arXiv:1701.02459v4 [] 27 Jul 2017

The Congruence Subgroup Problem
for the Free Metabelian group on n ≥ 4
generators
David El-Chai Ben-Ezra
March 30, 2018
Abstract
The congruence subgroup problem for a finitely generated group Γ
\
asks whether the map Aut
(Γ) → Aut(Γ̂) is injective, or more generally,
what is its kernel C (Γ)? Here X̂ denotes the profinite completion of X. It
n
is well known that for finitely
 generated free abelian groups C (Z ) = {1}
2
for every n ≥ 3, but C Z = F̂ω , where F̂ω is the free profinite group on
countably many generators.
Considering Φn , the free metabelian group on n generators, it was also
proven that C (Φ2 ) = F̂ω and C (Φ3 ) ⊇ F̂ω . In this paper we prove that
C (Φn ) for n ≥ 4 is abelian. So, while the dichotomy in the abelian case
is between n = 2 and n ≥ 3, in the metabelian case it is between n = 2, 3
and n ≥ 4.

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 properties of IA (Φn ) and its subgroups

6

3 The main theorem’s proof

8

4 Some elementary elements of hIA (Φn )m i
10
4.1 Elementary elements of type 1 . . . . . . . . . . . . . . . . . . . 10
4.2 Elementary elements of type 2 . . . . . . . . . . . . . . . . . . . 13

1

5 A main lemma
14
5.1 Reducing Lemma 3.1’s proof . . . . . . . . . . . . . . . . . . . . 14
5.2 A technical lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.3 Finishing Lemma 3.1’s proof . . . . . . . . . . . . . . . . . . . . . 23
6 Remarks and problems for further research

1

25

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. Equivalently, it asks whether the natural map Ĝ →
GLn (Ẑ) is injective, where Ĝ and Ẑ are the proﬁnite completions of the group
G and the ring Z, respectively. More generally, the CSP asks what is the kernel
of this map. It is a classical 19th century result that the answer is negative for
n = 2. Moreover (but not so classical, cf. [Mel], [L]), the kernel in this case is
F̂ω - the free proﬁnite group on a countable number of generators. On the other
hand, it was proved in 1962 by Mennicke [Men] and Bass-Lazard-Serre [BLS]
that for n ≥ 3 the answer is aﬃrmative, and the kernel is therefore trivial.
By the observation GLn (Z) ∼
= Aut (Zn ) = Out (Zn ), the CSP can be generalized as follows: Let Γ be a group and G ≤ Aut (Γ) (resp. G ≤ Out (Γ)). For
a ﬁnite index characteristic subgroup M ≤ Γ denote:
G (M ) =
(resp. G (M ) =

ker (G → Aut (Γ/M ))
ker (G → Out (Γ/M ))).

Such a G (M ) will be called a “principal congruence subgroup” 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. In some sense, the CSP tries to understand
whether every ﬁnite quotient of G comes from a ﬁnite quotient of Γ.
One can easily see that the CSP is equivalent to the question: Is the congruence map Ĝ = ←
lim
−G/U → lim
←−G/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 ) = limm∈N G/G (Mm ) ≤ limm∈N Aut(Γ/Mm )
←−
←−
←−
≤ Aut(lim
(Γ/M
))
= Aut(Γ̂) (resp. Out(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) (resp. Ĝ → Out(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 results mentioned above can therefore
be reformulated as C Z2 = F̂ω while C (Zn ) = {e} for n ≥ 3. So the ﬁnite
quotients of GLn (Z) are closely related to the ﬁnite quotients of Zn when n ≥ 3,
but the ﬁnite quotients of GL2 (Z) are far of being understandable by the ﬁnite
quotients of Z2 .
Very few results are known when Γ is non-abelian. Most of the results
are related to Γ = πg,n , the fundamental group of Sg,n , the closed surface of
genus g with n punctures. In these cases one can take G = P M od (Sg,n ), the
pure mapping class group of Sg,n , and can naturally view it as a subgroup of
Out (πg,n ) (cf. [FM], chapter 8). Considering this cases, it is known that:
Theorem 1.1. For g = 0, 1, 2 and every n ≥ 0, 1, 0 respectively, we have:
C (P M od (Sg,n ) , πg,n ) = {1}.
Z2 and P M od (S1,0 ) ∼
Note that when g = 1 and n = 0, π1,0 ∼
= SL2 (Z),
=

2
so: C (P M od (S1,0 ) , π1,0 ) = C SL2 (Z) , Z = F̂ω . The cases for g = 0 were
proved in [DDH] (see also [Mc]), the cases for g = 1 were proved in [A] (see also
[Bo1], [BER]), and the cases for g = 2 were proved in [Bo1] (see also [Bo2] for the
speciﬁc case where g = 2 and n = 0). In particular, as P M od (S1,1 ) is isomorphic
to the special outer-automorphism group of F2 , we have an aﬃrmative answer
for the full outer-automorphism group of F2 , and by some standard arguments
it shows that actually C (F2 ) is trivial (see [BER], [BL]). Note that for every
n > 0, πg,n ∼
= F2g+n−1 = the free group on 2g + n − 1 generators. Hence, the
above solved cases give an aﬃrmative answer for various subgroups of the outerautomorphism group of ﬁnitely generated free groups, while the CSP for the full
Aut (Fd ) when d ≥ 3 is still unsettled, and so is the situation with P M od (Sg,n )
when g ≥ 3.
All the above settled cases have a common property which plays a crucial
role in the proof of Theorem 1.1: There is an intrinsic description of G by
iterative extension process by virtually free groups (groups which have a ﬁnite
index free subgroup). Actually, in these cases, in some sense, we do understand
the ﬁnite quotients of G, and the CSP tells us that these quotients are closely
related to the ﬁnite quotients of Γ. This situation changes when we pass to
G = Aut (Fd ) for d ≥ 3 or P M od (Sg,n ) for g ≥ 3. In these cases we do not
have a description of G that can help to understand the ﬁnite quotients of G.
So in some sense, all the known cases do not give us a new understanding of
the ﬁnite quotients of G. Considering the abelian case, what makes the result
of Mennicke and Bass-Lazard-Serre so special is that it not only shows that the
ﬁnite quotients of GLn (Z) are related to the ﬁnite quotients of Zn , but also
gives us a description of the ﬁnite quotients of GLn (Z), which we have not
known without this result.
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
3

C (Φ2 ) = F̂ω . In addition, it was proven there that C (Φ3 ) ⊇ F̂ω . So, the
ﬁnite quotients of Aut (Φ2 ) and Aut (Φ3 ) are far of being connected to the ﬁnite
quotients of Φ2 and Φ3 , respectively.
Here comes the main theorem of this paper:
\
Theorem 1.2. For every n ≥ 4, C (IA (Φn ) , Φn ) is central in IA
(Φn ), where:
IA (Φn ) = ker (Aut (Φn ) → Aut (Φn /Φ′n ) = GLn (Z)) .
u
Using the commutative exact diagram (see 6):
\
\
\
IA
(Φn ) → Aut
(Φn ) → GL
n (Z)
ց
↓
↓
Aut(Φ̂n ) → GLn (Ẑ)

