1,t in GLn−1 (R), we have:
Er,s,t ∈ GLn−1 (R, σt R) , λ ∈ SLn−1 (R, Hm2 ) .
Note that when considering λ ∈ IGL′n−1,n ֒→ GLn−1 (R), i.e. when considering
λ ∈ GLn−1 (R) through the embedding of IGL′n−1,n in GLn−1 (R), we have λ ∈
GLn−1 (R, σn Hm2 ) ≤ GLn−1 (R). However, when we consider λ ∈ IGL′n−1,t ֒→
GLn−1 (R) we do not necessarily have: λ ∈ GLn−1 (R, σn Hm2 ), but we still
have λ ∈ GLn−1 (R, Hm2 ).
Thus, by similar arguments as in the ﬁrst case:
[Er,s,t , λ]

∈

[GLn−1 (R, σt R) , SLn−1 (R, Hm2 )]

⊆ En−1 (R, Hm2 ) ∩ GLn−1 (R, σt R)
= IEn−1,t (Hm2 ) ⊆ hIA (Φn )m i ⊆ IAm .
\
This ﬁnishes the argument which shows that Ci are central in IA
(Φn ).

23

6

Some elementary elements of hIA (Φn )m i
m

In this section we compute some elements in hIA (Φn ) i, which needed through
the proof of the main lemma. Additionally to the previous notations, on the
section, and also later on, we will use the notation:
σr,m = xm
r − 1 for 1 ≤ r ≤ n.
Proposition 6.1. Let n ≥ 4, 1 ≤ u ≤ n and m ∈ N. Denote by ~ei the i-th
row standard vector. Then, the elements of IA (Φn ) of the form (the following
notation means that the matrix is similar to the identity matrix, except the
entries in the u-th row):


Iu−1
0
0
 au,1 · · · au,u−1 1 au,u+1
· · · au,n  ← u-th row
0
0
In−u

when (au,1 , . . . , au,u−1 , 0, au,u+1 , . . . , au,n ) is a linear combination of the vectors:
1. {m (σi~ej − σj ~ei ) | i, j 6= u, i 6= j}
2. {σ k,m (σi~ej − σj ~ei ) | i, j, k 6= u, i 6= j}
3. {σu σu,m (σi~ej − σj ~ei ) | i, j 6= u, i 6= j}
m

with coefficients in Rn , belong to hIA (Φn ) i.
Proof. Without loss of generality, we assume that u = 1. Observe now that for
every ai , bi ∈ Rn for 2 ≤ i ≤ n one has:


 

1 a2 · · · an
1 b2 · · · bn
1 a2 + b 2 · · · an + b n
=
.
0
In−1
0
In−1
0
In−1
Hence, it is enough to prove that the elements of the following forms belong to
m
hIA (Φn ) i (when we write a~ei we mean that the entry of the i-th column in
the ﬁrst row is a):


1 mf (σi~ej − σj ~ei )
i, j 6= 1, i 6= j, f ∈ Rn
1.
0
In−1


1 σ k,m f (σi~ej − σj ~ei )
i, j, k 6= 1, i 6= j, f ∈ Rn
2.
0
In−1


1 σ1 σ1,m f (σi~ej − σj ~ei )
i, j 6= 1, i 6= j, f ∈ Rn
3.
0
In−1
We start with the elements of form 1. Here we have:
m
 

1 f (σi~ej − σj ~ei )
1 mf (σi~ej − σj ~ei )
m
∈ hIA (Φn ) i .
=
0
In−1
0
In−1

24

We pass to the elements of form 2. In this case we have:
"
−1 
m #
1 f (σi~ej − σj ~ei )
xk −σ1~ek
m
hIA (Φn ) i ∋
,
0
In−1
0
In−1


1 σ k,m f (σi~ej − σj ~ei )
.
=
0
In−1
We ﬁnish with the elements of form 3. The computation here is more complicated than in the previous cases, so we will demonstrate it for the private
case: n = 4, i = 2, j = 3. It is clear that symmetrically, with similar argument,
the same holds in general when n ≥ 4 for every i, j 6= 1, i 6= j. By similar
arguments as in the previous case we get:


1
0
0
0
 0
1
0
0 
m
.
hIA (Φ4 ) i ∋ 
 0
0
1
0 
0 σ3 σ1,m f −σ2 σ1,m f 1
Therefore, we also have:

x4 0 0 −σ1
 0 1 0
0
m

hIA (Φ4 ) i ∋ 
 0 0 1
0
0 0 0
1

1 −σ3 σ1 σ1,m f
 0
1
= 
 0
0
0
0

7

 

1
  0
,
  0
0

0
1
0
σ3 σ1,m f

σ2 σ1 σ1,m f 0
0
0 
.
1
0 
0
1

0
0
1
−σ2 σ1,m f


0

0 

0 
1

A main lemma

In this section we prove the main lemma which asserts that:
Lemma 7.1. For every n ≥ 4, m ∈ N and 1 ≤ i ≤ n we have:

m
IEn−1,i Hn,m2 ⊆ hIA (Φn ) i .

For simplifying the proof and the notations, we will prove the lemma for
the private case i = n, and symmetrically, all the arguments are valid for every
1 ≤ i ≤ n. Also in this section n will be constant, so we
Pncan use the notations:
m
IAm = hIA (Φn ) i, R = Rn and Hm = Hn,m =
r=1 Ur,m + Om where:
Ur,m = σr,m R and Om = mR.
In addition, using the isomorphism IGLn−1,n ∼
= GLn−1 (R, σn R) (Proposition 3.4), we will identify IGLn−1,n with GLn−1 (R, σn R), and IEn−1,n (Hm )
25

with GLn−1 (R, σn R)∩E n−1 (R, Hm ). So the goal of this section is proving that
GLn−1 (R, σn R) ∩ En−1 (R, Hm2 ) ⊆ IAm .
Through the proof we use also elements of IGL′ n−1,n , so as GLn−1 (R, σn R)∩
En−1 (R, Hm2 ) ≤ GLn−1 (R, σn R) ≤ IGL′ n−1,n ≤ GLn−1 (R) (Proposition
3.11), in this section, we will make all the computations in GLn−1 (R), and
omit the n-th row and column from each matrix.

7.1

Decomposing the proof

In this subsection we will start the proof of Lemma 7.1. In the end of the
subsection, there will be remained a few tasks, which will be accomplished in
the forthcoming subsections. We start with the following deﬁnition:
Definition 7.2. For every m ∈ N, deﬁne the following ideal of R:
Tm =

n
X

σr2 Ur,m +

r=1

n
X

2
σr Om + Om
.

r=1

By the observation that for every x ∈ R we have
one can have the following computation:
m2

x

−1 =

(x − 1)

2
m
−1
X

xj = (x − 1)

m−1
X
j=0

j=0

∈
⊆

Pm−1
j=0

xj

xj ∈ (x − 1)R + mR,

m−1
X

xjm

j=0

(x − 1) ((x − 1)R + mR) ((xm − 1)R + mR)
(x − 1)2 (xm − 1)R + (x − 1)2 mR + (x − 1)m2 R

which shows that Hm2 ⊆ Tm . Hence, it is enough to prove that:
GLn−1 (R, σn R) ∩ En−1 (R, Tm ) ⊆ IAm .
Equivalently, it is enough to prove that the group:
(GLn−1 (R, σn R) ∩ En−1 (R, Tm )) · IAm /IAm
is trivial.
We continue with the following lemma, which is actually a lemma of Suslin
(Corollary 1.4 in [Su]) with some elaborations of Bachmuth and Mochizuki [BM]:
Lemma 7.3. Let R be a commutative ring, d ≥ 3, and H ⊳ R ideal. Then,
Ed (R, H) is generated by the matrices of the form:
(Id − f Ei,j ) (Id + hEj,i ) (Id + f Ei,j )

(7.1)

for h ∈ H, f ∈ R and 1 ≤ i 6= j ≤ d.
Proof. By Suslin’s proof to Corollary 1.4 in [Su], and the remark which follows
Proposition 3.5 in [BM], Ed (R, H) is generated by the matrices of the form:
t

