# Bounded reduction for Chevalley groups of types $E_{6}$ and $E_{7}$

Pavel Gvozdevsky  Department of Mathematics, Bar-Ilan University, 5290002
Ramat Gan, ISRAEL<EMAIL_ADDRESS>
###### Abstract.

We prove that an element from the Chevalley group of type $E_{6}$ or $E_{7}$
over a polynomial ring with coefficients in a small-dimensional ring can be
reduced to an element of certain proper subsystem subgroup by a bounded number
of elementary root elements. The bound is given explicitly. This result is an
effective version of the early stabilisation of the corresponding
$K_{1}$-functor. We also give part of the proof of similar hypothesis for
$E_{8}$.

###### Keywords:

Chevalley groups, surjective stability of $K_{1}$, bounded reduction,
polynomial rings

###### 2020 Mathematics Subject Classification:

19B14(primary), 19B10, 20G35 and 20G41(secondary)

The paper is written as part of the author’s post-doctoral fellowship at Bar-
Ilan University, Department of Mathematics; and is supported by ISF grant
1994/20.

##  1\. Introduction

This paper deals with the bounded reduction in exceptional Chevalley groups
over polynomial rings.

Chevalley groups over certain rings have bounded width with respect to the
elementary generators. For example this holds for Dedekind domains of
arithmetic type, see [6], [7], [29], [49], [50],[56],[9],[10], [24]. Results
on such bounded generation are of great value, for example they are connected
to the congruence subgroup property, see [26],[30]; to Margulis–Zimmer
conjecture, see [37]; and have applications in studying strong boundedness,
see [52], [51], [54], [53], [55]. However bounded generation occurs very
rarely in the sense that classes of rings for which it is known to hold are
pretty narrow. Nevertheless, for some applications it is enough to have a
weaker result, such as: bounded length of conjugates of elementary generators
(see [46]), bounded length of commutators (see [38],[45],[17]), or bounded
generation with respect to a larger set of generators. Bounded reduction is a
variation of the last property.

A given Chevalley group $G$ over a given ring is said to have bounded
reduction if any element of $G$ can be decomposed as a product of bounded
number of elementary generators and one (not necessarily elementary) element
from a certain subsystem subgroup. In other words, it means that one can
reduce any element to the subsystem subgroup by bounded number of elementary
transformations. Without requirement for the number of elementary
transformations to be bounded this property is called the surjective stability
of the $K_{1}$-functor. In papers [42], [43], [31], [32], [33], and [15] this
problem is considered for rings that satisfy certain conditions on stable
rank, absolute stable rank, or other similar conditions. Actually, from the
proofs of the theorems in these papers one can recover the bound on the
required number of elementary transformations, despite the fact that this
bound is not stated in papers explicitly. Therefore, these are results on
bounded reduction.

However, conditions on stable rank are still very strong. Even though small
Jacobson dimension implies small stable rank, rings with large Jacobson
dimension usually fail to have small stable rank. In the present paper, we
consider another important class of rings. Namely we take a polynomial ring in
arbitrary number of variables with coefficients in a small-dimensional ring.
Here we use Krull dimension because the techniques require for dimension to
behave well with respect to adding an independent variable.

Without the bound on the number of elementary transformations similar result
for classical groups is known as early surjective stability of the
$K_{1}$-functor. For the special linear group this was proved by Suslin in
[47]. Similar result for the orthogonal group follows from [48], and for the
symplectic group it is proven in [20], see also [14], [21]. Note that if the
ring of coefficient is a Dedekind domain or a smooth algebra over a field,
then this result for all Chevalley groups follows from the homotopy invariance
of the non-stable $K_{1}$-functor, see [1],[41],[40].

In the case of special linear and symplectic groups, there are similar results
for Laurent polynomial rings, see [22], [23].

In the paper [58], Vaserstein obtained the effective version of the Suslin
result, i.e. he proved the bounded reduction for the special linear group over
a polynomial ring, and gave this bound explicitly. From this result he deduced
that the elementary subgroup of the general linear group over an arbitrary
finitely generated commutative ring has Kazhdan’s property (T).

In [35], the basic connection between bounded generation and property (T) has
been established and used to estimate the Kazhdan Constants for
$\mathop{\mathrm{SL}}\nolimits_{n}(\mathbb{Z})$. Later the bounds for these
constants were improved in [19]. In order to deduce property (T) from the
Vaserstein’s result one needs to refer to [36].

In fact, property (T) for Chevalley groups and groups similar to them has
already been studied by other methods, see [11],[12]. However, we believe that
the bounded reduction has an independent value, and we aim to study this
question for other Chevalley groups. It was noted in the concluding remarks of
[58] that the bounded reduction for the symplectic group follows formally from
the case of special linear group. In [16] the bounded reduction for orthogonal
groups was established, therefore, closing the problem for classical groups.

In the present paper we deal with exceptional groups. We prove bounded
reduction for the groups of types $E_{6}$ and $E_{7}$; and we give part of the
proof for the group $E_{8}$.

The main result of the present paper is the following theorem.

######  Theorem 1.1.

Let $C$ be a commutative Noetherian ring and $\dim C=D$. Let
$A=C[x_{1},\ldots,x_{n}]$. Let $\Delta\leqslant\Phi$ be one of the following
embeddings of root systems

  1. (a)

$D_{5}\leqslant E_{6}$;

  2. (b)

$E_{6}\leqslant E_{7}$;

Assume that

$D\leqslant\begin{cases}3&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\ 4&\text{
for }E_{6}\leqslant E_{7}\text{.}\end{cases}$

For the case $E_{6}\leqslant E_{7}$ assume additionally that $C$ is a Jacobson
ring.

Then every element of the group $G(\Phi,A)$ can be reduced to the subgroup
$G(\Delta,A)$ by multiplication from the left by $N$ elementary root elements,
where

$N=\begin{cases}36n^{2}+(72D+80)n+92&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 52n^{2}+(104D+249)n+244&\text{ for }E_{6}\leqslant
E_{7}\text{.}\end{cases}$

Therefore, this theorem is an extension of [58] and [16] to the groups of
types $E_{6}$ and $E_{7}$.

The paper is organised as follows. In Section 2, we give all necessary
preliminaries and introduce basic notation. In Section 3, we recall the notion
of an absolute flexible stable rank introduced in [16]. In Sections 4, 5 6,
and 7 we give the proof of the main result.

##  2\. Preliminaries and notation

### Rings, ideals and dimensions

By a ring we always mean associative and commutative ring with unity.

If $R$ is a ring, then by $R^{*}$ we denote the set of invertible elements in
$R$.

For the elements $r_{1}$,$\ldots$,$r_{k}\in R$, we denote by $\langle
r_{1},\ldots,r_{k}\rangle$ the ideal in $R$ generated by these elements.

In the present paper we use three different notions of a ring dimension.

  * •

By $\dim R$ we denote Krull dimension of the ring $R$. That is the supremum of
the lengths of all chains of prime ideals.

  * •

By $\operatorname{Jdim}R=\dim\operatorname{Max}(R)$ we denote the dimension of
the maximal spectrum $\operatorname{Max}(R)$ of the ring $R$. It is equal to
the supremum of the lengths of all chains of such prime ideals that coincide
with its Jacobson radical.

  * •

By $\operatorname{BSdim}R$ we denote the Bass–Serre dimension of a ring $R$.
That is the minimal $\delta$ such that $\operatorname{Max}(R)$ is a finite
union of irreducible Noetherian subspaces of dimension not greater than
$\delta$.

Obviously, for a Noetherian ring $R$ we have

$\operatorname{BSdim}R\leqslant\operatorname{Jdim}R\leqslant\dim R\text{.}$

The following property of Bass–Serre dimension is well known; see Lemma 4.17
in [2].

######  Lemma 2.1.

Let $R$ be a ring with $\operatorname{BSdim}R=d<\infty$. Then it has a finite
collection $P_{1}$,$\ldots$,$P_{m}$ of maximal ideals, such that for any $s\in
R\setminus\bigcup_{i}P_{i}$ we have $\operatorname{BSdim}R/(s)<d$. In case
where $d=0$, this means that $s\in R^{*}$.

### Chevalley groups

Let $\Phi$ be a reduced irreducible root system, let $G(\Phi,-)$, be a simply
connected Chevalley–Demazure group scheme over $\mathbb{Z}$ of type $\Phi$
(see [8]), and let $T(\Phi,-)$, be a split maximal torus in it. If $R$ is a
commutative ring with unit, the group $G(\Phi,R)$ is called the simply
connected Chevalley group of type $\Phi$ over $R$.

For a subset $X$ of a group we denote by $\langle X\rangle$ the subgroup
generated by $X$.

To each root $\alpha\in\Phi$ there correspond root unipotent elements
$x_{\alpha}(\xi)$, $\xi\in R$, elementary with respect to $T$. The group
generated by all these elements

$E(\Phi,R)=\left\langle x_{\alpha}(\xi)\colon\alpha\in\Phi\text{,}\,\xi\in
R\right\rangle\leqslant G(\Phi,R)\text{.}$

is called the elementary subgroup of $G(\Phi,R)$. For any $N\in\mathbb{N}$ we
denote by $E(\Phi,R)^{\leqslant N}$ the subset of $E(\Phi,R)$ consisting of
elements that can be expressed as the product of no more than $N$ elementary
root elements.

Any inclusion of root systems $\Delta\subseteq\Phi$ induces the homomorphisms
$G(\Delta,R)\to G(\Phi,R)$, and $E(\Delta,R)\to E(\Phi,R)$ taking elementary
root elements to elementary root elements.

By $U=U(\Phi,R)$ we denote the subgroup of $E(\Delta,R)$ generated by
elementary root elements with positive roots, i.e. the unipotent radical of
the standard Borel subgroup.

### Basic representations and weight diagrams

Let us fix an order on $\Phi$, and let $\Phi^{+}$, $\Phi^{-}$ and
$\Pi=\\{\alpha_{1}\text{,}\ldots\text{,}\alpha_{l}\\}$ be the sets of
positive, negative, and fundamental roots, respectively. Our numbering of the
fundamental roots follows that of [5] and [34]. By
$\varpi_{1}$,$\ldots$,$\varpi_{l}$ one denotes the corresponding fundamental
weights. Let $W=W(\Phi)$ be the Weyl group of the root system $\Phi$.

Recall that an irreducible representation $\pi$ of the complex semisimple Lie
algebra $L$ is called basic (see [28]) if the Weyl group $W=W(\Phi)$ acts
transitively on the set $\Lambda^{*}(\pi)$ of nonzero weights of the
representation $\pi$. The set of all weights of the representation $\pi$ we
denote by $\Lambda(\pi)$.

In case, where zero is not a weight of the basic representation $\pi$, such
representation is called microweight or minuscule representations, and the
list of these representations is classically known (see [4]).

Let $L$ be a simple Lie algebra of type $\Phi$. To each complex representation
$\pi$ of the algebra $L$ there corresponds a representation $\pi$ of the
Chevalley group $G=G(\Phi,R)$ on the free $R$-module
$V_{\pi}=V_{\pi}(R)=V_{\pi}(\mathbb{Z})\otimes_{\mathbb{Z}}R$ (see [28],[44]).
If $\pi$ is faithful, then we can identify $G$ with its image under this
representation. Thus, for an $g\in G$ and $v\in V$ we write $gv$ for the
action of $g$ on $v$.

Decompose the module $V=V_{\pi}$ into the direct sum of its weight submodules

$V=V^{0}\oplus\bigoplus_{\lambda\in\Lambda^{*}(\pi)}V^{\lambda}\text{.}$

Here all the modules $V^{\lambda}$, $\lambda\in\Lambda^{*}(\pi)$, are one-
dimensional. Matsumoto, [28, Lemma 2.3], has shown that there is a special
base of weight vectors $e_{\lambda}\in V^{\lambda}$,
$\lambda\in\Lambda^{*}(\pi)$, $v^{0}_{\alpha}$,
$\alpha\in\Delta(\pi)=\Pi\cap\Lambda^{*}(\pi)$ such that the action of root
unipotents $x_{\alpha}(\xi)$, $\alpha\in\Phi$, $\xi\in R$, is described by the
a following simple formulas:

$\begin{array}[]{lllll}\text{i.}&\text{ if
}&\lambda\in\Lambda^{*}(\pi)\text{,}\,\lambda+\alpha\notin\Lambda(\pi)\text{,}&\text{
then }&x_{\alpha}(\xi)e^{\lambda}=e^{\lambda}\text{,}\\\ \text{ii.}&\text{ if
}&\lambda,\lambda+\alpha\in\Lambda^{*}(\pi)\text{,}&\text{ then
}&x_{\alpha}(\xi)e^{\lambda}=e^{\lambda}\pm\xi e^{\lambda+\alpha}\text{,}\\\
\text{iii.}&\text{ if }&\alpha\notin\Lambda^{*}(\pi)\text{,}&\text{ then
}&x_{\alpha}(\xi)v^{0}=v^{0},\text{ for any }v^{0}\in V^{0}\text{,}\\\
\text{iv.}&\text{ if }&\alpha\in\Lambda^{*}(\pi)\text{,}&\text{ then
}&x_{\alpha}(\xi)v^{-\alpha}=v^{-\alpha}\pm\xi
v^{0}(\alpha)\pm\xi^{2}v^{\alpha}\text{,}\\\
&&&&x_{\alpha}(\xi)v^{0}=v^{0}\pm\xi\alpha_{*}(v^{0})v^{\alpha}\text{.}\end{array}$

Weight diagram of the representation $\pi$ is a graph whose vertices
correspond to the elements of $\Lambda^{*}(\pi)\sqcup\Delta(\pi)$; and whose
edges labeled be the numbers of fundamental roots shows the action of the
corresponding elementary root elements on the weight basis. These diagrams
serve as a great visual aid for calculations in Chevalley groups. The details
of how to construct and operate with weight diagrams can be found in [34] (see
also [42], [33] and [59]).

For a fundamental weight $\varpi$, we may consider the basic representation
with the highest weight $\varpi$. For simplicity, we call it the
representation $\varpi$.

Recall that in the present paper we study bounded reduction for the following
embedding of root systems: $D_{5}\leqslant E_{6}$, $E_{6}\leqslant E_{7}$, and
$E_{7}\leqslant E_{8}$. We denote by $\Phi$ the bigger system, and by $\Delta$
the smaller one. For the rest of the paper we fix the following representation
$\varpi$ of the group $G(\Phi,R)$:

  1. (a)

$\varpi=\varpi_{1}$ for $D_{5}\leqslant E_{6}$;

  2. (b)

$\varpi=\varpi_{7}$ for $E_{6}\leqslant E_{7}$;

  3. (c)

$\varpi=\varpi_{8}$ for $E_{7}\leqslant E_{8}$.

