$V:=T\cap U$. According to Lemma 6.17, $V\subset T$ and $V\subset U$ are
t-closed finite extensions such that $M=(V:S)=(V:T)=(V:U)\in\mathrm{Max}(V)$.
Set $k:=V/M,\ K:=T/M,\ L:=U/M\neq K,\ S^{\prime}:=S/M$ and $N^{\prime}:=N/M$.
Since $S/N\cong K\cong L$, we get that $R/P\subset K$ and $R/P\subset L$ are
finite separable field extensions, and so are $k\subset K$ and $k\subset L$.
Then, there exists $\alpha\in K$ such that $K=k[\alpha]$. Moreover, $K\subset
S^{\prime}$ and $L\subset S^{\prime}$ are distinct minimal ramified
extensions. We are in the situation of Lemma 6.27; so that, there exists $x\in
S^{\prime},\ x\neq 0$ such that $x^{2}=0$ with $S^{\prime}=K[x]=L[x]$, where
$N^{\prime}=Kx=Lx$ is the maximal ideal of $S^{\prime}$. In particular,
$N^{\prime 2}=0$. Furthermore, there exists a $k$-isomorphism of $k$-algebras
$\varphi:K\to L$ defined, for any $a\in K$, by $\varphi(a)=b$, where $b$ is
the unique element of $L$ such that $ax=bx$. Set $\beta:=\varphi(\alpha)$,
which gives $L=k[\beta]$. Since $\alpha x=\beta x$ in $S^{\prime}$, we have
$(\alpha-\beta)x=0$; so that, $y:=\alpha-\beta\in N^{\prime}$ satisfies
$y^{2}=0$ with $y\neq 0$. Let $Q(X):=\sum_{i=0}^{n}a_{i}X^{i}\in k[X]$ be the
minimal monic polynomial of $\alpha$, which is also the minimal monic
polynomial of $\beta$ through $\varphi$. Then,
$Q(\alpha)=Q(y+\beta)=0=\sum_{i=0}^{n}a_{i}(y+\beta)^{i}=\sum_{i=0}^{n}a_{i}\beta^{i}+\sum_{i=0}^{n}ia_{i}y\beta^{i-1}=Q(\beta)+yQ^{\prime}(\beta)=yQ^{\prime}(\beta)$,
with $Q^{\prime}(\beta)\in k[\beta]=L$, which is a field. It follows that
$Q^{\prime}(\beta)=0$ since $y\neq 0$. But $Q(X)$ is a separable polynomial, a
contradiction with $Q(X)$ the minimal polynomial of $\beta$. To conclude, our
assumptions do not hold and there do not exist $T,U\in[R,S],\ T\neq U$ such
that $T\subset S$ and $U\subset S$ are minimal ramified with the same crucial
maximal ideal. ∎

Contrary to Proposition 6.30, Example 6.29 shows that there exist
$T,U\in[R,S],\ T\neq U$ such that $T\subset S$ and $U\subset S$ are minimal
decomposed with the same crucial maximal ideal, where $(R,P)$ is a local ring,
$\mathrm{Max}(S)=\\{M_{1},M_{2}\\}$ and $R/P\subset S/M_{i}$ is a finite
separable field extension for $i=1,2$. It is enough to take $R:=\mathbb{Q},\
T:=K$ and $U:=L$; so that, $S/M_{i}\cong K\cong L$ for $i=1,2,\
R/P=\mathbb{Q}$ and $R/P\subset S/M_{i}$ is a finite separable field extension
for $i=1,2$.

##  7\. Cardinality of the lattices of integral FIP extensions

Let $R\subset S$ be an integral FIP extension. We are going to give
information about $|[R,S]|$. Because $R\subset S$ is a
$\mathcal{B}$-extension, it is enough to assume that $R$ is a local ring. In
order to examine the elements of $[R,S]$, we use the different closures (the
t-closure and the seminormalization) we studied in the previous sections to
get bounds for $|[R,S]|$.

Let $R\subset S$ be an integral FCP extension and $T$ its seminormalization
(resp.; t-closure). When it exists let $T^{\prime}$ be its co-subintegral
(resp.; co-infra-integral) closure. Define the map
$\varphi:[R,S]\to[R,T]\times[T,S]$ by $\varphi(U):=(T\cap U,TU)$ for any
$U\in[R,S]$. Recall that $T^{o}$ denotes the complement of $T$ in $[R,S]$ when
it exists and is unique. The following Proposition gives relations between
$\varphi$, the existences of a complement of $T$ and $T^{\prime}$.

######  Proposition 7.1.

Let $R\subset S$ be an integral FCP. Consider the following conditions:

  1. (1)

$\varphi$ is bijective.

  2. (2)

$\varphi$ is surjective.

  3. (3)

$T$ admits a complement.

  4. (4)

$T^{o}$ exists.

  5. (5)

$T^{\prime}$ exists.

We have the following implications: (1) $\Rightarrow$ (2) $\Rightarrow$ (3),
(1) $\Rightarrow$ (4) $\Rightarrow$ (3) and (3)+(5) $\Rightarrow$ (4) with
$T^{\prime}=T^{o}$ in this last case.

###### Proof.

We make the proof for $T={}_{S}^{+}R$.

(1) $\Rightarrow$ (2) is obvious.

(2) $\Rightarrow$ (3): Since $\varphi$ is surjective, there exists $U\in[R,S]$
such that $\varphi(U)=(R,S)=(T\cap U,TU)$; so that, $T\cap U=R$ and $TU=S$,
which shows that $U$ is a complement of $T$.

(1) $\Rightarrow$ (4): Since $\varphi$ is bijective, the previous proof shows
that $T$ has a unique complement.

(3) $+$ (5) $\Rightarrow$ (4): Assume that $T$ has a complement $V$ and that
the co-subintegral closure $T^{\prime}$ exists. By definition, $TV=S$ implies
that $V\subseteq S$ is subintegral; so that, $T^{\prime}\subseteq V$. Assume
that $T^{\prime}\neq V$. Since $R\subseteq V$ is seminormal, so is
$T^{\prime}\subseteq V$ and $T^{\prime}\subseteq V$ is subintegral because so
is $T^{\prime}\subseteq S$. Then, $T^{\prime}=V$, and $T$ has a unique
complement $T^{o}=T^{\prime}$.

When $T={}_{S}^{t}R$, it is enough to replace in the previous proof seminormal
with t-closed and subintegral with infra-integral. ∎

We begin to consider integral u-closed extensions.

######  Lemma 7.2.

Let $R\subset S$ be an integral FCP u-closed extension where $(R,M)$ is a
local ring such that $R\neq{}_{S}^{+}R\neq S$. Assume that $R\subseteq
S^{\prime}$ is t-closed for any $S^{\prime}\in[R,S]$ such that $MS^{\prime}$
is the maximal ideal of $S^{\prime}$. Then, ${}_{S}^{+}R$ has a complement.
If, in addition, this complement is unique, then it is the co-subintegral
closure of $R\subset S$, that is $({}_{S}^{+}R)^{o}=S^{+}_{R}$.

###### Proof.

Recall that ${}_{S}^{+}R={}_{S}^{t}R$ and
$\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is a bijection by Proposition 5.2.
Moreover, $R\subset S$ is unbranched and only ramified and inert minimal
extensions appear in a maximal chain of extensions composing $R\subset S$. Set
$\mathcal{F}:=\\{U\in[R,S]\mid U\subseteq S$ is subintegral$\\}$. We have
$\mathcal{F}\neq\emptyset$ because $S\in\mathcal{F}$. Since $R\subset S$ has
FCP, there exists some $U_{0}\in\mathcal{F}$ such that $\ell[U_{0},S]$ is
maximal; so that, there does not exists any $V\in[R,U_{0}]$ such that
$V\subset U_{0}$ is minimal ramified. As $R\subset S$ is not subintegral,
$R\neq U_{0}$. Then, according to [6, Lemma 17], $MU_{0}$ is a radical ideal
of $U_{0}$, that is an intersection of maximal ideals of $U_{0}$. But, $U_{0}$
is a local ring; so that, $MU_{0}$ is the maximal ideal of $U_{0}$, giving
that $R\subseteq U_{0}$ is t-closed by assumption. Then, $U_{0}\neq S$ because
$R\subset S$ is not t-closed. As $U_{0}\subset S$ is subintegral, we get that
${}_{S}^{+}R\cap U_{0}=R$ and ${}_{S}^{+}RU_{0}=S$; so that, $U_{0}$ is a
complement of ${}_{S}^{+}R$.

Assume, in addition, that this complement is unique and let $T$ be this
complement. Then, $T\in\mathcal{F}$. Of course, ${}_{S}^{+}R\cap T=R$ and
${}_{S}^{+}RT=S$ show that $R\subseteq T$ is t-closed and $T\subseteq S$ is
subintegral. Set $T^{\prime}:=\cap[U\in\mathcal{F}]$; so that,
$T^{\prime}\subseteq T$. If $T^{\prime}\neq T$, there exists some
$V\in\mathcal{F}$ with $V\not\in[T,S]$. Choose such $V\not\in[T,S]$ satisfying
that $\ell[V,S]$ is maximal. As we proved in the beginning of the proof for
$U_{0}$, we get that $V$ is a complement of ${}_{S}^{+}R$, a contradiction
with the uniqueness of this complement. Then, $T^{\prime}=T$ implies that
$T^{\prime}\subseteq S$ is subintegral; so that, $T^{\prime}$ is the co-
subintegral closure of $R\subset S$ by Theorem 6.15. ∎

######  Remark 7.3.

Lemma 6.17 shows that when ${}_{S}^{t}R$ has more than one complement, then
the co-subintegral closure of $R\subset S$ does not exists.

Contrary to Proposition 6.30, where separable field extensions appear, here is
an example illustrating Lemma 7.2 in the context of radicial field extensions.

######  Example 7.4.

Let $k$ be a field with $\mathrm{c}(k)=2$ and let $K:=k[t]$ be a radicial
field extension of degree 2 such that $t^{2}=a\in k$. Set
$S:=K[X]/(X^{2})=K[x]$, where $x$ is the class of $X$ in $S$, so that
$K\subset S$ is a minimal ramified extension and set $\alpha:=t+x$. Then,
$\alpha^{2}=t^{2}=a\in k$; so that, $L:=k[\alpha]$ is a radicial field
extension of $k$ of degree 2, with $Y^{2}-a$ the minimal polynomial of
$\alpha$ and $L$ is a field different from $K$ since $x\not\in K$. Moreover,
$L\subset S$ is a minimal ramified extension and $k\subset S$ has FCP. This
example illustrates also Lemma 6.27. The maximal ideal of $S$ is $Kx=Lx$.
Since $k$ is a field, its maximal ideal is $0$, and for any
$S^{\prime}\in[R,S]$ such that $0S^{\prime}=0$ is the maximal ideal of
$S^{\prime}$, we have that $S^{\prime}$ is a field; so that, $k\subseteq
S^{\prime}$ is t-closed. Then, $k\subseteq S$ satisfies the assumptions of
Lemma 7.2. In particular, $k^{\prime}:={}_{S}^{t}k=k+Kx=k+Lx$ because
$k^{\prime}/Lx\cong k$ shows that $k\subset k^{\prime}$ is subintegral, while
$k^{\prime}\subset S$ is t-closed because $(k^{\prime}:S)=Lx$, with $k\cong
k^{\prime}/Lx\subset S/Lx\cong L$ a field extension such that $L\cong K$. To
end, we remark that $K$ and $L$ are both complements of ${}_{S}^{t}k$.

In case $R\subset S$ is an integral u-closed FIP extension, the following
Proposition shows that when ${}_{S}^{+}R$ has a complement, it is unique and
equal to the co-subintegral closure of $R\subset S$ when this last one exists.

######  Proposition 7.5.

Let $R\subset S$ be an integral u-closed FIP extension. Assume that $(R,M)$ is
a local ring and $R\neq{}_{S}^{+}R\neq S$.