→ 1

\
and the fact that GL
n (Z) → GLn (Ẑ) is injective for n ≥ 3, we obtain that
C (IA (Φn ) , Φn ) is mapped onto C (Φn ). Therefore we deduce that:
Theorem 1.3. For every n ≥ 4, C (Φn ) is abelian.
This is dramatically diﬀerent from the cases of n = 2, 3 described above.
Theorem 1.3 tells us that when n ≥ 4 the situation changes, and the ﬁnite
quotients of Aut (Φn ) are closely related to the ﬁnite quotients of Φn in the
following manner:
Corollary 1.4. Let n ≥ 4. Then, for every finite index subgroup H ≤ G =
Aut (Φn ), there exists a finite index characteristic subgroup M ≤ Φn and r ∈ N
such that G (M )′ G (M )r ⊆ H.
Note that by a theorem of Bachmuth and Mochizuki [BM], Aut (Fn ) →
Aut (Φn ) is surjective for every n ≥ 4, and thus G = Aut (Φn ) is ﬁnitely generated. Hence, the principal congruence subgroups of the form G (M ) are ﬁnitely
generated, and thus, the subgroups of the form G (M )′ G (M )r are also of ﬁnite
′
r
index in Aut (Φn ). Therefore, the quotients of the form Aut (Φn ) /G (M ) G (M )
describe all the ﬁnite quotients of Aut (Φn ). In particular, our theorem gives
us a description of the ﬁnite quotients of Aut (Φn ) when n ≥ 4 - just like the
theorem of [Men] and [BLS] gives for GLn (Z) when n ≥ 3. Corollary 1.4 obviously does not hold for n = 2, 3. So, the picture is that while the dichotomy in
the abelian case is between n = 2 and n ≥ 3, in the metabelian case we have a
dichotomy between n = 2, 3 and n ≥ 4.
d
In [KN], Kassabov and Nikolov showed that ker(SL\
n (Z [x]) → SLn (Z [x])) is
central and not ﬁnitely generated, when n ≥ 3. In [Be2] we use their techniques
and an interesting surjective representations:
 x→1
IA (Φn ) ։ ker(GLn−1 Z[x±1 ] −→ GLn−1 (Z))

to show also that:

4

Theorem 1.5. For every n ≥ 4, C (IA (Φn ) , Φn ) is not finitely generated.
We remark that despite the result of the latter theorem, we do not know
whether C (Φn ) is also not ﬁnitely generated.uIn fact we can not even prove at
this point that it is not trivial (for more, see 6).
The main line of the proof of Theorem 1.2 is as follows: For G = IA (Φn )
we ﬁrst take the principal congruence subgroups G (Mn,m ) where Mn,m =
′
′ m m
(Φ′n Φm
(Φ /M ), and thus we deduce that
n ) (Φn Φn ) . By [Be1], Φ̂n = lim
←− n n,m
the subgroups of the form G Mn,m4 are enough to represent the congruence
subgroups of IA(Φn ) in the sense that every congruence subgroup contains one
of these principal congruence subgroups. Then, we follow the steps of the theorem of Bachmuth and Mochizuki [BM], showing that Aut (Fn ) → Aut (Φn ) is
m
surjective for n ≥ 4, and we try to build G Mn,m4 with elements of hIA (Φn ) i.
This process, combined with some classical results from algebraic K-theory enables us to show that

m
m
hIA (Φn ) i G Mn,m4 / hIA (Φn ) i

is ﬁnite and central in IA (Φn ) / hIA (Φn )m i, and thus, hIA (Φn )m i is of ﬁnite
index in IA (Φn ). In particular, as every normal subgroup of index m in IA (Φn )
m
m
contains hIA (Φn ) i, we deduce that the groups of the form hIA (Φn ) i are
enough to represent the ﬁnite index subgroups of IA (Φn ). From here, it follows
\
easily that C (IA (Φn ) , Φn ) is central in IA
(Φn ).
We hope that the solution of the free metabelian case will help to understand
some more cases of non-abelian groups, such as the automorphism group of a free
group and the mapping class group of a surface. The immediate next challenges
are the automorphism groups of free solvable
groups.
u
Let us point out that, as remarked in 5 in [BL], one can deduce from Theorem 1.3 that for every n ≥ 4, Aut (Φn ) is not large, i.e does not contain a ﬁnite
index subgroup which can be mapped onto a free group. This is in contrast
with Aut (Φ2 ) and Aut (Φ3 ) which are large.
u
The paper is organized as follows: In 2 we present some needed notations
and
lemma, in
u discuss IA (Φn ) and some of its subgroups. Then, up to a main
u
3 we prove the main theorem of the paper, Theorem 1.2. In 4 we compute
m
some
u elements of hIA (Φn ) i which we use in the proof of the main lemma.
In 5 we prove the main reminded lemma. We end the paper with the proof
of Theorem 1.3, and some remarks on the problem of computing C (Φn ) and
C (IA (Φn ) , Φn ).
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.

5

2

Some properties of IA (Φn) and its subgroups

Let G = IA (Φn ) = ker (Aut (Φn ) → Aut (Φn /Φ′n ) = GLn (Z)). We start with
recalling some of the properties of G = IA (Φn ) and its subgroups, as presented
in Section 3 in [Be2]. We also refer the reader to [Be2] for the proofs of the
statements in this section. 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 ) .

• IGn,m = G(Mn,m ) = ker (IA (Φn ) → Aut (Φn,m )) .
m

• IAm
n = hIA (Φn ) i.
±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.
P
• An = ni=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
n
ai (xi − 1)
| g ∈ Z , 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 [Ba], [Bi]):
IA (Φn ) ֒→
deﬁned by α

7→

GLn (Rn )

a1,1 · · ·
 ..
 .
an,1

6

···


a1,n
.. 
. 
an,n

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

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

One can ﬁnd the proof of the following proposition in [Be2] (Propositions
3.1 and 3.2):
Proposition 2.1. Let In + A ∈ IA (Φn ). Then:
• If one denotes the P
entries of A by ak,l for 1 ≤ k, l ≤ n, then for every
n
1 ≤ k, l ≤ n, ak,l ∈ l6=i=1 σi Rn ⊆ An .
Q
• det (In + A) is of the form: det (In + A) = nr=1 xsrr for some sr ∈ Z.

Consider now the map:



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



Pn
g a 1 t1 + . . . + a n tn
| g ∈ Znm , ai ∈ Zm [Znm ], 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 2.2. 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 )
The following proposition is proven in [Be2] (Proposition 3.4):
Proposition 2.3. For every 1 ≤ i ≤ n we have: IGLn−1,i ∼
= GLn−1 (Rn , σi Rn ).
We recall the following deﬁnitions from Algebraic K-Theory:
Definition 2.4. Let R be a commutative ring (with identity), H ⊳ R an ideal,
and d ∈ N. Then:
7

• Ed (R) = hId + rEi,j | r ∈ R, 1 ≤ i 6= j ≤ di ≤ SLd (R) where Ei,j is the
matrix which has 1 in the (i, j)-th entry and 0 elsewhere.
• 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.
Under the above identiﬁcation of IGLn−1,i with GLn−1 (Rn , σi Rn ), for every
1 ≤ i ≤ n we deﬁne:
Definition 2.5. Let H ⊳ Rn . Then:
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) .

3

The main theorem’s proof

Using the above notations we prove in

u

5 the following main lemma:

Lemma 3.1. For every n ≥ 4 and m ∈ N one has:
IGn,m2

⊆ IAm
n ·

n
Y

ISLn−1,i (σi Hn,m )

i=1

= IAm
n · ISLn−1,1 (σ1 Hn,m ) · . . . · ISLn−1,n (σn Hn,m ) .
Observe that it follows that when n ≥ 4, then for every m ∈ N:
IGn,m4

⊆

2

IAm
n ·

n
Y

ISLn−1,i σi Hn,m2

i=1

⊆
⊆

IAm
n ·
IAm
n ·

n
Y

i=1
n
Y

i=1

ISLn−1,i σi Hn,m2

ISLn−1,i Hn,m2 .





The following Lemma is proved in [Be2], using classical results from Algebraic
K-theory (Lemma 7.1 in [Be2]):
Lemma 3.2. For every n ≥ 4, 1 ≤ i ≤ n and m ∈ N one has:

IEn−1,i Hn,m2 ⊆ IAm
n.