By $\lambda_{1}$,$\ldots$,$\lambda_{\dim\varpi}$ we denote the weights of this
representation with multiplicities, where numbering of weights for
$(E_{6},\varpi_{1})$ and $(E_{7},\varpi_{7})$ follows that of [15] and [33].
We do not need to fix a numbering for $(E_{8},\varpi_{8})$, but we agree that
the highest weight has number $1$, and the lowest weight has number $-1$.

By $e_{1}$,$\ldots$,$e_{\dim\varpi}$ we denote the corresponding weight basis.
Thus $\lambda_{1}$ is the highest weight and $e_{1}$ is the highest weight
vector. For $b\in V_{\varpi}$ we denote by $b_{i}$ or $b_{\lambda_{i}}$ the
corresponding coordinate of $b$ in the basis
$e_{1}$,$\ldots$,$e_{\dim\varpi}$. We will identify $b$ with the column vector
with entries $b_{i}$.

By $V_{\varpi}=V_{\varpi}(R)$ we denote the underlying module of the
representation $\varpi$. By $\operatorname{Um}_{\varpi}R$ we denote the set of
unimodular vectors in $V_{\varpi}(R)$, i.e. the set of such vectors $b\in
V_{\varpi}(R)$ that the elements $b_{i}$ generate the unit ideal in $R$. By
$\operatorname{Eq}_{\varpi}$ we denote the set of equations that determine the
orbit of the highest weight vector of the representation $\varpi$ (see [25]
and [59]). By $\operatorname{Orb}_{\varpi}R$ we denote the set of vectors from
$V_{\varpi}(R)$ that satisfy the equations from $\operatorname{Eq}_{\varpi}$.
Further set
$\operatorname{Um}^{\prime}_{\varpi}R=\operatorname{Um}_{\varpi}R\cap\operatorname{Orb}_{\varpi}R$;
and $\operatorname{Um}^{\prime\prime}_{\varpi}R=G(\Phi,R)e_{1}$.

Let $\Sigma_{1}\leqslant\Phi$ be the set roots that have positive coefficient
in simple root $\alpha_{i}$, $i=1$ for $E_{6}$ and $i=7$ for $E_{7}$, and
$i=8$ for $E_{8}$. Therefore, $\Delta\cup\Sigma_{1}$ is a parabolic set of
roots with $\Delta$ being the symmetric part, and $\Sigma_{1}$ being the
special part. Let $U_{1}$ be the unipotent radical of the corresponding
parabolic subgroup, and $U_{1}^{-}$ be the unipotent radical of the opposite
parabolic subgroup.

The following Lemmas can be derived from the proof of Chevalley–Matsumoto
decomposition theorem (see [8],[28],[42]).

######  Lemma 2.2.

Let $b\in\operatorname{Um}^{\prime}R$ be such that $b_{1}=1$. Then there
exists $u\in U_{1}^{-}$ such that $b=ue_{1}$.

######  Lemma 2.3.

Let $g\in G(\Phi,R)$ be such that $(ge_{1})_{1}=1$. Then

$g\in U_{1}^{-}\cdot G(\Delta,R)\cdot U_{1}=U_{1}^{-}U_{1}G(\Delta,R)\text{.}$

We also need the following lemma.

######  Lemma 2.4.

If the ring $R$ is semilocal, then
$\operatorname{Um}^{\prime}R=\operatorname{Um}^{\prime\prime}R$.

###### Proof.

Let $b\in\operatorname{Um}^{\prime}R$. By Lemma 2.2, it is enough to prove
that there exists $g\in G(\Phi,R)$ such that $(gb)_{1}$ is invertible (we may
then make it 1 by a toric element). Let $J$ be the Jacobson radical of the
ring $R$. The reduction map $E(\Phi,R)\to E(\Phi,R/J)$ is surjective; hence it
is enough to find $g\in E(\Phi,R/J)$ such that $(g\overline{b})_{1}$ is
invertible in $R/J$, where $\overline{b}$ is the reduction of $b$ (in fact, we
have $G=E$ for both $R$ and $R/J$). So we may assume that $J=0$, so $R$ is a
product of fields. Moreover, we can look for such $g$ separately for each
factor, so we may assume that $R$ is a field. If $b$ has at least one nonzero
entry in position that correspond to a nonzero weight, then we can take $g$ to
be the element of the extended Weyl group that shifts this weight to the
highest weight. That concludes the proof for $E_{6},\varpi_{1}$ and
$E_{7},\varpi_{7}$ because these representations are minuscule. It remains to
consider the case for $E_{8},\varpi_{8}$, where $b\in V^{0}$. It follows
easily from the fact that the lattice $E_{8}$ is self-dual that we have
$x_{\alpha_{i}}(1)b\notin V^{0}$ for at least one simple root $\alpha_{i}$; so
the problem is reduced to the previous case. ∎

### Branching tables

From the weight diagram it is immediate to read off the branching of the
corresponding representation with respect to a subsystem subgroup. In the case
where $\Delta=\langle\Pi\setminus\\{\alpha_{h}\\}\rangle$ is the symmetric
part of the maximal parabolic subset obtained by dropping the $h$-th
fundamental root the procedure is particularly easy. Then the restriction of
$\pi$ to $G(\Phi,R)$ looks as follows: one has to cut the diagram of $\pi$
through the bonds with the label $h$.

Given a representation $\pi$ of the group $G(\Phi,R)$ and two fundamental
roots $\alpha_{h_{1}}$,$\alpha_{h_{2}}\in\Pi$, by "branching table, where
vertical lines correspond to cuting through the bonds marked with $h_{1}$, and
horizontal lines correspond to cuting through the bonds marked with $h_{2}$",
we mean the table build as follows: at the upper right corner we write the
representation $\pi$; at the remaining cells of the upper row we write the
components of restriction of $\pi$ to the group
$G(\langle\Pi\setminus\\{\alpha_{h_{1}}\\}\rangle,-)$; at the remaining cells
of the left column we write the components of restriction of $\pi$ to the
group $G(\langle\Pi\setminus\\{\alpha_{h_{2}}\\}\rangle,-)$; and in all the
remaining cells we write the intersection of the corresponding restrictions.
When this intersection is zero we leave the cell blank; and when the
intersection or a component is one-dimensional, we denote it by $\circ$, which
refer to the node of the weight diagram. The columns of the table, except the
left one, are denoted by bold letters, and the rows except the upper one, are
denoted by bold numbers.

Here is the example: the branching table for $(E_{6},\varpi_{1})$, where
vertical lines correspond to cuting through the bonds marked with 1, and
horizontal lines correspond to cuting through the bonds marked with 6.

|  | a | b | c
---|---|---|---|---
| $E_{6},\varpi_{1}$ | $\circ$ | $D_{5},\varpi_{5}$ | $D_{5},\varpi_{1}$
1) | $\mathop{\mathrm{D}}\nolimits_{5},\varpi_{1}$ | $\circ$ | $D_{4},\varpi_{1}$ | $\circ$
2) | $D_{5},\varpi_{5}$ |  | $D_{4},\varpi_{4}$ | $D_{4},\varpi_{3}$
3) | $\circ$ |  |  | $\circ$

### ASR-condition

Recall that a commutative ring $R$ satisfies the absolute stable rank
condition $\mathrm{ASR}_{d}$ if for any row $(b_{1},\ldots,b_{d})$ with
coordinates in $R$, there exist elements $c_{1}$,$\ldots$,$c_{d-1}\in R$ such
that every maximal ideal of $R$ containing the ideal $\langle
b_{1}+c_{1}b_{d},\ldots,b_{d-1}+c_{d-1}b_{d}\rangle$ contains already the
ideal $\langle b_{1},\ldots,b_{d}\rangle$. This notion was introduced in [13]
and used in [42], [43] and then in [31],[32], [33], and [15] to study
stability problems.

If we assume that a row $(b_{1},\ldots,b_{d})$ is unimodular, then the
absolute stable rank condition boils down to the usual stable rank condition
$\mathrm{SR}_{d}$ (see [3],[57]).

Absolute stable rank satisfies the usual properties, namely for every ideal
$I\unlhd R$ condition $\mathrm{ASR}_{d}$ for $R$ implies $\mathrm{ASR}_{d}$
for the quotient $R/I$, and if $d\geqslant d^{\prime}$, then
$\mathrm{ASR}_{d^{\prime}}$ implies $\mathrm{ASR}_{d}$. Finally, it is well
known that if the maximal spectrum of $R$ is a Noetherian space of dimension
$\operatorname{Jdim}R=d-2$, then both conditions $\mathrm{ASR}_{d}$ and
$\mathrm{SR}_{d}$ are satisfied (see [13],[27],[42]).

##  3\. Absolute flexible stable rank

In this section, we recall the definition and the basic properties of absolute
flexible stable rank introduced in [16]. Here is the definition.

######  Definition 3.1.

A commutative ring $A$ satisfies the absolute flexible stable rank condition
$\mathrm{AFSR}_{d}$ if for any row $(b_{1},\ldots,b_{d})$ with coordinates in
$A$, there exists an element $c_{1}\in A$ such that for any invertible element
$\varepsilon_{1}\in A^{*}$, there exists $c_{2}\in A$ such that for any
$\varepsilon_{2}\in A^{*}$, $\ldots$, there exists $c_{d-1}\in A$ such that
for any $\varepsilon_{d-1}\in A^{*}$, every maximal ideal of $A$ containing
the ideal $\langle
b_{1}+\varepsilon_{1}c_{1}b_{d},\ldots,b_{d-1}+\varepsilon_{d-1}c_{d-1}b_{d}\rangle$
contains already the ideal $\langle b_{1},\ldots,b_{d}\rangle$.

One can think of it as follows. Two players are playing a game. Player 1
chooses a row $(b_{1},\ldots,b_{d})$ with coordinates in $A$. Then they take
turns starting with Player 2. Player 2 in his $i$-th turn chooses element
$c_{i}\in A$; after that Player 1 in his turn chooses invertible element
$\varepsilon_{i}\in A^{*}$. Player 2 wins if after $d$ turns every maximal
ideal of $A$ containing the ideal $\langle
b_{1}+\varepsilon_{1}c_{1}b_{d},\ldots,b_{d-1}+\varepsilon_{d-1}c_{d-1}b_{d}\rangle$
contains already the ideal $\langle b_{1},\ldots,b_{d}\rangle$. A commutative
ring $A$ satisfies the absolute flexible stable rank condition
$\mathrm{AFSR}_{d}$ if Player 2 has a winning strategy.

The following Lemma shows that the condition $\mathrm{AFSR}_{d}$ holds for
small-dimensional rings. That generalises the result of [13].

######  Lemma 3.2.

(Lemma 3.2 in [16]). Let $A$ be a commutative ring. Assume that
$\operatorname{Max}(A)$ is Noetherian and $\operatorname{Jdim}A\leqslant d-2$.
Then $A$ satisfies $\mathrm{AFSR}_{d}$.

Now the following lemma shows how one can use the $\mathrm{AFSR}$ condition.

######  Lemma 3.3.

(Lemma 3.3 in [16]) Let $A$ be a commutative ring, and $S$ be a multiplicative
system in $A$. Assume that the localisation $A[S^{-1}]$ satisfies
$\mathrm{AFSR}_{d}$. Then for any row $(b_{1},\ldots,b_{d})$ with coordinates
in $A[S^{-1}]$ and for any $s\in S$, there exist $c_{1}$,$\ldots$,$c_{d-1}\in
sA$ such that every maximal ideal of $A[S^{-1}]$ containing the ideal $\langle
b_{1}+c_{1}b_{d},\ldots,b_{d-1}+c_{d-1}b_{d}\rangle$ contains already the
ideal $\langle b_{1},\ldots,b_{d}\rangle$.

##  4\. Reduction of Theorem 1.1 to propositions 4.2, 4.3 and 4.4

In this section, we divide the proof of Theorem 1.1 into three steps. One of
the steps, namely Proposition 4.4, will be formulated and then proved also for
the case $E_{7}\leqslant E_{8}$, so that if a proof of Propositions 4.2 and
4.3 are found for this case, it will finish the proof of bounded reduction for
Chevalley groups of type $E$.

Recall that by $U_{1}$ we denote the unipotent radical that correspond to the
set $\Sigma_{1}\leqslant\Phi$, which is the special part of the parabolic
subset of roots $\Delta\cup\Sigma_{1}$.

Note that

$|\Sigma_{1}|=\begin{cases}16&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\
27&\text{ for }E_{6}\leqslant E_{7}\text{.}\end{cases}$

Therefore, theorem 1.1 follows trivially from the following result and Lemma
2.3.

######  Theorem 4.1.

Under the condition of Theorem 1.1, for every column
$b\in\operatorname{Um}^{\prime}_{\varpi}A$ there exists a column

$b^{\prime}\in E(\Phi,A)^{\leqslant N}b\text{,}$

where

$N=\begin{cases}36n^{2}+(72D+80)n+60&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 52n^{2}+(104D+249)n+190&\text{ for }E_{6}\leqslant
E_{7}\text{,}\end{cases}$

such that $b^{\prime}_{1}=1$.

Consider the lexicographic order on the monomials in variables
$x_{1}$,$\ldots$,$x_{n}$. That is the order where $x_{1}^{k_{1}}\ldots
x_{n}^{k_{n}}$ is bigger than $x_{1}^{l_{1}}\ldots x_{n}^{l_{n}}$ if for some
$m$ we have $k_{i}=l_{i}$ for $i<m$, and $k_{m}>l_{m}$. A polynomial in
$A=C[x_{1},\ldots,x_{n}]$ is called lexicographically monic if its leading
coefficient in lexicographic order is equal to one.

Further we reduce Theorem 4.1 to the following three propositions.

######  Proposition 4.2.

Under the condition of Theorem 1.1, assuming $n=0$ (i.e. $A=C$), for every
column $b\in\operatorname{Um}^{\prime}_{\varpi}A$ there exists a column

$b^{\prime}\in E(\Phi,A)^{\leqslant N}b\text{,}$

where

$N=\begin{cases}60&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\ 190&\text{ for
}E_{6}\leqslant E_{7}\text{.}\end{cases}$

such that $b^{\prime}_{1}=1$.

