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\title{Classification of all connected subgroup schemes of a reductive group containing
a split maximal torus}
\author{Ekaterina Sopkina\\
{\it Department of Mathematics and Mechanics}\\
{\it St.~Petersburg University}\\
and \\
{\it Fakult\"at f\"ur Mathematik}\\
{\it Universit\"at Bielefeld}\\
E-mail: sopkina@yandex.ru
}
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\maketitle

\noindent {\bf Abstract.} The main result of the paper is a classification
of all connected subgroup schemes of a reductive group that contain
a split maximal torus, over an arbitrary field. The classification is expressed
in terms of functions on the root system.

\medskip

\noindent {\bf 2000 Mathematical Subject Classifications:} 20G15, 20G35, 14L15. 

\medskip

\noindent {\bf Keywords:} reductive groups, non-reduced affine group schemes.

\section{Introduction}
\indent

In the paper we completely describe connected subgroup schemes of
a reductive group that contain a split maximal torus over an arbitrary field.
This description is stated in purely combinatorial terms and depends only on
the corresponding root system. This is a decisive step towards
a description of {\bf all} (not necessarily connected) subgroup schemes
containing a torus. We intend to return to this problem in a subsequent paper. 

The main theorem is a broad generalization of classical results describing
algebraic subgroups of a split reductive group (see A.~Borel's and J.~Tits's
paper \cite{BT} and S\'eminaire de G\'eom\'etrie Alg\'ebrique \cite{SGA}).
In context of our problem, algebraic subgroups correspond to reduced
subgroup schemes. We briefly recall the results on algebraic subgroups
in Section \ref{extract}.

These results were generalized to groups over some classes
of rings, including semi-local ones, by Z.~Borewicz, N.~Vavilov, and others.
On the other hand, over a finite field G.~Seitz described overgroups
of all not necessarily split maximal tori. Surveys of research in
these directions and further references can be found in the papers
\cite{Vavobz,Vavob}.

In the present paper we generalize in a different direction.
Namely, we include not necessarily reduced subgroup schemes. Recall that
by Cartier's theorem (see, for example, \cite{Wat} 11.4)
any affine group scheme over a field of characteristic zero is reduced.
But in positive characteristic, subgroup schemes of a reductive group are not
exhausted by reduced ones.

Non-reduced subgroup schemes of reductive groups were considered,
in particular, by C.~Wenzel (see \cite{Wen1,Wen2}) and F.~Knop
(see \cite{Knop}). Namely, in the special case of parabolic subgroup schemes,
an analogous problem was addressed by Ch.Wenzel. He gave a complete description
of parabolic subschemes under some mild restriction on characteristic.
Futhermore, F.~Knop for \cite{Knop} described all subgroup schemes of $SL_2$.
One of our initial motivations was exactly to generalize these results. 

In the preceding paper \cite{Sop} we classified all (not necessarily reduced)
overgroup schemes of a maximal torus in $GL_n$. The result of the present paper
generalizes simultaneously the classical result on reduced subgroup schemes,
the result by Ch.~Wenzel on parabolic schemes, and the above result for $GL_n$.
In the proof we develop
the approach applied to the case of $GL_n$ in \cite{Sop}. The proof uses
standard technics and tools from algebraic geometry: reduction to
an algebraically closed field, Frobenius morphism, functorial properties.
But the most difficult step, Main Lemma \ref{mlemma},
concerns groups over rings and involves ideas from the works by Z.~Borewicz
and N.~Vavilov \cite{BV}, \cite{Vav}.

The rest of the article is organized as follows. In Section 2, we fix
notation and recall some results on the structure of reductive groups.
In Section 3, we formulate the main theorem. In Section 4, we discuss
the concept of quasi-closed sets. Section 5 contains a review of
the description of connected intermediate algebraic subgroups in a reductive
group. In Section 6, we discuss a Frobenius morphism. In Section 7,
we specialize and start proving Theorem 1, namely, we construct all
connected intermediate group schemes. In Section 8, we reduce Theorem 1
to the case of an algebraically closed field. In Section 9,
we consider the reduced subscheme of an intermediate group scheme.
Finally, in Section 10, we prove Main Lemma on intermediate subgroups
in the group of $R$-points of a reductive group and complete the proof
of Theorem 1.

\section*{Acknowledgements}
\indent

This research was partially supported by
INTAS 04-83-3000 Young Scientist Fellowship,
INTAS 03-51-3251, DAAD-program ``Michail Lomonosov'',
and{ \-INTAS 00-566}.

It is a special plesure for me to thank Prof.~N.~Vavilov and Prof.~A.~Bak
for their attention to the work and support in the preparing of this article.


\section{Preliminaries}
\indent

In this section we fix notation and recall necessary results concerning
reductive groups. We follow the books \cite{Jan}, \cite{Sp},
and \cite{SGA} Exp. XXII.

Let $G$ be a reductive group containing a split maximal torus $T$ over
a field $K$. Let $(X,\Phi,X\check{\ },\Phi\check{\ })$
be the root data of $(G,T)$. In particular, $X$ is the character group of $T$,
$X\check{\ }$ is the cocharacter group of $T$, i.e. the group of all homomorphisms
from $G_m$ to $T$, and the groups $X$ and $X\check{\ }$ are in duality by
a pairing ${\langle {\ }, {\ } \rangle: X\times X\check{\ } \to \mathbb{Z}}$.
For any root $\beta \in \Phi$, any $K$-algebra $R$, and any element
$t \in T(R)$ we have by definition $$(-\beta)(t)=\beta(t)^{-1}.$$

Consider an adjoint representation of $G$ on its Lie algebra $L$.
The $T$-module $L$ admits a decomposition
$$L=L_0\oplus(\underset{\beta \in \Phi}\oplus L_\beta)$$
Each root subspace $L_\beta$ has dimension 1.

For each root $\beta \in \Phi$ there is a homomorphism $x_\beta:G_a \to G$
such that the corresponding tangent map induces an isomorphism
of the Lie algebra of $G_a$ onto $L_\beta$. Any element $t$
of the torus $T(R)$ acts on the root subgroup
corresponding to a root $\beta$ by
$$tx_\beta(a)t^{-1}=x_\beta(\beta(t)a).$$
In particular, for $t=\lambda(\varepsilon)$, where $\lambda$ is a cocharacter
and $\varepsilon$ is an invertible element of $R$, we have
$$\lambda(\varepsilon)x_\beta(a)\lambda(\varepsilon)^{-1}=
x_\beta(\varepsilon^{\langle\beta,\lambda\rangle}a).$$
Such a root homomorphism is unique up to multiplication by an element of $K^*$.
For each root $\beta$ we fix a homomorphism $x_\beta$ that comes from
a homomorphism over $\mathbb{Z}$.

Fix a Weyl basis $\{e_\beta, \beta \in \Phi, h_\lambda, 1 \leq \lambda \leq \dim T\}$
in $L$, where $\{e_\beta, \beta \in \Phi \}$ is a Chevalley system in $L$. 
Then the adjoint action of a root subgroup $x_\alpha(G_a)$ on a basis element
$e_\beta$ with $\beta\neq -\alpha$ is given by
$$\Ad(x_\alpha(a))e_\beta=e_\beta+\sum_{r>0}M_{\alpha\beta r}a^r e_{r\alpha+\beta},$$
where $M_{\alpha\beta r}$ are integral coefficients.

For root subgroups corresponding to roots $\alpha$ and $\beta$
with $\beta\neq -\alpha$, we have the Chevalley commutator formula 
$$[x_\alpha(a),x_\beta(b)]=\prod_{r,s>0}x_{r\alpha+s\beta}(N_{\alpha\beta r s}a^rb^s),$$
where $N_{\alpha\beta r s}$ are integral coefficients. We take the values
of $M_{\alpha\beta r}$ and $N_{\alpha\beta r s}$ from the paper \cite{VP}.

In case of $\beta=-\alpha$, but one of $a$ and $b$ is nilpotent, we use the formula 
$$x_\alpha(a)x_{-\alpha}(b)=
x_{-\alpha}(\frac b{1+ab})
\alpha\check{\ }(1+ab)
x_\alpha(\frac a{1+ab})$$
(see \cite{SGA} Exp.XX 2.1).