Let us now quote the following proposition (see [Be2], Corollary 2.3):

Proposition 3.3. Let R be a commutative ring, H ⊳ R ideal of finite index
and d ≥ 3. Assume also that Ed (R) = SLd (R). Then:
SK1 (R, H; d) = SLd (R, H) /E d (R, H)
is a finite group which is central in GLd (R) /E d (R, H).
8

Now, according to Proposition 3.3 and the fact that Ed (Rn ) = SLd (Rn ) for
every d ≥ 3 [Su], we obtain that for every n ≥ 4:
SLn−1 (Rn , Hn,m ) /E n−1 (Rn , Hn,m ) = SK1 (R, Hn,m ; n − 1)
is a ﬁnite group. Thus
ISLn−1,i (Hn,m ) /IEn−1,i (Hn,m ) ≤ SLn−1 (Rn , Hn,m ) /E n−1 (Rn , Hn,m )
is also a ﬁnite group. Hence, the conclusion from Lemmas 3.1 and 3.2 is that
for every m ∈ N, one can cover IGn,m4 with ﬁnite number of cosets of IAm
n . As
IGn,m4 is obviously a ﬁnite index subgroup of IA (Φn ) we deduce that IAm
n is
also a ﬁnite index subgroup of IA (Φn ). Therefore, as every normal subgroup
of IA (Φn ) of index m cotains IAm
n we deduce that one can write explicitely
m
\
IA (Φn ) = lim
←− (IA (Φn ) /IAn ). On the other hand, it is proven in [Be1] that
Φ̂n = ←
lim
−Φn,m , and thus:
Corollary 3.4. For every n ≥ 4:


m
C (IA (Φn ) , Φn ) = ker lim
(IA
(Φ
)
/IA
)
→
lim
(IA
(Φ
)
/IG
)
n
n
n,m
n
←−
←−


m
4
IA
(Φ
)
/IG
= ker lim
(IA
(Φ
)
/IA
)
→
lim
n
n
n,m
n
←−
←−

m
m
= lim IAn · IGn,m4 /IAn .
←−
Now, Proposition 3.3 gives us also that for every m ∈ N and n ≥ 4, the subgroup SK1 (Rn , Hn,m ; n − 1) is central in GLn−1 (Rn ) /E n−1 (Rn , Hn,m ). This
fact is used in [Be2] to prove that (see the arguments in Section 5 in [Be2]) :
Proposition 3.5. For every n ≥ 4, m ∈ N and 1 ≤ i ≤ n the subgroup:

m
IAm
n · ISLn−1,i σi Hn,m2 /IAn

is central in IA (Φn ) /IAm
n.

Corollary 3.6. For every n ≥ 4 and m ∈ N the elements of the set
IAm
n ·

n
Y

i=1


ISLn−1,i σi Hn,m2 /IAm
n

belong to the center of IA (Φn ) /IAm
n.

The conclusion from the latter corollary is that for every n ≥ 4 and m ∈ N,
the set
n
Y

ISLn−1,i σi Hn,m2 /IAm
IAm
n
n ·
i=1

is an abelian group which contained in the center of IA (Φn ) /IAm
n . In particm
m
4 /IA
·
IG
ular, IAm
is
contained
in
the
center
of
IA
(Φ
)
/IA
n
n,m
n
n , and thus,
n
\
C (IA (Φn ) , Φn ) is in the center of IA (Φn ). This ﬁnishes, up to the proof of
Lemma 3.1, the proof of Theorem 1.2.
So it remains to prove Lemma 3.1. But before we start to prove this lemma,
we need to compute some elements of IAm
n . We will do this in the following
section.
9

4

Some elementary elements of hIA (Φn )m i
m

In this section we compute some elements of IAm
n = hIA (Φn ) i which needed
through the proof of Lemma 3.1. As one can see below, we separate the elementary elements to two types. In addition, we separate the treatment of the
elements of type 1, to two parts. We hope this separations will make the process
clearer.
Additionally to the previous notations, on the section, and also later on, we
will use the notation:
µr,m =

m−1
X

xir for 1 ≤ r ≤ n.

i=0

4.1

Elementary elements of type 1

Proposition 4.1. Let n ≥ 3, 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. {σk µk,m (σi~ej − σj ~ei ) | i, j, k 6= u, i 6= j}
3. {σk µi,m (σi~ej − σj ~ei ) | i, j, k 6= u, i 6= j, k 6= j}
with coefficients in Rn , belong to IAm
n.

Before proving this proposition, we present some more elements of this type.
Note that for the following proposition we assume n ≥ 4:
Proposition 4.2. Let n ≥ 4,
IA (Φn ) of the form:

Iu−1
 au,1 · · · au,u−1
0

1 ≤ u ≤ n and m ∈ N. Then, the elements of
0
1
0

au,u+1

0
···
In−u



au,n  ← u-th

row

when (au,1 , . . . , au,u−1 , 0, au,u+1 , . . . , au,n ) is a linear combination of the vectors:
 2
1.
σu µu,m (σi~ej − σj ~ei ) | i, j 6= u, i 6= j
2. {σu σj µi,m (σi~ej − σj ~ei ) | i, j 6= u, i 6= j}
with coefficients in Rn , belong to IAm
n.
10

Proof. (of Proposition 4.1) 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 + b 2 · · ·
an + b n
1 b2 · · ·
bn
1 a2 · · ·
an
.
=
0
In−1
0
In−1
0
In−1
Hence, it is enough to prove that the elements of the following forms belong to
IAm
ei we mean that the entry of the i-th column in the ﬁrst
n (when we write a~
row is a):


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


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


1 σk µi,m f (σi~ej − σj ~ei )
i, j, k 6= 1, i 6= j, k 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 )
=
∈ IAm
n.
0
In−1
0
In−1
We pass to the elements of form 2. In this case we have:
"
m #
−1 
1 f (σi~ej − σj ~ei )
xk −σ1~ek
m
IAn ∋
,
0
In−1
0
In−1


1 σk µk,m f (σi~ej − σj ~ei )
.
=
0
In−1
We ﬁnish with the elements of form 3. If k = i, it is a private case of the
previous case, so we assume k 6= i. So we assume that i, j, k are all diﬀerent
from each other and i, j, k 6= 1 - observe that this case is interesting only when
n ≥ 4. 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, k = 4. It
is clear that symmetrically, with similar argument, the same holds in general
when n ≥ 4 for every i, j, k 6= 1 which diﬀerent from each other. So:
−m 
 

1
0
0 0
1 0 −σ4 f σ3 f


 0 1
1
0 0 
0
0 
 
, 0

IAm
∋ 
4




 0 0
0 −σ3 x2 0
1
0
0
0
0 1
0 0
0
1


1 −σ4 f µ2,m σ3 σ4 f σ2 µ2,m 0
 0
1
0
0 
.
= 
 0
0
1
0 
0
0
0
1

11

We pass now to the proof of Proposition 4.2.
Proof. (of Proposition 4.2) Also here, without loss of generality, we assume that
u = 1. Thus, all we need to show is that also the elements of the following forms
belong to IAm
n:


1 σ12 µ1,m f (σi~ej − σj ~ei )
i, j 6= 1, i 6= j, f ∈ Rn
1.
0
In−1


1 σ1 σj µi,m f (σi~ej − σj ~ei )
i, j 6= 1, i 6= j, f ∈ Rn .
2.
0
In−1

Also here, to simplify the notations, we will demonstrate the proof in the private
case: n = 4, i = 2, j = 3. We start with the ﬁrst form. From Proposition 4.1
we have (an element of form 2 in Proposition 4.1):


1
0
0
0
 0
1
0
0 
.