######  Proposition 4.3.

Let $j$ be a number of a vertex on a weight diagram of the representation
$\varpi$. Under the condition of Theorem 1.1, for every column
$b\in\operatorname{Um}^{\prime}_{\varpi}A$ there exists a column

$b^{\prime}\in E(\Phi,A)^{\leqslant N}b\text{,}$

where

$N=\begin{cases}116&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\ 301&\text{ for
}E_{6}\leqslant E_{7}\text{,}\end{cases}$

such that its entry $b^{\prime}_{j}$ is lexicographically monic.

We state and prove the third proposition also for the case $E_{7}\leqslant
E_{8}$.

######  Proposition 4.4.

Let $B$ be a commutative ring such that $\operatorname{BSdim}B=d<\infty$. Let
$A=B[y]$, $b\in\operatorname{Um}^{\prime}_{\varpi}A$ such that its entry
$b_{j}$ is monic, where $j=24$ for $E_{6}$, $j=-1$ for $E_{7}$ and $E_{8}$.
Then

$E(\Phi,A)^{\leqslant
N}b\cap\operatorname{Um}_{\varpi}(B)\neq\varnothing\text{,}$

where

$N=\begin{cases}72d&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\ 104d&\text{
for }E_{6}\leqslant E_{7}\text{,}\\\ 291d&\text{ for }E_{7}\leqslant
E_{8}\text{.}\end{cases}$

First we need the following lemma.

######  Lemma 4.5.

Let $f\in C[x_{1},\ldots,x_{n}]$ be a lexicographically monic polynomial. Then
there exists an invertible change of variables

$x_{1},\ldots,x_{n}\leftrightarrow y_{1},\ldots,y_{n}\text{,}$

such that $f$ becomes monic in $y_{n}$.

###### Proof.

Take $K>\deg f$. Set $x_{i}=y_{i}+y_{n}^{K^{n-i}}$, $i=1$,$\ldots$,$n-1$, and
$x_{n}=y_{n}$. ∎

Now we deduce Theorem 4.1 from Propositions 4.2, 4.3 and 4.4.

Take $b\in\operatorname{Um}^{\prime}_{\varpi}A$. By Proposition 4.3 there
exists a column

$b^{\prime}\in E(\Phi,A)^{\leqslant N_{1}^{\prime}}b\text{,}$

where

$N_{1}^{\prime}=\begin{cases}116&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\
301&\text{ for }E_{6}\leqslant E_{7}\text{,}\end{cases}$

such that its entry $b^{\prime}_{j}$ is lexicographically monic, where $j$ is
as in Proposition 4.4. Applying Lemma 4.5, we change variables to
$y_{1}$,$\ldots$,$y_{n}$ so that $b^{\prime}_{j}$ is now monic in $y_{n}$. Now
we apply Proposition 4.4 to $B=C[y_{1},\ldots,y_{n-1}]$. Note that
$\operatorname{BSdim}B\leqslant\dim B=D+n-1$. Hence we can obtain a column
from

$\displaystyle E(\Phi,A)^{\leqslant
N_{1}^{\prime\prime}}b^{\prime}\cap\operatorname{Um}_{\varpi}B\leqslant
E(\Phi,A)^{\leqslant N_{1}}b\cap\operatorname{Um}_{\varpi}B\text{,}$

where

$N_{1}^{\prime\prime}=\begin{cases}72(D+n-1)&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 104(D+n-1)&\text{ for }E_{6}\leqslant
E_{7}\text{,}\end{cases}$

and $N_{1}=N_{1}^{\prime}+N_{1}^{\prime\prime}$.

Repeating this argument $n$ times we can obtain a column from

$E(\Phi,A)^{\leqslant N_{n}}b\cap\operatorname{Um}_{\varpi}C\text{,}$

where

$\displaystyle N_{n}=\begin{cases}72n(2D+n-1)/2+116n&\text{ for
}D_{5}\leqslant E_{6}\text{,}\\\ 104n(2D+n-1)/2+301n&\text{ for
}E_{6}\leqslant E_{7}\text{.}\\\ \end{cases}\text{.}$

Now Proposition 4.2 implies that there exists

$b^{\prime\prime}\in E(\Phi,A)^{\leqslant N}b\text{,}$

where

$\displaystyle N=\begin{cases}72n(2D+n-1)/2+116n+60&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 104n(2D+n-1)/2+301n+190&\text{ for }E_{6}\leqslant
E_{7}\text{,}\end{cases}$

such that $b^{\prime\prime}_{1}=1$.

##  5\. Bounded reduction for low dimensional rings

In this Section, we prove Proposition 4.2. Note that the case $E_{6}$ easily
follows from the proof of Lemma 2 in [15]. Similarly, the proof for the case
for the case $E_{7}$ can be obtain from the proof of the main theorem in [33].
In order to do so, we must estimate how many elementary root element it takes
to apply Lemma 2 in [33]. That proof starts with picking an element $e\in
E(E_{6},R)$ such that $(ae)_{\lambda_{1}}\equiv 1\mod\mathfrak{u}$ and
$(ae)_{\lambda_{i}}\equiv 0\mod\mathfrak{u}$ for $i\neq 1$. Note that it is
enough to require $(ae)_{\lambda_{1}}$ to be invertible modulo $\mathfrak{u}$
and not necessarily congruent to 1. Therefore, the element $e$ can be taken
from $XU_{1}$ where

$X=\\{x_{-\delta_{E_{6}}}(\xi_{1})x_{-\delta_{A_{5}}}(\xi_{2})x_{-\delta_{D_{5}(6)}}(\xi_{3})x_{-\alpha_{1}}(\xi_{4})x_{-\delta_{D_{5}(1)}}(\xi_{5})x_{-\alpha_{2}-\alpha_{3}-\alpha_{4}}(\xi_{6})x_{-\alpha_{2}}(\xi_{7})x_{-\alpha_{3}}(\xi_{8})\\\
x_{-\alpha_{4}}(\xi_{9})x_{-\alpha_{5}}(\xi_{10})x_{-\alpha_{6}}(\xi_{11})\colon\xi_{i}\in
R\\}\text{,}$

where $\delta_{E_{6}}$ is the maximal root of the system $E_{6}$;
$\delta_{A_{5}}$ is the maximal root of the system generated by
$\alpha_{1}$,$\alpha_{3}$,$\alpha_{4}$,$\alpha_{5}$ and $\alpha_{6}$;
$\delta_{D_{5}(6)}$ is the maximal root of the system generated by
$\alpha_{1}$,$\alpha_{2}$,$\alpha_{3}$,$\alpha_{4}$ and $\alpha_{5}$;
$\delta_{D_{5}(1)}$ is the maximal root of the system generated by
$\alpha_{2}$,$\alpha_{3}$,$\alpha_{4}$,$\alpha_{5}$ and $\alpha_{6}$.

The next step in the proof uses the element $e_{1}\in U_{1}$, hence $ee_{1}\in
XU_{1}\leqslant E^{\leqslant 27}(E_{6},R)$. Now it is easy to count that the
proof of Lemma 2 takes 67 elementary root elements, and the whole proof of the
main theorem in [33] takes 190 elementary root elements.

##  6\. Obtaining a monic polynomial

In this section, we give the proof of Proposition 4.3. First we need some
preparation.

The following Lemma was proved in [16].

######  Lemma 6.1.

(Lemma 5.2 in [16]). Let $C$ be a Noetherian ring, $A=C[x_{1},\ldots,x_{n}]$.
Let $S$ be a multiplicative system of lexicographically monic polynomials in
$A$. Then we have $\dim A[S^{-1}]\leqslant\dim C$.

Now we recall a definition from [58].

######  Definition 6.2.

Let $A$ be an associative ring with 1, $s$ be a central element of $A$,
$l\geqslant 2$, $v\in A^{l-1}$ (a column over $A$), $u\in\vphantom{,}^{l-1}A$
(a row over $A$). We define an $l$ by $l$ matrix over $A$ by

$\mu(u,s,v)=\begin{pmatrix}1_{l-1}+vsu&vs^{2}\\\
-uvu&1-uvs\end{pmatrix}\text{.}$

This matrix is invertible with $\mu(u,s,v)^{-1}=\mu(u,s,-v)$. If $s\in A^{*}$,
then

$\mu(u,s,v)=\begin{pmatrix}1_{l-1}&0\\\
-u/s&1\end{pmatrix}\begin{pmatrix}1_{l-1}&vs^{2}\\\
0&1\end{pmatrix}\begin{pmatrix}1_{l-1}&0\\\ u/s&1\end{pmatrix}$

The following lemma was proved in [58].

######  Lemma 6.3.

(Lemma 2.2 in [58]) When $l\geqslant 3$, the matrix $\mu(u,s,v)$ is a product
of $7l-3$ elementary transvections in $\mathop{\mathrm{GL}}\nolimits(l,R)$.

Now let $S$ be a multiplicative system of lexicographically monic polynomials
in $A$. It follows from Lemmas 6.1 and 3.2 that the ring $A[S^{-1}]$ satisfies
$\mathrm{AFSR}_{5}$ for the case $E_{6}$ resp. $\mathrm{AFSR}_{6}$ for the
case $E_{7}$, and so does any quotient of $A[S^{-1}]$.

######  Lemma 6.4.

Let $l\geqslant 3$, let $\mathfrak{I}$ be an ideal in $A$, and suppose that
$\dim\operatorname{Max}A/\mathfrak{I}[S^{-1}]\leqslant l-2$. Let $b\in
V_{D_{l},\varpi_{1}}A$ be such that it becomes unimodular in
$A/\mathfrak{I}[S^{-1}]$. Then there exists a column

$b^{\prime}\in E(D_{l},A)^{\leqslant 11l-7}b\text{,}$

such that $b_{1}^{\prime}$ is congruent to a lexicographically monic
polynomial modulo $\mathfrak{I}$.

###### Proof.

We perform the following steps.

Figure 1. $(D_{l},\varpi_{1})$

Step 1. Make the row $(b_{2},\ldots,b_{-1})$ unimodular in
$A/\mathfrak{I}[S^{-1}]$ by $l-1$ elementary elements.

Let $\mathfrak{A}=\mathfrak{I}+\langle b_{-l},\ldots,b_{-1}\rangle\unlhd
A[S^{-1}]$. Since $A[S^{-1}]/\mathfrak{A}$ satisfies $\mathrm{AFSR}_{l}$ and
the row $(b_{1},\ldots,b_{l})$ is unimodular in $A[S^{-1}]/\mathfrak{A}$, it
follows from Lemma 3.3 that there exist $c_{2}$,$\ldots$,$c_{l}\in A$ such
that the row $(b_{2}+c_{2}b_{1},\ldots,b_{l}+c_{l}b_{1})$ is unimodular in
$A[S^{-1}]/\mathfrak{A}$. Thus by applying the elements
$x_{-\alpha_{1}-\ldots-\alpha_{i-1}}(\pm c_{i})$ for $i=2$,$\ldots$,$l$, we
make the row $(b_{2},\ldots,b_{l})$ unimodular in $A[S^{-1}]/\mathfrak{A}$
without changing the ideal $\mathfrak{A}$. Thus the row
$(b_{2},\ldots,b_{-1})$ becomes unimodular in $A/\mathfrak{I}[S^{-1}]$.

Step 2. Make the row $(b_{1},b_{-l}\ldots,b_{-1})$ unimodular in
$A/\mathfrak{I}[S^{-1}]$ by $l-1$ elementary elements.

Since the row $(b_{2},\ldots,b_{-1})$ unimodular in $A/\mathfrak{I}[S^{-1}]$,
it follows that the ideal generated by $\mathfrak{I}$ and
$(b_{2},\ldots,b_{-1})$ in $A$ contains a lexicographically monic polynomial.
So for some $f_{2}$,$\ldots$,$f_{-1}\in A$, and $f\in\mathfrak{I}$ the
polynomial

$f+\sum_{i=2}^{-1}f_{i}b_{i}$

is lexicographically monic. Multiplying polynomials $f$ and $f_{i}$ by a large
enough power of $x_{1}$, we may assume that the polynomial

$b_{1}+f+\sum_{i=2}^{-1}f_{i}b_{i}$

is also lexicographically monic.

Let us now apply the elements $x_{\alpha_{1}+\ldots+\alpha_{i-1}}(f_{i})$ for
$i=2$,$\ldots$,$l$. Then the ideal generated by $\mathfrak{I}$, the new
$b_{1}$, and old $b_{-l}$,$\ldots$,$b_{-1}$ contains a lexicographically monic
polynomial. However, these elements do not change the ideal generated by
$b_{-l}$,$\ldots$,$b_{-1}$. Hence we actually achieve that the ideal generated
by $\mathfrak{I}$, and new $b_{1}$,$b_{-l}$,$\ldots$,$b_{-1}$ contains a
lexicographically monic polynomial. Thus the row $(b_{1},b_{-l}\ldots,b_{-1})$
becomes unimodular in $A/\mathfrak{I}[S^{-1}]$.

Step 3. Make the row $(b_{1},b_{-l}\ldots,b_{-2})$ unimodular in $A[S^{-1}]$
by $7l-3$ elementary elements.

Let $\mathfrak{A}=\mathfrak{I}+\langle b_{-l},\ldots,b_{-1}\rangle\unlhd
A[S^{-1}]$. Since $b_{1}$ is invertible in $A[S^{-1}]/\mathfrak{A}$, it
follows that there exist $\xi_{2}$,$\ldots$,$\xi_{l}\in A[S^{-1}]$ such that
$b_{i}-\xi_{i}b_{1}\in\mathfrak{A}$ for $i=2,\ldots,l$. Let $s$ be a common
denominator of $\xi_{i}$. Set

$g_{1}=\prod_{2\leqslant i\leqslant
l}x_{-\alpha_{1}-\ldots-\alpha_{i-1}}(\pm\xi_{i})\text{,}$

where signs are such that $(g_{1}b)_{i}=b_{i}-\xi_{i}b_{1}\in\mathfrak{A}$ for
$1\leqslant i\leqslant l$.