Fix an arbitrary ordering on $\Phi$ that is compatible with root heights,
i.e. if $ht(\alpha) < ht(\beta)$, then $\alpha$ goes before $\beta$.
When we consider a product like $\prod_{\beta\in\Phi}$, we take the factors
in this order.
\section{Statement of main results}
\indent

In this section we formulate a main result of the paper.

Let $K$ be a field of characteristic $p>0$. Let $\Phi$ be a root system,
$G(\Phi)$ be a reductive group of type $\Phi$ over the field $K$,
$T$ be a split maximal torus in $G(\Phi)$.

\begin{theorem} \label{theorem}
There is a one-to-one correspondence
between all {\bf connected} subgroup schemes of $G(\Phi)$ that contain $T$,
and all functions
$\varphi : \Phi \to \mathbb{N}\cup \{0, \infty \}$
satisfying 
$$\varphi(r\alpha+s\beta) \ge \min(\varphi(\alpha)-\log_pr,\varphi(\beta)-\log_ps) \eqno (*)$$
for any roots $\alpha$, $\beta$, $r\alpha+s\beta \in \Phi$ such that $p \not\vert N_{\alpha\beta r s}$.
\end{theorem}

\noindent{\bf Remark.} Suppose that the following restriction on characteristic holds:

$\bullet$ if $\Phi \supseteq B_2$, then $p > 2$;

$\bullet$ if $\Phi \supseteq G_2$, then $p > 3$.

\vspace{0.1cm}
\noindent Then $p \not\vert N_{\alpha\beta 1 1}$ for any two roots $\alpha$ and $\beta$
such that $\alpha+\beta$ is a root.
So in this case the condition $(*)$ is equivalent to the following one:
$$\varphi(\alpha+\beta) \ge \min(\varphi(\alpha),\varphi(\beta))$$
for any roots $\alpha$, $\beta\in\Phi$ such that $\alpha+\beta\in\Phi$. 

\noindent{\bf Examples.} 1. A function $\varphi$ satisfying $(*)$
corresponds to a reduced subgroup scheme if and only if all values
of $\varphi$ are zero and infinity. In the notation of Section~\ref{extract},
the function $\varphi$ corresponds to the scheme $G(S)$, where
$S$ is a quasi-closed set, if and only if
$$\varphi(\beta)=\begin{cases}
&\infty,\text{ if } \beta \in S;\\
&0, \text{ otherwise.}
\end{cases}$$

2. The product of the torus $T$ and the kernel $E_N$ of the $N$-th power of a Frobenius morphism
is a non-reduced connected subgroup scheme. All values of the corresponding
function are equal to $N$.

3. Suppose that $\Phi=A_{n-1}$, $G=\GLn$, and $T$ is the group of
diagonal matrices. $\GLn$ is represented by
$K[\GLn]={K[x_{11}, \dots, x_{nn},t]\slash (t{\rm det}(x_{ij})=1)}$.
In this case the subscheme corresponding to a function
$\varphi:A_{n-1} \to \mathbb{N}\cup \{0, \infty \}$
is represented by ${K[\GLn]\slash (x_{ij}^{p^{\varphi(ij)}}=0)}$,
where $\varphi(ij)=\varphi(e_i-e_j)$. See \cite{Sop} for details.

\section{Quasi-closed sets of roots}
\indent

In this section we introduce and discuss a notion of quasi-closed
sets of roots, and an equivalent definition of the property $(*)$.
The notion of quasi-closed set is taken from the paper \cite{BT} by A.~Borel
and J.~Tits, but the definition we use is distinct from the initial one.

\noindent{\bf Definition.} Let $S$ be a subset of a root system $\Phi$.
Suppose that for any two roots $\alpha$ and $\beta$ in $S$ and
positive integers $r$ and $s$ such that $r\alpha+s\beta$ is a root and
$p \not\vert \ N_{\alpha\beta r s}$, the root $r\alpha+s\beta$ lies in $S$.
Then the subset $S$ is called {\it quasi-closed}.

\noindent{\bf Remark.}
Any closed set of roots is quasi-closed. Conversely, any
quasi-closed set is closed under the restriction

$\bullet$ if $\Phi \supseteq B_2$, then $p > 2$;

$\bullet$ if $\Phi \supseteq G_2$, then $p > 3$.

\vspace{0.1cm}

\begin{lemma} \label{MandN}
Let $\Phi$ be a root system, $p$ be a prime number.

1. The condition $(*)$ is equivalent to the following one:
$$\varphi(r\alpha+\beta) \ge \min(\varphi(\alpha)-\log_pr,\varphi(\beta)) \eqno (**)$$
for any roots $\alpha$, $\beta$, $r\alpha+\beta \in \Phi$ such that $p \not\vert M_{\alpha\beta r}$.

2. A set $S$ is quasi-closed if and only if the following condition holds:
for any two roots $\alpha$ and $\beta$ in $S$ and
positive integer $r$ such that $r\alpha+\beta$ is a root and
$p \not\vert \ M_{\alpha\beta r}$, the root $r\alpha+\beta$ lies in $S$.

\end{lemma}
\begin{proof}
1. If $s=1$, then $N_{\alpha\beta r s}=M_{\alpha\beta r}$ and the conditions
$(*)$ and $(**)$ are the same. The case $r=1$ is similar.
The only case, when $r,s>1$ and
$r\alpha+s\beta$ is a root, is as follows:
$\alpha$ and $\beta$ are simple roots of $G_2$, $r=2$, and $s=3$
(or, symmetrically, $r=3$ and $s=2$).
In this case we have $N_{\alpha\beta 3 1}=1$ and
$N_{3\alpha+\beta, \beta 1 1}=1$ (see \cite{VP}), so
\begin{align*}
\varphi(3\alpha+2\beta) &\geq \min(\varphi(3\alpha+\beta),\varphi(\beta))
\geq \min(\varphi(\alpha)-\log_p3, \varphi(\beta))\\
&\geq\min(\varphi(\alpha)-\log_p3, \varphi(\beta)-\log_p2).
\end{align*}

2. The condition holds if and only if the function $\varphi_S$ satisfies $(**)$, where
$$\varphi_S(\beta)=\begin{cases}
&\infty,\text{ if } \beta \in S;\\
&0, \text{ otherwise.}
\end{cases}$$
The set $S$ is quasi-closed if and only if the function
$\varphi_S$ satisfies $(*)$. Apply the first assertion of the lemma to
the function $\varphi_S$.
\end{proof}

\section{Connected smooth subgroup schemes \\containing a torus}
\indent
\label{extract}

This section is based on \cite{SGA} Exp. XXII.
In this section we consider connected smooth subgroup schemes
containing a torus and show that they correspond to
quasi-closed subsets of roots.

\noindent{\bf Definition.}(\cite{SGA} Exp. XXII 5.2.1) \label{type(R)}
Let $X$ be a prescheme, $G$ be a smooth finitely represented group $X$-prescheme
with connected fibres, $H$ be a subgroup prescheme of $G$.
We say that $H$ is {\it of type} $(R)$ if the following conditions hold:

1. $H$ is smooth finitely represented $X$-prescheme with connected fibres;

2. for any $x \in X$, $H_{\bar{x}}$ contains a Cartan subgroup of $G_{\bar{x}}$.

Assume now that $G=G(\Phi)$ is a split reductive group over a field.
In this case any connected subgroup scheme containing a maximal torus $T$
is of type $(R)$ if and only if the scheme is smooth. By \cite{SGA} Exp. XXII 5.4.1,
any subgroup scheme of type (R) containing $T$ is uniquely determined
by its Lie algebra, which has a form
$$L_S=L_0\oplus(\underset{\beta \in S}\oplus L_\beta)$$
for some subset $S$ of $\Phi$.

\noindent{\bf Definition.}(\cite{SGA} Exp. XXII 5.4.2) \label{set(R)}
A subset $S$ of $\Phi$ is called a set {\it of type} $(R)$ if
$L_S$ is the Lie algebra of a subgroup scheme of type $(R)$ containing $T$.