  1. (1)

If ${}_{S}^{+}R$ has a complement, then $MS\not\in\mathrm{Max}(S)$. Moreover,
this complement is unique. If, in addition, $S^{+}_{R}$ exists, then
$({}_{S}^{+}R)^{o}=S^{+}_{R}$.

  2. (2)

The map $\varphi:[R,S]\to[R,{}_{S}^{+}R]\times[{}_{S}^{+}R,S]$ defined by
$\varphi(U)=(U\cap{}_{S}^{+}R,{}_{S}^{+}RU)$ for $U\in[R,S]$ is injective. In
particular,
$|[R,{}_{S}^{+}R]|+|[{}_{S}^{+}R,S]|-1\leq|[R,S]|\leq|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$.

###### Proof.

(1) Let $T^{\prime}\in[R,S]$ be a complement of ${}_{S}^{+}R$. Since
$R=T^{\prime}\cap{}_{S}^{+}R$ and $S={}_{S}^{+}RT^{\prime}$, we have
$T^{\prime}\neq R,S$ because $R\subset S$ is neither subintegral nor t-closed.
In particular, $T^{\prime}\subset S$ is subintegral; so that, there exists
some $U\in[T^{\prime},S]$ such that $U\subset S$ is minimal ramified. Then,
[6, Lemma 17] shows that $MS$ is not a radical ideal of $S$, and in
particular, $MS\not\in\mathrm{Max}(S)$.

Let $N$ be the maximal ideal of $S$, which is a local ring because $R\subset
S$ is u-closed. Assume now that there exist another complement
$T^{\prime\prime}\in[R,S]$ of ${}_{S}^{+}R$. Set
$T:=T^{\prime}T^{\prime\prime}\subseteq S$. Hence, $M$ is also the maximal
ideal of $T^{\prime}$ and $T^{\prime\prime}$ since $R\subset T^{\prime}$ and
$R\subset T^{\prime\prime}$ are t-closed, and also an ideal of $T$, a local
ring since $R\subset S$ is an i-extension. Set $k:=R/M,\
K^{\prime}:=T^{\prime}/M,\ K^{\prime\prime}:=T^{\prime\prime}/M$ and
$K:=T/M=K^{\prime}K^{\prime\prime}$; so that, $k,K^{\prime}$ and
$K^{\prime\prime}$ are fields, with $K^{\prime},K^{\prime\prime}\cong S/N$,
because $T^{\prime}\subset S$ and $T^{\prime\prime}\subset S$ are subintegral.
Moreover, $k\subset K$ has FIP, since so has $R\subset S$, and $K$ is a local
Artinian ring. We claim that $K$ cannot be a field, because $K^{\prime}\neq
K^{\prime\prime}$. Otherwise, if $K$ is a field and $K^{\prime}\neq
K^{\prime\prime}$, then $K^{\prime}\subset K$, and $T^{\prime}\subset T$; so
that, $M=(T^{\prime}:T)\in\mathrm{Max}(T)$, which gives that
$T^{\prime}\subset T$ is t-closed, a contradiction because $T^{\prime}\subset
S$ is subintegral. Using [1, Theorem 3.8], we have to look at two cases.
Assume first that $|k|=\infty$. Then, $K=k[\alpha]$, for some $\alpha\in K$
which satisfies $\alpha^{3}=0$. In this case, $\dim_{k}(K)\leq 3$ and is a
prime number. But $\dim_{k}(K)=\dim_{K^{\prime}}(K)[K^{\prime}:k]$ gives that
either $\dim_{K^{\prime}}(K)=1$ or $[K^{\prime}:k]=1$. This last condition
cannot happen since $k\neq K^{\prime}$; so that, $\dim_{K^{\prime}}(K)=1$,
which gives $K=K^{\prime}=K^{\prime\prime}$, a contradiction. At last, if $k$
is a finite field, so are $K^{\prime}$ and $K^{\prime\prime}$. Moreover, they
are isomorphic as $k$-algebras and $k\subset K^{\prime},K^{\prime\prime}$ are
Galois. According to [7, Proposition 7.8], we get a contradiction since
$T^{\prime}\subset S$ is subintegral. Then, $T^{\prime}=T^{\prime\prime}$.

Now $T^{\prime}=S^{+}_{R}$ by Proposition 7.1 when $S^{+}_{R}$ exists.

(2) We next consider $\varphi:[R,S]\to[R,{}_{S}^{+}R]\times[{}_{S}^{+}R,S]$
defined by $\varphi(T)=(T\cap{}_{S}^{+}R,{}_{S}^{+}RT)$ for $T\in[R,S]$. Set
$R^{\prime}:=T\cap{}_{S}^{+}R$ and $S^{\prime}:={}_{S}^{+}RT$, then
${}_{S}^{+}R={}_{S^{\prime}}^{+}R^{\prime}$. Let $T^{\prime}\in[R,S]$ be such
that $\varphi(T)=\varphi(T^{\prime})$. Then $T$ and $T^{\prime}$ are both
complements of ${}_{S^{\prime}}^{+}R^{\prime}$ in $[R^{\prime},S^{\prime}]$.
It follows that $T=T^{\prime}$, applying (1) to the extension
$R^{\prime}\subseteq S^{\prime}$; so that, $\varphi$ is injective. So we get
$|[R,S]|\leq|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$ and
$|[R,{}_{S}^{+}R]|+|[{}_{S}^{+}R,S]|-1\leq|[R,S]|$ comes from
$[R,{}_{S}^{+}R]\cup[{}_{S}^{+}R,S]\subseteq[R,S]$ with
$[R,{}_{S}^{+}R]\cap[{}_{S}^{+}R,S]=\\{{}_{S}^{+}R\\}$. ∎

######  Corollary 7.6.

Let $R\subset S$ be an integral u-closed FIP extension. Assume that $(R,M)$ is
a local ring and $R\neq{}_{S}^{+}R\neq S$.

If $R^{\prime}\subseteq S^{\prime\prime}$ is t-closed for any
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{+}R[\times]{}_{S}^{+}R,S]$ and for any
$S^{\prime\prime}\in[R^{\prime},S^{\prime}]$ such that
$M^{\prime}S^{\prime\prime}\in\mathrm{Max}(S^{\prime\prime})$, where
$\mathrm{Max}(R^{\prime})=\\{M^{\prime}\\}$, then
$|[R,S]|=|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$.

###### Proof.

First observe that
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{+}R]\times[{}_{S}^{+}R,S]$ implies that
${}_{S^{\prime}}^{+}R^{\prime}={}_{S}^{+}R$. Since $R\subset S$ is an
i-extension, any element of $[R,S]$ is a local ring. Using the notation of the
proof of Proposition 7.5, we have $|[R,S]|=|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$
if and only if $\varphi$ is surjective, if and only if for any
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{+}R]\times[{}_{S}^{+}R,S]$, there exists
$T\in[R,S]$ such that $R^{\prime}=T\cap{}_{S}^{+}R$ and
$S^{\prime}={}_{S}^{+}RT\ (*)$. If $R^{\prime}={}_{S}^{+}R$, it is enough to
take $T=S^{\prime}$, and if $S^{\prime}={}_{S}^{+}R$, it is enough to take
$T=R^{\prime}$. Then $(*)$ holds also.

Assume that for any
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{+}R[\times]{}_{S}^{+}R,S]$ and for any
$S^{\prime\prime}\in[R^{\prime},S^{\prime}]$ such that
$M^{\prime}S^{\prime\prime}\in\mathrm{Max}(S^{\prime\prime})$, where
$\mathrm{Max}(R^{\prime})=\\{M^{\prime}\\}$, this implies that
$R^{\prime}\subseteq S^{\prime\prime}$ is t-closed. We have
$R^{\prime}\neq{}_{S^{\prime}}^{+}R^{\prime}={}_{S^{\prime}}^{t}R^{\prime}\neq
S^{\prime}$. Applying Lemma 7.2 to the extension $R^{\prime}\subseteq
S^{\prime}$, we get that there exists
$T\in[R^{\prime},S^{\prime}]\subseteq[R,S]$ such that
$R^{\prime}=T\cap{}_{S}^{+}R$ and $S^{\prime}={}_{S}^{+}RT$ and $(*)$ holds.
It follows that $|[R,S]|=|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$. ∎

######  Proposition 7.7.

Let $R\subset S$ be a seminormal integral FCP extension such that $(R,M)$ is a
local ring and ${}_{S}^{t}R\neq R,S$. The following statements hold:

  1. (1)

$S$ is a semilocal ring whose maximal ideals are $N_{1},\ldots,N_{p}$. Then,
$(R:S)=M=\cap_{i\in\mathbb{N}_{p}}N_{i}$.

  2. (2)

${}_{S}^{t}R$ has a complement in $[R,S]$ if and only if $S/N_{i}\cong
S/N_{j}$, as $R/M$-algebras, for each $i,j\in\mathbb{N}_{p}$.

  3. (3)

If, in addition, this complement is unique, then it is the co-infra-integral
closure of $R\subset S$.

###### Proof.

(1) $S$ is a semilocal ring because $R\subset S$ is a finite extension. Let
$N_{1},\ldots,N_{p}$ be its maximal ideals. By Proposition 2.8, $(R:S)$ is an
intersection of finitely many maximal ideals of $S$ and of $R$; so that,
$(R:S)=M=\cap_{i\in\mathbb{N}_{p}}N_{i}$.

(2) Since $(R:S)=M$ is an ideal of $S$, let $\rho:S\to S/M$ be the canonical
map. As $M=\cap_{i\in\mathbb{N}_{p}}N_{i}$, we get that
$S^{\prime}:=S/M\cong\prod_{i\in\mathbb{N}_{p}}S/N_{i}$. Set $k:=R/M$ and
$K_{i}:=S/N_{i}$ for each $i\in\mathbb{N}_{p}$, which are field extensions of
$k$. Moreover, ${}_{S^{\prime}}^{t}k=({}_{S}^{t}R)/M$ by [17, Proposition
2.10].

Let $T\in[R,S]$; so that,
$T^{\prime}:=T/M\in[k,\prod_{i\in\mathbb{N}_{p}}K_{i}]$, up to a $k$-algebra
isomorphism. It follows that $T$ is a complement of ${}_{S}^{t}R$ in $[R,S]$
if and only if $T^{\prime}$ is a complement of ${}_{S^{\prime}}^{t}k$ in
$[k,S^{\prime}]$ if and only if $T^{\prime}\cap{}_{S^{\prime}}^{t}k=k$ and
$T^{\prime}{}_{S^{\prime}}^{t}k=S^{\prime}$. Assume that this condition holds.
Then, $k\subset T^{\prime}$ is t-closed; so that, $T^{\prime}$ is a field
extension of $k$ by [17, Lemme 3.10] and $T^{\prime}\subset S^{\prime}$ is
seminormal infra-integral, giving that $K_{i}\cong T^{\prime}$, as
$k$-algebras, for each $i\in\mathbb{N}_{p}$; so that, $S/N_{i}\cong S/N_{j}$,
as $R/M$-algebras, for each $i,j\in\mathbb{N}_{p}$.

Conversely, assume that $S/N_{i}\cong S/N_{j}$, as $R/M$-algebras, for each
$i,j\in\mathbb{N}_{p}$; so that, $K_{i}\cong K_{j}$, as $k$-algebras, for each
$i,j\in\mathbb{N}_{p}$. Set, for example, $K:=K_{1}$; so that,
$S^{\prime}\cong K^{p}$. Let $T^{\prime}\in[k,S^{\prime}]$ be such that
$T^{\prime}\cong K$ through the diagonal map; so that, $k\subset T^{\prime}$
is t-closed and $T^{\prime}\subset S^{\prime}$ is seminormal infra-integral.
There exists $T\in[R,S]$ such that $\rho(T)=T^{\prime}$ and satisfying
$R\subset T$ is t-closed and $T\subset S$ is seminormal infra-integral. Then,
$T\cap{}_{S}^{t}R=R$ and $T{}_{S}^{t}R=S$, which shows that $T$ is a
complement of ${}_{S}^{t}R$.