Since $A/\mathfrak{I}[S^{-1}]$ satisfies $\mathrm{AFSR}_{l}$, it follows from
Lemma 3.3 that there exist $c_{-l}$,$\ldots$,$c_{-2}\in s^{2}A$ such that
every maximal ideal of $A/\mathfrak{I}[S^{-1}]$ containing the ideal
$\langle(g_{1}b)_{-l}+c_{-l}(g_{1}b)_{-1},\ldots(g_{1}b)_{-2}+c_{-2}(g_{1}b)_{-1}\rangle$
contains already the ideal
$\langle(g_{1}b)_{-l},\ldots,(g_{1}b)_{-1}\rangle=\mathfrak{A}$. Set

$g_{2}=\prod_{2\leqslant i\leqslant l}x_{\alpha_{1}+\ldots+\alpha_{i-1}}(\pm
c_{-i})\text{,}$

where signs are such that $(g_{2}g_{1}b)_{i}=(g_{1}b)_{i}+c_{i}b_{-1}$ for
$-l\leqslant i\leqslant-2$.

We claim that the elements
$(g_{2}g_{1}b)_{1}$,$(g_{2}g_{1}b)_{-l}$,$\ldots$,$(g_{2}g_{1}b)_{-2}$
generate the unit ideal in $A/\mathfrak{I}[S^{-1}]$. Let us prove that.

Assume that some maximal ideal $\mathfrak{M}$ of the ring $A[S^{-1}]$ contains
$\mathfrak{I}$ and all the elements
$(g_{2}g_{1}b)_{1}$,$(g_{2}g_{1}b)_{-l}$,$\ldots$,$(g_{2}g_{1}b)_{-2}$

Since applying $g_{1}$ does not change the ideal generated by
$b_{-l}$,$\ldots$,$b_{-1}$, by choice of $c_{i}$ we have
$\mathfrak{A}\leqslant\mathfrak{M}$. Hence $(g_{1}b)_{i}\in\mathfrak{M}$ for
$2\leqslant i\leqslant l$. Thus $b_{1}=(g_{2}g_{1}b)_{1}+\sum_{2\leqslant
i\leqslant 7}\pm c_{-i}(g_{1}b)_{i}\in\mathfrak{M}$. However, by previous
step, $b_{1}$ and $\mathfrak{A}$ generate a unit ideal. This is a
contradiction.

Since applying $g_{1}^{-1}$ does not change the ideal generated by elements
$b_{1}$, $b_{-l}$,$\ldots$,$b_{-2}$, we obtain that the elements
$(g_{1}^{-1}g_{2}g_{1}b)_{i}$, where $i=1,-l,\ldots,-2$, generate the unit
ideal in $A[S^{-1}]$.

It remains to notice that the element $g_{1}^{-1}g_{2}g_{1}$ is the image of
the matrix $\mu(u,s,v)$ for certain $u$ and $v$ under the embedding
$G(A_{l-1},A)\to G(D_{l},A)$ as a subsystem subgroup. Therefore, by Lemma 6.3,
$g_{1}^{-1}g_{2}g_{1}\in E(D_{l},A)^{\leqslant 7l-3}$.

Step 4. Make the row $(b_{-l}\ldots,b_{-2})$ unimodular in
$A/\mathfrak{I}[S^{-1}]$ by $l-1$ elementary elements.

Since $A/\mathfrak{I}[S^{-1}]$ satisfies $\mathrm{AFSR}_{l}$ and the row
$(b_{1},b_{-l}\ldots,b_{-2})$ is unimodular in $A/\mathfrak{I}[S^{-1}]$, it
follows from Lemma 3.3 that there exist $c_{-l}$,$\ldots$,$c_{-2}\in A$ such
that the row $(b_{-l}+c_{-l}b_{1},\ldots,b_{-2}+c_{-2}b_{1})$ is unimodular in
$A/\mathfrak{I}[S^{-1}]$. Thus by applying the elements
$x_{\alpha_{2}+\ldots+\alpha_{i-1}-\delta}(\pm c_{-i})$ for
$i=2$,$\ldots$,$l$, we make the row $(b_{-l}\ldots,b_{-2})$ unimodular in
$A/\mathfrak{I}[S^{-1}]$.

Step 5. Make $b_{1}$ congruent to a lexicographically monic polynomial modulo
$\mathfrak{I}$ by $l-1$ elementary elements.

Since the row $(b_{-l}\ldots,b_{-2})$ is unimodular in
$A/\mathfrak{I}[S^{-1}]$, it follows that the ideal generated by
$\mathfrak{I}$ and $(b_{-l},\ldots,b_{-2})$ in $A$ contains a
lexicographically monic polynomial. So for some
$f_{1}$,$f_{-l}$,$\ldots$,$f_{-3}\in A$ and $f\in\mathfrak{I}$, the polynomial

$f+\sum_{-l\leqslant i\leqslant-2}f_{i}b_{i}$

is lexicographically monic. Multiplying polynomials $f_{i}$ and $f$ by a large
enough power of $x_{1}$, we may assume that the polynomial

$b_{-2}+f+\sum_{-l\leqslant i\leqslant-2}f_{i}b_{i}$

is also lexicographically monic.

Now applying the elements $x_{\delta-\alpha_{2}-\ldots-\alpha_{i-1}}(\pm
f_{-i})$ for $i=2$,$\ldots$,$l$, we achieve that $b_{1}$ is congruent to a
lexicographically monic polynomial modulo $\mathfrak{I}$. ∎

######  Remark 6.5.

One can notice that the proof above repeats the proof of Proposition 4.2 in
[16], which on its turn basically repeats the proof of stability theorem for
$K_{1}$-functor given in [42].

Now we prove Proposition 4.3 for the case $(E_{6},\varpi_{1})$.

###### Proof.

Consider the branching table for $(E_{6},\varpi_{1})$, where vertical lines
correspond to cuting through the bonds marked with 1, and horizontal lines
correspond to cuting through the bonds marked with 6.

|  | a | b | c
---|---|---|---|---
| $E_{6},\varpi_{1}$ | $\circ$ | $D_{5},\varpi_{5}$ | $D_{5},\varpi_{1}$
1) | $\mathop{\mathrm{D}}\nolimits_{5},\varpi_{1}$ | $\circ$ | $D_{4},\varpi_{1}$ | $\circ$
2) | $D_{5},\varpi_{5}$ |  | $D_{4},\varpi_{4}$ | $D_{4},\varpi_{3}$
3) | $\circ$ |  |  | $\circ$

Take $b\in\operatorname{Um}^{\prime}_{(E_{6},\varpi_{1})}A$. We need to obtain
a lexicographically monic polynomial by 116 elementary elements. Since the
Weyl group acts transitively on weights, it does not matter in which position
to obtain a lexicographically monic polynomial. Let us make it with $b_{1}$.
We perform the following steps.

Figure 2. $(E_{6},\varpi_{1})$

Step 1. Make the row that consists of elements in all the cells except a1
unimodular in $A[S^{-1}]$ by 4 elementary elements.

Let $\mathfrak{A}\unlhd A[S^{-1}]$ be the ideal generated by all the elements
$b_{i}$ except for $b_{1}$,$\ldots$,$b_{4}$,$b_{6}$. Since
$A[S^{-1}]/\mathfrak{A}$ satisfies $\mathrm{AFSR}_{5}$, and the row
$(b_{1},\ldots,b_{4},b_{6})$ is unimodular in $A[S^{-1}]/\mathfrak{A}$, it
follows from Lemma 3.3 that there exist $c_{2}$,$c_{3}$,$c_{4}$,$c_{6}\in A$
such that he row $(b_{2}+c_{2}b_{1},\ldots,b_{6}+c_{6}b_{1})$ is unimodular in
$A[S^{-1}]/\mathfrak{A}$. Thus by applying the elements $x_{\alpha_{1}}(\pm
c_{2})$,$\ldots$,$x_{\alpha_{1}+\alpha_{3}+\alpha_{4}+\alpha_{5}}(\pm c_{6})$,
we make the row $(b_{2},b_{3},b_{4},b_{6})$ unimodular in
$A[S^{-1}]/\mathfrak{A}$ without changing the ideal $\mathfrak{A}$. Thus the
row that consists of elements in all the cells except a1 becomes unimodular in
$A[S^{-1}]$.

Step 2. Make the row that consists of elements in columns a and c unimodular
in $A[S^{-1}]$ by 16 elementary elements.

Since the row $(b_{2},\ldots,b_{27})$ unimodular in $A[S^{-1}]$, it follows
that the ideal generated by $(b_{2},\ldots,b_{27})$ in $A$ contains a
lexicographically monic polynomial. So for some $f_{2}$,$\ldots$,$f_{27}\in
A$, the polynomial

$\sum_{i=2}^{27}f_{i}b_{i}$

is lexicographically monic. Multiplying polynomials $f_{i}$ by a large enough
power of $x_{1}$, we may assume that the polynomial

$b_{1}+\sum_{i=2}^{27}f_{i}b_{i}$

is also lexicographically monic.

Let us now apply the elements $x_{\lambda_{1}-\lambda_{i}}(\pm f_{i})$ for all
$\lambda_{i}$ from the column b. Then the ideal generated by the new $b_{1}$
and old $b_{\lambda}$, where $\lambda$ is from the column c, contains a
lexicographically monic polynomial. However, these elements do not change the
ideal generated by $b_{\lambda}$, where $\lambda$ is from the column c. Hence
we actually achieve that the ideal generated by new $b_{1}$, and
$b_{\lambda}$, where $\lambda$ is from the column c contains a
lexicographically monic polynomial. Thus the row of elements in columns a and
c becomes unimodular in $A[S^{-1}]$.

Step 3. Make the row that consists of elements in cells a1 and c1 unimodular
in $A[S^{-1}]$ by 48 elementary elements.

Apply Lemma 6.4 to the column c and the ideal generated by $b_{1}$.

Step 4. Make the element $b_{1}$ lexicographically monic by 48 elementary
elements.

Apply Lemma 6.4 to the row 1 and the zero ideal.

∎

######  Remark 6.6.

One can notice that the proof above basically repeats the proof of stability
theorem for $K_{1}$-functor given in [15].

Before we prove Proposition 4.3 for $(E_{7},\varpi_{7})$, we need one more
Lemma.

######  Lemma 6.7.

Under the condition of Theorem 1.1 in case $E_{6}\leqslant E_{7}$. Let
$\mathfrak{I}$ be an ideal in $A$. Let
$b\in\operatorname{Orb}_{E_{6},\varpi_{1}}A/\mathfrak{I}$ be such that it
becomes unimodular in $A/\mathfrak{I}[S^{-1}]$. Then there exists a column
vector

$b^{\prime}\in E(E_{6},A)^{\leqslant 91}b\text{,}$

such that the row $(b_{1}^{\prime},b_{18}^{\prime})$ is unimodular in
$A/\mathfrak{I}[S^{-1}]$.

###### Proof.

Let us choose in each irreducible component of
$\operatorname{Max}(A/\mathfrak{I}[S^{-1}])$ a maximal ideal
$\mathfrak{u}_{i}$, $i\in I$. Next denote by $\widetilde{\mathfrak{u}_{i}}$
the preimage of $\mathfrak{u}_{i}$ in $A$.

For each $i$ choose a maximal ideal $\mathfrak{v}_{i}\in\operatorname{Max}A$
such that it contains $\widetilde{\mathfrak{u}_{i}}$, and $b$ is unimodular in
$A/\mathfrak{v}_{i}$. Let us show that we can do it. The column $b$ is
unimodular in $A/\mathfrak{I}[S^{-1}]$; hence the ideal in $A$ generated by
$\mathfrak{I}$ and entries of $b$ contains a lexicographically monic
polynomial $f$. Clearly, the ideal $\widetilde{u_{i}}$ is prime and
$f\notin\widetilde{\mathfrak{u}_{i}}$. Since $C$ is a Jacobson ring and $A$ is
finitely generated over $C$, it follows that $A$ is a Jacobson ring; hence
there exists $\mathfrak{v}_{i}\in\operatorname{Max}A$ such that
$\widetilde{\mathfrak{u}_{i}}\leqslant\mathfrak{v}_{i}$ and
$f\notin\mathfrak{v}_{i}$. Then $b$ is unimodular in $A/\mathfrak{v}_{i}$.

Set $I_{1}=\\{i\in I\colon\mathfrak{v}_{i}=\widetilde{\mathfrak{u}_{i}}\\}$,
and $I_{2}=\\{i\in
I\colon\mathfrak{v}_{i}\neq\widetilde{\mathfrak{u}_{i}}\\}$, so $I=I_{1}\sqcup
I_{2}$.

Now we perform the following steps.

Step 1. Achive that $b_{1}\notin\bigcup_{i\in I}\mathfrak{v}_{i}$ by 11
elementary elements.

In other words, we should make $b_{1}$ invertible in $A/(\bigcap_{i\in
I}\mathfrak{v}_{i})$. The ring $A/(\bigcap_{i\in I}\mathfrak{v}_{i})$ is
semilocal, so we can do it in the same way we did in Section 5.

Step 2. Without changing $b_{1}$, make the row $(b_{2},\ldots,b_{27})$ is
unimodular in $A/\mathfrak{I}[S^{-1}]$, and achive that elements
$b_{5}$,$b_{7}$,$b_{8}$,$b_{9}$, $b_{11}$,$\ldots$,$b_{27}$ belong to
$\bigcap_{i\in I_{1}}\mathfrak{v}_{i}$ by element from $U_{1}$, i.e by 16
elementary elements.

It follows from Lemma 2.2 that for some $b^{\prime}=u_{1}b$, where $u_{1}\in
U_{1}$. we have $b^{\prime}_{\lambda}\in\bigcap_{i\in I_{1}}\mathfrak{v}_{i}$
for all $\lambda\neq\lambda_{1}$. Now let $\mathfrak{A}\unlhd A[S^{-1}]$ be
the ideal generated by $\mathfrak{I}$ and the elements
$b^{\prime}_{5}$,$b^{\prime}_{7}$,$b^{\prime}_{8}$,$b^{\prime}_{9}$,
$b^{\prime}_{11}$,$\ldots$,$b^{\prime}_{27}$ except for
$b^{\prime}_{1}$,$\ldots$,$b^{\prime}_{4}$,$b^{\prime}_{6}$,$b^{\prime}_{10}$.
Then $\mathfrak{A}\leqslant\bigcap_{i\in I_{1}}\mathfrak{v}_{i}$. Since
$A[S^{-1}]/\mathfrak{A}$ satisfies $\mathrm{AFSR}_{6}$, and the row
$(b^{\prime}_{1},\ldots,b^{\prime}_{4},b^{\prime}_{6},b^{\prime}_{10})$ is
unimodular in $A[S^{-1}]/\mathfrak{A}$, it follows from Lemma 3.3 that there
exist $c_{2}$,$c_{3}$,$c_{4}$,$c_{6}$,$c_{10}\in A$ such that he row
$(b^{\prime}_{2}+c_{2}b^{\prime}_{1},\ldots,b^{\prime}_{10}+c_{10}b^{\prime}_{1})$
is unimodular in $A[S^{-1}]/\mathfrak{A}$. Thus by applying the elements
$x_{\alpha_{1}}(\pm
c_{2})$,$\ldots$,$x_{\alpha_{1}+\alpha_{3}+\ldots+\alpha_{6}}(\pm c_{10})$, we
make the row
$(b^{\prime}_{2},b^{\prime}_{3},b^{\prime}_{4},b^{\prime}_{6},b^{\prime}_{10})$
unimodular in $A[S^{-1}]/\mathfrak{A}$ without changing the ideal
$\mathfrak{A}$. Thus the row $(b^{\prime}_{2},\ldots,b^{\prime}_{27})$ becomes
unimodular in $A/\mathfrak{I}[S^{-1}]$.