By $G(S)$ denote the subgroup scheme of type $(R)$ uniquely determined by the set $S$.

\begin{lemma} \label{cap}
$$x_\beta(G_a) \cap G(S)=\begin{cases}
&x_\beta(G_a),\text{ if } \beta \in S;\\
&1, \text{ otherwise.}
\end{cases}$$
\end{lemma}
\begin{proof}
See \cite{SGA} Exp. XXII 5.4.3.
\end{proof}

\begin{lemma}
Any quasi-closed set is of type (R). And conversely, any set of type (R) is
quasi-closed.
\end{lemma}
\begin{proof}
We show first that any set of type $(R)$ is quasi-closed. By Lemma \ref{MandN}
it is enough to check that
for any two roots $\alpha$ and $\beta$ in $S$ and
positive integer $r$ such that $r\alpha+\beta$ is a root and
$p \not\vert \ M_{\alpha\beta r}$, the root $r\alpha+\beta$ lies in $S$.

Since $\alpha \in S$ and $\beta \in S$, we have $e_\beta \in L_S$ and
$x_\alpha(G_a) \subseteq G(S)$ by Lemma~\ref{cap}. Therefore the element
$\Ad(x_\alpha(a))e_\beta=e_\beta+\sum_{r>0}M_{\alpha\beta r}a^r e_{r\alpha+\beta}$
lies in $L_S$ for any $a$. Since $L_S=L_0\oplus(\underset{\beta \in S}\oplus L_\beta)$
and $p \not\vert \ M_{\alpha\beta r}$, we conclude
$L_{r\alpha+\beta} \subseteq L_S$, and therefore the root $r\alpha+\beta$ lies in $S$.

Now we need to prove the converse. The assertion that any closed set is of type
$(R)$ is established in Theorem \cite{SGA} Exp. XXII 5.4.7. In fact,
the proof of this theorem uses only properties of quasi-closed sets 
and is therefore a valid proof for our assertion.
\end{proof}

The above lemmas imply the following corrolary.

\noindent
\begin{corollary} \label{G(S)}
The correspondence $S \mapsto G(S)$ gives a bijection
between all quasi-closed subsets of $\Phi$ and all
connected smooth subgroup schemes containing the torus $T$ in $G(\Phi)$.
\end{corollary}

\section{Frobenius morphism}
\indent

In this section we discuss a Frobenius morphism and introduce a scheme $E_N$.

Let $G$ be an affine scheme over a perfect field $K$.
Consider a Frobenius morphism $F:G \to F(G)$ corresponding to
the natural embedding $$K[F(G)]=K[G]^p \hookrightarrow K[G].$$

\begin{lemma} \label{red_frob}
Let $H$ and $H'$ be closed subschemes of any algebraic affine scheme $G$
over a perfect field $K$. If the subschemes $F^N(H)$ and $F^N(H')$ of $F^N(G)$
coincide, then the schemes $H\red$ and $H'\red$ coincide.
\end{lemma}
\begin{proof}
Let the scheme $G$ be represented by $K[G]$, and let the subschemes
$H$ and $H'$ be represented by $K[G]/I$ and $K[G]/I'$
for some ideals $I$ and $I'$ respectively. Then the subscheme $F^N(H)$
(respectively $F^N(H')$) is represented by the algebra
$(K[G]/I)^{p^N}=K[G]^{p^N}/(K[G]^{p^N} \cap I)$
(respectively $K[G]^{p^N}/(K[G]^{p^N} \cap I')$). Since  the subscheme $F^N(H)$
equals to the subscheme $F^N(H')$, the corresponding ideals $K[G]^{p^N} \cap I$
and $K[G]^{p^N} \cap I'$ coincide.
So we get $\sqrt[p^N]{I}=\sqrt[p^N]{I'}$, and
therefore the ideals $\sqrt{I}$ and $\sqrt{I'}$ coincide. Thus
we conclude
$$K[H\red]=K[G]/\sqrt{I}=K[G]/\sqrt{I'}=K[H'\red]$$ and $H\red=H'\red$.
\end{proof}

Let $F^N:G(\Phi) \to F^N(G(\Phi))$ be the $N$-th power of a Frobenius morphism.
Denote the kernel of $F^N$ by $E_N$.

\begin{lemma} \label{EGauss}
$E_N=T_N \prod_{\beta \in \Phi} x_\beta(\alpha_{p^N})$, \\
where $T_N=\Ker(F^N) \cap T$, and this product is direct.
\end{lemma}
\begin{proof}
See \cite{Jan}, II.3.2.
\end{proof}

\begin{lemma} \label{G(S)_N}
$G(S) \cap E_N=T_N\prod_{\beta \in S}x_\beta(\alpha_{p^N})$, \\
where $T_N=\Ker(F^N) \cap T$, and this product is direct.
\end{lemma}
\begin{proof}
Since the group schemes $G(S)$ and $E_N$ contain $T_N$ and
$x_\beta(\alpha_{p^N})$ for each $\beta \in S$,
the intersection $G(S) \cap E_N$ contains the right-hand member
of the equality. So it is enough to prove the other inclusion.

By Lemma \ref{EGauss}, for any $K$-algebra $R$
and any element $g \in G(S,R) \cap E_N(R)$
there is an {\it fppf} $R$-algebra $R'$ such that
$g=t\prod_{\beta \in \Phi} x_\beta(a_\beta)$,
where $t \in T_N(R')$ and $a_\beta \in R'$ with
$a_\beta^{p^N}=0$. Since $t$ lies in $G(S,R') \cap E_N(R')$,
the intersection $G(S,R') \cap E_N(R')$ contains the product
$\prod_{\beta \in \Phi} x_\beta(a_\beta)$.

Note that $G(S,R') \cap E_N(R')$ is a subgroup of $G(\Phi,R')$
and contains $T(R)$. So we may apply Main Lemma \ref{mlemma},
which is proven in Section \ref{tech}. Thus we get
$x_\beta(a_\beta) \in G(S,R')$ for each $\beta \in \Phi$.
Lemma \ref{cap} implies $a_\beta=0$ for each $\beta \notin S$,
and this completes the proof. 
\end{proof}

\section{Construction of $H_\varphi$} \label{Hvarphi}
\indent

In this section we construct a subgroup scheme $H_\varphi$ corresponding
to a function $\varphi$ that satisfies the inequality $(*)$.

For a given function $\varphi: \Phi \to \mathbb{N}\cup \{0, \infty \}$
define a subset $$S=\{\beta \in \Phi \mid \varphi(\beta)=\infty \}$$ of $\Phi$.
If the function $\varphi$ satisfies the condition $(*)$, then the set $S$ is
quasi-closed by definition. Fix any positive integer $N$
such that $N > \max\{\varphi(\beta)\mid \varphi(\beta) < \infty\}$.

Assume at first that the field $K$ is perfect.
Set $\psi(\beta)=\min(\varphi(\beta),N)$. Obviously
if $\varphi$ satisfies the condition $(*)$ then $\psi$ satisfies too. Consider a subscheme
$E_{\varphi,N}=T_N \prod_{\beta \in \Phi} x_\beta(\alpha_{p^{\psi(\beta)}})$
of $E_N$.

\begin{lemma} \label{Egroup}
If $\varphi$ satisfies $(*)$, then $E_{\varphi,N}$ is a subgroup scheme. 
\end{lemma}
\begin{proof} To prove this lemma it is enough to check that for any
$k$-algebra $R$ the set $E_{\varphi,N}(R)$ is a subgroup
of $G(\Phi,R)$. Obviously $E_{\varphi,N}(R)$ contains a unit, and for any
$t \in T_N(R)$ we have $t^{-1} \in T_N(R)$, and $x_\beta(a)^{-1}=x_\beta(-a)$.
So it is enough to show that an arbitrary product of elements of $T_N(R)$ and
$x_\beta(\alpha_{p^{\psi(\beta)}}(R))$ can be written in the form
$t\prod_{\beta \in \Phi} x_\beta(a_\beta)$, where
$t \in T_N(R)$ and each $a_\beta^{p^{\psi(\beta)}}=0$.