IAm
4 ∋ 0
0
1
0 
0 σ3 σ1 µ1,m f −σ2 σ1 µ1,m f 1
Therefore, we also have:

x4 0 0 −σ1


0
 0 1 0
∋ 
IAm
4
 0 0 1
0
0 0 0
1

2
1 −σ3 σ1 µ1,m f
 0
1
= 
 0
0
0
0

 

1
0
  0
1
,
  0
0
0 σ3 σ1 µ1,m f

σ2 σ12 µ1,m f 0
0
0 
.
1
0 
0
1

0
0
1
−σ2 σ1 µ1,m f


0

0 

0 
1

We pass to the elements of form 2. From Proposition 4.1 we have (an element
of form 3 in Proposition 4.1):


1
0
0
0
 0
1
0
0 


IAm
4 ∋ 0
0
1
0 
0 σ1 σ3 µ2,m f −σ1 σ2 µ2,m f 1

and therefore, we have:

1 0 σ4 −σ3
 0 1 0
0
m


IA4 ∋ 
0 0 1
0
0 0 0
1

1 −σ1 σ32 µ2,m f
 0
1
= 
 0
0
0
0

 

1
0
  0
1
,
  0
0
0 σ1 σ3 µ2,m f

σ3 σ1 σ2 µ2,m f 0
0
0 
.
1
0 
0
1
12

0
0
1
−σ1 σ2 µ2,m f


0

0 


0 
1

4.2

Elementary elements of type 2

Proposition 4.3. Let n ≥ 4, 1 ≤ u < v ≤ n and m ∈ N. Then, the elements
of IA (Φn ) of the form:


Iu−1
0
0
0
0
 0
1 + σu σv f
0
−σu2 f
0 

 ← u-th row
 0

0
I
0
0
v−u−1


 0
σv2 f
0
1 − σu σv f
0  ← v-th row
0
0
0
0
In−v
for f ∈ Hn,m , belong to IAm
n.

Proof. As before, to simplify the notations we will demonstrate the proof in the
case: n = 4, u = 1 and v = 2, and it will be clear from the computation that
the same holds in the general case, provided n ≥ 4.
First observe that for every f, g ∈ Rn we have:



1 + σ1 σ2 g
−σ12 g
0 0
1 + σ1 σ2 f
−σ12 f
0 0


σ22 g
1 − σ1 σ2 g 0 0 
σ22 f
1 − σ1 σ2 f 0 0 






0
0
1 0 
0
0
1 0
0
0
0 1
0
0
0 1


1 + σ1 σ2 (f + g)
−σ12 (f + g)
2

σ2 (f + g)
1 − σ1 σ2 (f + g)
=

0
0
0
0

0
0
1
0


0
0 

0 
1

so it is enough to consider the cases f ∈ mR4 and f ∈ σr µr,m R4 for 1 ≤ r ≤ 4,
separately. Consider now the following computation. For an arbitrary f ∈ Rn
we have:

 
−1 
1
0
0 0
x4 0 0 −σ1
 0

 
1
0 0 

 ,  0 x4 0 −σ2  
 0
0
1 0   0
0 1
0  
−σ2 f σ1 f 0 1
0
0 0
1

 

1
0
0 0
1 + σ1 σ2 f
−σ12 f
0 0


0
1
0 0 
1 − σ1 σ2 f 0 0 
σ22 f
=
.
·

0
0
1 0  
0
0
1 0 
−σ4 σ2 f σ4 σ1 f 0 1
0
0
0 1
Therefore, we conclude that

1
0
0

0
1
0


0
0
1
−σ4 σ2 f σ4 σ1 f 0

if:

 
1
0
 0
0 
, 
0   0
−σ2 f
1
13

0
1
0
σ1 f

0
0
1
0


0
0 
 ∈ IAm
4
0 
1

then also:



1 + σ1 σ2 f

σ22 f


0
0

−σ12 f
1 − σ1 σ2 f
0
0


0
0 
 ∈ IAm
4 .
0 
1

0
0
1
0

Thus, the cases f ∈ mR4 and f ∈ σr µr,m R4 for r 6= 4, are obtained immediately
from Proposition 4.1. Hence, it remains to deal with the case f ∈ σr µr,m R4 for
r = 4. However, it is easy to see that by switching the roles of 3 and 4, the
remained case is also obtained by similar arguments.

5

A main lemma

In this section we prove Lemma 3.1 which states that for every n ≥ 4 and m ∈ N
we have:
IGn,m2

⊆ IAm
n ·

n
Y

ISLn−1,i (σi Hn,m )

i=1

= IAm
n · ISLn−1,1 (σ1 Hn,m ) · . . . · ISLn−1,n (σn Hn,m ) .
The proof will be presented in a few stages - each of which will have a
separated subsection. In this sections n ≥ 4 will be constant, so we will make
notations simpler and write:
R = Rn , A = An , Hm = Hn,m , IAm = IAm
n , IGm = IGn,m .
We will also use the following notations:
Om = mR, Ur,m = µr,m R when µr,m =

Pm−1
i=0

xir for 1 ≤ r ≤ n .

Pn
Notice that it follows from the deﬁnitions, that: Hm = r=1 σr Ur,m + Om
(we note that in [Be2] we used the notation Ur,m for σr µr,m R).

5.1

Reducing Lemma 3.1’s proof

We start this subsection with introducing the following objects:
Definition 5.1. Let m ∈ N. Deﬁne:
R ⊲ Jm

=

n
X

2
σr3 Ur,m + A2 Om + AOm

r=1

Jm

=


In + A |

In + A ∈ IA (Φn ) ∩ GLn (R, Jm )
Qn
2
det (In + A) = r=1 xrsr m , sr ∈ Z

Proposition 5.2. For every m ∈ N we have:

IGm2 = IA (Φn ) ∩ GLn (R, Hm2 ) ⊆ Jm .
14



.

Pm−1
Proof. Let x ∈ R. Notice that i=0 xi ∈ (x − 1) R + mR. In addition, by
P
mi
replacing x by xm we obtain: m−1
∈ (xm − 1) R + mR. Hence:
i=0 x
m2

x

−1

2
m
−1
X

xi = (x − 1)

m−1
X

xi

m−1
X

xmi

=

(x − 1)

∈

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

⊆

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

i=0

i=0

i=0

P
2
2
2
Thus, we obtain that Hm2 = nr=1 (xm
r − 1)R + m R ⊆ Jm + Om . Now, let
In +A ∈ IGm2 = IA (Φn )∩GLn (R, Hm2 ). From the above observation and
 from
2
∩A =
Proposition 2.1, it follows that every entry of A belongs to Jm + Om
J
.
In
addition,
by
Proposition
2.1,
the
determinant
of
I
+
A
is
of
the
form
m
n
Qn
m2
sr
.
On
the
other
hand,
we
know
that
under
the
projection
x
→
7
1
and
x
r
r
r=1
Qn
m2 7→ 0 one has: In + A 7→ In and thus also r=1 xsrr = det (In + A) 7→ 1.
Qn
2
Therefore, det (In + A) is of the form r=1 xrm sr , as required.
Corollary 5.3. Let n ≥ 4 and m ∈ N. Then, for proving Lemma 3.1 it suffices
to prove that:
n
Y
ISLn−1,i (σi Hm ) .
Jm ⊆ IAm ·
i=1

We continue with deﬁning the following objects:
Definition 5.4. For 0 ≤ u ≤ n and 1 ≤ v ≤ n, deﬁne the following ideals of
±1
R = Rn = Z[x±1
1 , . . . , xn ]:
Ãu

=

n
X

σr R

r=u+1

J˜m,u,v

=



Pu
2
Ãu

r=1 Aσr Ur,m + AOm + Om +


P
 n
3
r=u+1 σr Ur,m

Pu
2

Ãu

r=1 Aσr Ur,m + AOm + Om +

P
 n
3
2
v6=r=u+1 σr Ur,m + Aσv Uv,m

v≤u
v>u

and for 0 ≤ u ≤ n deﬁne the groups: Ãu = IA (Φn ) ∩ GLn (R, Ãu ), and:
)
(
Qn
2
det (In + A) = i=1 xsi i m , every entry in
.
J̃m,u = In + A ∈ IA (Φn ) |
the v-th colmun of A belongs to J˜m,u,v
Remark 5.5. If In + A ∈ J̃m,u , the entries of the columns of A may belong to
diﬀerent ideals in R, so it is not obvious that J̃m,u is indeed a group, i.e. closed
under matrix multiplication and the inverse operation. However, showing that
J̃m,u is a group is not diﬃcult and we leave it to the reader.