Id + h (f1~ei + f2~ej ) (f2~ei − f1~ej )
26

for f1 , f2 ∈ R, h ∈ H and 1 ≤ i 6= j ≤ d. So it is enough to show that these
matrices are generated by the matrices of the form 7.1. We will show it for the
case i, j, d = 1, 2, 3 and it will be clear that the general argument is similar. So
we have the matrix:


1 + hf1 f2
−hf12
0
Id + h (f1~e1 + f2~e2 )t (f2~e1 − f1~e2 ) = 
hf22
1 − hf1 f2 0 
0
0
1
for some f1 , f2 ∈ R and h ∈ H,

1
0
1
= 0
f2 −f1


1
0
0
1 0
1
0  0 1
= 0
f2 −f1 1
0 0
As the matrix


1
 0
0

which is equals to:


−hf1
1
0 hf1
−hf2   0
1 hf2 
1
−f2 f1
1


−hf1
1
0 0
1 0
−hf2   0
1 0  0 1
1
−f2 f1 1
0 0

 
0 hf1
1 0
1 hf2  =  0 1
0 1
0 0

is generated by the matrices of the


1
0
0
1
 0
1
0  0
f2 −f1 1
0
=


hf1
hf2  .
1


0
hf2 
1


hf1
1 0
0  0 1
0 0
1

form 7.1, it remains to show that:


0 −hf1
1
0 0
1 −hf2   0
1 0 
0
1
−f2 f1 1






1
0
0
1 0 −hf1
1
0 0
 0
1
0  0 1
0  0
1 0 
f2 −f1 1
0 0
1
−f2 f1 1




1
0
0
1
0 0
1 0
0
1
0   0 1 −hf2   0
1 0 
· 0
0 0
1
f2 −f1 1
−f2 f1 1

is generated by the matrices of the


1
0
0
1
 0
1
0  0
f2 −f1 1
0


1 + hf1 f2
0
=
hf1 f22

form 7.1. Now:

0 −hf1
1
1
0  0
0
1
−f2
−hf12
1
−hf12 f2

27

0
1
f1


−hf1

0
1 − hf1 f2


0
0 
1



1
= 0
0


−hf12
1
−hf12 f2

1
0
1
=  0
0 −hf12 f2

1 0 0
·  0 1 0
f2 0 1


0
1 + hf1 f2
0
0 
hf1 f22
1

0
0 
1

1
 0
0

is generated by the matrices of the


1
0
0
1
 0
1
0  0
f2 −f1 1
0

1 −hf12
0
1
0
0
0
1
0

0
1
0


−hf1

0
1 − hf1 f2


0
0 
1


1
−hf1
0  0
1
−f2


0
1 0 
0 1

form 7.1, and by similar computation:


0
0
1
0 0
1 −hf2   0
1 0 
0
1
−f2 f1 1

is generated by these matrices as well.

We proceed with the followimg lemma:
Lemma 7.4. Let n ≥ 4. Then, every element of GLn−1 (R, σn R)∩En−1 (R, Tm )
can be decomposed as a product of elements of the following four forms:
1.

A−1 (In−1 + hEi,j ) A

h ∈ σn Om

2.

A−1 (In−1 + hEi,j ) A

h ∈ σn2 Un,m , σn σr2 Ur,m
for 1 ≤ r ≤ n − 1

3.

A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A

2
h ∈ Ōm
, f ∈ σn R

4.

A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A

h ∈ σr2 Ūr,m , σr Ōm for
1 ≤ r ≤ n − 1, f ∈ σn R

when A ∈ GLn−1 (R), i 6= j, and Ūr,m , Ōm are the projections of Ur,m , Om
±1
to Rn−1 = Z[x±1
1 , . . . , xn−1 ], induced by the projection σn 7→ 0.
Remark 7.5. Notice that as GLn−1 (R, σn R) is normal in GLn−1 (R), every
element of the above forms is an element of GLn−1 (R, σn R) ∼
= IGLn−1,n ≤
IA (Φn ).
Proof. (of Lemma 7.4) Let B ∈ GLn−1 (R, σn R) ∩ En−1 (R, Tm ). We ﬁrst claim
that for proving the lemma, it is enough to show that B can be decomposed as
a product of the elements in the lemma (Lemma 7.4), and arbitrary elements in
GLn−1 (Rn−1 ). Indeed, assume that we can write B = C1 D1 · · · Cn Dn for some
Di of the forms in the lemma and Ci ∈ GLn−1 (Rn−1 ) (notice that C1 or Dn
might be equal to In−1 ). Observe now that we can therefore write:
B = C1 D1 C1−1 · . . . · (C1 · . . . · Cn ) Dn (C1 · . . . · Cn )−1 (C1 · . . . · Cn )
28

and by deﬁnition, the conjugations of the Di -s can also be considered as of the
forms in the lemma. On the other hand, we have:
−1

(C1 · . . . · Cn ) Dn−1 (C1 · . . . · Cn )

· . . . · C1 D1−1 C1−1 B = C1 · . . . · Cn

and as the matrices of the forms in the lemma are all in GLn−1 (R, σn R) (by
Remark 7.5) we deduce that:
C1 · . . . · Cn ∈ GLn−1 (R, σn R) ∩ GLn−1 (Rn−1 ) = {In−1 }
i.e. C1 · . . . · Cn = In−1 . Hence:
−1

B = C1 D1 C1−1 · . . . · (C1 · . . . · Cn ) Dn (C1 · . . . · Cn )

i.e. B is a product of matrices of the forms in the lemma, as required.
So let B ∈ GLn−1 (R, σn R) ∩ En−1 (R, Tm ). According to Lemma 7.3, as
B ∈ En−1 (R, Tm ) and n − 1 ≥ 3, we can write B as a product of elements of
the form:
(In−1 − f Ei,j ) (In−1 + hEj,i ) (In−1 + f Ei,j )
Pn
Pn
2
2
for some f ∈ R, h ∈ Tm =
r=1 σr Ur,m +
r=1 σr Om + Om and 1 ≤ i 6=
j ≤ n − 1. We will show now that every element of the above form can be
writen as a product of the elements of the forms in the lemma and elements of
GLn−1 (Rn−1 ). Observe ﬁrst that as:
Tm

=

n
X

σr2 Ur,m +

r=1

⊆ σn

n
X

2
σr Om + Om

r=1

n−1
X

σr2 Ur,m

+ σn Un,m + Om

r=1

!

+

n−1
X

σr2 Ūr,m +

r=1

n−1
X

2
σr Ōm + Ōm

r=1

Pn−1

we can decompose h = σn h1 + h2 for some: h1 ∈ r=1 σr2 Ur,m + σn Un,m + Om
Pn−1
Pn−1
2
. Therefore, we can write:
and h2 ∈ r=1 σr2 Ūr,m + r=1 σr Ōm + Ōm
(In−1 − f Ei,j ) (In−1 + hEj,i ) (In−1 + f Ei,j )
= (In−1 − f Ei,j ) (In−1 + σn h1 Ej,i ) (In−1 + f Ei,j )
· (In−1 − f Ei,j ) (In−1 + h2 Ej,i ) (In−1 + f Ei,j ) .
Thus, as the matrix (In−1 − f Ei,j ) (In−1 + σn h1 Ej,i ) (In−1 + f Ei,j ) is clearly a
product of elements of forms 1 and 2 in the lemma, it is enough to deal with
the matrix:
(In−1 − f Ei,j ) (In−1 + h2 Ej,i ) (In−1 + f Ei,j )
Pn−1
Pn−1 2
2
. Let us now write: f = σn f1 + f2
σr Ōm + Ōm
when h2 ∈ r=1 σr Ūr,m + r=1
for some f1 ∈ R and f2 ∈ Rn−1 , and write:
(In−1 − f Ei,j ) (In−1 + h2 Ej,i ) (In−1 + f Ei,j )
= (In−1 − f2 Ei,j ) (In−1 − σn f1 Ei,j )
· (In−1 + h2 Ej,i ) (In−1 + σn f1 Ei,j ) (In−1 + f2 Ei,j ) .
29

Now, as (In−1 ± f2 Ei,j ) ∈ GLn−1 (Rn−1 ), it is enough to deal with the element:
(In−1 − σn f1 Ei,j ) (In−1 + h2 Ej,i ) (In−1 + σn f1 Ei,j )
which can be writen as a product of elements of the form:
(In−1 − σn f1 Ei,j ) (In−1 + kEj,i ) (In−1 + σn f1 Ei,j )
2
k ∈ Ōm
, σr2 Ūr,m , σr Ōm , for 1 ≤ r ≤ n − 1.