Consider an element $h$ of $G(\Phi,R)$, which is equal to a product of elements
$t_i \in T_N(R)$ and $x_\beta(a_{\beta,j})$ with
$a_{\beta,j}^{p^{\psi(\beta)}}=0$ in some order.
Let $I$ be the ideal of $R$ generated by all $a_{\beta,j}$. The ideal $I$ is
finitely generated and is contained in the nilradical $\Nil(R)$,
so we have $I^M=0$ for some positive integer $M$. We prove that $h$
modulo $I^r$ can be written in the required form, by induction.
Obviously the case $r=M$ completes the proof.

If $r$ is equal to $1$, this is trivial, because $a_{\beta,j} \in I$.
Now we prove that the case $r$ imlies the result for $r+1$.

First of all we note that we anytime can permute any element $t_i$ with
any element of $x_\beta(a_{\beta,j})$, because
$x_\beta(a_{\beta,j})t_i=t_ix_\beta(\beta(t_i^{-1})a_{\beta,j})$.

Suppose that $h$ is equal to a product of
$t_i$ and $x_\beta(a_{\beta,j})$ with the following property:
if we omit the factors $x_\beta(a_{\beta,j})$
with $a_{\beta,j} \in I^r$, then we get a product like
$t\prod_{\beta \in \Phi} x_\beta(a_\beta)$. Consider an omitted factor
$x_\alpha(a'_{\alpha})$ with $a'_{\alpha} \in I^r\backslash I^{r+1}$.
We can permute it with $t$, with $x_\beta(a_{\beta,j})$, where $\beta \neq
\pm \alpha$, and with $x_{-\alpha}(a_{-\alpha,j})$, as follows.

When we permute 
$x_\alpha(a'_\alpha)$ and $t$, we use the formula
$$x_\alpha(a'_\alpha)t=tx_\alpha(\alpha(t^{-1})a'_\alpha).$$

When we permute $x_\alpha(a'_\alpha)$ and $x_\beta(a_{\beta,j})$, where
$\beta \neq \pm \alpha$, we use the Chevalley commutator formula 
$$x_\alpha(a'_\alpha)x_\beta(a_{\beta,j})=
x_\beta(a_{\beta,j})x_\alpha(a'_\alpha)
\prod_{r,s>0}x_{r\alpha+s\beta}(N_{\alpha\beta r s}{a'_\alpha}^ra_{\beta,j}^s).$$
By the inequality $(*)$, we have
$(N_{\alpha\beta r s}{a'_\alpha}^ra_{\beta,j}^s)^{p^{\psi(r\alpha+s\beta)}}=0$.
So each new factor $x_{\gamma}(a'_\gamma)$
satisfies ${a'_\gamma}^{p^{\psi(\gamma)}}=0$ and $a'_\gamma \in I^{r+1}$.

When we permute $x_\alpha(a'_\alpha)$ and $x_{-\alpha}(a_{-\alpha,j})$,
we use the formula
$$x_\alpha(a'_\alpha)x_{-\alpha}(a_{-\alpha,j})=
x_{-\alpha}(\frac{a_{-\alpha,j}}{1+a'_\alpha a_{-\alpha,j}})
\alpha\check{\ }(1+a'_\alpha a_{-\alpha,j})
x_\alpha(\frac{a'_\alpha}{1+a'_\alpha a_{-\alpha,j}})$$
for a homomorphism $\alpha\check{\ }:G_m \to T$ (see \cite{SGA} Exp.XX 2.1).
Obviously we have $\alpha\check{\ }(1+a'_\alpha a_{-\alpha,j}) \in T_N(R)$.

Using these permutations, we put $x_\alpha(a'_\alpha)$ near the factor
$x_\alpha(a_\alpha)$ and eliminate the former by the formula
$$x_\alpha(a_\alpha)x_\alpha(a'_\alpha)=x_\alpha(a_\alpha+a'_\alpha).$$

Proceeding this process for each omitted factor
$x_\alpha(a'_{\alpha})$ with $a'_{\alpha} \in I^r\backslash I^{r+1}$, we get
a required product modulo $I^{r+1}$.
\end{proof}

\begin{lemma} \label{Norm}
$G(S)$ normalizes $E_{\varphi,N}$.
\end{lemma}
\begin{proof}
Let us show that conjugation by $G(S)$ leaves $T_N$ inside $E_{\varphi,N}$.
In particular, then conjugation by $x_\gamma(G_a)$ leaves $T_N$ inside
$E_{\varphi,N}$ for any $\gamma \in S$.
Since $G(S)$ contains $T$ and since $E_N$ is a normal
subgoup scheme of $G(\Phi)$, we have the inclusion
$$T_N^{G(S)} =(T\cap E_N)^{G(S)}\subseteq G(S)\cap E_N.$$
Using Lemma \ref{G(S)_N}, we conclude
$$T_N^{G(S)} \subseteq T_N \prod_{\beta \in S} x_\beta(\alpha_{p^N})
\subseteq E_{\varphi,N}.$$

Let us show that the commutator of subgroup schemes $x_\beta(\alpha_{p^{\psi(\beta)}})$
and $x_\gamma(G_a)$ with $\gamma \in S$ lies inside $E_{\varphi,N}$.
Suppose that $\gamma \neq -\beta$. Then by the Chevalley commutator formula we have
$$[x_\beta(\alpha_{p^{\psi(\beta)}}),x_\gamma(G_a)]\subseteq
\prod_{r,s>0}x_{r\beta+s\gamma}(N_{\beta \gamma r s}(\alpha_{p^{\psi(\beta)}})^r).$$
By the inequality $(**)$, if $p \not\vert N_{\beta \gamma r s}$, then 
\begin{align*}
\varphi(r\beta+s\gamma)
&\geq\min(\varphi(\beta)-\log_pr,\varphi(\gamma)-\log_ps)\\
&=\min(\varphi(\beta)-\log_pr,\infty)=\varphi(\beta)-\log_pr,
\end{align*}
and therefore $\psi(r\beta+s\gamma)\geq \psi(\beta)-\log_pr$. So we have the inclusion
$$N_{\beta \gamma r s}(\alpha_{p^{\psi(\beta)}})^r \subseteq 
\alpha_{p^{\psi(r\beta+s\gamma)}},$$ and therefore the following inclusion holds:
$$[x_\beta(\alpha_{p^{\psi(\beta)}}),x_\gamma(G_a)]
\subseteq T_N \prod_{\beta \in \Phi} x_\beta(\alpha_{p^{\psi(\beta)}})
\subseteq E_{\varphi,N}.$$

Now suppose that $\gamma=-\beta$.
Then by the commutation formula
\begin{align*}
[x_\beta(a),x_{-\beta}(b)]&=x_\beta(a)x_{-\beta}(b)x_\beta(-a)x_{-\beta}(-b)\\
&=x_\beta(a)x_\beta(-\frac a{1-ab})\beta\check{\ }(1-ab)
x_{-\beta}(\frac b{1-ab})x_{-\beta}(-b)\\
&=x_\beta(a\frac{-ab}{1-ab})\beta\check{\ }(1-ab)x_{-\beta}(b\frac{ab}{1-ab}),
\end{align*}
where $a$ is nilpotent, we get the inclusion
$$[x_\beta(\alpha_{p^{\psi(\beta)}}),x_{-\beta}(G_a)]\subseteq
x_\beta(\alpha_{p^{\psi(\beta)}})\cdot T_N\cdot x_{-\beta}(\alpha_{p^N})\subseteq E_{\varphi,N}.$$

Since $E_{\varphi,N}$ is a product of subgroups $T_N$ and
$x_\beta(\alpha_{p^{\psi(\beta)}})$, the above assertions imply that
each root subgroup $x_\gamma(G_a)$ with
$\gamma \in S$ normalizes $E_{\varphi,N}$. Obviously $T$ also normalizes
$T_N$ and $x_\beta(\alpha_{p^{\psi(\beta)}})$, so $T$ normalizes the whole
$E_{\varphi,N}$.