(3) Assume, in addition, that the complement of ${}_{S}^{t}R$ is unique. We
are going to show that it is the co-infra-integral closure of $R\subset S$. We
keep the previous notation with $T$ the complement of ${}_{S}^{t}R$, and, to
make simpler the proof, assume that $S^{\prime}=K^{p}$ (which is satisfied up
to an isomorphism). Let $U\in[R,S]$ be such that $U\subseteq S$ is infra-
integral. Then, $U$ is a semilocal ring. Set
$\mathrm{Max}(U):=\\{M_{1},\ldots,M_{n}\\}$; so that,
$M=\cap_{j\in\mathbb{N}_{n}}M_{j}$. Since $U\subset S$ is infra-integral, we
have $U/M_{j}\cong S/N_{i}$ for any $N_{i}$ lying over $M_{j}$. But
$S/N_{i}\cong K$, as $R/M$-algebra, for each $i\in\mathbb{N}_{p}$ leads to
$U/M_{j}\cong K$, as $R/M$-algebra, for each $j\in\mathbb{N}_{n}$. Then,
$U/M\cong K^{n}$. In particular, there exists $V\in[R,U]$ such that
$\rho(V)\cong K$. This shows that $R\subseteq V$ is t-closed and $V\subseteq
S$ is seminormal infra-integral. This means that $V$ is a complement of
${}_{S}^{t}R$. By uniqueness of this complement, it follows that $V=T$, giving
$T\subseteq U$. Then, $T$ is the co-infra-integral closure of $R\subset S$. ∎

######  Remark 7.8.

Let $R\subset S$ be an FIP seminormal extension such that ${}_{S}^{t}R\neq
R,S$. Assume that $(R,M)$ is a local ring. Then, $S$ is semilocal since
$R\subset S$ is a finite extension. Let $N_{1},\ldots,N_{p}$ be the maximal
ideals of $S$. As in Corollary 7.6, we can define a map
$\varphi:[R,S]\to[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$ by
$\varphi(T):=(T\cap{}_{S}^{t}R,{}_{S}^{t}RT)$. But, contrary to the results of
this corollary, we are going to show that $\varphi$ is not always injective
and never surjective.

Assume first that there exists $T$ such that $R\subset T$ is t-closed and
$T\subset S$ is infra-integral; so that, $\varphi(T):=(R,S)$ and $T$ is a
complement of ${}_{S}^{t}R$. Then $M\in\mathrm{Max}(T)$. Each $N_{i}$ lies
over $M$ in $T$ and $M=\cap_{i=1}^{p}N_{i}=(R:S)$ by Proposition 2.8.
Moreover, $S/N_{i}\cong T/M$ for each $i=1,\ldots,p$. Set $k:=R/M,\ K:=T/M$
and $L:=S/M\cong\prod_{i=1}^{p}S/N_{i}\cong K^{p}$. Assume that the Galois
group of the field extension $k\subset K$ has more than one element and let
$\sigma$ be such an element different from the identity. Since $R\subset T$
has FIP, so has $k\subset K$, and there exists $x\in K$, algebraic over $k$,
such that $K=k[x]$ in view of the Primitive Element Theorem. In particular,
$\sigma(x)\neq x$. Set $y:=(\sigma(x),x,\ldots,x)\in K^{p}$ and
$K^{\prime}:=k[y]$. Then $K$ and $K^{\prime}$ are $k$-isomorphic distinct
$k$-subalgebras of $K^{p}$, and the inverse image $T^{\prime}$ of $K^{\prime}$
under the canonical map $S\to S/M$ is an $R$-subalgebra of $S$ such that
$R=T^{\prime}\cap{}_{S}^{t}R$ and $S={}_{S}^{t}RT^{\prime}$, with $T\neq
T^{\prime}$. It follows that $\varphi(T)=\varphi(T^{\prime})$ and $\varphi$ is
not injective. See also Example 6.29, where $\varphi(K)=\varphi(L)$ with
$K\neq L$. But, if $k\subset K$ is a radicial extension, its Galois group has
only one element, and $K^{p}$ has no $k$-subalgebra isomorphic to $K$ distinct
from $K$. Then there is a unique complement of ${}_{S}^{t}R$. This reasoning
still holds for any
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$. In this
case, $\varphi$ is injective.

Now, we claim that $\varphi$ is never surjective. For each
$i\in\mathbb{N}_{p}$, set $M_{i}:=N_{i}\cap{}_{S}^{t}R$. Then
${}_{S}^{t}R/M_{i}\cong R/M$. Let $R_{1}\in[{}_{S}^{t}R,S]$ be such that
${}_{S}^{t}R\subset R_{1}$ is minimal (inert) with conductor
$M_{1}\in\mathrm{Max}({}_{S}^{t}R)$ (after a suitable reordering), and set
$N^{\prime}_{j}:=N_{j}\cap R_{1}$ for $j\in\\{2,\ldots,p\\}$. It follows that
${}_{R_{1}}^{t}R={}_{S}^{t}R$ in view of Proposition 2.6 and $R\subset R_{1}$
is an FIP seminormal extension. Moreover,
$R_{1}/N^{\prime}_{j}\cong{}_{S}^{t}R/M_{j}\cong R/M$ for each
$j\in\\{2,\ldots,p\\}$. But $R_{1}/M_{1}\not\cong{}_{S}^{t}R/M_{1}$; so that,
$R_{1}/M_{1}\not\cong R_{1}/N^{\prime}_{j}$ for each $j\in\\{2,\ldots,p\\}$.
It follows by Proposition 7.7 that ${}_{R_{1}}^{t}R$ does not admits a
complement in $[R,R_{1}]$; so that, there does not exist $T$ such that
${}_{S}^{t}RT=R_{1}$ and $R={}_{S}^{t}R\cap T$. Then, $\varphi$ is not
surjective.

######  Proposition 7.9.

Let $R\subset S$ be an infra-integral FCP extension. The map
$\varphi:[R,S]\to[R,{}_{S}^{+}R]\times[{}_{S}^{+}R,S]$ defined by
$\varphi(T):=(T\cap{}_{S}^{+}R,{}_{S}^{+}RT)$ is injective.

In particular, if $R\subset S$ has FIP, then
$|[R,S]|\leq|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$.

###### Proof.

Let $U,U^{\prime}\in[R,S]$ be such that $\varphi(U)=\varphi(U^{\prime})$. Set
$R^{\prime}:=U\cap{}_{S}^{+}R$ and $S^{\prime}:={}_{S}^{+}RU$. Then,
${}_{S}^{+}R={}_{S^{\prime}}^{+}R^{\prime}$, with $R^{\prime}\subseteq
S^{\prime}$ infra-integral. Moreover, $U$ and $U^{\prime}$ are both
complements of ${}_{S^{\prime}}^{+}R^{\prime}$. As Theorem 6.6 asserts that
$R^{\prime}\subseteq S^{\prime}$ admits a co-subintegral closure $T$, it
follows from Proposition 7.1 that $T=U=U^{\prime}$; so that, $\varphi$ is
injective. Then, $|[R,S]|\leq|[R,{}_{S}^{+}R]||[{}_{S}^{+}R,S]|$ if $R\subset
S$ has FIP. ∎

The following proposition gives us another way to calculate $|[R,S]|$, when
$R\subseteq S$ is an integral FIP extension with $R$ local.

######  Proposition 7.10.

Let $R\subset S$ be an FIP integral extension, over the local ring $(R,M)$ and
set $\mathrm{Max}(S):=\\{N_{1},\ldots,N_{p}\\}$. Let $n[R,S]$ be the number of
complements of ${}_{S}^{t}R$ in $[R,S]$. The following conditions hold:

  1. (1)

There exists a greatest $R$-subalgebra $R^{\prime}$ of $S$ such that $M$ is an
ideal of $R^{\prime}$.

  2. (2)

If $S/N_{i}\not\cong S/N_{j}$ as $R/M$-algebras, for some $i\neq
j\in\mathbb{N}_{p}$, then $n[R,S]=0$.

  3. (3)

Assume that $n[R,S]\neq 0$. Then, any complement $T$ of ${}_{S}^{t}R$ is in
$[R,R^{\prime}]$ and satisfies $S/N_{i}\cong T/M$ for any
$i\in\mathbb{N}_{p}$. Let $P(X)\in(R/M)[X]$ be the minimal monic polynomial of
some $y\in T/M$ such that $T/M=(R/M)[y]$. For any $i\in\mathbb{N}_{p}$, there
exists some $y_{i}\in S/N_{i}$ such that $S/N_{i}=(R/M)[y_{i}]$ where $P(X)$
is also the minimal monic polynomial of $y_{i}$. Moreover,
$n[R,S]=|\\{(R/M)[x]\mid x\in R^{\prime}/M,\ P(x)=0\\}|$.

###### Proof.

First observe that $S$ is semilocal by Propositions 2.6 and 2.8.

(1) Set $\mathcal{R}:=\\{T\in[R,S]\mid M$ is an ideal of $T\\}$. Then
$\mathcal{R}$ is a finite nonempty set. It has maximal elements. Assume that
$R^{\prime}$ and $R^{\prime\prime}$ are two maximal elements of $\mathcal{R}$.
If $R^{\prime}\neq R^{\prime\prime}$, then $R^{\prime}\subset
R^{\prime}R^{\prime\prime}$ with $R^{\prime}R^{\prime\prime}\in\mathcal{R}$, a
contradiction. Then $\mathcal{R}$ has a unique maximal element, which is its
greatest element. Let $R^{\prime}$ be this element.

(2) Assume that $S/N_{i}\not\cong S/N_{j}$ as $R/M$-algebras, for some $i\neq
j\in\mathbb{N}_{p}$ and that $n[R,S]\neq 0$. Then ${}_{S}^{t}R$ admits some
complement $T$ such that $R\subseteq T$ is t-closed; so that, $T$ is local
with maximal ideal $M$ and $T\subseteq S$ is infra-integral. It follows that
$S/N_{i}\cong T/M$ for each $i\in\mathbb{N}_{p}$. In particular, $S/N_{i}\cong
S/N_{j}$ as $R/M$-algebras, for each $i\neq j\in\mathbb{N}_{p}$, a
contradiction, whence $n[R,S,]=0$.

(3) According to (2), $n[R,S]\neq 0$ implies that $S/N_{i}\cong S/N_{j}$ as
$R/M$-algebras, for each $(i,j)\in\mathbb{N}_{p}$. Assume that ${}_{S}^{t}R$
admits some complement $T$; so that, $R\subseteq T$ is t-closed. Therefore,
$T$ is local with maximal ideal $M$, whence $M=(R:T)$, giving
$T\in[R,R^{\prime}]$. Then $R/M\subseteq T/M$ is a finite FIP field extension;
so that, there exists some $y\in T/M$ such that $T/M=(R/M)[y]$. Since
$T\subseteq S$ is infra-integral; so that, $S/N_{i}\cong T/M$ for each
$i\in\mathbb{N}_{p}$, let $\varphi_{i}:T/M\to S/N_{i}$ be the $R/M$-algebra
isomorphism, and set $y_{i}:=\varphi_{i}(y)$ for each $i\in\mathbb{N}_{p}$. It
follows that
$S/N_{i}=\varphi_{i}(T/M)=\varphi_{i}(R/M[y])=R/M[\varphi_{i}(y)]=R/M[y_{i}]$.
Let $P(X)\in(R/M)[X]$ be the minimal monic polynomial of $y$, then $P(X)$ is
also the minimal monic polynomial of $y_{i}$ and $T/M\cong(R/M)[X]/(P(X))\cong
S/N_{i}$ for each $i\in\mathbb{N}_{p}$.