15

Notice now the extreme cases:
1. For u = 0 we have (for every v and m): Ã0 = A, and Jm ⊆ J˜m,0,v . Hence,
we have Jm ⊆ J̃m,0 .
2. For u = n we have (for every v and m): Ãn = J˜m,n,v = 0. Hence, we also
have J̃m,n = {In }.
Corollary 5.6. For proving Lemma 3.1, it is enough to prove that for every
1 ≤ u ≤ n:
J̃m,u−1 ⊆ IAm · ISLn−1,u (σu Hm ) · J̃m,u .
Proof. Using that IAm is normal in IA (Φn ) and the latter observations, under
the above assumption, one obtains that:
Jm ⊆ J̃m,0

⊆
⊆
⊆

IAm · ISLn−1,1 (σ1 Hm ) · J̃m,1
...
n
Y
(IAm · ISLn−1,u (σu Hm )) · J̃m,n

u=1

=

IAm

n
Y

ISLn−1,u (σu Hm )

u=1

which is the requirement of Corollary 5.3.
We continue with deﬁning the following objects:
Definition 5.7. For 0 ≤ u ≤ n and 1 ≤ v ≤ n, deﬁne the following ideals of R:
 Pu

2
A

r=1 Aσr Ur,m + AOm + Om +

Pn

3
v≤u
r=u+1 σr Ur,m

Pu
Jm,u,v =
2

+
Aσ
U
+
AO
+
O
A
r
r,m
m

m

Pn r=1
3
2
v>u
v6=r=u+1 σr Ur,m + Aσv Uv,m
and for 0 ≤ u ≤ n deﬁne the group:


Qn
si m2
, every entry in .
Jm,u = In + A ∈ IA (Φn ) | det (In + A) = i=1 xi
the v-th colmun of A belongs to Jm,u,v
It follows from the deﬁnitions that for every 1 ≤ u ≤ n we have:
1. Jm,u−1,v ⊆ Jm,u,v , but Ãu−1 ⊇ Ãu . Thus, we have also
2. Jm,u−1 ⊆ Jm,u , but Ãu−1 ⊇ Ãu .
Here comes the connection between the latter objects to the objects deﬁned in
Deﬁnition 5.4.
Proposition 5.8. For every 0 ≤ u ≤ n and 1 ≤ v ≤ n we have: Jm,u,v ∩ Ãu =
J˜m,u,v , and hence: Jm,u ∩ Ãu = J̃m,u .
16

Proof. It is clear from the deﬁnitions that we have: J˜m,u,v ⊆ Jm,u,v ∩ Ãu , so we
have to show an inclusion on the opposite way. Let a ∈ Jm,u,v ∩ Ãu . As:
(P
n
σ 3 Ur,m
v≤u
˜
Jm,u,v ⊇ Pnr=u+1 r 3
2
v6=r=u+1 σr Ur,m + Aσv Uv,m v > u


Pu
2
we can assume that: a ∈ A
r=1 Aσr Ur,m + AOm + Om ∩ Ãu .
Observe now that by dividing an element b ∈ R by σu+1 , . . . , σn (with
residue), one can present b as a summand of an element of Ãu with an ele±1
ment of Ru = Z[x±1
1 , . . . , xu ]. Hence, R = Ãu + Ru and A = Ãu + Au , where
Au is the augmentation ideal of Ru . Hence:
a

∈

(Ãu + Au )2

u
X

σr µr,m (Ãu + Ru )

r=1
2

⊆

+ (Ãu + Au ) m(Ãu + Ru ) + (Ãu + Au )m2 (Ãu + Ru )
u
X
J˜m,u,v + A2u
σr µr,m Ru + A2u mRu + Au m2 Ru .
r=1


Pu
Hence, we can assume that a ∈ A2u r=1 σr µr,m Ru + A2u mRu + Au m2 Ru ∩
Ãu = {0}, i.e. a = 0 ∈ J˜m,u,v , as required.
Due to the above, we can now reduce Lemma 3.1’s proof as follows.
Corollary 5.9. For proving Lemma 3.1 it suffices to show that given 1 ≤ u ≤ n,
for every α ∈ J̃m,u−1 there exist β ∈ IAm ∩ Jm,u and γ ∈ ISLn−1,u (σu Hm ) ∩
Jm,u such that γαβ ∈ Ãu .
Proof. As clearly Jm,u ⊇ Jm,u−1 ⊇ J̃m,u−1 , we obtain from Proposition 5.8 that:
γαβ ∈ Ãu ∩ Jm,u = J̃m,u . Thus:
α ∈ ISLn−1,u (σu Hm ) · J̃m,u · IAm = IAm · ISLn−1,u (σu Hm ) · J̃m,u .
This yields that J̃m,u−1 ⊆ IAm ·ISLn−1,u (σu Hm )·J̃m,u which is the requirement
of Corollary 5.6.

5.2

A technical lemma

I this section we will prove a technical lemma, which will help us in subsection
5.3 to prove Lemma 3.1. In the following subsections 1 ≤ u ≤ n will be constant.
Under this statement, we will use the following notations:
• For a ∈ R we denote its image in Ru under the projection xu+1 , . . . , xn 7→ 1
by ā. In addition, we denote its image in Ru−1 under the projection
¯.
xu , . . . , xn 7→ 1 by ā
• For α ∈ GLn (R) we denote its image in GLn (Ru ) under the projection
xu+1 , . . . , xn 7→ 1 by ᾱ.
17

• Similarly, we will use the following notations for every m ∈ N:
P
– Ā = Au = ui=1 σi P
Ru , Ūr,m = µr,m Ru for 1 ≤ r ≤ u, Ōm = mRu
u
and H̄m = Hu,m = r=1 σr Ūr,m + Ōm .
P
u−1
¯ =A
¯
– Ā
u−1 =
i=1 σi Ru−1 , Ūr,m = µr,m Ru−1 for 1 ≤ r ≤ u − 1 and
¯ = mR
Ō
.
m

u−1

Now, let α = In + A ∈ J̃m,u−1 , and denote the entries of A by ai,j . Consider
the u-th row of A. Under the above assumption, for every v we have:

P

u−1
2

+
Aσr Ur,m + AOm + Om
Ãu−1

r=1


Pn σ 3 U
v<u
r=u r r,m

au,v ∈
Pu−1
2

Ãu−1

r=1 Aσr Ur,m + AOm + Om +


Pn
3
2
v ≥ u.
v6=r=u σr Ur,m + Aσv Uv,m

Hence we have:

 P
u−1
2

Ū
+
Ā
Ō
+
Ō
+ Āσu2 Ūu,m
Āσ
σ
r,m
m
r
u

m
r=1




= σu Pu Āσr Ūr,m + ĀŌm + Ō2
m
r=1
P

āu,v ∈
u−1
2

+ σu3 Ūu,m
Āσr Ūr,m + ĀŌm + Ōm
σu

r=1


P


u−1
= σ
2
2
Ū
+
σ
Ū
+
Ā
Ō
+
Ō
Āσ
r,m
u,m
m
r
u
u
m
r=1

v=u

(5.1)

v 6= u.