Since $G(\Phi)$ is a group scheme over a field,
the normalizer of the subgroup scheme $E_{\varphi,N}$ in $G(\Phi)$
is an affine subgroup scheme (see \cite{SGA} Exp. VIII 6.7).
The reduced subscheme of the normalizer
is a closed subgroup of $G(\Phi)$ and contains $T$ and $x_\gamma(G_a)$ with
$\gamma \in S$. On the other hand, the product
$T\prod_{\gamma \in S}x_\gamma(G_a)$ is an open dense set in $G(S)$
(see \cite{SGA} Exp. XXII 5.4.4).
Thus the whole $G(S)$ is contained in the normalizer,
i.e. $G(S)$ normalizes $E_{\varphi,N}$.
\end{proof}

Define a scheme $H_\varphi$ as a product $G(S)E_{\varphi,N}$.
Since the scheme $G(S)$ normalizes $E_{\varphi,N}$ by Lemma \ref{Norm},
$H_\varphi$ is a group scheme.
Since $H_\varphi$ is defined as a product of subgroup schemes,
we can consider $H_\varphi$ as a functor on $k$-algebras given by
\begin{align*}
H_\varphi(R)=\{&g \in G(\Phi,R) \mid \text{there are
an {\it fppf} $R$-algebra $R'$ and elements}\\
&h \in G(S,R'), f \in E_{\varphi,N}(R') \text{ with }
hf=g \text{ in } G(\Phi,R')\}.
\end{align*}
See \cite{Jan}, I.6.2 for details.

This construction is invariant under field extension,
so we can define $H_\varphi$ over any (not necessarily perfect) field $K$
as usual by $K[(H_\varphi)_K]=K\otimes \mathbb{F}_p[(H_\varphi)_{\mathbb{F}_p}]$,
since any field $K$ of characteristic $p$ is an extension of the perfect
field $\mathbb{F}_p$. This new definition is compatible with the previous one.

\begin{lemma} \label{check}
The scheme $H_\varphi$ does not depend on the choice of $N$ and
satisfies all the conditions of Theorem \ref{theorem},
i.e. it is a connected subgroup scheme of $G(\Phi)$ containing $T$. Moreover
$(H_\varphi)\red=G(S)$.
\end{lemma}
\begin{proof}

Let us show that $H_\varphi$ does not depend on the choice of $N$.
Let $N'$ be greater than $N$, $N > \max\{\varphi(\beta)\mid \varphi(\beta)=\infty\}$.
Obviously $E_{\varphi,N'}$ contains $E_{\varphi,N}$,
so we get the inclusion $G(S)E_{\varphi,N'}\supseteq G(S)E_{\varphi,N}$.
On the other hand $T_N$ is contained in $G(S)$,
$x_\beta (\alpha_{p^{N'}})$ is contained in $G(S)$ for $\beta \in S$,
and $x_\beta (\alpha_{p^{\varphi(\beta)}})$ is contained in $E_{\varphi,N}$ for
$\beta \notin S$. Thus we get the inclusion
$$E_{\varphi,N'}=T_{N'}\prod_{\beta \in \Phi}
x_\beta(\alpha_{p^{\min(\varphi(\beta), N')}}) \subseteq G(S)E_{\varphi,N},$$
and therefore $G(S)E_{\varphi,N'} \subseteq G(S)E_{\varphi,N}$.
So the group schemes $G(S)E_{\varphi,N'}$ and $G(S)E_{\varphi,N}$ coincide,
and $H_\varphi$ does not depend on the choice of $N$.

Let us prove that the reduced subscheme of the scheme $H_\varphi$ is equal to $G(S)$.
Assume at first that the field $K$ is perfect.
By definition we get $F^N(H_\varphi)=F^N(G(S))$. So by Lemma \ref{red_frob}
we get $(H_\varphi)\red=G(S)$. Any field extension preserves this equality,
so it holds for any (not necessarily perfect) field $K$.

To prove that the scheme $H_\varphi$ is connected it is enough
to check that $(H_\varphi)\red$ is connected,
because for any non-trivial idempotent $f \in K[H_\varphi]$
the element $\bar{f} \in (K[H_\varphi]/\Nil)$
is also a non-trivial idempotent. But the subscheme $(H_\varphi)\red$
is equal to $G(S)$ and therefore is connected.

Since the subscheme $(H_\varphi)\red$ contains the torus $T$, the scheme
$H_\varphi$ does.
\end{proof}

\begin{lemma} \label{Hcap}
$x_\beta(G_a)\cap H_\varphi =x_\beta(\alpha_{p^{\varphi(\beta)}})$. \\
In particular, all schemes $H_\varphi$ are distinct.
\end{lemma}
\begin{proof}
Assume at first that the field $K$ is perfect.

Suppose that $\varphi(\beta)=\infty$. Then $\beta$ lies in $S$ by definition,
and by Lemma \ref{cap} we get the inclusion $x_\beta(G_a) \subseteq G(S)$.
So $x_\beta(G_a) \subseteq H_\varphi$ and
$x_\beta(G_a)\cap H_\varphi =x_\beta(G_a)$.

If $\varphi(\beta) <\infty$, then $\varphi(\beta) < N$ by the choice of $N$.
So we have $$x_\beta(\alpha_{p^{\varphi(\beta)}})=
x_\beta(\alpha_{p^{\min(\varphi(\beta),N)}}) \subseteq E_{\varphi,N} \subseteq
H_\varphi.$$

We know that $F^N(E_{\varphi,N})=1$, so
$F^N(H_\varphi)=F^N(G(S))$. 
Since $\varphi(\beta)$ is finite, $\beta \notin S$, and by Lemma \ref{cap}
we have $x_\beta(G_a) \cap G(S)=1$.
Since $F$ preserves root data, we get
\begin{align*}
F^N(x_\beta(G_a)\cap H_\varphi) &\subseteq F^N(x_\beta(G_a))\cap F^N(H_\varphi)\\
&=F^N(x_\beta(G_a))\cap F^N(G(S))=F^N(x_\beta(G_a)\cap G(S)),
\end{align*}
i.e. $F^N(x_\beta(G_a)\cap H_\varphi)=1$, so
$x_\beta(G_a)\cap H_\varphi \subseteq x_\beta(\alpha_{p^N})$.

Consider an element $x_\beta(a) \in H_\varphi(R)$. By definition of $H_\varphi$
there exist an {\it fppf} $R$-algebra $R'$,
an element $g \in G(S,R')$, and an element $h \in E_{\varphi,N}(R')$
such that $x_\beta(a)=gh$. We know that $a^{p^N}=0$, so $F^N(x_\beta(a))=1$.
Also we know that $F^N(h)=1$. Thus $F^N(g)=1$, or, in other words,
the element $g$ lies in the intersection $G(S,R') \cap E_N(R')$. By Lemma \ref{G(S)_N},
we have the inclusion ${G(S,R') \cap E_N(R') \subseteq E_{\varphi,N}(R')}$.
So we get $x_\beta(a) \in E_{\varphi,N}(R')$, and therefore
$x_\beta(a) \in E_{\varphi,N}(R)$. By the uniqueness of the decomposition
in Lemma \ref{EGauss} and by definition of $E_{\varphi,N}(R)$ we get
$a \in \alpha_{p^{\varphi(\beta)}}(R)$,
so $x_\beta(G_a)\cap H_\varphi \subseteq x_\beta(\alpha_{p^{\varphi(\beta)}})$.

We proved the assertion for a perfect field $K$.
Since any field extension preserves the equality required,
it holds for an arbitrary field $K$. 
\end{proof}

\section{Reduction to an algebraically closed field} \label{alg}
\indent

Now to prove Theorem \ref{theorem} it is enough to show that any connected
subgroup scheme $H$ of $G(\Phi)$ containing $T$ over $K$ is a scheme
$H_\varphi$ for some function $\varphi$ satisfying the inequality $(*)$.
In this section we reduce the general situation to the case of an algebraically
closed field $K$.

Let $H$ be a connected intermediate group scheme $T \le H \le G(\Phi)$.
Let $H_{\overline{K}}$ be the scheme over the algebraic closure $\overline{K}$
represented by $\overline{K}[H]=\overline{K} \otimes K[H]$. Then $H_{\overline{K}}$
is a subgroup scheme of $G(\Phi)_{\overline{K}}$ containing the split
maximal torus $T_{\overline{K}}$. The scheme $H_{\overline{K}}$
is also connected, because field extension preserves connectedness of group schemes,
see \cite{SGA}, exp. VIa 2.4.