Let $T^{\prime}$ be another complement of ${}_{S}^{t}R$. As we have just seen
before, $T^{\prime}\in[R,R^{\prime}]$ and is local with maximal ideal $M$ and
$T^{\prime}\subseteq S$ is infra-integral, which implies that $S/N_{i}\cong
T^{\prime}/M$ for each $i\in\mathbb{N}_{p}$. Then, there exists an
$R/M$-algebra isomorphism $\theta:T\to T^{\prime}$. Setting $x:=\theta(y)$, we
get that $P(X)$ is also the minimal monic polynomial of $x$.

Conversely, let $x^{\prime}\in R^{\prime}/M$ be such that $P(x^{\prime})=0$
and let $T^{\prime}\in[R,R^{\prime}]$ be the inverse image of
$(R/M)[x^{\prime}]$ by the canonical map $\rho:R^{\prime}\to R^{\prime}/M$.
Then $T^{\prime}$ is an $R$-subalgebra of $S$ with maximal ideal $M$, because
$T^{\prime}/M=(R/M)[x^{\prime}]\cong(R/M)[X]/(P(X))$, which is a field; so
that, $R\subseteq T^{\prime}$ is t-closed, since $M=(R:T^{\prime})$. As we
have $T^{\prime}/M\cong(R/M)[X]/(P(X))\cong S/N_{i}$ for each
$i\in\mathbb{N}_{p}$, it follows that $T^{\prime}\subseteq S$ is infra-
integral. Then $T^{\prime}$ is a complement of ${}_{S}^{t}R$. To end, let
$x,x^{\prime}\in R^{\prime}/M$ be such that $P(x)=P(x^{\prime})=0$. If
$(R/M)[x]=(R/M)[x^{\prime}]$, their inverse images by $\rho$ give the same
$T$. Then $n[R,S]=|\\{(R/M)[x]\mid x\in R^{\prime}/M,\ P(x)=0\\}|$. ∎

We deduce the following theorem.

######  Theorem 7.11.

Let $R\subset S$ be an FIP integral extension, over the local ring $(R,M)$.
For each $(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$, let
$n[R^{\prime},S^{\prime}]$ be the number of complements of
${}_{S^{\prime}}^{t}R^{\prime}$ in $[R^{\prime},S^{\prime}]$. Then
$|[R,S]|=\sum_{(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]}n[R^{\prime},S^{\prime}]$.

###### Proof.

Let $(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$. As we
already have seen, we have ${}_{S}^{t}R={}_{S^{\prime}}^{t}R^{\prime}$.

Let $T\in[R,S]$. Set $R^{\prime}:=T\cap{}_{S}^{t}R$ and
$S^{\prime}:={}_{S}^{t}RT$; so that,
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$. Then,
$T\in[R^{\prime},S^{\prime}]$ and is a complement of
${}_{S^{\prime}}^{t}R^{\prime}={}_{S}^{t}R$ in $[R^{\prime},S^{\prime}]$.
Moreover, for any
$(R^{\prime\prime},S^{\prime\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$
such that $(R^{\prime\prime},S^{\prime\prime})\neq(R^{\prime},S^{\prime})$, we
get that $T$ is not a complement of
${}_{S^{\prime\prime}}^{t}R^{\prime\prime}={}_{S}^{t}R$ in
$[R^{\prime\prime},S^{\prime\prime}]$. Consider the map
$\varphi:[R,S]\to[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$ defined by
$\varphi(T):=(T\cap{}_{S}^{t}R,{}_{S}^{t}RT)$. Since
$n[R^{\prime},S^{\prime}]=|\\{T\in[R^{\prime},S^{\prime}]\mid
R^{\prime}=T\cap{}_{S^{\prime}}^{t}R^{\prime},\
S^{\prime}={}_{S^{\prime}}^{t}R^{\prime}T\\}|$, we get that
$n[R^{\prime},S^{\prime}]=|\\{T\in[R,S]\mid\varphi(T)=(R^{\prime},S^{\prime})\\}|$.
It follows that
$|[R,S]|=\sum_{(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]}n[R^{\prime},S^{\prime}]$,
because $\varphi(T)=(R^{\prime},S^{\prime})$ implies that
$T\in[R^{\prime},S^{\prime}]$. We may remark that it can be that
$n[R^{\prime},S^{\prime}]=0$ for some
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$ according to
Proposition 7.10(2). ∎

######  Remark 7.12.

In the preceding theorem, we may define $n[R^{\prime},S^{\prime}]$ even if
$R^{\prime}$ is not a local ring. But, since $R^{\prime}\subseteq S^{\prime}$
is an integral FIP extension, it is a $\mathcal{B}$-extension. Moreover, for
any $M\in\mathrm{MSupp}_{R^{\prime}}(S^{\prime}/R^{\prime})$, we have
${}_{S^{\prime}_{M}}^{t}R^{\prime}_{M}=({}_{S^{\prime}}^{t}R^{\prime})_{M}$;
so that, for some $T\in[R,S]$, we have
$R^{\prime}=T\cap{}_{S^{\prime}}^{t}R^{\prime}$ and
$S^{\prime}={}_{S^{\prime}}^{t}R^{\prime}T$ if and only if
$R^{\prime}_{M}=T_{M}\cap{}_{S^{\prime}_{M}}^{t}R^{\prime}_{M}$ and
$S^{\prime}_{M}={}_{S^{\prime}_{M}}^{t}R^{\prime}_{M}T_{M}$ for any
$M\in\mathrm{MSupp}_{R^{\prime}}(S^{\prime}/R^{\prime})$. It follows that
$n[R^{\prime},S^{\prime}]=\prod_{i=1}^{n}n[R^{\prime}_{M_{i}},S^{\prime}_{M_{i}}]$,
which we can calculate using Proposition 7.10.

We now consider some types of integral FIP extensions. We still work with an
extension $R\subset S$ such that $(R,M)$ is a local ring since $R\subset S$ is
a $\mathcal{B}$-extension. In the following Proposition, we recall and
generalize a result of [5], using the Nagata rings $R(X)$ and $S(X)$.

######  Proposition 7.13.

Let $R\subset S$ be an FIP subintegral extension. Assume that $(R,M)$ is a
local ring.

  1. (1)

[5, Proposition 4.13] If $|R/M|=\infty$, then $R\subset S$ is chained and
$|[R,S]|=\ell[R,S]+1$.

  2. (2)

Assume that $|R/M|<\infty$ and let $n$ be the nilpotency index of $M/(R:S)$.
If $R+SM^{2}\subseteq S$ is chained and ${\mathrm{L}}_{R}(MS/M)=n-1$, then
$R\subset S$ is chained and $|[R,S]|=\ell[R,S]+1={\mathrm{L}}_{R}(N/M)$, where
$N$ is the maximal ideal of $S$.

###### Proof.

Since $R$ is a local ring and $R\subset S$ is an FIP subintegral extension,
$S$ is also a local ring.

(2) Assume that $|R/M|<\infty,\ R+SM^{2}\subseteq S$ is chained and
${\mathrm{L}}_{R}(MS/M)=n-1$. By [5, Corollary 3.31], we get that
$R(X)\subseteq S(X)$ has FIP. Moreover, $(R(X),MR(X))$ is a local ring such
that $|R(X)/MR(X)|=\infty$, and $R(X)\subset S(X)$ is a finite subintegral
extension [5, Lemma 3.15]. Then, (1) gives that $R(X)\subset S(X)$ is chained.
But the map $\varphi:[R,S]\to[R(X),S(X)]$, defined by $\varphi(T):=T(X)$ for
each $T\in[R,S]$, is an order-preserving and order-reflecting injection [5,
Lemma 3.1(d)]; so that, $R\subset S$ is also chained, giving
$|[R,S]|=\ell[R,S]+1={\mathrm{L}}_{R}(N/M)+1$ by [4, Lemma 5.4]. ∎

######  Remark 7.14.

(1) When $(R,M)$ is a local ring with $|R/M|<\infty$ such that $R(X)\subseteq
S(X)$ has not FIP, the previous reasoning does not hold, as well as the
equalities of Proposition 7.13 (2). We use [5, Example 3.12] to illustrate
this situation. Let $K$ be a finite field with $n:=|K|$. Set
$T:=K[Y]/(Y^{4})=K[y]$, where $y$ is the class of $Y$ in $T$; so that,
$y^{4}=0$. Then, $T$ is a finite local Artinian ring with maximal ideal
$N:=yT$. Moreover, $K\subset T$ is finite subintegral. Let $S\in[K,T]$ be such
that $S\subset T$ is minimal (ramified). Then, $S$ is also a finite local
Artinian ring. Let $M$ be its maximal ideal. Then, $N^{2}\subseteq M$ gives
$y^{2},y^{3}\in M$; so that, $K[y^{2},y^{3}]\subseteq S\subset T$ implies
$K[y^{2},y^{3}]=S$ by considering the dimensions of the different $K$-vector
spaces. Obviously, $K\subset K^{\prime}:=K[y^{3}]$ and $K\subset
K_{\alpha}:=K[y^{2}+\alpha y^{3}]$ are minimal extensions, all distinct, for
any $\alpha\in K$. Moreover, $K^{\prime}\subset S$ and $K_{\alpha}\subset S$
are also minimal extensions. Then,
$[K,T]=\\{K,K^{\prime},K_{\alpha},S,T\mid\alpha\in K\\}$; so that,
$|[K,T]|=n+4$. As $K\subset T$ is not chained, $K(X)\subset T(X)$ has not FIP,
according to [5, Corollary 3.31].

(2) We can add another property about $|[R,S]|$ for an FIP subintegral
extension $R\subseteq S$ when $(R,M)$ is a finite local ring. Let $N$ be the
maximal ideal of $S$ and let $T\in[R,S]$ be such that $T\subset S$ is a
minimal (ramified) extension. Then, $P:=(T:S)$ is the maximal ideal of $T$ and
an $N$-primary ideal of $S$ such that $P$ and $N$ are adjacent in $S$
(equivalently, $\dim_{R/M}(N/P)=1$ by Theorem 1.3). Assume that there exists
$T^{\prime}\in[R,S]$ such that $T^{\prime}\subset S$ is minimal with
$(T^{\prime}:S):=P$. Then, $T=T^{\prime}=R+P$. Conversely, if $P$ is an
$N$-primary ideal of $S$ such that $P$ and $N$ are adjacent in $S$ and
$M=P\cap R$, setting $S^{\prime}:=R+P$, we get that $S^{\prime}\subset S$ is a
minimal ramified extension. Then, the map $\psi:\\{P$ an $N$-primary ideal of
$S\mid P$ and $N$ adjacent, $M=P\cap R\\}\to\\{S^{\prime}\in[R,S]\mid
S^{\prime}\subset S$ minimal$\\}$ defined by $\psi(P):=R+P$ is a bijection.
Let $m$ be the number of $N$-primary ideals $P$ of $S$ such that $P$ and $N$
are adjacent and $M=P\cap R$. In particular, if ${\mathrm{L}}_{R}(N/M)=2$,
then $\ell[R,S]=2$ by [4, Lemma 5.4] and $|[R,S]|=m+2$. It follows by [22,
Theorem 6.1 (6)] that either $|[R,S]|=3\ (*)$ or $|[R,S]|=|R/M|+3\ (**)$. In
case $(*)$, we get that $R\subseteq S$ is chained and $m=1$. In case $(**)$,
we get that $|R/M|=m-1$.

######  Proposition 7.15.

Let $R\subset S$ be an FIP integral t-closed extension with $C:=(R:S)$. Then
$|[R,S]|=\prod_{M\in\mathrm{V}_{R}(C)}|[S/MS,R/M]|$. In fact, $|[S/MS,R/M]|$
is the number of $R/M$-subalgebras of $S/SM$.