We can state now the technical lemma:

Lemma 5.10. Let α = In + A ∈ J̃m,u−1 . Then, there exists δ ∈ IAm ∩ J̃m,u−1
such that for every v 6= u, the (u, v)-th entry of αδ −1 belongs to σu2 H̄m .
We will prove the lemma in two steps. Here is the ﬁrst step:
Proposition 5.11. Let α = In + A ∈ J̃m,u−1 . Then, there exists δ ∈ IAm ∩
J̃m,u−1 such that for every v < u, the (u, v)-th entry of αδ −1 belongs to σu2 H̄m .
Proof. So let α = In + A ∈ J̃m,u−1 , and observe that for every 1 ≤ v ≤ u − 1 one
Pu−1
2
.
can write āu,v = σu b̄u,v for some: b̄u,v ∈ r=1 Āσr Ūr,m + σu2 Ūu,m + ĀŌm + Ōm
In addition, as it is easy to see that:
u−1
X
r=1

Āσr Ūr,m =

u−1
X

¯ )⊆σ
¯ )σ (σ Ū + Ū
(σu Ru + Ā
r,m
u
r u r,m

r=1

u−1
X

σr Ūr,m +

r=1

u−1
X

¯
¯ σ Ū
Ā
r r,m

r=1

2
¯2
¯ + Ō
¯ Ō
¯ )2 ⊆ σ Ō + Ā
¯ ) + (σ Ō + Ō
¯ )(σ Ō + Ō
ĀŌm + Ōm
= (σu Ru + Ā
m
m
u m
m
u m
u m
m
¯
for every 1 ≤ v ≤ u − 1, for some:
one can write b̄ = σ c̄ + b̄
u,v

¯b̄
u,v

u,v

u u,v

∈

u−1
X

¯¯
¯ σ Ū
¯2
¯
Ā
r r,m + ĀŌm + Ōm

r=1

c̄u,v

∈

u−1
X

σr Ūr,m + σu Ūu,m + Ōm = H̄m .

r=1

18

Notice, that as A satisﬁes the condition A~σ = ~0 we have the equality σ1 au,1 +
. . . + σn au,n = 0, which yields the following equalities as well:
σ1 āu,1 + . . . + σu−1 āu,u−1 + σu āu,u

=

0

σ1 b̄u,1 + . . . + σu−1 b̄u,u−1 + āu,u

⇓
=

0

σ1¯b̄u,1 + . . . + σu−1¯b̄u,u−1

⇓
=

0.

Observe now that for every 1 ≤ v ≤ u − 1 we have:
!
u−1
X
2
¯
¯
¯
¯
¯
¯
⊆ J˜m,u−1,v
Āσr Ūr,m + ĀŌm + Ō
σu b̄u,v ∈ σu
m

r=1

and thus, if we deﬁne:


δ =  σu¯b̄u,1

Iu−1
···
0

σu¯b̄u,u−1


0
0
1
0  ← u-th row
0 In−u

then δ ∈ J̃m,u−1 . We claim now that we also have δ ∈ IAm . We will prove this
claim soon, but assuming this claim, we can now multiply α from the right by
δ −1 ∈ J̃m,u−1 ∩ IAm and obtain an element in J̃m,u−1 such that the image of
its (u, v)-th entry for 1 ≤ v ≤ u − 1, under the projection xu+1 , . . . , xn 7→ 1, is:
āu,v − σu¯b̄u,v (1 + āu,u )

= σu2 c̄u,v − σu¯b̄u,v āu,u
∈

σu2 H̄m

+

σu2

u−1
X

¯ σ Ū
¯¯
¯
¯2
Ā
r r,m + ĀŌm + Ōm

r=1

= σu2 H̄m

!

as required.
So it remains to prove the following claim:
Pu−1 ¯ ¯
¯ 2 for
¯ + Ō
¯ Ō
σr Ūr,m + Ā
Claim 5.12. Let n ≥ 4, 1 ≤ u ≤ n, and ¯b̄u,v ∈ r=1 Ā
m
m
1 ≤ v ≤ u − 1 which satisfy the condition:
σ1¯b̄u,1 + . . . + σu−1¯b̄u,u−1 = 0.
Then:


u-th row →  σu¯b̄u,1

Iu−1
···
0

σu¯b̄u,u−1

19


0
0
1
0  ∈ IAm .
0 In−u

(5.2)

Proof. It will be easier to prove a bit more - we will prove that if for every
1 ≤ v ≤ u − 1:
u−1
X
¯b̄ ∈
¯
¯2 ¯
¯
¯ σ Ū
Ā
r r,m + Ā Ūv,m + Ōm
u,v
v6=r=1

then the vector: ~b = (¯b̄u,1 , . . . , ¯b̄u,u−1 , 0, . . . , 0) is a linear combination of the
vectors:


σk µk,m (σi~ej − σj ~ei )
, m (σi~ej − σj ~ei ) | i, j, k ≤ u − 1, i 6= j
σk µi,m (σi~ej − σj ~ei )

with coeﬃcients in Ru−1 . This will show that σu (¯b̄u,1 , . . . , ¯b̄u,u−1 , 0, . . . , 0) is a
linear combination of the vectors in Propositions 4.1 and 4.2, so the claim will
follow.
We start with expressing ¯b̄u,1 explicitly by writing:
¯b̄ =
u,1

u−1
X u−1
X

σi σr µr,m pi,r +

r=2 i=1

u−1
X

σi σj µ1,m qi,j + mr

i,j=1

for some pi,r , qi,j , r ∈ Ru−1 . Now, Equation 5.2 gives that under the projection
Pu−1
¯
σ2 , . . . , σu−1 7→ 0, ¯b̄u,1 7→ 0. It follows that ¯b̄u,1 ∈
i=2 σi Ru−1 ⊆ Ā. In
¯ , so we can write:
particular, r ∈ Ā
¯b̄ =
u,1

u−1
X u−1
X

σi σr µr,m pi,r +

r=2 i=1

u−1
X

σi σj µ1,m qi,j +

u−1
X

σi mri

i=1

i,j=1

for some pi,r , qi,j , ri ∈ Ru−1 .
Observe now that by dividing r1 by σ2 , . . . , σu−1 (with residue) we can write
Pu−1
r1 = r1′ + i=2 σi ri′ where r1′ depends only on x1 . Therefore, by replacing r1
by r1′ and ri by ri + σ1 ri′ for 2 ≤ i ≤ n, we can assume that r1 depends only on
x1 . Similarly, by dividing q1,1 by σ2 , . . . , σu−1 , we can assume that q1,1 depends
only on x1 . Now, by replacing ~b with:
~b

−
−

u−1
X u−1
X

r=2 i=1
u−1
X u−1
X

σi µr,m pi,r (σr ~e1 − σ1~er )
σj µ1,m qi,j (σi~e1 − σ1~ei ) −

u−1
X

σ1 µ1,m q1,j (σj ~e1 − σ1~ej )

j=2

i=2 j=1

−

u−1
X

mri (σi~e1 − σ1~ei )

i=2

we can assume that ¯b̄u,1 is a polynomial which depends only on x1 . On the
Pu−1
other hand, we already saw that Equation 5.2 yields that ¯b̄u,1 ∈ i=2 σi Ru−1 ,
so we can actually assume that ¯b̄u,1 = 0.
20

We continue in this manner by induction. In the 1 ≤ v ≤ u − 1 stage we
assume that ¯b̄u,1 = . . . ¯b̄u,v−1 = 0. Then we write:
¯b̄ =
u,v

u−1
X u−1
X

u−1
X

σi σr µr,m pi,r +

σi σj µv,m qi,j + mr

i,j=1

v6=r=1 i=1

for some pi,r , qi,j , r ∈ Ru−1 . The condition ¯b̄u,1 = . . . = ¯b̄u,v−1 = 0 and
Equation 5.2 give that σv ¯b̄u,v +σv+1¯b̄u,v+1 +. . .+σu−1¯b̄u,u−1 = 0 and thus, under
Pu−1
¯ . In
the projection σv+1 , . . . , σu−1 7→ 0, ¯b̄u,v 7→ 0, so ¯b̄u,v ∈ i=v+1 σi Ru−1 ⊆ Ā
¯
particular, r ∈ Ā, so we can write:
¯b̄ =
u,v

u−1
X u−1
X

σi σr µr,m pi,r +

u−1
X

σi σj µv,m qi,j +

σi mri

i=1

i,j=1

v6=r=1 i=1

u−1
X

for some pi,r , qi,j , ri ∈ Ru−1 .
Now, as we explained previously, by dividing pi,r , qi,j , ri for 1 ≤ i, j, r ≤
v by σv+1 , . . . , σu−1 , we can assume that these polynomials depend only on
x1 , . . . , xv . Thus, by replacing ~b with:
~b