Any subgroup scheme of a given group scheme is closed
(see \cite{SGA} Exp. VIb 1.4.2). 

Suppose now that Theorem \ref{theorem} is proved in the case of
an algebraically closed field. Then the following lemma implies
Theorem \ref{theorem} in the general case.

\begin{lemma} \label{alg_closed}
Let $H$ and $H'$ be closed subschemes of any algebraic affine scheme $G$
over a field $K$. If the schemes $H_{\overline{K}}$ and $H'_{\overline{K}}$
coincide as subschemes of the scheme $G_{\overline{K}}$ over
the algebraic closure $\overline{K}$, then the schemes $H$ and $H'$ coincide.
\end{lemma}
\begin{proof}Let the scheme $G$ be represented by $K[G]$, and let the subschemes
$H$ and $H'$ be represented by $K[G]/I$ and $K[G]/I'$
for some ideals $I$ and $I'$ respectively. Then the subscheme $H_{\overline{K}}$ is
represented by
$\overline{K} \otimes (K[G]/I)=\overline{K}[G]/\overline{K}[G]I$,
and similarly the subscheme $H'_{\overline{K}}$ is represented by
$\overline{K}[G]/\overline{K}[G]I'$.
Since the schemes $H_{\overline{K}}$ and $H'_{\overline{K}}$ coincide, we get
$\overline{K}[G]I=\overline{K}[G]I'$. Thus we can conclude 
$$I=K[G] \cap \overline{K}[G]I=K[G] \cap \overline{K}[G]I'=I',$$
because the embedding $K[G]\hookrightarrow \overline{K}[G]$ is faithfully flat.
\end{proof}

Now we may assume the field $K$ is algebraically closed in
the proof of Theorem~\ref{theorem}.

\section{Reduced subscheme}
\indent

In this section we assume that the field $K$ is algebraically closed.

Let $H$ be a connected subgroup scheme of $G(\Phi)$ containing
a split maximal torus $T$ over an algebraically closed field $K$.
Consider the reduced subscheme $H\red$. It is represented by
$K[H]/\Nil(K[H])$. The scheme $H\red$ contains the torus $T$,
because the torus $T$ is reduced. The reduced subscheme of
a group scheme over a perfect (e.g. over an algebraically closed)
field is a group scheme (see \cite{SGA} Exp. VIa 0.2).
The reduced subscheme of a connected group scheme is also connected.
Any reduced group scheme over an algebraically closed field is smooth
(see \cite{Wat} 11.6). So $H\red$ is a connected smooth subgroup
scheme of $G(\Phi)$ containing the torus $T$. By Corollary \ref{G(S)},
the scheme $H\red$ is equal to $G(S)$ for some quasi-closed subset $S$
of $\Phi$.

Consider a nilradical $\Nil(K[H])$ of the algebra $K[H]$.
It is a finitely generated nilpotent ideal, so there is a positive
integer $N$ such that $\Nil(K[H])^{p^N}=0$. Obviously we have
$F^N(H)=F^N(H\red)=F^N(G(S))$.

Consider any $K$-algebra $R$ and any element $g \in H(R)$.
Since the element $F^N(h)$ lies in the group $F^N(G(S))(R)$,
there is an {\it fppf} $R$-algebra $R'$ such that $F^N(g)=F^N(h)$
for some element $h \in G(S,R')$. Then the element $f=h^{-1}g$ satisfies
$F^N(f)=1$, and therefore $f\in E_N(R')$. Thus the element $h$ is equal
to the product $hf$, where $h \in G(S,R')$ and $f \in E_N(R')$. By definition
this means that $H$ is a subscheme of $G(S)E_N$. The inclusions
$G(S) \subseteq H \subseteq G(S)E_{\varphi_N}$ imply
$H=G(S)\cdot(H \cap E_N)$.

The above assertions establish the following lemma.

\begin{lemma} \label{HGauss}
Let $H$ be a connected subgroup scheme of $G(\Phi)$ containing $T$
over an algebraically closed field $K$.
Then the following assertions hold.

1. $H\red=G(S)$ for some quasi-closed set $S$;

2. $F^N(H\red)=F^N(G(S))$ for some positive integer $N$;

3. $H=G(S)(H\cap E_N)$, where $S$ and $N$ as above.
\end{lemma}

\section{Subgroups of reductive groups over rings} \label{tech}
\indent

In this section we prove Main Lemma \ref{mlemma}, which is a useful technical result concerning
overgroups of the maximal torus in $G(\Phi,R)$.

Let $G$ be any affine scheme over a field $K$, $R$ any $K$-algebra, $I$ any ideal in
$R$, $h$ and $h'$ elements of $G(R)$. We write $h \equiv h' \mod I$, if
the images of $h$ and $h'$ under the natural map $G(R) \to G(R/I)$ coincide.
Obviously this gives an equivalence relation on $G(R)$,
and if $G$ is a group scheme, then multiplication preserves the relation.

\begin{lemma} \label{relation}
$x_\alpha(a)x_\beta(b) \equiv x_\beta(b)x_\alpha(a) \mod ab$.
\end{lemma}
\begin{proof}
It is enough to prove Lemma for  the algebra $K[a,b]$, because we can apply
the functor $G$ to the commutative diagram
$$
\begin{CD}
K[a,b] @>>> K[a,b]/(ab)\\
@VVV @VVV \\
R@>>> R/(ab),\\
\end{CD}
$$
and if the images of the left and right parts of the equivalence coincide
in $K[a,b]/(ab)$, then they coincide in $R/(ab)$.

The ideals $(a)$ and $(b)$ are coprime in $K[a,b]$,
i.e. the product $(a)(b)$ is equal to the intersection $(a)\cap(b)$.
Therefore the natural map $$K[a,b]/(ab) \to K[a,b]/(a)\oplus K[a,b]/(b)$$
is injective. So the map
$$G(K[a,b]/(ab)) \to G(K[a,b]/(a)\oplus K[a,b]/(b))$$
is injective,
and it is enough to prove that the images of $x_\alpha(a)x_\beta(b)$
and $x_\beta(b)x_\alpha(a)$ in
$G(K[a,b]/(a)\oplus K[a,b]/(b))=G(K[a,b]/(a))\times G(K[a,b]/(b))$
coincide. In other words, it is enough to check that
$x_\alpha(a)x_\beta(b)$ is equal to $x_\beta(b)x_\alpha(a)$ modulo
$a$ and modulo $b$, that is obvious.
\end{proof}

\begin{mlemma} \label{mlemma}
Let $R$ be a $K$-algebra, where $K$ is an algebraically closed field.
Let $H$ be a subgroup of $G(\Phi,R)$ containing $T(R)$,
$h$ be an element of $H$ such that 
$h=\prod_{\beta \in \Phi} x_\beta(a_\beta)$, where
$a_\beta \in \Nil(R)$. Then for each root $\beta$
the element $x_\beta(a_\beta)$ lies in the group $H$.
\end{mlemma}

To prove Main Lemma \ref{mlemma} we use several sublemmas.