The composition of $u_{1}$ with the elements as above is then the required
element of $U_{1}$.

Step 3. Preserving the fact that the image of $b_{1}$ in
$A/\mathfrak{I}[S^{-1}]$ does not belong to $\bigcup_{i\in
I}\mathfrak{u}_{i}$, make the row that consists of elements in columns a and c
(we use the same branching table as in the proof above) unimodular in
$A/\mathfrak{I}[S^{-1}]$ by 16 elementary elements.

Since the row $(b_{2},\ldots,b_{27})$ unimodular in $A/\mathfrak{I}[S^{-1}]$,
it follows that the ideal generated by $\mathfrak{I}$ and elements
$b_{2}$,$\ldots$,$b_{27}$ in $A$ contains a lexicographically monic
polynomial. So for some $f_{2}$,$\ldots$,$f_{27}\in A$ and $f\in I$, the
polynomial

$f+\sum_{k=2}^{27}f_{k}b_{k}$

is lexicographically monic. Clearly for any $i\in I_{2}$ the ideal
$\mathfrak{v}_{i}$ contains some $h_{i}\in S$. Multiplying polynomials $f$ and
$f_{k}$ by $\prod_{i\in I_{2}}h_{i}$ and then by a large enough power of
$x_{1}$, we may assume that the, firstly all the $f_{k}$ belong to
$\bigcap_{i\in I_{2}}\mathfrak{v}_{i}$, and secondly, that the polynomial

$b_{1}+f+\sum_{i=2}^{27}f_{i}b_{i}$

is lexicographically monic.

Let us now apply the elements $x_{\lambda_{1}-\lambda_{i}}(\pm f_{i})$ for all
$\lambda_{i}$ from the column b. Clearly we preserve the fact that
$b_{1}\notin\bigcup_{i\in I_{2}}\mathfrak{v}_{i}$; hence we preserve the fact
that the image of $b_{1}$ in $A/\mathfrak{I}[S^{-1}]$ does not belong to
$\bigcup_{i\in I_{2}}\mathfrak{u}_{i}$. Further the ideal generated by
$\mathfrak{I}$, the new $b_{1}$, and old $b_{\lambda}$, where $\lambda$ is
from the column c, contains a lexicographically monic polynomial. However,
these elements do not change the ideal generated by $b_{\lambda}$, where
$\lambda$ is from the column c. Hence we actually achieve that the ideal
generated by new $b_{1}$, and $b_{\lambda}$, where $\lambda$ is from the
column c contains a lexicographically monic polynomial. Thus the row of
elements in columns a and c becomes unimodular in $A[S^{-1}]$. In addition,
the ideal generated by $b_{\lambda}$, where $\lambda$ is from the column c is
contained in $\bigcap_{i\in I_{1}}\mathfrak{v}_{i}=\bigcap_{i\in
I_{1}}\widetilde{u_{i}}_{i}$, because we make it so at step 2, and it was not
changed. Hence the image of new $b_{1}$ in $A/\mathfrak{I}[S^{-1}]$ does not
belong to $\bigcup_{i\in I_{2}}\mathfrak{u}_{i}$.

Step 4. Make the row $(b_{1},b_{18})$ unimodular in $A/\mathfrak{I}[S^{-1}]$
by 48 elementary elements.

Since the image of $b_{1}$ in $A/\mathfrak{I}[S^{-1}]$ does not belong to
$\bigcup_{i\in I}\mathfrak{u}_{i}$, it follows that
$\dim\operatorname{Max}A/(\mathfrak{I}+\langle
b_{1}\rangle)[S^{-1}]\leqslant\dim\operatorname{Max}A/\mathfrak{I}[S^{-1}]-1\leqslant
3$. Therefore, we can apply Lemma 6.4 to the column c and the ideal
$\mathfrak{I}+\langle b_{1}\rangle$. ∎

######  Remark 6.8.

One can notice that the proof above basically repeats the proof of Lemma 2 in
[33].

######  Remark 6.9.

The Lemma above is the only place, where we use the assumption that $C$ is a
Jacobson ring. It is easy to see that this assumption can be lifted, if we
assume that $\dim C\leqslant 3$.

Now we prove Proposition 4.3 for the case $(E_{7},\varpi_{7})$.

###### Proof.

Consider the branching table for $(E_{7},\varpi_{7})$, where vertical lines
correspond to cuting through the bonds marked with 1, and horizontal lines
correspond to cuting through the bonds marked with 7.

|  | a | b | c
---|---|---|---|---
| $E_{7},\varpi_{7}$ | $D_{6},\varpi_{1}$ | $D_{6},\varpi_{6}$ | $D_{6},\varpi_{1}$
1) | $\circ$ | $\circ$ |  |
2) | $E_{6},\varpi_{6}$ | $D_{5},\varpi_{1}$ | $D_{5},\varpi_{5}$ | $\circ$
3) | $E_{6},\varpi_{1}$ | $\circ$ | $D_{5},\varpi_{4}$ | $D_{5},\varpi_{1}$
4) | $\circ$ |  |  | $\circ$

Take $b\in\operatorname{Um}^{\prime}_{(E_{7},\varpi_{7})}A$. We need to obtain
a lexicographically monic polynomial by 301 elementary elements. Since the
Weyl group acts transitively on weights, it does not matter in which position
to obtain a lexicographically monic polynomial. Let us make it with $b_{1}$.
We perform the following steps.

Figure 3. $(E_{7},\varpi_{7})$

Step 1. Make the row that consists of elements in all the cells except a1
unimodular in $A[S^{-1}]$ by 5 elementary elements.

Let $\mathfrak{A}\unlhd A[S^{-1}]$ be the ideal generated by all the elements
$b_{i}$ except for $b_{1}$,$\ldots$,$b_{5}$,$b_{7}$. Since
$A[S^{-1}]/\mathfrak{A}$ satisfies $\mathrm{AFSR}_{6}$, and the row
$(b_{1},\ldots,b_{5},b_{7})$ is unimodular in $A[S^{-1}]/\mathfrak{A}$, it
follows from Lemma 3.3 that there exist $c_{2}$,$\ldots$,$c_{5}$,$c_{7}\in A$
such that he row $(b_{2}+c_{2}b_{1},\ldots,b_{7}+c_{7}b_{1})$ is unimodular in
$A[S^{-1}]/\mathfrak{A}$. Thus by applying the elements $x_{\alpha_{7}}(\pm
c_{2})$,$\ldots$,$x_{\alpha_{7}+\ldots+\alpha_{3}}(\pm c_{7})$, we make the
row $(b_{2},\ldots,b_{5},b_{7})$ unimodular in $A[S^{-1}]/\mathfrak{A}$
without changing the ideal $\mathfrak{A}$. Thus the row that consists of
elements in all the cells except a1 becomes unimodular in $A[S^{-1}]$.

Step 2. Make the row that consists of elements in the rows 1, 3, and 4
unimodular in $A[S^{-1}]$ by 27 elementary elements.

Since the row $(b_{2},\ldots,b_{-1})$ unimodular in $A[S^{-1}]$, it follows
that the ideal generated by $(b_{2},\ldots,b_{-1})$ in $A$ contains a
lexicographically monic polynomial. So for some $f_{2}$,$\ldots$,$f_{-1}\in
A$, the polynomial

$\sum_{i=2}^{-1}f_{i}b_{i}$

is lexicographically monic. Multiplying polynomials $f_{i}$ by a large enough
power of $x_{1}$, we may assume that the polynomial

$b_{1}+\sum_{i=2}^{-1}f_{i}b_{i}$

is also lexicographically monic.

Let us now apply the elements $x_{\lambda_{1}-\lambda_{i}}(\pm f_{i})$ for all
$\lambda_{i}$ from the row 2. Then the ideal generated by the new $b_{1}$ and
old $b_{\lambda}$, where $\lambda$ is from the rows 3 and 4, contains a
lexicographically monic polynomial. However, these elements do not change the
ideal generated by $b_{\lambda}$, where $\lambda$ is from the rows 3 and 4.
Hence we actually achieve that the ideal generated by new $b_{1}$, and
$b_{\lambda}$, where $\lambda$ is from the rows 3 and 4 contains a
lexicographically monic polynomial. Thus the row of elements in the rows 1, 3,
and 4 becomes unimodular in $A[S^{-1}]$.

Step 3. Make the row that consists of elements in the cells a1, b3, c3, and c4
unimodular in $A[S^{-1}]$ by 5 elementary elements.

Let $\mathfrak{A}\unlhd A[S^{-1}]$ be the ideal generated by all the elements
$b_{\lambda}$ for $\lambda$ in rows 1, 3, and 4, except for
$b_{-28}$,$\ldots$,$b_{-23}$. Since $A[S^{-1}]/\mathfrak{A}$ satisfies
$\mathrm{AFSR}_{6}$, and the row $(b_{-28},\ldots,b_{-23})$ is unimodular in
$A[S^{-1}]/\mathfrak{A}$, it follows from Lemma 3.3 that there exist
$c_{-27}$,$\ldots$,$c_{-23}\in A$ such that he row
$(b_{-27}+c_{-27}b_{-28},\ldots,b_{-23}+c_{-23}b_{-28})$ is unimodular in
$A[S^{-1}]/\mathfrak{A}$. Thus by applying the elements $x_{-\alpha_{1}}(\pm
c_{-27})$,$\ldots$,$x_{-\alpha_{1}-\alpha_{3}-\ldots-\alpha_{6}}(\pm
c_{-23})$, we make the row $(b_{-27},\ldots,b_{-23})$ unimodular in
$A[S^{-1}]/\mathfrak{A}$ without changing the ideal $\mathfrak{A}$. Thus the
row that consists of elements in the cells a1, b3, c3, and c4 becomes
unimodular in $A[S^{-1}]$.

Step 4. Make the row that consists of elements in the cells a1, a3, c3, and c4
unimodular in $A[S^{-1}]$ by 16 elementary elements.

Let $\Gamma$ be the set of weights in cells a1, b3, c3, and c4 Since the row
that consists of elements $\\{b_{\lambda}\colon\lambda\in\Gamma\\}$ is
unimodular in $A[S^{-1}]$, it follows that the ideal generated by
$\\{b_{\lambda}\colon\lambda\in\Gamma\\}$ in $A$ contains a lexicographically
monic polynomial. So for some $f_{\lambda}\in A$, where $\lambda\in\Gamma$ the
polynomial

$\sum_{\lambda\in\Gamma}f_{\lambda}b_{\lambda}$

is lexicographically monic. Multiplying polynomials $f_{\lambda}$ by a large
enough power of $x_{1}$, we may assume that the polynomial

$b_{-28}+\sum_{\lambda\in\Gamma}f_{\lambda}b_{\lambda}$

is also lexicographically monic.

Let us now apply the elements $x_{\lambda_{-28}-\lambda}(\pm f_{\lambda})$ for
all $\lambda_{i}$ from the cell b3. Then the ideal generated by the new
$b_{-28}$ and old $b_{\lambda}$, where $\lambda$ is from the cells a1, c3, and
c4, contains a lexicographically monic polynomial. However, these elements do
not change the ideal generated by $b_{\lambda}$, where $\lambda$ is from the
cells a1, c3, and c4. Hence we actually achieve that the ideal generated by
new $b_{1}$, and $b_{\lambda}$, where $\lambda$ is from the cells a1, c3, and
c4, contains a lexicographically monic polynomial. Thus the row of elements in
in the cells a1, a3, c3, and c4 becomes unimodular in $A[S^{-1}]$.

Step 5. Make the row that consists of elements in the cells a1, a2, a3, and c2
unimodular in $A[S^{-1}]$ by 59 elementary elements.

Apply Lemma 6.4 to the column c and the ideal generated by elements from the
column a.

Step 6. Make the row that consists of elements in the cells a1, a2, and c2
unimodular in $A[S^{-1}]$ by 39 elementary elements.

Let $\mathfrak{A}\unlhd A[S^{-1}]$ be the ideal generated by elements from
column a. Since $b_{28}$ is invertible in $A[S^{-1}]/\mathfrak{A}$, it follows
that there exist $\xi_{-7}$,$\ldots$,$\xi_{-11}\in A[S^{-1}]$ such that
$b_{i}-\xi_{i}b_{28}\in\mathfrak{A}$ for $i=-7$,$\ldots$,$-11$. Let $s$ be a
common denominator of $\xi_{i}$. Set

$g_{1}=\prod_{-11\leqslant
i\leqslant-7}x_{\lambda_{i}-\lambda_{28}}(\pm\xi_{i})\text{,}$

where signs are such that $(g_{1}b)_{i}=b_{i}-\xi_{i}b_{28}\in\mathfrak{A}$
for $-11\leqslant i\leqslant-7$.

Since $A[S^{-1}]$ satisfies $\mathrm{AFSR}_{6}$, it follows from Lemma 3.3
that there exist $c_{7}$,$\ldots$,$c_{11}\in s^{2}A$ such that every maximal
ideal of $A[S^{-1}]$ containing the ideal
$\langle(g_{1}b)_{7}+c_{7}(g_{1}b)_{-28},\ldots(g_{1}b)_{11}+c_{11}(g_{1}b)_{-28}\rangle$
contains already the ideal
$\langle(g_{1}b)_{7},\ldots,(g_{1}b)_{11},(g_{1}b)_{-28}\rangle=\mathfrak{A}$.
Set

$g_{2}=\prod_{7\leqslant i\leqslant 11}x_{\lambda_{i}-\lambda_{-28}}(\pm
c_{i})\text{,}$

where signs are such that $(g_{2}g_{1}b)_{i}=(g_{1}b)_{i}+c_{i}b_{-28}$ for
$7\leqslant i\leqslant 11$.