−

u−1
X
X u−1

σi µr,m pi,r (σr ~ev − σv ~er ) −

u−1
X u−1
X

σj µv,m qi,j (σi~ev − σv ~ei ) −

u−1
X

v
u−1
X
X

σi µv,m qi,j (σj ~ev − σv ~ej )

i=1 j=v+1

i=v+1 j=1

−

σr µr,m pi,r (σi~ev − σv ~ei )

r=1 i=v+1

r=v+1 i=1

−

v−1
X
X u−1

mri (σi~ev − σv ~ei )

i=v+1

we can assume that ¯b̄u,v is a polynomial which depends only on x1 , . . . , xv ,
without changing the assumption that ¯b̄u,w = 0 for w < v. But we saw that in
Pu−1
this situation Equation 5.2 yields that ¯b̄u,v ∈ i=v+1 σi Ru−1 , so we can actually
assume that ¯b̄u,v = 0, as required.
Here is the second step of the technical lemma’s proof:
Proposition 5.13. Let α = In + A ∈ J̃m,u−1 such that for every v < u,
āu,v ∈ σu2 H̄m . Then, there exists δ ∈ IAm ∩ J̃m,u−1 such that for every v 6= u,
the (u, v)-th entry of αδ −1 belongs to σu2 H̄m .
Proof. So let α = In + A ∈ J̃m,u−1 such that for every v < u, , āu,v ∈
σu2 H̄
m . We remined that by Equation 5.1,for every v > u we have: āu,v ∈
P
u−1
2
2
σu
r=1 Āσr Ūr,m + σu Ūu,m + ĀŌm + Ōm . Hence, we can write explicitly:
āu,v = σu

u−1
u
XX

σi σr µr,m pr,i + σu2 µu,m q +

r=1 i=1

u
X
i=1

21

mσi ri + m2 s

!

2
for some: pr,i , q, ri , s ∈ Ru . Clearly, as ĀŌm ⊇ ĀŌm
, by dividing s by σi for
1 ≤ i ≤ u (with residue), we can assume that s ∈ Z. Consider now the following
element:
m2

IAm ∋ (In + σv Eu,u − σu Eu,v )

= In + σv µv,m2 Eu,u − σu µv,m2 Eu,v = δ ′ .

By the computation in the proof of Proposition 5.2, we obtain that:
2
µv,m2 ∈ σv2 Uv,m + σv Om + Om

and thus (we remind that v > u):
σv µv,m2
σu µv,m2


2
∈ σv σv2 Uv,m + σv Om + Om
⊆ J˜m,u−1,u

2
2
∈ σu σv Uv,m + σv Om + Om ⊆ J˜m,u−1,v .
2

′
In addition, the determinant of δ ′ is xm
v . Therefore, δ ∈ J̃m,u−1 . Observe now
that as v > u, under the projection σu+1 , . . . , σn 7→ 0, xv 7→ 1, and δ is therefore
maped to:
δ̄ ′ = In − m2 σu Eu,v .

Thus, if we multiply α from the right by δ ′s we obtain that the value of the
entries in the u-th row under the projection σu+1 , . . . , σn 7→ 0 does not change,
besides the value of the entry in the v-th colmun, which changes to (see Equation
5.1 for the ideal which contains āu,u ):
!
u−1
X
Āσr Ūr,m + σu2 Ūu,m + ĀŌm
āu,v − sm2 σu (1 + āu,u ) ∈ σu
+ σu2
=

σu

r=1
u
X

r=1
u−1
X

Āσr Ūr,m + ĀŌm +

Āσr Ūr,m +

r=1

σu2 Ūu,m

2
Ōm

!

+ ĀŌm

!

.


Pu
Pu−1
Hence, we can assume that āu,v ∈ σu i=1 σi fi + σu2
r=1 σr Ūr,m + Ōm =
P
Pu−1
u−1
σu i=1 σi fi + σu2 H̄m , for some fi ∈
r=1 σr Ūr,m + Ōm . Deﬁne now (the
coeﬃcient of ~ev is the value of the (u, v)-th entry):


Iu−1
0 
0

Pu−1


δv =  −σv σu f1 · · · −σv σu fu−1 1
σu i=1 σi fi ~ev  ∈ J̃m,u−1 .
0
0
In−u
By proposition 4.1, we obviously have: δv ∈ IAm . In addition, as v > u, under
the projection σu+1 , . . . , σn 7→ 0 we have:


Iu−1 0

P 0


u−1
ev  .
1 σu
δ̄v =  0
i=1 σi fi ~
0
0
In−u
22

Thus, by multiplying α from the right by δ̄v−1 we obtain that the value of the
entries in the u-th row under the projection σu+1 , . . . , σn 7→ 0 does not change,
besides the value of the entry in the v-th colmun, which changes to:
!
!
u
u−1
X
X
2
2
2
Āσr Ūr,m + AŌm + Ōm
σi fi (1 + āu,u ) ∈ σu H̄m + σu
āu,v − σu
r=1

i=1

= σu2 H̄m .

Qn
Thus, deﬁninig δ = v=u+1 δv ﬁnishes the proof of the proposition, and hence,
also the proof of the technical lemma.

5.3

Finishing Lemma 3.1’s proof

We remind that we have a constant 1 ≤ u ≤ n. We remind also that by Corollary
5.9, it suﬃces to show that given α ∈ J̃m,u−1 there exist β ∈ IAm ∩ Jm,u and
γ ∈ ISLn−1,u (σu Hm ) ∩ Jm,u such that γαβ ∈ Ãu .
So let α = In + A ∈ J̃m,u−1 . By the above technical lemma, there exists
δ ∈ IAm ∩ J̃m,u−1 ⊆ IAm ∩ Jm,u such that for every v 6= u, the (u, v)-th entry
of αδ −1 belongs to σu2 H̄m . Thus, by replacing α with αδ −1 , with out loss of
generality, we can assume that we have āu,v ∈ σu2 H̄m for every v 6= u. I.e. for
every v 6= u one can write: āu,v = σu2 b̄u,v for some b̄u,v ∈ H̄m .
Now, for every v 6= u deﬁne the matrix:


σ1 b̄u,v (σv ~eu − σu~ev )
 σ2 b̄u,v (σv ~eu − σu~ev ) 


δv = In + 
 ∈ Jm,u
..


.
σn b̄u,v (σv ~eu − σu~ev )

which is equals, by direct computation, to the multiplication of the matrices:


0
Jm,u ∋ εv,k = In +  σk b̄u,v (σv ~eu − σu~ev )  ← k-th row
0

for k 6= u, v and the matrix (the following is an example for v > u):


0
 σu b̄u,v (σv ~eu − σu~ev )  ← u-th row



0
Jm,u ∋ ηv = In + 


 σv b̄u,v (σv ~eu − σu~ev )  ← v-th row
0

Qn
i.e. δv = ηv · u,v6=k=1 εv,k (observe that the matrices εv,k commute, so the
product is well deﬁned). One can see that by Propositions 4.1 and 4.2, εv,k ∈
IAm for every k 6= u, v. Moreover, by Proposition 4.3, ηv ∈ IAm . Hence,

23

Pn
δv ∈ IAm ∩ Jm,u . Now, as for every 1 ≤ i ≤ n we have j=1 ai,j σj = 0 (by the
Qn
condition A~σ = ~0), α · u6=v=1 δv is equals to:






σ1 b̄u,v (σv ~eu − σu~ev )
a1,1 · · · a1,n
n




 ..
..  Y I +  σ2 b̄u,v (σv ~eu − σu~ev ) 
I
+

 n 
 n  .
..
. 