\begin{sublemma} \label{tor}
For any element $\varepsilon \in K^*$ and any root $\beta \in \Phi$
there is an element $t \in T(K)$ such that $\beta(t)=\varepsilon$.
\end{sublemma}
\begin{proof} Since $\langle {\ },\rangle$ is non-degenerated, we can choose
a cocharacter $\lambda$ such that $\langle\beta,\lambda\rangle\neq 0$.
Set $t=\lambda(\varepsilon^{\frac 1{\langle\beta,\lambda\rangle}})$.
\end{proof}

\begin{sublemma} \label{2root}
Let $\alpha$ and $\gamma$ be non-collinear roots. Then
there is an element $t \in T(K)$ such that $\alpha(t)\neq 1$, $\gamma(t)=1$.
\end{sublemma}
\begin{proof}
Since $\langle {\ },\rangle$ is non-degenerated, we can choose
a cocharacter $\lambda$ such that $\langle\alpha,\lambda\rangle\neq 0$
and $\langle\gamma,\lambda\rangle=0$. Set $t=\lambda(\varepsilon)$,
where $\varepsilon^{\langle\gamma,\lambda\rangle}\neq1$.
\end{proof}

\begin{sublemma} \label{sublemma1}
In the settings of Main Lemma \ref{mlemma}
let $I \le \Nil(R)$ be an ideal.
Suppose that $a_\alpha \in I^r$ for some root $\alpha$, and
$a_\beta \in I^s$ for $\beta \neq \alpha$,
$1 \leq r \leq s$. Then there is an element $h'$ in $H$ such that
$h'=\prod_{\beta \in \Phi} x_\beta(a'_\beta)$, where
$a'_\alpha \equiv a_\alpha \mod I^{r+1}$, $a'_{-\alpha} \in I^{s+1}$,
and $a'_\beta \in I^{s}$ for $\beta \neq \pm \alpha$. 
\end{sublemma}
\begin{proof}
Since the field $K$ is algebraically closed, there are elements
$\varepsilon$ and $\xi$ in $K^*$ such that $\varepsilon+\xi+1 \in K^*$
and $\varepsilon^{-1}+\xi^{-1}+1=0$.
By Sublemma \ref{tor}, there are $t_1, t_2, t_3 \in T(K)$
such that $\alpha(t_1)=\varepsilon$, $\alpha(t_2)=\xi$, and
$\alpha(t_3)=(\varepsilon+\xi+1 )^{-1}$.

Since $h \equiv 1 \mod \Nil(R)$, we have
$t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}\equiv 1 \mod \Nil(R)$, so
the element $t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}$ lies in
$\Ker(F^M: G \to F^M(G))(R)$ for some $M$. Then, by Lemma \ref{EGauss},
the element $t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}$ lies in the big cell,
and therefore we have a decomposition
$$t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}=t_4\prod_{\beta \in \Phi} x_\beta(a'_\beta)$$
for some $t_4 \in T(R)$ and $a'_\beta \in R$.
Consider an element
$$h'=t_4^{-1}t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}=\prod_{\beta \in \Phi} x_\beta(a'_\beta).$$
It lies in the group $H$, because $H$ contains $h$ and $T(R)$.
On the other hand, we have
\begin{align*}
h'&=t_4^{-1}t_3((t_1ht_1^{-1})(t_2ht_2^{-1})h)t_3^{-1}\\
&=t_4^{-1}\prod_{\beta \in \Phi} x_\beta(\beta(t_1t_3)a_\beta)
\prod_{\beta \in \Phi} x_\beta(\beta(t_2t_3)a_\beta)
\prod_{\beta \in \Phi} x_\beta(\beta(t_3)a_\beta)\\
&\equiv t_4^{-1}\prod_{\beta \in \Phi} x_\beta(\beta(t_3)(\beta(t_1)+\beta(t_2)+1)a_\beta)
\mod I^{r+s}
\end{align*}
by Lemma \ref{relation}. Since Gauss decomposition is unique for the big cell
of the group $G(\Phi, R/I^{r+s})$, we have
$a'_\beta \equiv \beta(t_3)(\beta(t_1)+\beta(t_2)+1)a_\beta\mod I^{r+s}$ for each root
$\beta$. In particular, since $I^{r+1} \subseteq I^{r+s}$ and
$I^{s+1} \subseteq I^{r+s}$, we have $a'_\alpha \equiv a_\alpha\mod I^{r+1}$,
$a'_{-\alpha} \in I^{s+1}$, and $a'_\beta \in I^s$
for each $\beta \neq \pm \alpha$.
\end{proof}

\begin{sublemma} \label{sublemma2}
Under the assumptions of Sublemma \ref{sublemma1},
there is an element $h'$ in $H$ such that
$h'=\prod_{\beta \in \Phi} x_\beta(a'_\beta)$, where
$a'_\alpha \equiv a_\alpha \mod I^{r+1}$ and
$a'_\beta \in I^{s+1}$ for $\beta \neq \alpha$. 
\end{sublemma}
\begin{proof}
By Sublemma \ref{sublemma1}, we may assume that
$a_{-\alpha} \in I^{r+1}$.
     
Fix any root $\gamma \neq \pm \alpha$. By Sublemma \ref{2root},
there is an element $t_1\in T(K)$
such that $\gamma(t_1)=1$ and $\alpha(t_1)=\eta\neq 1$. 
Since $(1-\eta) \in R^*$, Sublemma \ref{tor} gives
an element $t_2 \in T(K)$ such that $\alpha(t_2)=(1-\eta)^{-1}$.

Since $h \equiv 1 \mod \Nil(R)$, we have
$t_2[h,t_1]t_2^{-1}\equiv 1 \mod \Nil(R)$, so
the element $t_2[h,t_1]t_2^{-1}$ lies in the big cell, and therefore
we have a decomposition
$$t_2[h,t_1]t_2^{-1}=t_3\prod_{\beta \in \Phi} x_\beta(a''_\beta)$$
for some $t_3 \in T(R)$ and $a''_\beta \in R$.
Consider an element
$$h''=t_3^{-1}t_2[h,t_1]t_2^{-1}=\prod_{\beta \in \Phi} x_\beta(a''_\beta).$$
It lies in the group $H$, because $H$ contains $h$ and $T(R)$.
On the other hand, we have
\begin{align*}
h''&=t_3^{-1}t_2[h,t_1]t_2^{-1}=t_3^{-1}(t_2ht_2^{-1})(t_1t_2h^{-1}t_2^{-1}t_1^{-1})\\
&=t_3^{-1}\prod_{\beta \in \Phi} x_\beta(\beta(t_2)a_\beta)
\prod_{-\beta \in \Phi} x_\beta(-\beta(t_1t_2)a_\beta)\\
&\equiv t_3^{-1}\prod_{\beta \in \Phi} x_\beta(\beta(t_2)(1-\beta(t_1))a_\beta) \mod I^{r+s}
\end{align*}
by Lemma \ref{relation}. Since Gauss decomposition is unique for the big cell
of the group $G(\Phi, R/I^{r+s})$, we have
$a''_\beta \equiv \beta(t_2)(1-\beta(t_1))a_\beta\mod I^{r+s}$ for each root
$\beta$. In particular, since $I^{r+1} \subseteq I^{r+s}$ and
$I^{s+1} \subseteq I^{r+s}$, we have $a''_\alpha \equiv a_\alpha\mod I^{r+1}$,
$a''_\gamma \in I^{s+1}$,
and if $a_\beta \in I^{s+1}$, then $a''_\beta \in I^{s+1}$.

Proceeding this argument for all roots $\gamma$ such that $\gamma \neq \pm \alpha$,
we obtain an element $h'$ required.
\end{proof}

\begin{sublemma} \label{sublemma3}
Under the assumptions of Sublemma \ref{sublemma1}, suppose that
the ideal $I$ is finitely generated. Then
there is an element $a'_\alpha \in R$ such that $a_\alpha \equiv a'_\alpha \mod I^{r+1}$
and $x_\alpha(a'_\alpha) \in H$.
\end{sublemma}
\begin{proof}
This follows from Sublemma \ref{sublemma2} by induction on $s$,
since some power of the ideal $I$ is equal to zero.
\end{proof}

\begin{sublemma} \label{sublemma4}
Under the assumptions of Sublemma \ref{sublemma1},
suppose that the ideal $I$ is finitely generated and $r=s$.
Then for any positive integer $q$ there is an element
$a'_\alpha \in R$ such that $a_\alpha \equiv a'_\alpha \mod I^{r+q}$
and $x_\alpha(a'_\alpha) \in H$.
\end{sublemma}
\begin{proof}
Let us prove the sublemma by induction on $q$. Sublemma \ref{sublemma3} implies
the assertion for $q=1$. Suppose now that the assertion holds for $q$ and
for each root $\alpha$, and prove it for $q+1$.