Finally, as for every such k one can write:
(In−1 − σn f1 Ei,j ) (In−1 + kEj,i ) (In−1 + σn f1 Ei,j )
= (In−1 + kEj,i ) [(In−1 − kEj,i ) , (In−1 − σn f1 Ei,j )]
and (In−1 + kEj,i ) ∈ GLn−1 (Rn−1 ), we actually ﬁnished.
Corollary 7.6. For proving Lemma 7.1, it is enough to show that every element
in the forms of Lemma 7.4 is in IAm .
We start to deal here with the elements of form 1:
Proposition 7.7. The elements of the following form are in IAm :
A−1 (In−1 + hEi,j ) A, for A ∈ GLn−1 (R) , h ∈ σn Om and i 6= j.
Proof. In this case we can write h = σn mh′ for some h′ ∈ R. So as:
A−1 (In−1 + σn h′ Ei,j ) A ∈ GLn−1 (R, σn R) ≤ IA (Φn )
we obtain that:
A−1 (In−1 + hEi,j ) A =
=
as required.

A−1 (In−1 + σn mh′ Ei,j ) A =
m
A−1 (In−1 + σn h′ Ei,j ) A ∈ IAm

We will devote the remaining sections to deal with the elements of the three
other forms. In these cases the proof will be much more diﬃcult, and we will
need the help of the following computations.

7.2

Some needed computations

Proposition 7.8. For every f, g ∈ R we have

1 − f g −f g
 fg
1 + fg
0
0

30

the following equalities:

0
0 
1



1
0 0
=  fg 1 0
f g2 0 1

1 −f g
1
· 0
0 −f g 2



1
0
0
  0 1 + fg
−f 
2
0
fg
1 − fg


0
1 − fg 0
f
1 0
 0 1
0 
0
1
0
1
−f g 2 0 1 + f g
0 0


1
0 0
 fg
1 0 
−f g 2 0 1


1 −f g 0
1
0 
· 0
0 f g2 1

=



1
 0
f

1
· 0
0

=

=



1
0
0

0
1 + fg
−f g 2

1 − fg
0
f g2


0
1 − fg 0
0 
0
1
1
−f
0

0
0
1 + fg
f g2  
−f
1 − fg

0
1
f




1
0 0
1 − fg
 0
1 0 
0
−f −f 1
f


1
0
0
·  0 1 + f g −f g 2  
0
f
1 − fg

0
1
0


0

f
1 − fg

−f
1
 0
0
1 + fg
0


f g2
1 0
  fg 1
0
1 + fg
0 0

2
1 −f g −f g
0
1
0 
0
0
1

0 −f g 2
1
  fg
1
0
0 1 + fg
0

2
1 −f g f g
0
1
0 
0
0
1

Proof. Here is the computation for Equation 7.2:

 

1 − fg
−f g 0
1 0 f
 fg
1 + f g 0  =  0 1 −f  
0
0
1
g g 1



1 0 0
1 0 f
1
0 0
1 0
=  0 1 0   0 1 −f   0
g g 1
0 0 1
−g −g 1



−f
f 
1
(7.3)


0 f
1 −f 
0 1

0
−f g 2 
1
(7.4)


0
0
1 f g2 
0
1
(7.5)


1
0 −f
0
1
f 
−g −g 1


1 0 −f
 0 1 f 
0 0 1




1 0 0
1 0 0
1
0 0
1 0 
=  0 1 0   0 1 −f   0
g g 1
0 0 1
−g −g 1




1 0 0
1 0 f
1
0 0
1 0
1 0  0 1
· 0 1 0  0 1 0  0
g g 1
0 0 1
−g −g 1
0 0
31

(7.2)


−f
f 
1



1
=  fg
f g2
=


0
1 − fg
−f  
0
1 − fg
−f g 2

0
1 + fg
f g2


1
0 0
 fg 1 0
f g2 0 1

1 −f g
1
· 0
0 −f g 2

−f g
1
−f g 2


f
1
 0
0
1 + fg
0


0 −f
1 f 
0 1




1
0
0
  0 1 + fg
−f 
2
0
fg
1 − fg


0
1 − fg 0
f
1
 0
0 
0
1
0
1
−f g 2 0 1 + f g
0


0 −f
1 f .
0 1

Equation 7.3 is obtained Similarly by changing the signs of f and g simultaneously. Here is the computation for Equation 7.4:

 


1 − fg
−f g 0
1 0 g
1
0 −g
 fg
1 + f g 0  =  0 1 −g   0
1
g 
0
0
1
f f 1
−f −f 1





1 0 0
1 0 g
1
0 0
1 0 −g
1 0  0 1 g 
=  0 1 0   0 1 −g   0
f f 1
0 0 1
−f −f 1
0 0 1


=



1
 0
f

1
· 0
0

1
= 0
f
=




0
1 0
0  0 1
1
0 0

g
1
−g   0
1
0

1 − fg
  fg
−f

0
1
f

0
1
f

1
 0
f

1
· 0
0

0
1
0
0
0
1


g
1 0
−g   0 1
1
−f 0

0 0
1 0
1 0  0 1
−f 1
0 0


0
f g2
1
1 −f g 2   0
0 1 + fg
0


0
1 − fg 0
0 
0
1
1
−f
0

0
0
1 + fg
f g2  
−f
1 − fg

0
1
f


0
1
0  0
1
0

−g
g 
1
−f g
1 + fg
−f


f g2
1 0
  fg 1
0
1 + fg
0 0

2
1 −f g −f g
0
1
0 
0
0
1

0
1
0


−g
g 
1


−f g 2
f g2 
1 − fg

0
−f g 2 
1

and Equation 7.5 is obtained similarly by changing the signs of f and g simultaneously.
In the
corollary, a 3 × 3 matrix B ∈ GL3 (R) denotes the blocks
 following 
B
0
∈ GLn−1 (R).
matrix
0 In−4
32

Corollary 7.9. Let n ≥ 4, f ∈ σn

P

n−1
r=1

σr Ur,m + Un,m + Om



and g ∈ R.

Then, mod IAm we have the following equalities (the indices are intended to
help us later to recognize forms of matrices: form 7, form 12 etc.):


1 − f g −f g 0
 fg
1 + fg 0 
(7.6)
0
0
1 13
 


1
0
0
1 − fg 0
f

−f  
0
1
0
≡  0 1 + fg
2
2
0
fg
1 − fg 1
−f g
0 1 + fg 2

 

1
0
0
1 − fg 0
−f
 

f
0
1
0
≡  0 1 + fg
0 −f g 2 1 − f g 3
f g2
0 1 + fg 4

 

1 − fg 0
f g2
1
0
0
  0 1 + fg
0
1
0
f g2 
≡
−f
0 1 + fg 5
0
−f
1 − fg 6

 

2
1 − f g 0 −f g
1
0
0
  0 1 + f g −f g 2 
0
1
0
≡
f
0 1 + fg 7
0
f
1 − fg 8

and:





1 − fg
0
≡
f


fg
0
1 + fg 0 
0
1 14
 
2
−f g
1
0
  0 1 + fg
0
1 + fg 7
0
−f

1 − fg
 −f g
0

0
1
0

(7.7)

0
f g2  .
1 − fg 6

Moreover (the inverse of a matrix is denoted by the same index - one can
observe that the inverse of each matrix in these equations is obtained by changing
the sign of f ):