We claim that the elements $(g_{2}g_{1}b)_{\lambda}$, where $\lambda$ is in
the cells a1, a2, and c2, generate the unit ideal in $A[S^{-1}]$ Let us prove
that.

Assume that some maximal ideal $\mathfrak{M}$ of the ring $A[S^{-1}]$ contains
all the elements $(g_{2}g_{1}b)_{\lambda}$, where $\lambda$ is in the cells
a1, a2, and c2

Since applying $g_{1}$ does not change the ideal generated by elements from
column a, by choice of $c_{i}$ we have $\mathfrak{A}\leqslant\mathfrak{M}$.
Hence $(g_{1}b)_{i}\in\mathfrak{M}$ for $-11\leqslant i\leqslant-7$. Thus
$b_{28}=(g_{2}g_{1}b)_{28}+\sum_{7\leqslant i\leqslant 11}\pm
c_{i}(g_{1}b)_{-i}\in\mathfrak{M}$. However, by previous step, $b_{28}$ and
$\mathfrak{A}$ generate a unit ideal. This is a contradiction.

Since applying $g_{1}^{-1}$ does not change the ideal generated by elements
from the cells a1, a2, and c2, we obtain that the elements
$(g_{1}^{-1}g_{2}g_{1}b)_{\lambda}$, where $\lambda$ is in the cells a1, a2,
and c2, generate the unit ideal in $A[S^{-1}]$

It remains to notice that the element $g_{1}^{-1}g_{2}g_{1}$ is the image of
the matrix $\mu(u,s,v)$ for certain $u$ and $v$ under the embedding
$G(A_{5},A)\to G(E_{7},A)$ as a subsystem subgroup. Therefore, by Lemma 6.3,
$g_{1}^{-1}g_{2}g_{1}\in E(E_{7},R)^{\leqslant 39}$.

Step 7. Make the row that consists of elements in the cells a1, and a2
unimodular in $A[S^{-1}]$ by 91 elementary elements.

It follows from the proof of Lemma 1 in [33] that the elements in the row 2,
taken modulo the ideal $\langle b_{1}\rangle\unlhd A$, form an element of
$\operatorname{Orb}_{E_{6},\varpi_{6}}A/\langle b_{1}\rangle$. Therefore, we
can apply Lemma 6.7 to the row 2 and the ideal $\langle b_{1}\rangle$.

Step 8. Make the element $b_{1}$ lexicographically monic by 59 elementary
elements.

Apply Lemma 6.4 to the column a and the zero ideal.

∎

##  7\. Eliminating a variable

In this section, we give the proof of Proposition 4.4. First we need some
preparation.

For $\Phi=E_{6}$,$E_{7}$ or $E_{8}$, let $\Sigma_{2}\leqslant\Phi$ be the set
roots that have positive coefficient in simple root $\alpha_{2}$, and
$\Delta_{2}$ be the set roots that have zero coefficient in $\alpha_{2}$.
Therefore, $\Delta_{2}\cup\Sigma_{2}$ is a parabolic set of roots with
$\Delta_{2}$ being the symmetric part, and $\Sigma_{2}$ being the special
part. Let $U_{2}$ be the unipotent radical of the corresponding parabolic
subgroup, and $U_{2}^{-}$ be the unipotent radical of the opposite parabolic
subgroup.

Note that

$|\Sigma_{2}|=\begin{cases}21&\text{ for }E_{6}\text{,}\\\ 42&\text{ for
}E_{7}\text{,}\\\ 92&\text{ for }E_{8}\text{.}\end{cases}$

Further let $\Lambda$ be the set of weights of the representation $\varpi$.
Denote by $\Lambda_{i}\leqslant\Lambda$ the subset of such weights $\lambda$
that in the decomposition of $\lambda_{1}-\lambda$ in simple roots the
coefficient in $\alpha_{2}$ is equal to $i$. Therefore,

$\Lambda=\bigcup_{i=0}^{i_{\max}}\Lambda_{i}\text{,}$

where

$i_{max}=\begin{cases}2&\text{ for }E_{6}\text{,}\\\ 3&\text{ for
}E_{7}\text{,}\\\ 6&\text{ for }E_{8}\text{.}\end{cases}$

For an ideal $I\unlhd R$, set

$U_{2}^{-}(I)=\langle x_{\gamma}(\xi)\colon\gamma\text{ has negative
coefficient in }\alpha_{2}\text{,}\;\xi\in I\rangle\text{.}$

######  Lemma 7.1.

Let $R$ be a commutative ring. Let $0\leqslant r\leqslant i_{\max}-1$. Let
$b$,$b^{\prime}\in\operatorname{Orb}_{\varpi}R$ be such that for all
$0\leqslant i\leqslant r$ and for all $\lambda\in\Lambda_{i}$ we have
$b_{\lambda}=b^{\prime}_{\lambda}$. Let $I=\langle
b_{\lambda}-b^{\prime}_{\lambda}\colon\lambda\in\Lambda\rangle\unlhd R$.
Suppose that the elements $\\{b_{\lambda}\colon\lambda\in\Lambda_{0}\\}$
generate the unit ideal. Let $\gamma_{1}$,$\ldots$,$\gamma_{p}\in\Phi$ be all
the roots with coefficient in $\alpha_{2}$ being equal to $-(r+1)$. Then there
exists an element $u=x_{\gamma_{1}}(\xi_{1})\ldots x_{\gamma_{p}}(\xi_{p})$,
where all the $\xi_{j}$ are in $I$, such that for all $0\leqslant i\leqslant
r+1$ and for all $\lambda\in\Lambda_{i}$ we have
$(ub)_{\lambda}=b^{\prime}_{\lambda}$.

######  Remark 7.2.

In particular, this means that if $r+1$ is bigger than the maximal coefficient
in $\alpha_{2}$, then the assumptions on $b$ and $b^{\prime}$ implies that
$b=b^{\prime}$.

###### Proof.

Note that such an element $u$ does not change the elements $b_{\lambda}$ for
$\lambda\in\bigcup_{i\leqslant r}\Lambda_{i}$. Therefore, we must ensure the
equalities $(ub)_{\lambda}=b^{\prime}_{\lambda}$ only for
$\lambda\in\Lambda_{r+1}$. It is easy to see that this equalities are linear
equations in $\xi_{j}$. We must seek $\xi_{j}$ in form
$\xi_{j}=\sum_{\lambda}\zeta_{j,\lambda}(b_{\lambda}-b^{\prime}_{\lambda})$.
Therefore, we have a system of linear equations in $\zeta_{j,\lambda}$. For a
system of linear equations over a ring, existence of a solution is a local
property, see, for example, Proposition 1 in [18]. Hence it is enough to
consider the case where $R$ is a local ring.

Note that the system $\Delta_{2}$ has type $A_{|\Lambda_{0}|-1}$ and the
summand of the representation $\varpi$ that correspond to $\Lambda_{0}$ is the
vector representation of $G(\Delta_{2},-)$. Hence in the local case, since the
elements $\\{b_{\lambda}\colon\lambda\in\Lambda_{0}\\}$ generate the unit
ideal, there exists an element $g\in G(\Delta_{2},R)\leqslant G(\Phi,R)$ such
that $(gb)_{\lambda_{1}}=1$ and $(gb)_{\lambda}=0$ for all
$\lambda\in\Lambda_{0}\setminus\\{\lambda_{1}\\}$. The same equalities hold
for $gb^{\prime}$. It follows by Lemma 2.2 that $gb=u_{1}e_{1}$ and
$gb^{\prime}=u_{1}^{\prime}e_{1}$ for some $u_{1}$,$u_{1}^{\prime}\in
U_{1}^{-}$. Since $(gb)_{\lambda}=(gb^{\prime})_{\lambda}=0$ for all
$\lambda\in\Lambda_{0}\setminus\\{\lambda_{1}\\}$, it follows that
$u_{1}$,$u_{1}^{\prime}\in U_{2}^{-}$. We denote by $U_{2}^{-(j)}$ the
subgroup of $U_{2}^{-}$ generated by root subgroups for all roots that have
coefficient in $\alpha_{2}$ less or equal than $-(j+1)$. Similarly, we define
the subgroup $U_{2}^{-(j)}(I)$. Since $g\in G(\Delta_{2},R)$, the assumptions
on $b$ and $b^{\prime}$ imply that $(gb)_{\lambda}=(gb^{\prime})_{\lambda}$
for all $\lambda\in\bigcup_{i\leqslant r}\Lambda_{i}$; hence we have
$u_{1}\equiv u_{1}^{\prime}\mod U_{2}^{-(r)}$. Moreover, it is easy to see
that the ideal generated by elements $(gb)_{\lambda}-(gb^{\prime})_{\lambda}$
for all $\lambda\in\Lambda$ is equal to $I$. Therefore, $u_{1}\equiv
u_{1}^{\prime}\mod U_{2}^{-(r)}(I)$. Set
$\widetilde{u}=u_{1}^{\prime}u_{1}^{-1}\in U_{2}^{-(r)}(I)$. Then we have

$b^{\prime}=g^{-1}gb^{\prime}=g^{-1}u_{1}^{\prime}e_{1}=g^{-1}\widetilde{u}u_{1}e_{1}=g^{-1}\widetilde{u}gb\text{.}$

Since $g\in G(\Delta_{2},R)$, it follows that $g^{-1}\widetilde{u}g\in
U_{2}^{-(r)}(I)$. Therefore, $g^{-1}\widetilde{u}g=u\widehat{u}$, where
$u=x_{\gamma_{1}}(\xi_{1})\ldots x_{\gamma_{p}}(\xi_{p})$, where $\xi_{j}\in
I$, and $\widehat{u}\in U_{2}^{-(r+1)}$. Then we have
$(ub)_{\lambda}=(g^{-1}\widetilde{u}gb)_{\lambda}=b^{\prime}_{\lambda}$ for
$\lambda\in\Lambda_{r+1}$. ∎

######  Lemma 7.3.

Let $R$ be a commutative ring. Let
$b$,$b^{\prime}\in\operatorname{Orb}_{\varpi}R$ be such that for all
$\lambda\in\Lambda_{0}$ we have $b_{\lambda}=b^{\prime}_{\lambda}$. Let
$I=\langle
b_{\lambda}-b^{\prime}_{\lambda}\colon\lambda\in\Lambda\rangle\unlhd R$.
Suppose that the elements $\\{b_{\lambda}\colon\lambda\in\Lambda_{0}\\}$
generate the unit ideal. Then there exists an element $u\in U_{2}^{-}(I)$ such
that $ub=b^{\prime}$.

###### Proof.

Follows from Lemma 7.1 by induction. ∎

######  Lemma 7.4.

Let $R$ be a commutative Noetherian ring, $s\in R$. Then there exists
$k\in\mathbb{N}$ such that for any
$b$,$b^{\prime}\in\operatorname{Orb}_{\varpi}R$ that satisfy the following
conditions:

  * •

for all $\lambda\in\Lambda_{0}$ we have $b_{\lambda}=b^{\prime}_{\lambda}$,

  * •

$s\in\langle b_{\lambda}\colon\lambda\in\Lambda_{0}\rangle\unlhd R$,

  * •

$b_{\lambda}-b^{\prime}_{\lambda}$ is divisible by $s^{k}$ for all
$\lambda\in\Lambda$,

there exists an element $u\in U_{2}^{-}$ such that $ub=b^{\prime}$.

###### Proof.

The annihilators of the elements $s^{i}$, $l\in\mathbb{N}$ form a ascending
chain

$\operatorname{Ann}s\leqslant\operatorname{Ann}s^{2}\leqslant\ldots\text{.}$

Since the ring $R$ is Noetherian, it follows that for some $l\in\mathbb{N}$,
we have $\operatorname{Ann}s^{l+m}=\operatorname{Ann}s^{l}$ for any
$m\in\mathbb{N}$.

Now consider the ring
$\mathbb{Z}[\\{\widetilde{b}_{\lambda}\colon\lambda\in\Lambda\\}][\\{\widetilde{b^{\prime}}_{\lambda}\colon\lambda\in\Lambda\\}][\\{\widetilde{a}_{\lambda}\colon\lambda\in\Lambda_{0}\\}]$
of polynomials over $\mathbb{Z}$ in $2|\Lambda|+|\Lambda_{0}|$ variables. Set

$\widetilde{R}=\mathbb{Z}[\\{\widetilde{b}\colon\lambda\in\Lambda\\}][\\{\widetilde{b^{\prime}}\colon\lambda\in\Lambda\\}][\\{\widetilde{a}\colon\lambda\in\Lambda_{0}\\}]/\mathfrak{I}\text{,}$

where the ideal $\mathfrak{I}$ is generated by the following elements:
equations form $\operatorname{Eq}_{\varpi}$ for the column $\widetilde{b}$,
equations form $\operatorname{Eq}_{\varpi}$ for the column
$\widetilde{b^{\prime}}$, elements
$\widetilde{b}_{\lambda}-\widetilde{b^{\prime}}_{\lambda}$ for all
$\lambda\in\Lambda_{0}$. Then set

$\widetilde{s}=\sum_{\lambda\in\Lambda_{0}}\widetilde{a}_{\lambda}\widetilde{b}_{\lambda}\text{.}$

It follows by Lemma 7.3 that over the ring $\widetilde{R}[\widetilde{s}^{-1}]$
there exists an element $\widetilde{u}\in U_{2}^{-}$ such that
$\widetilde{u}\widetilde{b}=\widetilde{b^{\prime}}$ in
$\widetilde{R}[\widetilde{s}^{-1}]$. Let
$\widetilde{u}=x_{\gamma_{1}}(\widetilde{\xi_{1}})\ldots
x_{\gamma_{q}}(\widetilde{\xi_{q}})$, where $\gamma_{j}$ are roots that have
negative coefficient in $\alpha_{2}$. Moreover, we can choose
$\widetilde{\xi_{j}}$ to be in the ideal generated by
$\widetilde{b}_{\lambda}-\widetilde{b^{\prime}}_{\lambda}$ in
$\widetilde{R}[\widetilde{s}^{-1}]$. Let $\widetilde{k}\in\mathbb{N}$ be such
that for all $i$ elements $\widetilde{s}^{\widetilde{k}}\widetilde{\xi_{j}}$
belong to the ideal generated by
$\widetilde{b}_{\lambda}-\widetilde{b^{\prime}}_{\lambda}$ in $\widetilde{R}$,
i.e