The element $h=\prod_{\alpha \in \Phi} x_\alpha(a_\alpha)$ lies in $H$,
and for each root $\alpha$ there is an element $a' \in R$ such that
$a \equiv a'\ \mod I^{r+q}$ and $x_\alpha(a') \in H$.
Consider an element
$(\prod_{\alpha \in \Phi} x_\alpha(a_\alpha))(\prod_{\alpha \in \Phi} x_\alpha(a'_\alpha))^{-1}$.
For each root $\alpha$ we have $a_\alpha \in \Nil(R)$ and $a'_\alpha \in \Nil(R)$,
so this element lies in the big cell, and therefore we have a decomposition
$$(\prod_{\alpha \in \Phi} x_\alpha(a_\alpha))
(\prod_{\alpha \in \Phi} x_\alpha(a'_\alpha))^{-1}=
t\prod_{\alpha \in \Phi} x_\alpha(b_\alpha)$$
for some $t \in T(R)$ and $a'_\alpha \in R$. Consider an element
$$h'=t^{-1}(\prod_{\alpha \in \Phi} x_\alpha(a_\alpha))
(\prod_{\alpha \in \Phi} x_\alpha(a'_\alpha))^{-1}=
\prod_{\alpha \in \Phi} x_\alpha(b_\alpha).$$
It lies in the group $H$, because $H$ contains $h$, $T(R)$, and
$x_\alpha(a')$ for each root $\alpha$. On the other hand, we have 
$$h'=t^{-1}(\prod_{\alpha \in \Phi} x_\alpha(a_\alpha))
(\prod_{\alpha \in \Phi} x_\alpha(a'_\alpha))^{-1}
\equiv t^{-1}\prod_{\alpha \in \Phi} x_\alpha(a_\alpha-a'_\alpha) \mod I^{2r+q}$$
by Lemma \ref{relation}, because
$x_\alpha(a_\alpha)x_\alpha(a'_\alpha)^{-1}=x_\alpha(a_\alpha-a'_\alpha)$,
and we permute only elements of type
$x_\alpha(a_\alpha-a'_\alpha)$, where $a_\alpha-a'_\alpha \in I^{r+q}$,
and elements of type $x_\alpha(a'_\alpha)$, where
$a'_\alpha \in I^r$. Since Gauss decomposition is unique for the big cell
of the group $G(\Phi, R/I^{2r+q})$, we have
$b_\alpha \equiv a_\alpha-a'_\alpha \mod I^{2r+q}$ for each root
$\alpha$, and, in particular, $b_\alpha \equiv a_\alpha-a'_\alpha \mod I^{r+q+1}$
and $b_\alpha \in I^{r+q}$.
Apply Sublemma \ref{sublemma3} to the element $h'$. Then
for each root $\alpha$ there is an element $a''_\alpha$ such that
$a''_\alpha \equiv b_\alpha \mod I^{r+q+1}$
and $x_\alpha(a''_\alpha) \in H$. Thus we have
$$x_\alpha(a'_\alpha+a''_\alpha)=x_\alpha(a'_\alpha)x_\alpha(a''_\alpha) \in H$$
with $a'_\alpha+a''_\alpha \equiv a_\alpha \mod I^{r+q+1}.$
\end{proof}

\begin{proof}[Proof of Main Lemma \ref{mlemma}] This follows from
Sublemma \ref{sublemma4} for $r=s=1$,
$I=\langle a_\alpha \mid \alpha \in \Phi \rangle$, and for $q$ large enough,
because some power of $I$ is equal to zero.
\end{proof}

\section{Determining of a function $\varphi$} \label{varphi}
\indent

In this section we determine a function $\varphi$ corresponding
to a given connected subgroup scheme $H$ in $G(\Phi)$ that contains $T$
and check that $\varphi$ satisfies the condition $(*)$.

\begin{lemma} \label{decomp}
Under the assumptions of Lemma \ref{HGauss}, the following equality holds:
$H=G(S)\prod_{\beta \in \Phi}(H \cap x_\beta(\alpha_{p^N}))$.
\end{lemma}
\begin{proof}
By Lemmas \ref{HGauss} and \ref{EGauss}, it is enough to prove the equality
$$H\cap (T_N \prod_{\beta \in \Phi} x_\beta(\alpha_{p^N}))
=T_N\prod_{\beta \in \Phi}(H \cap x_\beta(\alpha_{p^N})),$$ which follows
immediately from Main Lemma \ref{mlemma}. 
\end{proof}

Any subgroup scheme of the scheme $\alpha_{p^N}$ is a scheme $\alpha_{p^M}$
for some $M \leq N$.
Define a function $\varphi: \Phi \to \mathbb{N}\cup \{0, \infty \}$ as follows:\\
\indent if $\beta \in S$, set $\varphi(\beta)=\infty$;\\
\indent otherwise $\varphi(\beta)$ is determined by $H\cap x_\beta(\alpha_{p^N})=x_\beta(\alpha_{p^{\varphi(\beta)}})$.

\begin{lemma} \label{trans}
$x_\beta(G_a) \cap H=x_\beta(\alpha_{p^{\varphi(\beta)}})$.
\end{lemma}
\begin{proof}
The inclusion $x_\beta(G_a) \cap H \supseteq x_\beta(\alpha_{p^{\varphi(\beta)}})$
is obvious. Let us prove the other inclusion. The case $\beta \in S$ (in other words,
$\varphi(\beta)=\infty$) is trivial, because we have the inclusions
$x_\beta(G_a) \subseteq G(S) \subseteq H$.

Assume that $\beta \notin S$. We have
$F^N(x_\beta(G_a) \cap H)\subseteq F^N(H)=F^N(G)$, so
$F^N(x_\beta(G_a) \cap H)\subseteq F^N(G) \cap F^N(x_\beta(G_a))$.
Since $F$ preserves root data and $\beta \notin S$,
we have $F^N(G) \cap F^N(x_\beta(G_a))=1$. Therefore
$x_\beta(G_a) \cap H$ lies in the kernel
$\Ker(F^N:G(\Phi) \to F^N(G(\Phi)))$. So
$x_\beta(G_a) \cap H \subseteq x_\beta(\alpha_{p^N})$.
By the setting of $\varphi$, this implies
$x_\beta(G_a) \cap H=x_\beta(\alpha_{p^{\varphi(\beta)}})$.
\end{proof}

\begin{lemma} \label{ineq}
The function $\varphi$ satisfies the inequality $(*)$.
\end{lemma}
\begin{proof} 
Consider the $K$-algebra $R=K[a,b]/(a^{p^{\varphi(\alpha)}},b^{p^{\varphi(\beta)}})$.
Elements $x_{\alpha}(a)$ and $x_{\beta}(b)$ lie in $H(R)$.
Since $H(R)$ is a group, the commutator
$$[x_{\alpha}(a),x_{\beta}(b)]=\prod_{r,s \geq 1} x_{r\alpha+s\beta}
(N_{\alpha \beta rs}a^rb^s)$$ lies in $H(R)$.
By Main Lemma \ref{mlemma}, we have
$x_{r\alpha+s\beta}(N_{\alpha \beta r s}ab) \in H(R)$
for any positive integers $r$ ans $s$ such that
$r\alpha+s\beta \in \Phi$. By Lemma \ref{trans},
this means that $(N_{\alpha \beta r s}ab)^{p^{\varphi(r\alpha+s\beta)}}=0$.
Obviously, this is equivalent to the condition $(*)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{theorem}]
In Section \ref{Hvarphi} we defined a scheme $H_\varphi$
that corresponds to a function $\varphi$ satisfying $(*)$.
In Lemma \ref{check} we prove that the schemes $H_\varphi$
satisfies all the conditions of Theorem \ref{theorem}.
Lemma \ref{Hcap} shows that all schemes $H_\varphi$ are distinct.
It remains to prove that any scheme $H$ satisfying the conditions of
Theorem \ref{theorem} is a scheme $H_\varphi$ for an appropriate function
$\varphi$ satisfying $(*)$. By Lemma \ref{alg_closed}, we may assume
$K$ is algebraically closed. In Section \ref{varphi}
we determine a function $\varphi$ corresponding to a given scheme $H$.
In Lemma \ref{ineq} we prove that the function $\varphi$ satisfies $(*)$.
By the definition of $\varphi$ we have $H=H_\varphi$,
and this completes the proof.
\end{proof}

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\end{document}