###### Proof.

Since $R\subset S$ is t-closed, $C=(R:S)$ is an intersection of maximal ideals
of $R$ and $S$ by Proposition 2.8. Since $R\subset S$ is an i-extension, for
each $M\in\mathrm{V}_{R}(C)$, there is one and only one maximal ideal $N$ of
$S$ lying over $M$. We claim that $N$ is of the form $N=SM$. First,
$SM\subseteq N$. But $MR_{M}=MS_{M}=MS_{N}=NS_{N}$ by [2, Proposition 2, page
40] and Proposition 2.6, since $S_{M}=S_{N}$. Moreover, for
$M^{\prime}\in\mathrm{Max}(R),\ M^{\prime}\neq M$, we have
$MS_{M^{\prime}}=S_{M^{\prime}}=NS_{M^{\prime}}$. Then, $N=SM$. Since
$R\subset S$ is a $\mathcal{B}$-extension, we have
$|[R,S]|=\prod_{M\in\mathrm{MSupp}(S/R)}|[R_{M},S_{M}]|$. But,
$MR_{M}=(R_{M}:S_{M})$; so that,
$|[R_{M}/(MR_{M}),S_{M}/(MR_{M})]|=|[R_{M},S_{M}]|$ by [4, Proposition 3.7].
Since $MR_{M}\in\mathrm{Max}(S_{M})$, it follows that $R_{M}/(MR_{M})\to
S_{M}/(MR_{M})$ is a field extension. At last, $R_{M}/(MR_{M})\cong R/M$ and
$S_{M}/(MR_{M})\cong S/MS$. ∎

We end this section with open questions. In Lemma 7.2, we proved that for a
u-closed integral extension $R\subset S$ where $R$ is a local ring, under some
special assumptions, if ${}_{S}^{+}R$ has a unique complement in $[R,S]$, then
it is the co-subintegral closure of $R\subset S$. We can ask if the same
result holds in the more general situation of Proposition 7.5. In a similar
way, Proposition 7.7 shows that for a seminormal integral extension $R\subset
S$ where $R$ is a local ring, under some special assumptions, if ${}_{S}^{t}R$
has a unique complement in $[R,S]$, then it is the co-infra-integral closure
of $R\subset S$. We can ask if the same result holds in the more general
situation of Proposition 7.10.

## References

  * [1] D. D. Anderson, D. E. Dobbs and B. Mullins, The primitive element theorem for commutative algebras, Houston J. Math., 25, (1999), 603–623.
  * [2] N. Bourbaki, Algèbre Commutative, Chs. 5–6, Hermann, Paris, 1964.
  * [3] D. E. Dobbs, B. Mullins, G. Picavet and M. Picavet-L’Hermitte, On the FIP property for extensions of commutative rings, Comm. Algebra, 33 (2005), 3091–3119.
  * [4] D. E. Dobbs, G. Picavet and M. Picavet-L’Hermitte, Characterizing the ring extensions that satisfy FIP or FCP, J. Algebra, 371 (2012), 391–429.
  * [5] D. E. Dobbs, G. Picavet and M. Picavet-L’Hermitte, Transfer results for the FIP and FCP properties of ring extensions, Comm. Algebra, 43 (2015), 1279–1316.
  * [6] D. E. Dobbs, G. Picavet and M. Picavet-L’Hermitte, When an extension of Nagata rings has only finitely many intermediate rings, each of those is a Nagata ring, Int. J. Math. Math. Sci., Vol 2014 (2012), Article ID 315919, 13 pp..
  * [7] D. E. Dobbs, G. Picavet, M. Picavet-L’Hermitte and J. Shapiro, On intersections and composites of minimal ring extensions, JP J. Algebra, Number Theory and Appl., 26 (2012), 103–158.
  * [8] D. Ferrand and J.-P. Olivier, Homomorphismes minimaux d’anneaux, J. Algebra, 16 (1970), 461–471.
  * [9] A. Grothendieck and J. Dieudonné, Eléments de Géométrie Algébrique I, Springer Verlag, Berlin, (1971).
  * [10] B. Iversen, Generic local structure of the morphisms in Commutative Algebra, Lecture Notes in Mathematics, vol. 310, Springer, 1973.
  * [11] M. Knebusch and D. Zhang, Manis Valuations and Prüfer Extensions I, Springer, Berlin (2002).
  * [12] K. Kubota, A note on radicial extensions of rings, Mem. Defense Acad., 16 (1976), no 3, 75–80.
  * [13] D. Lazard, Autour de la platitude, Bull. Soc. Math. France, 97, (1969), 81–128.
  * [14] M. Manaresi, Some properties of weakly normal varieties, Nagoya Math. J., 77 (1980), 61–74.
  * [15] G. Picavet, Anodality, Comm. Algebra, 26 (1998), 345–393.
  * [16] G. Picavet and M. Picavet-L’Hermitte, Etale extensions with finitely many subextensions, Boll. Unione Mat. Ital., 10, (2017), 549–573. DOI:10.1007/s40574-016-0088-7.
  * [17] G. Picavet and M. Picavet-L’Hermitte, Morphismes t-clos, Comm. Algebra, 21 (1993), 179–219.
  * [18] G. Picavet and M. Picavet-L’Hermitte, T-Closedness, pp. 369–386, in: Non-Noetherian Commutative Ring Theory, Math. Appl. 520, Kluwer, Dordrecht, 2000.
  * [19] G. Picavet and M. Picavet-L’Hermitte, Prüfer and Morita hulls of FCP extensions, Comm. Algebra, 43 (2015), 102–119.
  * [20] G. Picavet and M. Picavet-L’Hermitte, FIP and FCP products of ring morphisms, Palest. J. Math., 5  (Special Issue) (2016), 63–80.
  * [21] G. Picavet and M. Picavet-L’Hermitte, Quasi-Prüfer extensions of rings, pp. 307–336, in: Rings, Polynomials and Modules, Springer, 2017.
  * [22] G. Picavet and M. Picavet-L’Hermitte, Rings extensions of length two, J. Algebra Appl., 18 (2019 no 8), 1950174, 34pp..
  * [23] G. Picavet and M. Picavet-L’Hermitte, Boolean FIP ring extensions, Comm. Algebra, 48 (2020), 1821–1852.
  * [24] G. Picavet and M. Picavet-L’Hermitte, The Loewy series of an FCP (distributive) ring extension, Int. Electon. J. Algebra, 29, (2021), 15–49.
  * [25] G. Picavet and M. Picavet-L’Hermitte, Catenarian FCP ring extensions, to appear in J. Commut. Algebra, projecteuclid.org/euclid.jca/1599184821.
  * [26] G. Picavet and M. Picavet-L’Hermitte, Splitting ring extensions, arXiv:2107.04102v1 [math.AC] .
  * [27] M. Picavet-L’Hermitte, Weak normality and t-closedness, Comm. Algebra, 28 (2000), 2395–2422.
  * [28] M. Raynaud, Anneaux locaux Henséliens, Lecture Notes in Mathematics, vol. 169, Springer, Berlin, 1970.
  * [29] P. Samuel, La notion de place dans un anneau. Bull. Soc. Math. France, 85, (1957), 123–133.
  * [30] R. G. Swan, On seminormality, J. Algebra, 67 (1980), 210–229.
  * [31] H. Yanagihara, On an intrinsic definition of weakly normal rings, Kobe J. Math., 2 (1985), 89–98.


# Matrices of simple spectrum in irreducible representations of cyclic
extension of simple algebraic groups

A.E. Zalesski  Department of Physics, Mathematics and Informatics, National
Academy of Sciences of Belarus, 66 Prospekt Nezavisimosti, Minsk, Belarus
<EMAIL_ADDRESS>
###### Key words and phrases:

Algebraic group representations, matrices with simple spectrum

Abstract. Let $H$ be a linear algebraic group whose connected component $G\neq
1$ is simple, $H/G$ is cyclic and $Z(H)\subseteq Z(G)$. We determine the
irreducible projective representations $\phi$ of $H$ such that $\phi(G)$ is
irreducible and $\phi(h)$ has simple spectrum for some $h\in H$. The latter
means that all irreducible constituents of the group $\phi(\langle h\rangle)$
are of multiplicity 1. (Here $\langle h\rangle$ is the subgroup of $H$
generated by $h$.) This extends an earlier known result for $H=G$.

##  1\. Introduction

The aim of this paper is to extend the result of [4, §6], [11] on the
irreducible representations of simple algebraic groups whose image contains a
matrix of simple spectrum. We assume that the representation field $F$ is
algebraically closed; in this case, the spectrum of a square matrix $M$ over
$F$ is the set of its eigenvalues and the spectrum of $M$ is called simple if
every eigenvalue has multiplicity 1. In particular, a matrix with simple
spectrum is similar to a diagonal matrix with all diagonal entries are
distinct.

Let $G$ be a linear algebraic group, $\rho$ a rational representation and
$g\in G$. If $M=\rho(g)$ has simple spectrum then $g$ is a semisimple element
and all weights of $\rho$ are of multiplicity 1. In fact, this is equivalent
for $\rho(G)$ to contain a matrix with simple spectrum. For every simple
algebraic group the irreducible representations whose all weights are of
multiplicity 1 have been determined in [11], extending an earlier results by
Seitz [4, §6].

Recently R. Guralnick raised a question whether this result can be extended to
groups $H=\langle G,\sigma\rangle$, where $G$ is a simple algebraic group and
$\sigma$ is an outer automorphism of $G$. We prove

######  Theorem 1.1.

Let G be a simple algebraic group in defining characteristic $p\geq 0$, let
$\sigma$ be a non-trivial graph automorphism of G and $H=\langle
G,\sigma\rangle$. Let V be an H-module such that $V|_{G}$ is rational and
irreducible. Let $\omega$ be the highest weight of $V|_{G}$. Suppose that some
$h\in H$ has simple spectrum on V. Then one of the following holds:

$(1)$ There exists $g\in G$ whose spectrum on V is simple;

$(2)$ $G$ is of type $A_{2}$, $p\neq 2,3$ and
$\omega=p^{j}(\omega_{1}+\omega_{2})$ for some integer $j\geq 0;$

$(3)$ $G$ is of type $D_{4}$, $p=2$ and $\omega=2^{j}\omega_{2}$ for some
integer $j\geq 0.$

Moreover, in each case $(2),(3)$ there is an element $h\in(H\setminus G)$ with
simple spectrum on $V$.

It is well known that $|\sigma|\in\\{2,3\\}$. If $|\sigma|=p$ and $h$ is
diagonalizable then $h$ has simple spectrum on $V$ if and only if so has
$g=h^{p}$ so (1) holds. So we assume that $(|\sigma|,p)=1$. In this case
$h\in(H\setminus G)$ is diagonalizable if and only if so is $h^{|\sigma|}$.

Theorem 1.1 answers the original question by R. Guralnick, and is sufficient
to an application he held in mind. The assumption on $V|_{G}$ to be rational
is not essential due to a theorem of Borel and Tits [6, Theorem 42]. The
assumption “$V|_{G}$ is irreducible” is essential. In general, the problem of
describing the irreducible $FH$-modules $V$ containing a simple spectrum
matrix but reducible on $G$ seems to be more complex. We have the following
observations:

######  Proposition 1.2.

Let $G,H,\sigma$ be as in Theorem 1.1, $k=|\sigma|$, and let V be an
irreducible $FH$-module such that $V|_{G}$ is reducible. Let $h\in(H\setminus
G)$. Then the spectrum of h on V is simple if and only if $h^{k}$ has simple
spectrum on every irreducible constituent of $V|_{G}$. In particular, if the
spectrum of h on V is simple then the weights of every irreducible constituent
of V are of multiplicity $1$.

The converse is not true in general.

For finite groups of Lie type $G(q)$ the irreducible representations (over a
field of describing characteristic) whose image contains a matrix with simple
spectrum are determined in [8, 9]. The problem of extending the results of [8,
9] to cyclic extensions $H(q)=\langle G(q),h\rangle$, where $h\in H$
normalizes $G(q)$, remains open. We obtain some partial results in the
framework of Theorem 1.1, that is, in the case where $H(q)\subset H$. Then
Theorem 1.1 reduces the problem to the cases $(1)-(3)$ of it, and the case (1)
is sorted out in [8, 9].