.
u6
=
v=1
an,1 · · · an,n
σn b̄u,v (σv ~eu − σu~ev )




σ1 b̄u,v (σv ~eu − σu~ev )
a1,1 · · · a1,n
n



..  + X  σ2 b̄u,v (σv ~eu − σu~ev )  .
= In +  ...



.
.
..


u6
=
v=1
an,1 · · · an,n
σn b̄u,v (σv ~eu − σu~ev )
Qn
It is easy to see now that if we denote α · u6=v=1 δv = In + C, then for every
v 6= u, c̄u,v = 0, when ci,j is the (i, j)-th entry of C. Hence, we also have:
c̄u,u σu =

n
X

c̄u,v σ̄v = 0 =⇒ c̄u,u = 0.

v=1

Thus, we can write α ·
following properties:

Qn

u6=v=1 δv

= In + C̄ when the matrix C̄ has the

• The entries of the u-th row of C̄ are all 0.
• As ai,v ∈ J˜m,u−1,v for
5.1
Pevery i, v, by the computation for Equation

u−1
2
2
for every
we have: āi,v ∈ σu
r=1 Āσr Ūr,m + σu Ūu,m + ĀŌm + Ōm
i, v 6= u. Hence, for every i, v 6= u we have:
!
u−1
X
2
2
Āσr Ūr,m + σu Ūu,m + ĀŌm + Ōm + σu ĀH̄m
c̄i,v ∈ σu
r=1

=


2
.
σu ĀH̄m + Ōm

Qn
Qu
2
Now, as det(δv ) = 1 for every v 6= u, det(α · u6=v=1 δv ) = det(α) = i=1 xsi i m .
However, as the entries of C̄ have the above properties, Qthis determinant is
mapped to 1 under the projection σu 7→ 0. Thus, det(α · nu6=v=1 δv ) is of the
2

form xsuu m . Now, set i0 6= u, and denote:

ζ = In + σu µu,m2 Ei0 ,i0 − σi0 µu,m2 Ei0 ,u = (In + σu Ei0 ,i0 − σi0 Ei0 ,u )

m2

∈ IAm .

By the computation in the proof of Proposition 5.2, we obtain that:
2
µu,m2 ∈ σu2 Uu,m + σu Om + Om

and thus:
σu µu,m2
σi0 µu,m2



2
2
⊆ Jm,u,i0
⊆ σu ĀH̄m + Ōm
∈ σu σu2 Uu,m + σu Om + Om

2
2
∈ σi0 σu Uu,m + σu Om + Om ⊆ Jm,u,u
24

2

so ζ ∈ IAm ∩ Jm,u . In addition det (ζ) = xm
u . Therefore, α ·
writen as In + C̄, has the following properties:

Q

v6=u δv ζ

−su ,

• The entries of the u-th row of C̄ are all 0.

2
, so we can write
• For every i, v 6= u we have: c̄i,v ∈ σu ĀH̄m + Ōm
2
c̄i,v = σu di,v for some di,v ∈ ĀH̄m + Ōm
.
Pu
Pu−1
• For every 1 ≤ i ≤ n we have: k=1 σk c̄i,k = 0, so c̄i,u = − k=1 σk di,k .

• det In + C̄ = 1.

I.e.:



i=u
0 P
c̄i,j = − u−1
σ
d
j=u
k
i,k
k=1


σu di,j
i, j 6= u

2
for some di,j ∈ ĀH̄mQ+ Ōm , and det In + C̄ = 1.
Deﬁne now β = v6=u δv ζ −su and γ −1 by (γ −1 i,j is the (i, j)-th entry of γ −1 .
Notice that the sums in the u-th column run until n and not until u − 1 as in
In + C̄, as we need the following matrix to be an element of IA (Φn ). However,
clearly, this addition does not changes the value of the determinant, and it stays
to be 1):


i=u
0 P
−1
γ i,j = − nu6=k=1 σk di,k j = u


σu di,j
i, j 6= u.

m
2
⊆ Hm and det γ −1 =
So β ∈ IA  ∩ Jm,u . In addition, as di,j ∈ ĀH̄m + Ōm
det In + C̄ = 1, γ ∈ ISLn−1,u (σu Hm ). Moreover, γ ∈ Jm,u . Hence, we
obtained β ∈ IAm ∩ Jm,u and γ ∈ ISLn−1,u (σu Hm ) ∩ Jm,u such that γαβ = In ,
i.e. γαβ ∈ Ãu , as required.

6

Remarks and problems for further research

We will prove now Theorem 1.3, which asserts that C (Φn ) is abelian for every
n ≥ 4. But before, let us have the following proposition, which is slightly more
general than Lemma 2.1. in [BER], but proven by similar arguments:
α

β

Proposition 6.1. Let 1 → G1 → G2 → G3 → 1 be a short exact sequence of
groups. Assume also that G1 is finitely generated. Then:
α̂

β̂

1. The sequence Ĝ1 → Ĝ2 → Ĝ3 → 1 is also exact.
α̂
2. The kernel ker(Ĝ1 → Ĝ2 ) is central in Ĝ1 .
Proof. (of Theorem 1.3) By Proposition 6.1, the commutative exact diagram:
1 → IA (Φn ) → Aut (Φn )
ց
↓
Aut(Φ̂n )
25

→ GLn (Z) → 1
↓
→ GLn (Ẑ) .

gives rise to the commutative exact diagram:
\
\
\
IA
(Φn ) → Aut
(Φn ) → GL
n (Z)
ց
↓
↓
Aut(Φ̂n ) → GLn (Ẑ)

→ 1

\
Now, as n ≥ 4, by the CSP for GLn (Z), the map GL
n (Z) → GLn (Ẑ) is injec\
tive, so one obtains by diagram chasing, that C (IA (Φn ) , Φn ) = ker(IA
(Φn ) →
\
Aut(Φ̂n )) is mapped onto C (Φn ) = ker(Aut (Φn ) → Aut(Φ̂n )) through the map
\
\
IA
(Φn ) → Aut
(Φn ). In particular, as by Theorem 1.2 C (IA (Φn ) , Φn ) is cen\
tral in IA
(Φn ) for every n ≥ 4, it is also abelian, and thus C (Φn ) is an image
of abelian group, and therfore abelian, as required.
Problem 6.2. Is C (Φn ) not ﬁnitely generated? trivial?
We proved in [Be2] that C (IA (Φn ) , Φn ) is not ﬁnitely generated for every
n ≥ 4. This may suggest that also C (Φn ) is not ﬁnitely generated, or at list,
\
not trivial. Moreover, if C (IA (Φn ) , Φn ) was not central in IA
(Φn ), we could
use the fact that IA (Φn ) is ﬁnitely generated for every n ≥ 4 [BM], and by the
second part of Proposition 6.1 we could derive that the image of C (IA (Φn ) , Φn )
\
in Aut
(Φn ) is not trivial. However, we showed that C (IA (Φn ) , Φn ) is central
\
\
\
in IA (Φn ), so it is possible that C (IA (Φn ) , Φn ) ⊆ ker(IA
(Φn ) → Aut
(Φn ))
and thus C (Φn ) is trivial.
We saw in [Be2] that for every i there is a natural surjective map

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

These maps enabled us to show in [Be2] that for every n ≥ 4, C (IA (Φn ) , Φn )
can be written as
C (IA (Φn ) , Φn ) = (C

(IA (Φn ) , Φn ) ∩ni=1

ker ρ̂i ) ⋊

n
Y

Ci

i=1

where
Ci

∼
=
∼
=


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

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

\
are central in IA
(Φn ). Here we showed that also C (IA (Φn ) , Φn ) ∩ni=1 ker ρ̂i lie
\
in the center of IA
(Φn ) but we still do not know to determine whether:
Q
Problem 6.3. Is C (IA (Φn ) , Φn ) = ni=1 Ci or it contains more elements?

It seems that having the answer to Problem 6.3 will help to solve Problem
6.2.
26

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

28