$\widetilde{\xi_{j}}=\widetilde{s}^{-\widetilde{k}}\sum_{\lambda\in\Lambda}\widetilde{\zeta}_{j,\lambda}(\widetilde{b}_{\lambda}-\widetilde{b^{\prime}}_{\lambda})\text{,}\qquad\widetilde{\zeta}_{j,\lambda}\in\widetilde{R}\text{.}$

We claim that we can take $k=\widetilde{k}+l$. Let
$b$,$b^{\prime}\in\operatorname{Orb}_{\varpi}R$ satisfy the conditions. Then
there exists a ring homomorphism $\varphi\colon\widetilde{R}\to R$ that maps
$\widetilde{b}$ to $b$, $\widetilde{b^{\prime}}$ to $b^{\prime}$, and
$\widetilde{s}$ to $s$. Set
$\zeta_{j,\lambda}=\varphi(\widetilde{\zeta}_{j,\lambda})$. Let
$b_{\lambda}-b^{\prime}_{\lambda}=s^{k}c_{\lambda}$ for all
$\lambda\in\Lambda$. We claim that we can take
$u=x_{\gamma_{1}}(\xi_{1})\ldots x_{\gamma_{q}}(\xi_{q})$, where

$\xi_{j}=s^{l}\sum_{\lambda\in\Lambda}\zeta_{j,\lambda}c_{\lambda}\text{.}$

It is easy to see that the homomorphism $\widetilde{R}[\widetilde{s}^{-1}]\to
R[s^{-1}]$ induced by $\varphi$ send $\widetilde{u}$ to $u$. Therefore,
$ub=b^{\prime}$ over $R[s^{-1}]$, i.e. for any $\lambda\in\Lambda$ we have
$(ub)_{\lambda}-b^{\prime}_{\lambda}\in\operatorname{Ann}s^{m}$ for some $m$.
On the other hand, since all the $\xi_{j}$ are divisible by $s^{l}$, it
follows that $(ub)_{\lambda}-b_{\lambda}$ are divisible by $s^{l}$; hence
$(ub)_{\lambda}-b^{\prime}_{\lambda}$ are divisible by $s^{l}$. Let
$(ub)_{\lambda}-b^{\prime}_{\lambda}=s^{l}\theta_{\lambda}$. Then
$s^{m+l}\theta_{\lambda}=s^{m}((ub)_{\lambda}-b^{\prime}_{\lambda})=0$, i.e
$\theta_{\lambda}\in\operatorname{Ann}s^{m+l}=\operatorname{Ann}s^{l}$.
Therefore, $(ub)_{\lambda}-b^{\prime}_{\lambda}=s^{l}\theta_{\lambda}=0$, i.e
$ub=b^{\prime}$. ∎

Set $\Lambda_{0}^{\prime}=\Lambda_{0}\setminus\\{\nu\\}$, where $\nu$ is the
lowest weight in $\Lambda_{0}$.

######  Lemma 7.5.

Let $B$ be a commutative Noetherian ring, $A=B[y]$. Let
$b=b(y)\in\operatorname{Um}^{\prime}A$, and $s\in B\cup\langle
b(y)_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\rangle$. Then there exists
$m\in\mathbb{N}$ such that

$b(y+s^{m}z)\in E(\Phi,A[z])^{\leqslant N}b(y)\text{,}$

where

$\displaystyle N=\begin{cases}65&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\
94&\text{ for }E_{6}\leqslant E_{7}\text{,}\\\ 152&\text{ for }E_{7}\leqslant
E_{8}\text{.}\end{cases}$

###### Proof.

Take $k$ from Lemma 7.4 (for $R=A[z]$) and set $m=k+2$. Let
$b=(\begin{smallmatrix}b^{0}\\\ b^{1}\end{smallmatrix})$, where $b_{0}$ is a
column with entries $b_{\lambda}$ for $\lambda\in\Lambda_{0}$ and $b_{1}$ is a
column with entries $b_{\lambda}$ for $\lambda\notin\Lambda_{0}$. Recall that
the system $\Delta_{2}$ has type $A_{|\Lambda_{0}|-1}$ and the summand of the
representation $\varpi$ that correspond to $\Lambda_{0}$ is the vector
representation of $G(\Delta_{2},-)$. Therefore, it follows from Corollary 2.4
of [58] that there exists an element $g(z)\in E(\Delta_{2},A[z])^{\leqslant
8|\Lambda_{0}|-4}$ such that $g(z)b^{0}(y)=b^{0}(y+s^{2}z)$. Moreover, it
follows from the proof that $g$ is congruent to the identity element modulo
$z$, see the proof of Lemma 6.5 in [16] for details.

Therefore, for some $\widetilde{b}^{1}\in
A[z]^{|\Lambda\setminus\Lambda_{0}|}$ we have

$b^{\prime}=\begin{pmatrix}b^{0}(y+s^{m}z)\\\
\widetilde{b}^{1}\end{pmatrix}=g(s^{k}z)b\in E(\Phi,A[z])^{\leqslant
8|\Lambda_{0}|-4}b\text{.}$

In addition, $\widetilde{b}^{1}$ is congruent to $b^{1}(y)$, and hence to
$b^{1}(y+s^{m}z)$, modulo $s^{k}$. Now applying Lemma 7.4 to vectors
$b^{\prime}$ and $b(y+s^{m}z)$, we obtain that

$b(y+s^{m}z)\in E(\Phi,A[z])^{\leqslant|\Sigma_{2}|}b^{\prime}\subseteq
E(\Phi,A[z])^{\leqslant 8|\Lambda_{0}|-4+|\Sigma_{2}|}b\text{.}$

Here we used that $s\in B$, so the shift of the variable does not change the
fact that $s\in\langle b_{\lambda}\colon\lambda\in\Lambda_{0}\rangle$. ∎

######  Lemma 7.6.

There is an element $w\in W(E_{8})$ such that $w(\alpha_{2})=\alpha_{8}$,
$w(\alpha_{4})=\alpha_{7}$, $w(\alpha_{5})=\alpha_{6}$, and
$w(-\delta_{A_{8}})=\delta$, where $\delta_{A_{8}}$ is the maximal root of the
subsystem generated by
$\alpha_{1}$,$\alpha_{3}$,$\ldots$,$\alpha_{8}$,$\delta$.

###### Proof.

Let $\delta_{A_{7}}$ be is the maximal root of the subsystem generated by
$\alpha_{1}$,$\alpha_{3}$,$\ldots$,$\alpha_{8}$. Let us show that
$w^{\prime}\in W(E_{8})$ such that $w(\alpha_{2})=\alpha_{8}$,
$w(\alpha_{4})=\alpha_{7}$, $w(\alpha_{5})=\alpha_{6}$, and
$w(-\delta_{A_{7}})=\delta$, where $\delta_{A_{8}}$. Note that all the roots
in the condition belong to the subsystem of type $D_{8}$ generated by
$\alpha_{2}$,$\alpha_{3}$,$\ldots$,$\alpha_{8}$,$\delta$. We can realise this
$D_{8}$ in the Euclidean space with the orthonormal basis
$e_{1}$,$\ldots$,$e_{8}$ so that $\delta=e_{1}-e_{2}$,
$\alpha_{8}=e_{2}-e_{3}$, $\alpha_{7}=e_{3}-e_{4}$, $\alpha_{6}=e_{4}-e_{5}$,
$\alpha_{5}=e_{5}-e_{6}$, $\alpha_{4}=e_{6}-e_{7}$, $\alpha_{2}=e_{7}-e_{8}$,
$\alpha_{3}=e_{7}+e_{8}$, $\delta_{A_{7}}=e_{1}+e_{8}$. An element form
$W(D_{8})$ can perform any permutation of $e_{i}$ and in addition replace any
even number of $e_{i}$ with $-e_{i}$. So we can take $w^{\prime}\in W(D_{8})$
to be the element such that $w^{\prime}(e_{1})=e_{1}$,
$w^{\prime}(e_{2})=e_{8}$, $w^{\prime}(e_{3})=e_{7}$, $w(e_{4})=e_{6}$,
$w(e_{5})=-e_{5}$, $w(e_{6})=-e_{4}$, $w(e_{7})=-e_{3}$, $w(e_{8})=-e_{2}$.

Now we can take $w=w^{\prime}w_{\alpha_{1}}$, where $w_{\alpha_{1}}$ os the
reflection with respect to $\alpha_{1}$. ∎

######  Lemma 7.7.

Let $B$ be a commutative ring, $P_{1}$,$\ldots$,$P_{m}$ be distinct maximal
ideals in $B$, $A=B[y]$, $b=b(y)\in\operatorname{Um}^{\prime}_{\varpi}A$ such
that $b_{j}$ is monic, where $j=24$ for $E_{6}$, $j=-1$ for $E_{7}$ and
$E_{8}$. Then there exists a column vector

$b^{(1)}\in E(\Phi,A)^{\leqslant N}b$

such that $b^{(1)}_{j}$ is monic and

$\left(\langle
b^{(1)}_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\rangle\cap
B\right)\setminus\bigcup_{i=1}^{m}P_{i}\neq\varnothing\text{,}$

where

$\displaystyle N=\begin{cases}7&\text{ for }D_{5}\leqslant E_{6}\text{,}\\\
10&\text{ for }E_{6}\leqslant E_{7}\text{,}\\\ 139&\text{ for }E_{7}\leqslant
E_{8}\text{.}\end{cases}$

###### Proof.

Set

$R=B/\left(\bigcap_{i=1}^{m}P_{i}\right)=\prod_{i=1}^{m}B/P_{i}\text{.}$

First we show that the last condition on $b^{(1)}$ holds if $b^{(1)}_{1}$ is
monic and the elements
$\\{b^{(1)}_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\\}$ generate the
unit ideal in $R[y]$.

Let $c_{\lambda}\in A$, where $\lambda\in\Lambda_{0}^{\prime}$, be such that
$\sum_{\lambda\in\Lambda_{0}^{\prime}}c_{\lambda}b^{(1)}_{\lambda}\equiv 1\mod
P_{i}$ for every $i$.

Set
$f=\sum_{\lambda\in\Lambda_{0}^{\prime}\setminus\\{\lambda_{1}\\}}c_{\lambda}b^{(1)}_{\lambda}$.
Then $b^{(1)}_{1}$ and $f$ are coprime in $B/P_{i}[y]$ for every $i$.

Since $b^{(1)}_{1}$ is monic, it follows that the resultant
$\operatorname{res}(b^{(1)}_{1},f)$ modulo $P_{i}$ is equal to the resultant
of $b^{(1)}$ taken modulo $P_{i}$ and $f$ taken modulo $P_{i}$ (even if $f$
modulo $P_{i}$ has smaller degree).

Therefore, we have

$\operatorname{res}(b^{(1)}_{1},f)\in\left(\langle
b^{(1)}_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\rangle\cap
B\right)\setminus\bigcup_{i=1}^{m}P_{i}\text{.}$

Thus it remains to prove that a given column
$b\in\operatorname{Um}^{\prime}_{\varpi}A$, with $b_{j}$ being monic, can be
transformed by $N$ elementary elements so that $b_{1}$ becomes monic, $b_{j}$
remains monic, and the new elements
$\\{b_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\\}$ generate the unit
ideal in $R[y]$.

Proof for $(E_{6},\varpi_{1})$. Here we perform the following steps.

Figure 4. $(E_{6},\varpi_{1})$

Step 1. Make the polynomial $b_{3}$ monic and the row
$(b_{1},\ldots,b_{16},b_{18},\ldots b_{22},b_{24})$ unimodular in $R[y]$ by
the element $x_{\delta}(\xi)$.

Since $R$ is a product of fields and $b_{24}$ is monic, it follows that the
ring $R[y]/\langle b_{24}\rangle$ is semilocal; hence it is easy to see that
there exists $\widetilde{\xi}$ such that the row
$((x_{\delta}(\widetilde{\xi})b)_{1},\ldots,(x_{\delta}(\widetilde{\xi})b)_{16},(x_{\delta}(\widetilde{\xi})b)_{18},\ldots(x_{\delta}(\widetilde{\xi})b)_{22},(x_{\delta}(\widetilde{\xi})b)_{24})$
is unimodular in $R[y]$. Therefore, if we take

$\xi=\widetilde{\xi}+y^{K}b_{24}$

for some $K\in\mathbb{N}$, then we guarantee that the row
$(b_{1},\ldots,b_{16},b_{18},\ldots b_{22},b_{24})$ becomes unimodular in
$R[y]$. It remains to notice that if $K$ is large enough, then we also make
$b_{3}$ monic.

Step 2. Make the polynomial $b_{2}$ monic and the row $(b_{1},\ldots
b_{21},b_{23})$ unimodular in $R[y]$ by the element $x_{\alpha_{3}}(\xi)$.

This is done similarly to step 1.

Step 3. Make the polynomial $b_{1}$ monic and the row $(b_{1},\ldots b_{17})$
unimodular in $R[y]$ by the element $x_{\alpha_{1}}(\xi)$.

This is done similarly to step 1.

Step 4. Make the row $(b_{1},b_{2},b_{3},b_{4},b_{6})$ unimodular in $R[y]$ by
the element
$x_{\alpha_{2}}(\xi_{4})x_{\delta_{D_{4}}}(\xi_{3})x_{\alpha_{6}}(\xi_{2})x_{\delta_{D_{5}}}(\xi_{1})$,
where $\delta_{D_{5}}$ is the maximal root of the subsystem generated by
$\alpha_{2},\ldots\alpha_{6}$, and $\delta_{D_{4}}$ is the maximal root of the
subsystem generated by $\alpha_{2},\ldots\alpha_{5}$.

Existence of such $\xi_{1}$,$\ldots$,$\xi_{4}$ follows easily from the fact
that $R[y]/\langle b_{1}\rangle$ is semilocal.

Note that neither of steps change $b_{24}$; hence it remains monic. Also step
4 does not change $b_{1}$; hence it remains monic after being made so in step
3.

Proof for $(E_{7},\varpi_{7})$.

Consider the branching table for $(E_{7},\varpi_{7})$, where vertical lines
correspond to cuting through the bonds marked with 1, and horizontal lines
correspond to cuting through the bonds marked with 7.

|  | a | b | c
---|---|---|---|---
| $E_{7},\varpi_{7}$ | $D_{6},\varpi_{1}$ | $D_{6},\varpi_{6}$ | $D_{6},\varpi_{1}$
1) | $\circ$ | $\circ$ |  |
2) | $E_{6},\varpi_{6}$ | $D_{5},\varpi_{1}$ | $D_{5},\varpi_{5}$ | $\circ$
3) | $E_{6},\varpi_{1}$ | $\circ$ | $D_{5},\varpi_{4}$ | $D_{5},\varpi_{1}$
4) | $\circ$ |  |  | $\circ$

Now we perform the following steps, which are similar to those for $E_{6}$.