1 − f g 0 −f g


0
1
0
(7.8)
fg
0 1 + f g 15

 

1
0
0
1 − fg
f
0
f g 2   −f g 2 1 + f g 1 
≡  0 1 − fg
0
−f
1 + fg 8
0
0
0 9

 

1
0
0
1 − fg
−f
0
1 + fg 0 
≡  0 1 − f g −f g 2   f g 2
0
f
1 + fg 6
0
0
1 10




1 − fg
f g2
0
1
0
0
1 + fg 0   0 1 − fg
−f 
≡  −f
2
0
0
1 11
0
fg
1 + fg 3
33



and:

1 − fg
f
≡
0



and:

1 − fg
f
≡
0

 
0
1
0   0
1 12
0




0
fg

1
0
0 1 + f g 16
 
0
1
0
0   0 1 − fg
1 12
0
f g2

0
1 − fg
−f g 2

1 − fg

0
−f g

−f g 2
1 + fg
0


0

f
1 + fg 1
(7.9)

0
−f 
1 + fg 3





and:

−f g 2
1 + fg
0

1 − fg
0
≡
−f


1
0
0
 0 1 − fg
−f g 
0
fg
1 + f g 17
 
0
f g2
1 + f g −f g 2
 
1
0
f
1 − fg
0 1 + fg 5
0
0

(7.10)

0
1 
0 11





1 + fg
≡  −f g 2
0


1
0
0
 0 1 − fg
fg 
0 −f g 1 + f g 18
 
f
0
1 − fg 0
1 − fg 0  
0
1
0
1 10
f g2
0

(7.11)

−f

0
1 + fg 4

Remark 7.10. We remark that as f ∈ σn R, then every matrix which takes part
in the above epualities is indeed in GLn−1 (R, σn R) ∼
= IGLn−1,n ≤ IA (Φn ).

P
n−1
Proof. As f ∈ σn
r=1 σr Ur,m + Un,m + Om , Equation 7.6 is obtained by
applying Proposition 7.8 combined with Proposition 6.1. Equation 7.7 is obtained similarly by transposing all the computations which led to the ﬁrst part
of Equation 7.6. Similarly, by switching the roles of the second row and column
with the third row and column, one obtains Equations 7.8 and 7.9. By switching
one more time the roles of the ﬁrst row and column with the second row and
column, we obtain Equations 7.10 and 7.11 as well.

7.3

Elements of form 2

Proposition 7.11. The elements of the following form, belong to IAm :
A−1 (In−1 + hEi,j ) A
where A ∈ GLn−1 (R), h ∈ σn σr2 Ur,m , σn2 Un,m for 1 ≤ r ≤ n − 1 and i 6= j.

34



Notice that for every n ≥ 4, the groups En−1 σn2 Un,m and En−1 σn σr2 Ur,m
for 1 ≤ r ≤ n − 1 are normal
in GLn−1 (R), and
thus, all the above ele

ments are in En−1 σn2 Un,m and En−1 σn σr2 Ur,m . Hence, for proving Proposition 7.11, it is enough to show
 that for every 1 ≤ r ≤ n − 1 we have,
En−1 σn2 Un,m , En−1 σn σr2 Ur,m ⊆ IAm . Therefore, by Lemma 7.3, for proving Proposition 7.11, it is enough to show that the elements of the following
form are in IAm :
(In−1 − f Ej,i ) (In−1 + hEi,j ) (In−1 + f Ej,i )
when h ∈ σn σr2 Ur,m , σn2 Un,m for 1 ≤ r ≤ n − 1, f ∈ R and i 6= j. We will prove
it in a few stages, starting with the following lemma.
Lemma 7.12. Let h ∈ σn σr Ur,m , σn Un,m for 1 ≤ r ≤ n − 1 and f1 , f2 ∈ R.
Assume that the elements of the forms:
(In−1 ± f1 Ej,i ) (In−1 + hEi,j ) (In−1 ∓ f1 Ej,i )
(In−1 ± f2 Ej,i ) (In−1 + hEi,j ) (In−1 ∓ f2 Ej,i )
for every 1 ≤ i 6= j ≤ n − 1, belong to IAm . Then, the elements of the form:
(In−1 ± (f1 + f2 ) Ej,i ) (In−1 + hEi,j ) (In−1 ∓ (f1 + f2 ) Ej,i )
for 1 ≤ i 6= j ≤ n − 1, also belong to IAm .
Proof. Observe ﬁrst that by Proposition 6.1, all the matrices of the form In−1 +
hEi,j for h ∈ σn σr Ur,m , σn Un,m belong to IAm . We will use it in the following
computations. Without loss of generality, we will show that for i, j = 2, 1 we
have:
(In−1 − (f1 + f2 ) E1,2 ) (In−1 + hE2,1 ) (In−1 + (f1 + f2 ) E1,2 ) ∈ IAm
and the general argument is similar. In the following computation we use the
following notations:


B
0
∈ GLn−1 (R) is denoted by B ∈ GL3 (R).
- A matrix
0 In−4
- “=” denotes an equality between matrices in GLn−1 (R).
- “≡” denotes an equality in IA (Φn ) /IAm .
(In−1 − (f1 + f2 ) E1,2 ) (In−1 + hE2,1 ) (In−1 + (f1 + f2 ) E1,2 )


2
1 − h (f1 + f2 ) −h (f1 + f2 )
0
=
h
1 + h (f1 + f2 ) 0 
0
0
1



1
0
− (f1 + f2 )
1
0
(f1 + f2 )
 0

1
1
1
−1
= 0
−h −h (f1 + f2 )
1
h h (f1 + f2 )
1
35



1
=  0
−h

1
·  0
0


0
0 
−h (f1 + f2 ) 1

0 − (f1 + f2 )
1
 0
1
1
0
1
h
0
1

0
1
h (f1 + f2 )


0
1
0  0
1
0


0 (f1 + f2 )
 .
1
−1
0
1

As the product of the elements in the square brackets is a conjugation of an element of GLn−1 (R, σn R) it is also an element of GLn−1 (R, σn R) ∼
= IGLn−1,n ≤
IA (Φn ). In addition, by the remark in the beginning of the proof, the remained
element is in IAm . So mod IAm , the above expresion is equals to:




1 0 − (f1 + f2 )
1
0
0
1 0 f1 + f2
 0 1
 0
1
−1 
1
0  0 1
0 0
1
0 0
1
h h (f1 + f2 ) 1
=

=






1 0
0
1 0 −f1
1 0 −f2
 0 1
1  0 1
0  0 1
0 
0 0
1
0 0
1
0 hf2 1




1
0
0
1 0 f1
1 0 f2
· 0
1
0   0 1 −1   0 1 0 
h hf1 1
0 0 1
0 0 1



1 0 −f2
1
 0 1
0  0
0 0
1
0


1
1 0 f1
·  0 1 −1   0
h
0 0 1


1 0 −f1
1
1  0
· 0 1
0 0
1
h


−f1
1
0
0
1  0
1
0
1
0 hf2 1

0
0
1
0
1
0  0
1
hf1 1
−h −hf1

0 0
1 0 f1
1 0   0 1 −1
hf1 1
0 0 1

0
1
0



1
1 0 −f1
1 0 −f2
1  0
0  0 1
=  0 1
0 0
1
0 0
1
0


2
1
0
0
1 − hf1 −hf1
1
0 
h
1 + hf1
· 0
h hf1 1
0
0


36





0
0 
1


1
 0
0


0 0
1 0 
hf2 1

0
1 0
0  0 1
0 0
1


0 f2
1 0 
0 1


1 0 f1
0 1 −1 
0 0 1

f2
0 
1


1
1 0 −f2
0  0
=  0 1
0
0 0
1


1
0
0
1
1
0  0
· 0
0
h hf1 1





1 0 −f2
1
=  0 1
0  0
0 0
1
0


1
0
0
1
1
0  0
· 0
h hf1 1
0




1
0 0
0 −f1
1 0 f1
1
1  0
1 0   0 1 −1 
0
1
0 0 1
0 hf2 1


2
1 − hf1 −hf1
−hf1 f2
0 f2
h
1 + hf1
hf2 
1 0 
0 1
0
0
1


0 −f1
1
1
1  0
0
1
0

0 f2
1
1 0  0
0 1
0


0
1
0  0
1
0

−hf1 f2
hf2  
1

0
1
hf2
0
1
0


f1
−1 
1

0
1
0

1 − hf1
h
0

−hf12
1 + hf1
0

Notice now that by assumption, the right matrix in the above exspression
belongs to IAm , and the nearer matrix also belong to IAm by the remark in
the beginning of the proof. So, mod IAm , the latter expression is equals to:




1 0 −f2
1 0 −f1
1
0
0
 0 1
0  0 1
1  0
1
0 
0 0
1
0 0
1
0 hf2 1




1 0 f1
1
0
0
1 0 f2
1
0  0 1 0 
·  0 1 −1   0
0 0 1
h hf1 1
0 0 1





1 0 −f2
1 0 −f1
1
0 0
1
0  0 1
1  0
1 0  0
=  0 1
0 0
1
0 0
1
0 hf2 1
0




1 0 f2
1 0 −f2
1
0
0
1
0  0
1
0  0
· 0 1 0  0 1
0 0 1
0 0
1
h hf1 1
0
=

=




1 0 − (f2 + f1 )
1
0
 0 1
 0
1
1
0 0
1
0 hf2


1 − hf2 −hf1 f2 −hf22

0
1
0
·
h
hf1
1 + hf2



0
1
0  0
1
0

0
1
0
0
1
0


f1
−1 
1

f2
0 
1


0 f2 + f1
1
−1 
0
1



1
0
0
1 −hf2 (f1 + f2 ) hf2 (f1 + f2 )
 0 1 + hf2

−hf2   0
1
0
0
0
1
0
hf2
1 − hf2



1 −hf1 f2 0
1 − hf2 0 −hf22

1
0 
0
1
0
· 0
0
hf1
1
h
0 1 + hf2
37


0
0 .
1



1
0
≡  0 1 + hf2
0
hf2


0
−hf2  .
1 − hf2

Consider now the last part of Equation 7.6 in Corollary 7.9, and switch the roles
of f, g by h, f2 respectively. Then, by switching the roles of the ﬁrst row and
column with the third row and column in the equation, and transposing the
expresion, one deduces that:


1
0
0
 0 1 + hf2
−hf2 
0
hf2
1 − hf2



1 − hf2 −hf22 0
1 + hf2 0 −hf22
 ≡ In−1
h
1 + hf2 0  
0
1
0
≡
0
0
1
h
0 1 − hf2
and this ﬁnishes the proof of the lemma.
We pass to the next stage:
Proposition 7.13. The elements of the following form belong to IAm :
(In−1 − f Ej,i ) (In−1 + hEi,j ) (In−1 + f Ej,i )
where h ∈

σn σr2 Ur,m , σn2 Un,m

for 1 ≤ r ≤ n − 1, f ∈ Z and i 6= j.

Proof. According to Lemma 7.12, it is enough to prove the proposition for f =
±1. Without loss of generality, we will prove the proposition for r = 1, i.e.
h ∈ σn σ12 U1,m , and symmetrically, the same is valid for every 1 ≤ r ≤ n − 1.
The case h ∈ σn2 Un,m will be considered separately.
So let h ∈ σn σ12 U1,m and write: h = σ1 u for some u ∈ σn σ1 U1,m . We will
prove the proposition for i 6= j ∈ {1, 2, 3} - as one can see below, we will do it
simultaneously for all the options for i 6= j ∈ {1, 2, 3}. The treatment in the
other cases in which i 6= j ∈ {1, k, l} such that 1 < k 6= l ≤ n − 1 is obtained
symmetrically, so we get that the proposition is validfor every 1 ≤
 i 6= j ≤ n−1.
B
0
∈ GLn−1 (R)
As before, we denote a given matrix of the form
0 In−4
by B ∈ GL3 (R). Let have now the following computations (the indices of the
matrices below intended to recognize the appropriate form of matrix in Corollary
7.9, as will be explained below. We remind that the inverse of a matrix is denoted
by the same index, and one can observe that the inverse of each indexed matrix
is obtained by changing the sign of u). We remind that u ∈ σn σ1 U1,m ⊆ σn R.
Thus, by Proposition 6.1 we have:





1 − σ1 u −σ12 u 0
x2 −σ1 0
1
0 0

u
1 + σ1 u 0 
1
0   ux2 1 0 
=  0
0
0
1 12
0
0
1
0
0 1
 −1

−1
x2
x2 σ1 0
· 0
1
0  ∈ IAm
0
0
1
38



1 − σ1 u

0
u


0 −σ12 u

1
0
0 1 + σ1 u 7




0
0

1 + σ1 u
u
2
−σ1 u 1 − σ1 u 3




0
−σ12 u 
1 + σ1 u 6

1
 0
0

1
0
 0 1 − σ1 u
0
u



1
0 0
x3 0 −σ1
0  0
1 0 
=  0 1
0 0
1
ux3 0 1
 −1

−1
x3
0 x3 σ1
 ∈ IAm
· 0
1
0
0
0
1





0 0
1 0 
0 1

1
 0
−σ2

1
0
σ2




0 0
1 0 
0 1

1
  −σ3
0

1
σ3
0

1
=  uσ2
−uσ1 σ2

1 0 0
· 0 1 u
0 0 1
=

1
 −uσ1 σ3
uσ3

1 0 0
· 0 1 0
0 u 1


0
0
1
0 
−σ1 1

0 0
1 0  ∈ IAm
σ1 1
0
1
0


0
−σ1 
1


0 0
1 σ1  ∈ IAm .
0 1

By switching the signs of σ1 , σ2 and σ3 in the two latter computations we obtain
also that:

 

1
0
0
1
0
0
 0 1 − σ1 u
 ,  0 1 + σ1 u −σ12 u  ∈ IAm .
u
2
0 −σ1 u 1 + σ1 u 1
0
u
1 − σ1 u 8

Consider now the identities which we got in Corollary 7.9, and switch the roles
of f, g in the corollary by u,σ1 , respectively. Remember now that u ∈ σn σ1 U1,m .
Hence, as by the computations above the matrices which correspond to forms 7
and 8 belong to IAm , we obtain from the last part of Equation 7.6 that also form
13 belong to IAm . Thus, as we showed that forms 1, 3, 6 also belong to IAm ,
Equation 7.6 shows that also forms 2, 4, 5 belong to IAm . Similar arguments
show that Equations 7.6-7.11 give that all the 18 forms belong to IAm . In
particular, the matrices which correspond to forms 13 − 18 belong to IAm , and
these matrices (and their inverses) are precisely the matrices of the form (we
remind that h = σ1 u):
(In−1 ± Ej,i ) (In−1 + hEi,j ) (In−1 ∓ Ej,i ) , i 6= j ∈ {1, 2, 3} .
Clearly, by similar arguments, the proposition holds for every 1 ≤ i 6= j ≤
n − 1 and every h ∈ σn σr2 Ur,m for 1 ≤ r ≤ n − 1. The case h ∈ σn2 Un,m is a
bit diﬀerent, but easer. In this case one can consider the same computations we
built for r = 1, with the following ﬁttings: Firstly, write h ∈ σn2 Un,m as h = σn u
39

for some u ∈ σn Un,m . Secondly, change σ1 to σn , change σ2 , σ3 to 0 and change
x2 , x3 to 1 in the right side of the above equations. It is easy to see that in this
situation we obtain in the left side of the equations the same matrices, just that
instead of σ1 we will have σn . From here we continue exactly the same.
Proposition 7.14. The elements of the following form belong to IAm :
(In−1 − f Ej,i ) (In−1 + hEi,j ) (In−1 + f Ej,i )
where h ∈
i 6= j.