######  Theorem 1.3.

Let $G(q)=SL_{3}(q)$ or $SU_{3}(q)$, $(q,6)=1$ and $H(q)=\langle
G(q),\sigma\rangle$, where $\sigma$ is the transpose-inverse automorphism of
$G(q)$. Let $V$ be an irreducible $FH(q)$-module such that $V|_{G(q)}$ is
irreducible with highest weight $\omega_{1}+\omega_{2}$. Then $h$ has simple
spectrum on V for some $h\in(H(q)\setminus G(q))$.

The sentence “$V|_{G(q)}$ is irreducible with highest weight
$\omega_{1}+\omega_{2}$” means $V|_{G(q)}=\overline{V}|_{G(q)}$, where
$\overline{V}$ is an irreducible $SL_{3}(F)$-module with highest weight
$\omega_{1}+\omega_{2}$.

Similar results are obtained for $H(q)=\langle D_{4}(q)\cdot\sigma\rangle$ and
$H(q)=\langle{}^{3}D_{4}(q)\cdot\sigma\rangle$, where $q>8$ even and $\sigma$
is a triality automorphism of $D_{4}(q)$ or ${}^{3}D_{4}(q)$, see Sections 6.1
and 6.2. The cases of $q=4,8$ are still open. Note that the answer is opposite
for $q=2$ in both the cases, as for $h\in H(2)$ of odd order we have $|h|<d$,
where $d$ is the degree of any faithful $\sigma$-stable irreducible
representation $\phi$ of $G(q)$.

Notation We keep $G$ to denote a simply connected simple algebraic group
defined over an algebraically closed field $F$. All representations and
modules are assumed to be rational. For general facts of the theory of
algebraic groups and their representations we refer to [6] and [1, 2]; in
particular, the ordering of simple roots of $G$ is as in [1]. If $G$ is of
rank $n$ then $\omega_{1},\ldots,\omega_{n}$ are fundamental weights and
$\alpha_{1},\ldots,\alpha_{n}$ are simple roots of $G$. If $V$ is a $G$-module
and $X$ is a subgroup of $G$ then $V|_{X}$ denote the restriction of $V$ to
$X$. We write $|X|$ for the order of $X$, and also $|x|$ for the order of an
element $x\in G$; $\langle x\rangle$ means the group generated by $x$. The
normalizer and the centralizer of $X$ in $G$ are denoted by $N_{G}(X)$ and
$C_{G}(X)$, respectively. $Z(X)$ denotes the center of a group $X$. By
$\mathop{\mathbb{Z}}\nolimits$, $\mathop{\mathbb{Q}}\nolimits,$ and
$\mathop{\mathbb{F}}\nolimits_{q}$ we denote the ring of integers, the field
of rational numbers and the finite field of $q$ elements, respectively. A
diagonal matrix with diagonal entries $d_{1},\ldots,d_{n}$ is denoted by
$\mathop{\rm diag}\nolimits(d_{1},\ldots,d_{n})$.

##  2\. Reduction to representations with non-zero weights of
multiplicity $1$

The following lemma is obvious:

######  Lemma 2.1.

Let $h\in GL_{n}(F)$ be a diagonalizable element and $l\geq 1$ is an integer.
If h has simple spectrum then the multiplicity of every eigenvalue of $h^{l}$
does not exceed $l$.

######  Lemma 2.2.

Let $M$ be a vector space over an algebraically closed field and let
$M=M_{1}\oplus...\oplus M_{l}$, where $d:=\dim M_{1}>1$. Let $x\in GL(M)$ be a
semisimple element; suppose that $xM_{l}=M_{1}$, $xM_{i}=M_{i+1}$ for
$i=1,\ldots,l-1$, and that $x^{l}$ acts scalarly on $M_{1}$. Then each
eigenvalue of $x$ on $M$ is of multiplicity $d$.

###### Proof.

This is straightforward.∎

In the following statement we do not assume $V|_{G}$ to be irreducible.

######  Proposition 2.3.

Let $G,H,\sigma$ be as in Theorem 1.1. Let V be an H-module such that $V|_{G}$
is rational and $h\in H$. Suppose that h has simple spectrum on V. Then all
non-zero weights of G on V are of multiplicity $1$ and the multiplicity of
weight $0$ is at most $k:=|\sigma|$.

###### Proof.

If $h\in G$ then $h\in T$ for some maximal torus of $G$. Then obviously, all
$T$-weights are of multiplicity 1, as required. So we assume $h\notin G$. Set
$g=h^{k}$ so we have $g\in G$. Then $g\in G$ is semisimple, and hence $g\in
T$, where $T$ is some maximal torus of $G$. Moreover, $T$ can be chosen
$h$-stable. (Indeed, $C_{G}(g)^{0}$, the connected component of $C_{G}(g)$, is
an $h$-stable semisimple algebraic group on which $h$ acts as a semisimple
automorphism; by [5, Theorem 7.5], $C_{G}(g)^{0}$ has an $h$-stable maximal
torus, which remains maximal in $G$.) We fix this $T$, so $h\in N_{H}(T)$.)

Set $W_{H}=N_{H}(T)/T$ and $W=N_{G}(T)/T$. Then $W$ is the Weyl group of $G$
and $W_{H}=\langle\sigma,W\rangle$. The conjugation action of $N_{H}(T)$ on
$T$ induces the action of $W_{H}$ on $\Omega:={\rm
Hom}\,(T,F^{\times})\cong\mathop{\mathbb{Z}}\nolimits^{n}$, the set of
$T$-weights of $G$. Denote by $\omega(V)$ the set of $T$-weights of $V$. Then
$\omega(V)$ is stable under $W_{H}$. For $\lambda\in\omega(V)$ denote by
$V_{\lambda}$ the $\lambda$-weight space of $V$. Then
$hV_{\lambda}=V_{\sigma(\lambda)}$ for every $\lambda\in\Omega$, so $h$
permutes $V_{\lambda}$’s. (Here we define
$\sigma(\lambda)(t)=\lambda(\sigma(t))$ and $\sigma(t)=h^{-1}th$ for $t\in
T$.) Clearly, $g$ stabilizes every $V_{\lambda}$ and acts scalarly on it.

Suppose the contrary, and choose $0\neq\lambda\in\omega(V)$ so that $\dim
V_{\lambda}>1$. Then $y(\lambda)\in\omega(V)$ and $\dim V_{y(\lambda)}=\dim
V_{\lambda}$ for every $y\in W_{H}$.

(i) Suppose first that $hV_{w(\lambda)}\neq V_{w(\lambda)}$ for some $w\in W$.
Since $k$ is a prime, $\sum_{i=0}^{k-1}h^{i}V_{w(\lambda)}$ is a direct sum as
any sum of distinct weight spaces is a direct sum of them. By Lemma 2.2, $h$
has an eigenvalue of multiplicity at least 2 (in fact, the multiplicity in
question equals $\dim V_{\lambda}$). (Indeed, set $M_{1}=V_{w(\lambda)}$,
$M_{i}=h^{i}V_{w(\lambda)}$ in Lemma 2.2.) So we have a contradiction.

(ii) Suppose that $hV_{w(\lambda)}=V_{w(\lambda)}$ for all $w\in W$. Then
$V_{w(\lambda)}=hV_{w(\lambda)}=V_{\sigma(w(\lambda))}$ and hence $\sigma$
fixes $w(\lambda)$ or all $w\in W$.

Next let $R$ be the vector space over $\mathop{\mathbb{Q}}\nolimits$ spanned
by $\Omega.$ Then $W_{H}\subset GL(R)$, and $W$ is well known to be a (finite)
irreducible subgroup of $GL(R)$. Let $L$ be the
$\mathop{\mathbb{Q}}\nolimits$-span of the set $\\{w(\lambda):w\in W\\}$. Then
$L=R$ as $WL=L$, $L\neq 0$ and $W$ is irreducible. This means that $\sigma$
fixes every weight of $G$, and hence $\sigma(t)=t$ for every $t\in T$. In
addition, $\sigma(\alpha)=\alpha$ for every root $\alpha$ of $G$ (as
$\alpha\in R$). This implies that $\sigma$ is a trivial automorphism of $G$,
which is a contradiction.

Let $V_{0}=\\{v\in V:tv=v$ for all $t\in T\\}$, so $V_{0}$ is the zero weight
space on $V$. Then $hV_{0}=V_{0}$ and $h^{k}$ acts on $V_{0}$ trivially. Let
$h_{0}$ be the restriction of $h$ to $V_{0}$. Then $h_{0}$ has simple spectrum
on $V_{0}$, and the claim follows from Lemma 2.1.∎

Remark. In Proposition 2.3, If $V|_{G}$ is irreducible and tensor-decomposable
then the weight 0 has multiplicity 0 or 1. This follows from [10, Theorem
2(2)].

##  3\. Proof of Theorem 1.1

Below $F$ is an algebraically closed field of characteristic $p\geq 0$ and
$F^{\times}$ is the multiplicative group of $F$. Note that the simple
algebraic groups of types $A_{1},B_{n},n>1,C_{n},n>1,E_{7},E_{8},F_{4},G_{2}$
have no non-trivial graph automorphism. It is well known that $|\sigma|=2$ for
groups $G$ of type $A_{n}$, $n>1$, $D_{n},$ $n>4$ and $E_{6}$, and
$|\sigma|\in\\{2,3\\}$ for $G$ of type $D_{4}$.

######  Lemma 3.1.

Let $G=SL_{n}(F)$, $p\neq 2$, let $\sigma$ be the transpose-inverse
automorphism of $G$ and $H=\langle G,\sigma\rangle$. Let T be the group of
diagonal matrices in G and let $h\in(H\setminus G)$. Suppose that $h^{2}\neq
1$ is a semisimple element of G.

$(1)$ There is a conjugate $h_{1}$ of h such that $h_{1}(T)=T$ and
$h_{1}^{2}\in T$.

$(2)$ Let $\overline{h}_{1}$, $\overline{\sigma}$ be the projections of
$h_{1},\sigma$ into $W_{H}=N_{H}(T)/T$. Then
$\overline{h}_{1}=w\overline{\sigma}$, where $1\neq w\in W=N_{G}(T)/T$,
$w^{2}=1$ and $w\overline{\sigma}=\overline{\sigma}w$.

###### Proof.

We have seen in the proof of Lemma 2.3 that there exists a maximal torus
$T_{1}$ of $G$ such that $h(T_{1})=T_{1}$ and $h^{2}\in T_{1}$. As the maximal
tori of $G$ are conjugate, we have $gT_{1}g^{-1}=T$ for some $g\in G$, and
then $h_{1}=ghg^{-1}$ is as required in (1).