Step 1. Make the polynomial in the cell a3 monic and the row that consists of
elements in the cells a1, a2, a3, b2, b3, and c4 unimodular in $R[y]$ by the
element $x_{\delta}(\xi)$.

Step 2. Make the polynomial $b_{2}$ (highest weight in the cell a2) monic and
the row that consists of elements in the cells a1, a2, a3, b2, c2, the upper
half of the cell b3 with respect to cutting through the bonds marked with 6,
and the element that correspond to the highest weight in the cell c3
unimodular in $R[y]$ by the element $x_{\delta_{D_{6}}}(\xi)$, where
$\delta_{D_{6}}$ is the maximal root of the subsystem generated by
$\alpha_{2},\ldots\alpha_{7}$.

Step 3. Make the polynomial $b_{1}$ monic and the row that consists of
elements in the cells a1, a2, b2, and c2 unimodular in $R[y]$ by the element
$x_{\alpha_{7}}(\xi)$.

Step 4. Make the row $(b_{1},b_{2},b_{3},b_{4},b_{5},b_{7})$ unimodular in
$R[y]$ by the element
$x_{\alpha_{2}}(\xi_{7})x_{\alpha_{2}+\alpha_{3}+\alpha_{4}}(\xi_{6})x_{\delta_{D_{5}(1)}}(\xi_{5})x_{\alpha_{1}}(\xi_{4})x_{\delta_{D_{5}(6)}}(\xi_{3})x_{\delta_{A_{5}}}(\xi_{2})x_{\delta_{E_{6}}}(\xi_{1})$,
where $\delta_{E_{6}}$ is the maximal root of the system generated by
$\alpha_{1}$,$\alpha_{2}$,$\alpha_{3}$,$\alpha_{4}$,$\alpha_{5}$ and
$\alpha_{6}$; $\delta_{A_{5}}$ is the maximal root of the system generated by
$\alpha_{1}$,$\alpha_{3}$,$\alpha_{4}$,$\alpha_{5}$ and $\alpha_{6}$;
$\delta_{D_{5}(6)}$ is the maximal root of the system generated by
$\alpha_{1}$,$\alpha_{2}$,$\alpha_{3}$,$\alpha_{4}$ and $\alpha_{5}$;
$\delta_{D_{5}(1)}$ is the maximal root of the system generated by
$\alpha_{2}$,$\alpha_{3}$,$\alpha_{4}$,$\alpha_{5}$ and $\alpha_{6}$.

Proof for $(E_{8},\varpi_{8})$. Here we perform the following steps.

Step 0. Make the row $(b_{1},b_{-1})$ unimodular in $R[y]$ by the element
$u\in U$.

Since $R[y]/\langle b_{-1}\rangle$ is semilocal, by Lemma 2.4 there exists
$g\in G(E_{8},R[y]/\langle b_{-1}\rangle)$ such that $gb=e_{1}$ in
$R[y]/\langle b_{-1}\rangle$. By Theorem 1.1 in [39], we have $g=hu_{1}vu$,
where $h\in T$, $u,u_{1}\in U$, and $v\in U^{-}$. Therefore, we have
$ub=v^{-1}u_{1}^{-1}h^{-1}e_{1}$ in $R[y]/\langle b_{-1}\rangle$; hence
$(ub)_{1}$ is invertible in $R[y]/\langle b_{-1}\rangle$. Clearly $u$ can be
lifted to the element of $U(\Phi,B[y])$. Note that $(ub)_{-1}=b_{-1}$; hence
the row $((ub)_{1},(ub)_{-1})$ is unimodular in $R[y]$.

Now consider the subsystem $D_{8}\leqslant E_{8}$ generated by
$\alpha_{2}$,$\alpha_{3}$,$\ldots$,$\alpha_{8}$,$\delta$. If we restrict our
representation to the group $G(D_{8},-)$ one of the summands will be the
representation $(D_{8},\varpi_{8})$. Take $w\in W(E_{8})$ from Lemma 7.6. If
we move our subsystem $D_{8}$ with element $w$, then the highest weight of the
representation $(D_{8},\varpi_{8})$ become the highest weight of the entire
$(E_{8},\varpi_{8})$. In addition, three weights next to it become weights
from $\Lambda_{0}^{\prime}$ (Figure 5). It is clear that lowest weight of
$(D_{8},\varpi_{8})$ become the lowest weight of $(E_{8},\varpi_{8})$.

Figure 5. Part of $(E_{8},\varpi_{8})$ and action of $w$

Consider the weight diagram for $(D_{8},\varpi_{8})$. If we cut it through the
bonds marked with 2 (here marks refer to the numbering of simple root in
$D_{8}$ as shown at Figure 6), then we obtain the union of diagrams
$(D_{6},\varpi_{6})$, $(D_{6},\varpi_{5})\otimes(A_{1},\varpi_{1})$, and
$(D_{6},\varpi_{6})$.

Figure 6. Numbering of simple roots in $D_{l}$

Diagram for $(D_{6},\varpi_{5})$ differs from the diagram for
$(D_{6},\varpi_{6})$ by swaping two labels; so essentially we have 4 copies of
diagram $(D_{6},\varpi_{6})$. We give number 1 to the one containing the
highest weight, number 2 to the upper half of the component
$(D_{6},\varpi_{5})\otimes(A_{1},\varpi_{1})$, number 3 to its lower half, and
number 4 to the one containing the lowest weight. Now we give to every vertex
of the diagram $(D_{8},\varpi_{8})$ the number of the form $i/j$, where $i$ is
the number of weight in $(D_{6},\varpi_{6})$ according to Figure 7, and $j$ is
the number of the copy of $(D_{6},\varpi_{6})$.

Figure 7. $(D_{6},\varpi_{6})$

Now it remains to prove the following statement. For any column vector
$b=b(y)\in V_{(D_{8},\varpi_{8})}A$ such that it becomes unimodular in $R[y]$
and that $b_{32/4}$ is monic, there exists a column vector

$b^{(1)}\in E(\Phi,A)^{\leqslant 19}b$

such that $b^{(1)}_{32/4}$ and $b^{(1)}_{1/1}$ are monic and the row
$(b^{(1)}_{1/1},b^{(1)}_{2/1},b^{(1)}_{3/1},b^{(1)}_{5/1})$ is unimodular in
$R[y]$.

We prove this statement similarly to how we prove it for $(E_{6},\varpi_{1})$
and $(E_{7},\varpi_{7})$. Here we perform the following steps (numbering of
roots is as at Figure 6).

Step 1. Make the polynomial $b_{32/1}$ monic and simultaneously make the row
that consists of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
32,\,1\leqslant j\leqslant 3\\}\cup\\{b_{32/4}\\}$ unimodular in $R[y]$ by the
element $x_{\delta_{D_{8}}}(\xi)$.

Step 2. Make the polynomial $b_{8/1}$ monic and simultaneously make the row
that consists of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
24,\,1\leqslant j\leqslant 3\\}\cup\\{b_{32/1},b_{8/4}\\}$ unimodular in
$R[y]$ by the element $x_{\delta_{D_{6}}}(\xi)$, where $\delta_{D_{6}}$ is the
maximal root of the subsystem generated by $\alpha_{3},\ldots\alpha_{8}$.

Step 3. Make the row of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
16,\,1\leqslant j\leqslant 3\\}\cup\\{b_{32/1},b_{8/4}\\}$ unimodular in
$R[y]$ by the element $x_{\alpha_{3}}(\xi)$ (the polynomial $b_{8/1}$ remains
the same).

Step 4. Make the polynomial $b_{2/1}$ monic and simultaneously make the row
that consists of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
14,\,1\leqslant j\leqslant 3\\}\cup\\{b_{26/1},b_{2/4}\\}$ unimodular in
$R[y]$ by the element $x_{\delta_{D_{4}}}(\xi)$, where $\delta_{D_{6}}$ is the
maximal root of the subsystem generated by $\alpha_{5},\ldots\alpha_{8}$.

Step 5. Make the row of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
12,\,1\leqslant j\leqslant 3\\}\cup\\{b_{26/1},b_{2/4}\\}$ unimodular in
$R[y]$ by the element $x_{\alpha_{5}}(\xi)$ (the polynomial $b_{2/1}$ remains
the same).

Step 6. Make the row of elements $\\{b_{i/j}\colon 1\leqslant i\leqslant
8,\,1\leqslant j\leqslant 3\\}\cup\\{b_{22/1},b_{2/4}\\}$ unimodular in $R[y]$
by the element $x_{\alpha_{4}}(\xi)$ (the polynomial $b_{2/1}$ remains the
same).

Step 7. Make the row of elements $\\{b_{i/1}\colon 1\leqslant i\leqslant
8\\}\cup\\{b_{i,j}\colon 1\leqslant i\leqslant 7,\,2\leqslant j\leqslant
3\\}\cup\\{b_{22/1},b_{2/4}\\}$ unimodular in $R[y]$ by the element
$x_{\alpha_{7}}(\xi)$ (the polynomial $b_{2/1}$ remains the same).

Step 8. Make the row of elements $\\{b_{i/1}\colon 1\leqslant i\leqslant
8\\}\cup\\{b_{i,j}\colon 1\leqslant i\leqslant 6,\,2\leqslant j\leqslant
3\\}\cup\\{b_{22/1},b_{2/4}\\}$ unimodular in $R[y]$ by the element
$x_{\alpha_{6}+\alpha_{8}}(\xi)$ (the polynomial $b_{2/1}$ remains the same).

Step 9. Make the polynomial $b_{1/1}$ monic and simultaneously make the row
that consists of elements $\\{b_{i/1}\colon 1\leqslant i\leqslant
7\\}\cup\\{b_{i,j}\colon i\in\\{1,2,3,5\\},\,2\leqslant j\leqslant
3\\}\cup\\{b_{21/1},b_{1/4}\\}$ unimodular in $R[y]$ by the element
$x_{\alpha_{8}}(\xi)$.

Step 10. Make the row
$(b^{(1)}_{1/1},b^{(1)}_{2/1},b^{(1)}_{3/1},b^{(1)}_{5/1})$ unimodular in
$R[y]$ by the element
$x_{\alpha_{7}}(\xi_{10})x_{\alpha_{6}}(\xi_{9})x_{\alpha_{4}}(\xi_{8})x_{\alpha_{5}}(\xi_{7})x_{\alpha_{3}}(\xi_{6})x_{\alpha_{4}}(\xi_{5})x_{\alpha_{2}}(\xi_{4})x_{\alpha_{3}}(\xi_{3})x_{\alpha_{1}}(\xi_{2})x_{\alpha_{2}}(\xi_{1})$.
∎

Now we are ready to prove Proposition 4.4.

For simplicity, we will write

$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{N}$$\textstyle{w}$

instead of

$w\in E(\Phi,R)^{\leqslant N_{1}}v\text{,}$

where $v$ and $w$ are columns in $V_{\varpi}$.

Set

$\displaystyle N_{1}=\begin{cases}65&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 94&\text{ for }E_{6}\leqslant E_{7}\text{,}\\\ 152&\text{ for
}E_{7}\leqslant E_{8}\text{.}\end{cases}$

and

$\displaystyle N_{2}=\begin{cases}7&\text{ for }D_{5}\leqslant
E_{6}\text{,}\\\ 10&\text{ for }E_{6}\leqslant E_{7}\text{,}\\\ 139&\text{ for
}E_{7}\leqslant E_{8}\text{.}\end{cases}$

Applying Lemmas 2.1 and 7.7 $d$ times, we obtain elements
$s_{1}$,$\ldots$,$s_{d}\in B$ and columns
$b=b^{(0)}$,$b^{(1)}$,$\ldots$,$b^{(d)}\in\operatorname{Um}^{\prime}_{\varpi}A$
such that, firstly,

$\textstyle{b^{(i)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{N_{2}}$$\textstyle{b^{(i+1)}}$$\textstyle{i=0,\ldots,d-1\text{,}}$

secondly, $s_{i}\in\langle
b^{(i)}_{\lambda}\colon\lambda\in\Lambda_{0}^{\prime}\rangle$ for
$i=1,\ldots,d$, and thirdly, $\operatorname{BSdim}B/(s_{1},\ldots
s_{i+1})<\operatorname{BSdim}B/(s_{1},\ldots,s_{i})$ for $i=0,\ldots,d-1$. In
particular, the elements $s_{1}$,$\ldots$,$s_{d}$ generate the unit ideal.

By Lemma 7.5 we have

$\textstyle{b^{(i)}(y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{N_{1}}$$\textstyle{b^{(i)}(y+s_{i}^{m_{i}}z)}$

in $A[z]$.

Therefore, we have the following chain of transformations in
$A[z_{1},\ldots,z_{d}]$:

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
24.05574pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{b=b^{(0)}(y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
48.05574pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
48.05574pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{b^{(1)}(y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
102.54234pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
102.54234pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{b^{(1)}(y+s_{1}^{m_{1}}z_{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
184.62889pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
184.62889pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{b^{(2)}(y+s_{1}^{m_{1}}z_{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
266.71544pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
266.71544pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\ldots}$}}}}}}}\ignorespaces}}}}\ignorespaces\\\
\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
54.20982pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{b^{(d)}(y+s_{1}^{m_{1}}z_{1}+\ldots+s_{d-1}^{m_{d-1}}z_{d-1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
78.20982pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
78.20982pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{b^{(d)}(y+s_{1}^{m_{1}}z_{1}+\ldots+s_{d}^{m_{d}}z_{d})\text{.}}$}}}}}}}\ignorespaces}}}}\ignorespaces$

Thus we have

$\textstyle{b(y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!d(N_{1}+N_{2})}$$\textstyle{b^{(d)}(y+s_{1}^{m_{1}}z_{1}+\ldots+s_{d}^{m_{d}}z_{d})\text{.}}$

Since the elements $s_{1}$,$\ldots$,$s_{d}$ generate the unit ideal, it
follows that so do the elements $s_{1}^{m_{1}}$,$\ldots$,$s_{d}^{m_{d}}$.
Specializing the indeterminates $z_{i}$ to elements in $yB$, we make
$y+s_{1}^{m_{1}}z_{1}+\ldots+s_{d}^{m_{d}}z_{d}$ equal to zero; this concludes
the proof of Proposition 4.4.

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