σn2 Un,m , σn σr2 Ur,m

for 1 ≤ r ≤ n − 1, f ∈ σs R for 1 ≤ s ≤ n and

Proof. We will use again the result of Corollary 7.9, when we switch the roles
of f, g in the corollary by h, σs u respectively for some u ∈ R and 1 ≤ s ≤ n.
Similarly to proposition 7.13, we 
will prove it for s = 1, i 6= j ∈ {1, 2, 3}, and
B
0
∈ GLn−1 (R) by B ∈ GL3 (R).
denote a given matrix of the form
0 In−4
As h ∈ σn σr2 Ur,m , σn2 Un,m , we have also σ1 uh ∈ σn σr2 Ur,m , σn2 Un,m . Hence,
we obtain from the previous proposition, that the matrices of the forms 13 − 18
belong to IAm . In addition:





1
0
0
1
0 0
1
0
0
 0 1 − uσ1 h
 =  −huσ2
h
1 0  0
1
0 
0 −u2 σ12 h 1 + uσ1 h 1
−hu2 σ1 σ2 0 1
−uσ2 uσ1 1



1 0 0
1
0
0
1
0  ∈ IAm
· 0 1 h  0
0 0 1
uσ2 −uσ1 1



1
0
0
 0 1 − uσ1 h −u2 σ12 h 
0
h
1 + uσ1 h 6

=



1
0
 −hu2 σ1 σ3 1
huσ3
0


1 0 0
· 0 1 0 
0 h 1


0
1
0   uσ3
1
0
1
−uσ3
0

0
1
0


0
0
1 −uσ1 
0
1


0
uσ1  ∈ IAm
1

and by switching the signs of u and h simultaneously, we get also forms 3 and
8. So we easily conclude from Corollary 7.9 (Equations 7.6 and 7.8) that also
the matrices of the other eight forms are in IAm . In particular, the matrices of
the form:
(In−1 − σ1 uEj,i ) (In−1 + hEi,j ) (In−1 + σ1 uEj,i ) , i 6= j ∈ {1, 2, 3}

belong to IAm . The treatment for every i 6= j and 1 ≤ s ≤ n − 1 is similar, and
the treatment in the case s = n is obtained by replacing σ1 by σn and σ2 , σ3 by
0 in the above equations.
Pn
Corollary 7.15. As every f ∈ R can be decomposed as f = s=1 σs fs + f0 for
some f0 ∈ Z and fi ∈ R, we obtain from Lemma 7.12 and from the above two
propositions that we actually finised the proof of Proposition 7.11.
40

7.4

Elements of form 3

Proposition 7.16. The elements of the following form, belong to IAm :
A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A
2
where A ∈ GLn−1 (R), f ∈ σn R, h ∈ Ōm
and i 6= j, when Ōm is the projection
±1
±1
of Om to Rn−1 = Z[x1 , . . . , xn−1 ] which induced by the projection σn 7→ 0.

We will prove the proposition in the case i, j = 2, 1, and the same arguments
are valid for arbitrary i 6= j. In this case one can write: h = m2 h′ for some
h′ ∈ Rn−1 , and thus, our element is of the form:



1 − f m2 h ′
f
0
1 −f
0

2
0 A
A−1  −f m2 h′
1 + f m2 h ′
0  0 1
0 0 In−3
0
0
In−3

for some A ∈ GLn−1 (R), f ∈ σn R and h′ ∈ Rn−1 . The proposition will follow
easily from the following lemma:
Lemma
7.17.
 Let h1 , h2 ∈ R, f ∈ σn R and denote a given matrix

B
0
∈ GLn−1 (R) by B ∈ GL3 (R). Then:
0 In−4


1 − f m (h1 + h2 )
f
0
1 −f
2
A−1  −f (m (h1 + h2 )) 1 + f m (h1 + h2 ) 0   0 1
0 0
0
0
1



1 − f mh1
f
0
≡ A−1  −f (mh1 )2 1 + f mh1 0  
0
0
1


1 − f mh2
f
0
1
·  −f (mh2 )2 1 + f mh2 0   0
0
0
0
1

1 −f
0 1
0 0
−f
1
0


0
0 
1


of the form


0
0 A
1

0
0  A mod IAm
1

Now, if one proves the lemma, he can deduce that:



1 − f m2 h ′
f
0
1 −f
0

2
0 A
A−1  −f m2 h′
1 + f m2 h ′
0  0 1
0 0 In−3
0
0
In−3




 m
1 − f mh′
f
0
1 −f
0
≡ A−1  −f (mh′ )2 1 + f mh′
0  A
mod IAm
0  0 1
0 0 In−3
0
0
In−3
and as the latter element is obviously belong to IAm we obtain that:



1 −f
0
1 − fh
f
0
0  A ∈ IAm
0  0 1
A−1  −f h2 1 + f h
0 0 In−3
0
0
In−3
as required. So it is enough to prove Lemma 7.17.
41

Proof. (of Lemma 7.17) Notice through the computation the reminder (Remark
7.5) that as GLn−1 (R, σn R) is normal in GLn−1 (R) every conjugation of an
element of GLn−1 (R, σn R) ≤ IA (Φn ) by an element of GLn−1 (R), belongs to
GLn−1 (R, σn R) ≤ IA (Φn ). We remark also that through the computation, we
will use freely the result of Proposition 7.7 and the three notations which we
used in the proof
 of Lemma7.12:
B
0
∈ GLn−1 (R) is denoted by B ∈ GL3 (R).
- A matrix
0 In−4
- “=” denotes an equality between matrices in GLn−1 (R).
- “≡” denotes an equality in IA (Φn ) /IAm .



1 − f m (h1 + h2 )
f
0
1 −f 0
2
A−1  −f (m (h1 + h2 )) 1 + f m (h1 + h2 ) 0   0 1 0  A
0 0 1
0
0
1
=



1
0
−f
0
1 −f m (h1 + h2 )
A−1 
−m (h1 + h2 ) 1
1


1
0
f
0
1 f m (h1 + h2 )  
·
m (h1 + h2 ) −1
1




1
0 0
0
1 0 
= A−1 
−m (h1 + h2 ) 1 1


1
0 0
1
·
0
1 0  0
m (h1 + h2 ) −1 1
0

1
0
0

0
1
0




1 −f
0 1
0 0

−f
−f m (h1 + h2 )
1

0
f
1 f m (h1 + h2 )  
0
1


0
0 A
1



1 −f
0 1
0 0


0
0 A
1






1
0 0
1
0 0
1 0
−f
0
1 0 
0
1 0   0 1 −f mh1 
= A−1 
−mh2 0 1
0 0
1
−mh1 1 1




1
0 0
1 0
f
1 0
−f
1
0
1 0   0 1 f mh1   0 1 −f mh1   0
1
· 0
mh1 −1 1
0 0
1
0 0
1
mh2 0



1
0 0
1 0
0
1
0
0
1 0   0 1 −f mh2  
0
1
·A−1 
−m (h1 + h2 ) 1 1
0 0
1
m (h1 + h2 ) −1


 m


1 0
0
1 −f f
· A−1  0 1 f (h1 + h2 )  A A−1  0 1 0  A
0 0
1
0 0 1
42


0
0 A
1

0
0 A
1





1 − f mh1
f
0
1
0 0
0
1 0   −f (mh1 )2 1 + f mh1 0 
≡ A−1 
−mh2 0 1
0
0
1



1 0
−f
1
0 0
1 0 A
·  0 1 −f mh1   0
0 0
1
mh2 0 1




 m
1
0 0
1 0
0
1
0 0
· A−1 
0
1 0   0 1 −f h2  
0
1 0  A
−m (h1 + h2 ) 1 1
0 0
1
m (h1 + h2 ) −1 1