It is well known that $H=\langle G,\sigma\rangle$. As $\sigma(t)=t^{-1}$ for
every $t\in T$, it follows that $\sigma$ acts as $-\mathop{\rm Id}\nolimits$
on $\Omega={\rm Hom}(T,F^{\times})$. So $\sigma w=w\sigma$ for every $w\in W$.
As $W_{H}=\langle W,\overline{\sigma}\rangle$, we have
$\overline{h}_{1}=w\overline{\sigma}$ for some $w\in W$. Then
$\overline{h}_{1}^{2}=1$ implies $w^{2}=1$. If $w=1$ then $h_{1}=t\sigma$ for
some $t\in T$, whence $h_{1}^{2}=t\sigma t\sigma=t\cdot t^{-1}=1$ and
$h^{2}=1$, contrary the assumption. So (2) follows.∎

In view of Proposition 2.3, to prove Theorem 1.1 we have to examine the
irreducible representation of $G$ whose all non-zero weights are of
multiplicity 1. As above, let $V_{0}$ be the zero weight space in $V$. If
$\dim V_{0}\leq 1$ then all weight multiplicities are equal to 1; in this case
there exists $h\in G$ with simple spectrum (see for instance [7, Proposition
1]). So it suffices to deal with the irreducible representations of simple
algebraic groups whose all non-zero weights are of multiplicity 1 and $\dim
V_{0}>1$. Such representations are determined in [10], in particular, these
are tensor-indecomposable, and hence are Frobenius twists of those listed in
[10, Table 2]. This table is reproduced as Table 1 below; it also provides
$\dim V_{0}$. By Proposition 2.3, we ignore the cases with $\dim
V_{0}>|\sigma|$. In addition, if $V|_{G}$ is irreducible then the highest
weight of it is $\sigma$-invariant, that is, it remains unchanged under the
action of the graph automorphism $\tau$, say, of order $k=|\sigma|$, see [3,
Theorem 1.15.2]. In terms of highest weights, if $\sum a_{i}\omega_{i}$
($a_{i}\in\mathop{\mathbb{Z}}\nolimits$) is the highest weight of $V|_{G}$ and
$\tau(i)=j$ for the nodes $i,j$ at the Dynkin diagram then $\tau(\sum
a_{i}\omega_{i})=\sum a_{i}\omega_{\tau(i)}$.

So we have to delete the entries of Table 1 that do not satisfy the above
conditions.

Let $G=A_{n},n>1$. In this case $|H:G|=2=|\sigma|$, so we drop the cases with
$p=2$ and those with $\dim V_{0}>2$. By Table 1, this implies $n=2$,
$\omega=\omega_{1}+\omega_{n}$, $p\neq 2,3$ or $n=3$, $\omega=2\omega_{2}$,
$p\neq 2,3$.

Let $G=D_{n},n\geq 4$. If $n>4$ then $|H:G|=2=|\sigma|$ and there is no entry
with $\dim V_{0}\leq 2$. Let $n=4$. If $|\sigma|=2$ then $p\neq 2$, and then
$\omega=\omega_{2}$ or $2\omega_{2}$ in Table 1. The latter contradicts
Proposition 2.3. So $|a|=3$. Then $\omega=\omega_{2}$ by Table 1, and
Proposition 2.3 implies $p=2$.

Let $G=E_{6}$. Then $|H:G|=2=|\sigma|$ and $\dim V_{0}>2$ in Table 1, so we
may ignore this case.

Summarizing this we obtain:

######  Lemma 3.2.

In assumption of Theorem 1.1 one of the following holds:

$(1)$ There exists $g\in G$ whose spectrum on V is simple;

$(2)$ $G$ is of type $A_{2}$, $p\neq 2,3$ and
$\omega=p^{m}(\omega_{1}+\omega_{2});$

$(3)$ $G$ is of type $D_{4}$, $p=2$ and $\omega=2^{m}\omega_{2};$

$(4)$ $G$ is of type $A_{3}$, $p\neq 2,3$ and $\omega=2p^{m}\omega_{2},$
where $m\geq 0$ is an integer.

To prove the main statement of Theorem 1.1 we are left to rule out the case
(4). To prove the additional statement of the theorem, we have to construct
$h\in H$ with simple spectrum on $V$ in the cases (2), (3) of Lemma 3.2.

##  4\. The case $G=A_{3}$, $\omega=2\omega_{2}$, $p\neq 2,3$

In this section we deal with case (4) of Lemma 3.2. Let
$\omega_{1},\omega_{2},\omega_{3}$ be the fundamental weights of $G$.

######  Lemma 4.1.

Let $G=SL_{4}(F)$, $p\neq 2,3$, and let V be an irreducible G-module with
highest weight $2\omega_{2}$. Let $H=\langle G,\sigma\rangle$, where $\sigma$
is the transpose-inverse automorphism of G. Then V extends to an $FH$-module
and H has no element with simple spectrum on V.

###### Proof.

Suppose the contrary, and let $h\in H$ have simple spectrum on $V$. Then
$h\notin G$ as the zero weight of $V$ has multiplicity 2 (see Table 1). Let
$T$ be the group of diagonal matrices in $G$. By Lemma 3.1, we can assume that
$h^{2}\in T$ and $h=\sigma n_{w}t$, where $t\in T$ and $n_{w}\in N_{G}(T)$ is
such that $w$, the projection of $n_{w}$ in $N_{G}(T)/T$, is an involution. We
can choose $n_{w}$ up to a multiple in $T$, and we do this so that
$n_{w}=\mathop{\rm diag}\nolimits(\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},1,1)$ or $\mathop{\rm diag}\nolimits(\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},\begin{pmatrix}0&1\\\ -1&0\end{pmatrix})$. Then $\sigma
n_{w}\sigma=n_{w}$.

Note that $\sigma(\mu)=-\mu$ for every weight of $G$, in particular,
$\sigma(2\omega_{2})=-2\omega_{2}$. If $0\neq v\in V$ is a vector of weight
$2\omega_{2}$ then $t\sigma(v)=\sigma t^{-1}v$, so $\sigma(v)$ lies in the
$-2\omega_{2}$-weight space. In addition,
$w(2\omega_{2})=w(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3}-\varepsilon_{4})=2\omega_{2}$
[1, Planche I]. So $\sigma n_{w}(v)$ lies in the $-2\omega_{2}$-weight space.
Therefore, $h$ permutes the $2\omega_{2}$\- and $-2\omega_{2}$-weight spaces.
Let $E$ be the sum of them, so $hE=E$.

Let $t=\mathop{\rm diag}\nolimits(t_{1},t_{2},t_{3},(t_{1}t_{2}t_{3})^{-1})$.
Then $h^{2}=\sigma n_{w}t\sigma n_{w}t=n_{w}\sigma t\sigma
n_{w}t=n_{w}t^{-1}n_{w}t$. This equals to $\mathop{\rm
diag}\nolimits(-t_{1}t_{2}^{-1},-t_{1}^{-1}t_{2},1,1)$ or $\mathop{\rm
diag}\nolimits(-t_{1}t_{2}^{-1},-t_{1}^{-1}t_{2},-t_{1}^{-1}t_{2}^{-1}t_{3}^{2},-t_{1}t_{2}t_{3}^{-2})$,
respectively. So
$(2\omega_{2})(h^{2})=(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3}-\varepsilon_{4})(h^{2})=1=(-2\omega_{2})(h^{2})$
for both the choice of $w$. So the eigenvalues of $h$ on $E$ are the square
roots of 1, that is, 1 and $-1$.

As $h$ stabilizes $V_{0}=C_{V}(T)$, the zero weight space of $V$, and $h^{2}$
acts on $V_{0}$ trivially, it follows that the eigenvalues of $h$ on $V_{0}$
are square roots of unity. So either 1 or $-1$ is an eigenvalue of $h$ of
multiplicity at least 2, a contradiction. ∎

##  5\. The case $G=A_{2}$, $\omega=\omega_{1}+\omega_{n}$, $p\neq 2,3$.

In this case the module $V=V_{\omega}$ can be realized as the adjoint module
for $G=SL_{3}(F)$, whose underlying space $L$ is the space of $(3\times
3)$-matrices with trace 0, and the action of $G$ on $L$ is the conjugation
action $l\rightarrow glg^{-1}$ for $l\in L,g\in G$. Note that $\dim V=8$. We
can choose for $T$ the group of diagonal matrices of $G$ and for $\sigma$ the
transpose-inverse automorphism of $G$. Then $\sigma$ acts on $L$ as
$l\rightarrow-l^{tr}$, where $l^{tr}$ is the transpose of $l$, see [2, Ch.
VIII, §13.1]. For a basis of $L$ we choose the matrix units $E_{ij}$ ($1\leq
i\neq j\leq 3)$ and $E_{11}-E_{22},E_{22}-E_{33}$, the two latter span the
zero weight space.

Let $h=\sigma n_{w}t$, where $t=\mathop{\rm
diag}\nolimits(t_{1},t_{2},(t_{1}t_{2})^{-1})\in T$ and
$n_{w}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&-1\end{pmatrix}\in N_{G}(T)$, so
$n_{w}^{2}=1$ and $\sigma n_{w}\sigma=n_{w}$. We show that the spectrum of $h$
on $V$ is simple for a suitable choice of $t_{1},t_{2}\in F$. Set $g=h^{2}$.

Note that $g=\sigma n_{w}t\sigma n_{w}t=n_{w}\sigma t\sigma
n_{w}t=n_{w}t^{-1}n_{w}t=\mathop{\rm
diag}\nolimits(t_{1}t_{2}^{-1},t_{1}^{-1}t_{2},1)\in T$. Note that $E_{ij}$
$(1\leq j,j\leq 3)$ are eigenvectors for $g$, and the eigenvalues of $g$ on
$V=L$ are $t_{1}^{2}t_{2}^{-2}$, $t_{1}^{-2}t_{2}^{2}$, each of multiplicity
1, and 1,$t_{1}t_{2}^{-1}$, $t_{1}^{-1}t_{2}$, each of multiplicity 2. If we
show that $h$ is not scalar on each of the 2-dimensional $g$-eigenspaces of
$V$ then the eigenvalues of $h$ on them are distinct square roots of those of
$g$, and hence the spectrum of $h$ is simple.

We have $h(E_{12})=\sigma
n_{w}t(E_{12})t^{-1}n_{w}\sigma=t_{1}t_{2}^{-1}\sigma
E_{21}\sigma=-t_{1}t_{2}^{-1}E_{12}$, and similarly,
$h(E_{21})=-t_{1}^{-1}t_{2}E_{12}$. For the $g$-eigenspaces of multiplicity 2
we have $h(E_{22}-E_{33})=\sigma
n_{w}t((E_{22}-E_{33}))t^{-1}n_{w}\sigma=-(E_{11}-E_{33})$, so $h$ is non-
scalar on the 1-eigenspace of $h^{2}$. Similarly, $h(E_{13})=\sigma
n_{w}t(E_{13})t^{-1}n_{w}\sigma=t_{1}^{2}t_{2}\sigma
E_{23}\sigma=-t_{1}^{2}t_{2}E_{32}$ and $h(E_{23})=\sigma
n_{w}t((E_{23}))t^{-1}n_{w}\sigma=t_{1}t_{2}^{2}\sigma
E_{13}\sigma=-t_{1}t_{2}^{2}E_{31}$. So the eigenvalues of $h$ on $V$ are
$-t_{1}t_{2}^{-1},$ $-t_{1}^{-1}t_{2}$, $\pm 1,$ $\pm\sqrt{t_{1}t_{2}^{-1}}$,
$\pm\sqrt{t_{1}^{-1}t_{2}}$. Whence the claim.

For $t_{2}=1$ this implies:

######  Lemma 5.1.

Let $t=\mathop{\rm diag}\nolimits(t_{1},1,t_{1}^{-1})\in T$,
$n_{w}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&-1\end{pmatrix}\in N_{G}(T)$ and
$V,\sigma$ as above. Suppose that $\pm 1$, $-t_{1},$ $-t_{1}^{-1}$,
$\pm\sqrt{t_{1}}$, $\pm\sqrt{t_{1}^{-1}}$ are distinct. Then $h=\sigma n_{w}t$
has simple spectrum on V.

In particular, if $\zeta^{4}=-1$, $\zeta\in F$ and $t_{1}=\zeta^{2}$ then
these are $\pm 1$, $-\zeta^{\pm 2}$, $\pm\zeta^{\pm 1}$, which are all 8-roots
of unity. Therefore, for this choice of $h$ the spectrum of $h$ is simple.
Note that in the latter case $t=\mathop{\rm
diag}\nolimits(\zeta^{2},1,\zeta^{-2})$ so $|t|=4$.