1 −f f
·A−1  0 1 0  A
0 0 1


1
0
≡ A−1 
−mh2

1
0 0
1 0
· 0
mh2 0 1
=




1 − f mh1
0 0
1 0   −f (mh1 )2
0 1
0


1 −f f
 0 1 0 A
0 0 1

1
0
A−1 
−mh2

1
0 0
1 0
· 0
mh2 0 1

1
0 0
1 0
· 0
mh2 0 1

1
0 0
1 0
· 0
mh2 0 1

f
1 + f mh1
0


0
1 0


0 1
0
0 0
1



1 − f mh1
f
0
0 0
1 0   −f (mh1 )2 1 + f mh1 0 
0 1
0
0
1



1
0 0
1 0 −f

0
1 0  0 1 0 
−mh2 0 1
0 0 1



1
0 0
1 0
0

0
1 0   0 1 −f mh1 
−mh2 0 1
0 0
1


1 −f f
 0 1 0 A
0 0 1

43


−f
−f mh1 
1





 m

1 − f mh1
f
0
1
0
0
0
1
0  A
= A−1  −f (mh1 )2 1 + f mh1 0  A · A−1 
f mh1 h2 −f h2 1
0
0
1




1
0 0
1 0 −f
1
0 0
0
1 0  0 1 0  0
1 0 A
·A−1 
−mh2 0 1
0 0 1
mh2 0 1




 m
1
0 0
1 0
0
1
0 0
· A−1 
0
1 0   0 1 −f h1   0
1 0  A
−mh2 0 1
0 0
1
mh2 0 1


1 −f f
·A−1  0 1 0  A
0 0 1
≡



A−1 


1
· 0
0



1 − f mh1
f
0
1
0 0
0
1 0 
−f (mh1 )2 1 + f mh1 0  
−mh
0
0
1
2 0 1



0 −f
1
0 0
1 −f f
1 0  0
1 0  0 1 0 A
0 1
mh2 0 1
0 0 1



1 − f mh1
f
= A−1  −f (mh1 )2 1 + f mh1
0
0


1 f 0
1
0
·A−1  0 1 0  
0 0 1
−mh2


1
0 0
1
0
1 0  0
·A−1 
−mh2 0 1
0


1 − f mh1
= A−1  −f (mh1 )2
0


1
0
· A−1  0
1
0 f h2

1 − f mh2
−1 
0
·A
2
f (mh2 )


0
1 −f 0
0  0 1 0
0 0 1
1

0 0
1 −f
1 0  0 1
0 1
0 0

f −f
1
1 0  0
0 1
mh2

f
1 + f mh1
0
 m
0
0  A
1
f
1
−f mh2

44



A


0
1
0  0
1
mh2

0 0
1
1 0  0
0 1
0


0
1 −f
0  0 1
0 0
1


−f
1
 0
0
0
1 + f mh2


0 0
1 0 A
0 1

−f f
1 0 A
0 1


0
0 A
1

−f
1
0


f
0 A
1





1 − f mh1
f
0
1 −f 0
A−1  −f (mh1 )2 1 + f mh1 0   0 1 0  A
0 0 1
0
0
1




 m
1 f 0
1
0
0
1
0  A
·A−1  0 1 0  A · A−1  0
0 0 1
0 −f h2 1



1 − f mh2 0
−f
1 −f f
 0 1 0 A
0
1
0
·A−1 
2
0 0 1
f (mh2 )
0 1 + f mh2

≡






1 − f mh1
f
0
1 −f 0
1 f 0
≡ A−1  −f (mh1 )2 1 + f mh1 0   0 1 0   0 1 0 
0 0 1
0 0 1
0
0
1




1 − f mh2 0
−f
1 −f 0
1 0 f
 0 1 0  0 1 0 A
0
1
0
·
2
0 0 1
0 0 1
f (mh2 )
0 1 + f mh2
=

≡




1 − f mh1
f
0
A−1  −f (mh1 )2 1 + f mh1 0  
0
0
1


2
1
f mh2
0
1
0 A
·A−1  0
2
2
0 −f (mh2 ) 1


1 − f mh2 0
−f

0
1
0
·A−1 
2
f (mh2 )
0 1 + f mh2

1
0
0

−f
1
0

1 0
0 1
0 0


0
0 A
1


f
0 A
1





1 − f mh1
f
0
1 −f 0
A−1  −f (mh1 )2 1 + f mh1 0   0 1 0 
0 0 1
0
0
1



1 − f mh2 0
−f
1 0 f
  0 1 0  A.
0
1
0
·
2
0 0 1
f (mh2 )
0 1 + f mh2

Now, by similar computation as for Equation 7.5, we have (switch the roles
of f, g in the equation by f, mh2 respectively, and then switch the roles of the
ﬁrst row and column with the third row and column):

 

1
0
0
1 −f −f
 0 1 + f mh2
f mh2  =  0 1
0 
0 −f mh2
1 − f mh2
0 0
1
45




f
1
  f (mh2 )2
0
0
1 − f mh2

f
0
1
0
1 + f mh2 0  
2
0
1
f (mh2 )

1 + f mh2
0
·
−f (mh2 )2

1 − f mh2
·  −f (mh2 )2
0

0
1
0

Therefore, using Proposition 7.7, we have:



1
1
0
0
f mh2  A ≡ A−1  0
A−1  0 1 + f mh2
0
0 −f mh2
1 − f mh2


1 + f mh2
0
·
2
−f (mh2 )

Now, observe that:


1 − f mh2
0
f
  −f (mh2 )2
1
0
0
0 1 − f mh2




1
A−1  0
0

0
1 + f mh2
−f mh2



1
0
= A−1  0 1 + f h2
0 −f h2


0
0
1 f mh2 
0
1

0
0
1
0 .
−f mh2 1
−f
1
0

f
1 + f mh2
0


−f
0 
1


0
0  A.
1


0
f mh2  A
1 − f mh2

 m
0
f h2  A ∈ IAm .
1 − f h2

Hence we obtain that mod IAm we have:



1 + f mh2 0
f
1 −f −f

0
1
0
0 
A−1  0 1
2
0 0
1
−f (mh2 ) 0 1 − f mh2


1 − f mh2
f
0
·  −f (mh2 )2 1 + f mh2 0  A ≡ In−1 .
0
0
1

But as the inverse of every matrix in the above equation is obtained by replacing
f by −f , we have:



1 −f 0
1 0 −f
A−1  0 1 0   0 1 0  A
0 0 1
0 0 1


1 + f mh2
2
−1 
≡A
f (mh2 )
0

−f
1 − f mh2
0


0
1 − f mh2


0
0
1
f (mh2 )2

46


0
−f
A
1
0
0 1 + f mh2

and thus:

1 − f mh2 0
−f
1 0 f
 0 1 0
0
1
0
A−1 
2
0 0 1
f (mh2 )
0 1 + f mh2


1 − f mh2
f
0
1 −f
≡ A−1  −f (mh2 )2 1 + f mh2 0   0 1
0 0
0
0
1




A


0
0  A.
1

This ﬁnishes the proof of the lemma, and hence, also the proof of Proposition
7.16.

7.5

Elements of form 4

Proposition 7.18. The elements of the following form, belong to IAm :
A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A
where A ∈ GLn−1 (R), f ∈ σn R, h ∈ σr2 Ūr,m , σr Ōm for 1 ≤ r ≤ n− 1 and i 6= j,
±1
when Ūr,m , Ōm are the projections of Ur,m , Om to Rn−1 = Z[x±1
1 , . . . , xn−1 ]
which induced by the projection σn 7→ 0.
 through the subsection we denote a given matrix of the form
 As before,
B
0
∈ GLn−1 (R) by B ∈ GL3 (R). We start the proof of this propo0 In−4
sition with the following computation.
Lemma 7.19. Let f, g ∈ R and A ∈ GLn−1 (R). Then:


1 − fg
−f g 0
1 + fg 0  A
A−1  f g
0
0
1





1
0 0
1
0
0
−f  
= A−1  f g 1 0  AA−1  0 1 + f g
2
2
fg 0 1
0
fg
1 − fg





1 0 0
1 −f g 0
1
0  AA−1 
·A−1  0 1 −f  AA−1  0
2
0 0 1
0 −f g 1



1 − fg 0
f
1
0
0
 AA−1  f 2 g 2 1 −f 2 g
0
1
0
·A−1 
−f g 2 0 1 + f g
0
0
1


0
f A
1

1 0 0
0 1 f A
0 0 1



1 0 −f
 AA−1  0 1 0  A.
0 0 1

1 0
0 1
0 0

Proof. The lemma follows from the following computations, which based on
Proposition 7.8, Equation 7.2:


1 − f g −f g 0
 fg
1 + fg 0 
0
0
1
47



1
=  fg
f g2

1
· 0
0

1
· 0
0


0 0
1
0
1 0   0 1 + fg
0 1
0
f g2

0 0
1 −f g
1 −f   0
1
0 1
0 −f g 2

0 0
1 − fg 0
1 −f  
0
1
0 1
−f g 2 0


1
1
0 0
=  fg 1 0   0
0
f g2 0 1


1 0 0
1
·  0 1 −f   0
0 0 1
0

1 − fg 0
f
0
1
0
·
−f g 2 0 1 + f g


0
−f
1 − fg

0
0 
1



1
 0
0


0 0
1 f 
0 1


1 0 0
0 1 f 
0 0 1


f
1 0 0
1 0
 0 1 f  0 1
0
1 + fg
0 0 1
0 0

0
1 + fg
f g2
−f g
1
−f g 2


0
−f
1 − fg

0
0 
1

1
  f 2g2
0

0
1
0



1
 0
0

1 0
0 1
0 0


−f
0 
1


0 0
1 f 
0 1


0
f 
1


0
1 0
−f 2 g   0 1
1
0 0


−f
0 .
1

Observe now that if we have f ∈ σn R and instead of g we have h ∈
σr2 Ūr,m , σr Ōm for 1 ≤ r ≤ n − 1, then by Propositions 7.7 and 7.11, we have:


1 − f h −f h 0
1 + fh 0  A
A−1  f h
0
0
1




1 0 0
1 −f h 0
1 0 0
1
0   1 1 0  A ∈ IAm .
= A−1  −1 1 0   0
0 0 1
0
0
1
0 0 1

Therfore, by Propositions 7.7, 7.11 and the previous
GLn−1 (R) we have the following equation mod IAm :


1
0
0
1 0
−f   0 1
A−1  0 1 + f h
0
f h2
1 − fh
0 0


1 − fh
0
≡ A−1 
−f h2


0
f
1 0
 0 1
1
0
0 1 + fh
0 0

lemma, for every A ∈

0
f A
1

−1
−f
0  A.
1

I.e. we have the following corollary (notice the switching of the sign of f ):

48

Corollary 7.20. For every h ∈ σr2 Ūr,m , σr Ōm for 1 ≤ r ≤ n − 1, f ∈ σn R and
A ∈ GLn−1 (R) , the following is equals mod IAm :
A−1 [(In−1 + hE3,2 ) , (In−1 + f E2,3 )] A ≡ A−1 [(In−1 − f E1,3 ) , (In−1 + hE3,1 )] A.
Let have now the following proposition:
Proposition 7.21. Let h ∈ σ12 Ū1,m , σ1 Ōm and f ∈ σn R. Then:
[(In−1 + hE3,2 ) , (In−1 + f E2,3 )] ∈ IAm .
Proof. Denote h = σ1 u

1
0
1
IAm ∋  0
−σ2 u σ1 u
=

for some u ∈ σ1 Ū1,m , Ōm . Then:



0
1 0 0
1
0
0
1 0
0  0 1 f  0
1
0  0 1
1
0 0 1
σ2 u −σ1 u 1
0 0


0
−f 
1




1
0 0
 f σ2 u
1 0 
2
σ1 σ2 u f 0 1


1
0
0
1
1
0  0
· 0
0 σ1 u 1
0

and thus:


0 0
1
1 f  0
0 1
0


1
[(In−1 + hE3,2 ) , (In−1 + f E2,3 )] =  0
0

as required.

0
1
h


0
0
1
1
0  0
−σ1 u 1
0
 
0
1
0 , 0
1
0


0 0
1 −f 
0 1


0 0
1 f  ∈ IAm
0 1

We can now pass to the following proposition.
Proposition 7.22. Let h ∈ σ12 Ū1,m , σ1 Ōm , f ∈ σn R and A ∈ GLn−1 (R).
Then:
A−1 [(In−1 + hE3,2 ) , (In−1 + f E2,3 )] A ∈ IAm .
Proof. We will prove the proposition by induction. By result of Suslin [Su], as
n − 1 ≥ 3, SLn−1 (R) is generated by the elementary matrices of the form:
In−1 + rEl,k for r ∈ R, and 1 ≤ l 6= k ≤ n − 1.
Qn
So as the invertible elements of R are the elements of the form ± i=1 xsi i for
si ∈ Z (see [CF], chapter 8), GLn−1 (R) is generated by the elementary matrices
and the matrices of the form: In−1 + (±xi − 1) E1,1 for 1 ≤ i ≤ n. Therefore,
it is enough to show that if:
A−1 [(In−1 + hE3,2 ) , (In−1 + f E2,3 )] A ∈ IAm
49

and E is one of the above generators, then mod IAm we have:
A−1 E −1 [(In−1 + hE3,2 ) , (In−1 + f E2,3 )] EA
−1

≡A

(7.12)

[(In−1 + hE3,2 ) , (In−1 + f E2,3 )] A.

So if E is of the form In−1 + (±xi − 1) E1,1 , we obviously have property 7.12.
If E is an elementary matrix of the form In−1 + rEl,k such that l, k ∈
/ {2, 3} then
we also have property 7.12 in an obvious way. Consider now the case l, k = 2, 3.
In this case, by Corollary 7.20 we have:

 

1 0 0
1 0 0
A−1 E −1  0 1 0  ,  0 1 f  EA
0 h 1
0 0 1

 
 
 

1 0 0
1 0 −f
1 0 0
1 0 0
≡ A−1  0 1 −r   0 1 0  ,  0 1 0   0 1 r  A
0 0 1
0 0 1
h 0 1
0 0 1




2 2
2
1 0 0
1 − hf + h f
0 −hf
1 0 0
 0 1 r A
0
1
0
= A−1  0 1 −r  
2
0 0 1
−h f
0 1 + hf
0 0 1




2 2
2
1 − hf + h f
0 −hf
1
0
0
 AA−1  rh2 f 1 −rhf  A
0
1
0
= A−1 
2
−h f
0 1 + hf
0
0
1

 



1 0 −f
1 0 0
1
0
0
= A−1  0 1 0  ,  0 1 0  AA−1  rh2 f 1 −rhf  A.
0 0 1
h 0 1
0
0
1
So by applying Propositions 7.7, 7.11 and Corollary 7.20 once again on the
opposite way, we obtain that this element belongs to IAm by the induction
hypothesis. The other cases for l, k are treated by similar arguments: if l, k =
3, 2 we do exactly the same, and if l or k are diﬀerent from 2 and 3, then
the situation is easier - we use similar arguments, but without passing to
[(In−1 − f E1,3 ) , (In−1 + hE3,1 )] through Corollary 7.20.
Corollary 7.23. Let h ∈ σ12 Ū1,m , σ1 Ōm , f ∈ σn R and A ∈ GLn−1 (R). Then,
for every i 6= j we have:
A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A ∈ IAm .
Proof. Denote a permutation matrix, which its action on GLn−1 (R) by conjugation, moves 2 7→ j and 3 7→ i, by P . Then:
A−1 [(In−1 + hEi,j ) , (In−1 + f Ej,i )] A
= A−1 P −1 [(In−1 + hE3,2 ) , (In−1 + f E2,3 )] P A ∈ IAm .
Now, as one can see that symmetrically, the above corollary is valid for
every h ∈ σr2 Ūr,m , σr Ōm for 1 ≤ r ≤ n − 1, we actually ﬁnished the proof of
Proposition 7.18.
50

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Institute of Mathematics
The Hebrew University
Jerusalem, ISRAEL 91904
davidel-chai.ben-ezra@mail.huji.ac.il

53