######  Corollary 5.2.

Let $G(q)=SL_{3}(q)$, $(q,6)=1$ and $H(q)=\langle G(q),\sigma\rangle$, where
$\sigma$ is the transpose-inverse automorphism of $G(q)$. Let $V$ be an
irreducible $FH(q)$-module such that $V|_{G(q)}$ is irreducible with highest
weight $\omega_{1}+\omega_{2}$. Then $h$ has simple spectrum on V for some
$h\in(H(q)\setminus G(q))$.

###### Proof.

We show that the element $h=\sigma n_{w}t$ in Lemma 5.1 can be chosen in
$G(q)$. We take $t=\mathop{\rm diag}\nolimits(t_{1},1,t_{1}^{-1})$, where
$t_{1}\in F$ with $|t_{1}|=q-1$. As $q\neq 3$ is odd, the elements $1$,
$t_{1}^{2}$, $t_{1}$, $t_{1}^{-1}$ are distinct. As $t,n_{w}\in G(q)$, the
result follows from Lemma 5.1. ∎

Let $G_{1}=GL_{3}(q^{2})$. Let $V_{1}$ be the natural module for
$GL_{3}(q^{2})$. We choose a basis $B$ in $V_{1}$ and realize $G_{1}$ as a
matrix group with respect of $B$. For $x\in G_{1}$ we write $\overline{x}$ for
the matrix obtained from $x$ by applying the Galois automorphism of
$\mathop{\mathbb{F}}\nolimits_{q^{2}}/\mathop{\mathbb{F}}\nolimits_{q}$ to
every entry of $x$. Keep $\sigma$ for the transpose-inverse automorphism of
$G_{1}$. It is well known that $G(q)\cong SU_{3}(q)$ can be realized as
$\\{x\in SL_{3}(q^{2}):\sigma(x)=\overline{x}\\}$. Then $\sigma(G(q))=G(q)$.
Let $T$ be the group of diagonal matrices in $G(q)$ with respect to this
basis. It is easy to check that $T$ is of exponent $q+1$ and $|T|=(q+1)^{2}$.
For $t\in T$ we can write $t=\mathop{\rm
diag}\nolimits(t_{1},t_{2},(t_{1}t_{2})^{-1})$ with
$t_{1}^{q+1}=t_{2}^{q+1}=1$.

######  Corollary 5.3.

Let $G(q)=SU_{3}(q)$, $(q,6)=1$ and $H(q)=\langle G(q),\sigma\rangle$, where
$\sigma$ is the transpose-inverse automorphism of $G(q)$. Let $V$ be an
irreducible $FH(q)$-module such that $V|_{G(q)}$ is irreducible with highest
weight $\omega_{1}+\omega_{2}$. Then $h$ has simple spectrum on V for some
$h\in(H(q)\setminus G(q))$.

###### Proof.

We show that the element $h=\sigma wt$ in Lemma 5.1 can be chosen in $G(q)$.
We take $t_{2}=1$, $|t_{1}|=q+1$ so $t=\mathop{\rm
diag}\nolimits(t_{1},1,t_{1}^{-1})$. As $p\geq 5$, the elements $\pm 1$,
$-t_{1},$ $-t_{1}^{-1}$, $\pm\sqrt{t_{1}}$, $\pm\sqrt{t_{1}^{-1}}$ in Lemma
5.1 are distinct. We choose $n_{w}$ as in Lemma 5.1, that is,
$n_{w}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&-1\end{pmatrix}\in N_{G}(T)$ under
basis $B$. As $t,n_{w}\in G(q)$, the result follows from Lemma 5.1. ∎

Remark. If $G(q)$ and $V$ are as in Theorem 1.3 and $q$ is a 3-power then
there exists $g\in G$ that has simple spectrum on $V$. This is a special case
of [8, Theorem 0.4].

##  6\. The case $G=D_{4},$ $p=2$, $\omega=\omega_{2}$ and $|\sigma|=3$

The irreducible $G$-module $V$ with highest weight $\omega_{2}$ is a
constituent of the adjoint one, so $V$ is the only non-trivial irreducible
constituent of $L(G)$, where $L(G)$ is the Lie algebra of $G$ viewed as a
$G$-module. It is well known that $\dim V_{\omega_{2}}=26$.

Let $T$ be a maximal torus of $G$. Then $T$ is well known to be parameterized
as $(t_{1},t_{2},t_{3},t_{4})$ with $t_{1},t_{2},t_{3},t_{4}\in F^{\times}$,
so the weight lattice $\Omega={\rm Hom}\,(T,F^{\times})$ is of rank 4. Define
$\varepsilon_{i}\in\Omega$ by $\varepsilon_{i}(t_{1},t_{2},t_{3},t_{4})=t_{i}$
for $i=1,2,3,4$. If $\omega=\sum a_{i}\varepsilon_{i}\in\Omega$
$(a_{i}\in\mathop{\mathbb{Z}}\nolimits)$, then
$\omega(t_{1},t_{2},t_{3},t_{4})=t_{1}^{a_{1}}t_{2}^{a_{2}}t_{3}^{a_{3}}t_{4}^{a_{4}}$.
If $\omega$ is a linear combination of roots then $a_{i}$’s are integers; this
is the case for the weights of $V_{\omega_{2}}$. We denote by $R$ the
$\mathop{\mathbb{Q}}\nolimits$-span of $\Omega$, so $\dim R=4$. We assume that
$\sigma(T)=T$, so $\sigma$ acts on $R$ as a linear transformation and
$\sigma(\Omega)=\Omega$. Note that $\\{\varepsilon_{i}:i=1,2,3,4\\}$ is a
basis of $R$.

We denote by $\Phi$ the set of roots of $G$, which are
$\alpha_{1}=\varepsilon_{1}-\varepsilon_{2}$,
$\alpha_{2}=\varepsilon_{2}-\varepsilon_{3}$,
$\alpha_{4}=\varepsilon_{3}-\varepsilon_{4}$,
$\alpha_{4}=\varepsilon_{3}+\varepsilon_{4}$, $\alpha_{1}+\alpha_{2}$,
$\alpha_{2}+\alpha_{3}$, $\alpha_{2}+\alpha_{4}$,
$\alpha_{1}+\alpha_{2}+\alpha_{3}$, $\alpha_{1}+\alpha_{2}+\alpha_{4}$,
$\alpha_{2}+\alpha_{2}+\alpha_{3}$,
$\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}$,
$\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}$, and their negations, see [1,
Planche IV]. These are non-zero weights of $L(G)$ and of $V$. One can choose a
basis of $V$ to be root elements $X_{\alpha}$, $\alpha\in\Phi$, added by two
weight zero elements. The outer automorphism $\sigma$ of $G$ of order 3 acts
on the simple roots $\alpha_{i}$, $i=1,2,3,4$, by fixing $\alpha_{2}$ and
permuting $\alpha_{1},\alpha_{3},\alpha_{4}$, and we can assume that
$\sigma(\alpha_{1})=\alpha_{4}$, $\sigma(\alpha_{3})=\alpha_{1}$,
$\sigma(\alpha_{4})=\alpha_{3}$, see [2, Ch. VIII, §13.4]. In addition,
$X_{\alpha}$ can be chosen so that $\sigma(X_{\alpha})=X_{\sigma(\alpha)}$ for
every $\alpha\in\Phi$.

In particular, we have

$\sigma(X_{\alpha_{1}})=X_{\alpha_{4}}$,
$\sigma(X_{\alpha_{3}})=X_{\alpha_{1}}$,
$\sigma(X_{\alpha_{4}})=X_{\alpha_{3}}$,
$\sigma(X_{\alpha_{1}+\alpha_{2}})=X_{\alpha_{2}+\alpha_{4}}$,
$\sigma(X_{\alpha_{2}+\alpha_{4}})=X_{\alpha_{2}+\alpha_{3}}$,

$\sigma(X_{\alpha_{2}+\alpha_{3}})=X_{\alpha_{1}+\alpha_{2}}$,
$\sigma(X_{\alpha_{1}+\alpha_{2}+\alpha_{3}})=X_{\alpha_{1}+\alpha_{2}+\alpha_{4}}$,
$\sigma(X_{\alpha_{1}+\alpha_{2}+\alpha_{4}})=X_{\alpha_{2}+\alpha_{3}+\alpha_{4}}$,
$\sigma(X_{\alpha_{2}+\alpha_{3}+\alpha_{4}})=X_{\alpha_{1}+\alpha_{2}+\alpha_{3}}$,
$\sigma(X_{\alpha_{2}})=X_{\alpha_{2}}$,
$\sigma(X_{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}})=X_{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}}$,
$\sigma(X_{\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}})=X_{\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}}$.

The expressions $\alpha_{i}=\varepsilon_{i}-\varepsilon_{i+1}$ for $i=1,2,3$
and $\alpha_{4}=\varepsilon_{3}+\varepsilon_{4}$ yield the following formulae
for the eigenvalues of $t=(t_{1},t_{2},t_{3},t_{4})$ on $X_{\alpha_{i}}$ for
$i=1,2,3,4$:

$tX_{\alpha_{1}}t^{-1}=t_{1}t_{2}^{-1}X_{\alpha_{1}},\,tX_{\alpha_{2}}t^{-1}=t_{3}t_{3}^{-1}X_{\alpha_{2}},\,tX_{\alpha_{3}}t^{-1}=t_{3}t_{4}^{-1}X_{\alpha_{3}},\,tX_{\alpha_{4}}t^{-1}=t_{3}t_{4}X_{\alpha_{4}}.$
(1)

In addition, we have
$\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=\varepsilon_{1}+\varepsilon_{3}$,
$\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}=\varepsilon_{1}+\varepsilon_{2}$,
whence the eigenvalues of $t$ on the corresponding root elements
$X_{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}},X_{\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}}$
are $t_{1}t_{3}$ and $t_{1}t_{2}$.

Let $h=\sigma t$ for $t=(t_{1},t_{2},t_{3},t_{4})\in F$. Then
$X_{\alpha_{2}},X_{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}},X_{\alpha_{1}+2\alpha_{2}+\alpha_{3}+\alpha_{4}}$
are eigenvectors of $h$, as well as
$X_{-\alpha_{2}},X_{-\alpha_{1}-\alpha_{2}-\alpha_{3}-\alpha_{4}},X_{-\alpha_{1}-2\alpha_{2}-\alpha_{3}-\alpha_{4}}$.
So we have

$h^{3}(X_{\alpha_{1}})=h^{2}\sigma
t(X_{\alpha_{1}})=t_{1}t_{2}^{-1}h^{2}\sigma(X_{\alpha_{1}})=t_{1}t_{2}^{-1}h^{2}(X_{\alpha_{4}})=t_{1}t_{2}^{-1}h\sigma
t(X_{\alpha_{4}})=\\\
=t_{1}t_{2}^{-1}t_{3}t_{4}h\sigma(X_{\alpha_{4}})=t_{1}t_{2}^{-1}t_{3}t_{4}h(X_{\alpha_{3}})=t_{1}t_{2}^{-1}t_{3}t_{4}\sigma
t(X_{\alpha_{3}})=t_{1}t_{2}^{-1}t_{3}t_{4}t_{3}t_{4}^{-1}\sigma(X_{\alpha_{3}})=\\\
=t_{1}t_{2}^{-1}t_{3}^{2}(X_{\alpha_{1}})$. Similarly,

$h^{3}(X_{\alpha_{1}+\alpha_{2}})=h^{2}\sigma
t(X_{\alpha_{1}+\alpha_{2}})=h^{2}\sigma
t(X_{\varepsilon_{1}-\varepsilon_{3}})=t_{1}t_{3}^{-1}h^{2}h^{2}\sigma(X_{\alpha_{1}+\alpha_{2}})=t_{1}t_{3}^{-1}h^{2}(X_{\alpha_{2}+\alpha_{4}})=t_{1}t_{3}^{-1}h\sigma
t(X_{\alpha_{2}+\alpha_{4}})=t_{1}t_{3}^{-1}h\sigma
t(X_{\varepsilon_{2}+\varepsilon_{4}})=t_{1}t_{3}^{-1}t_{2}t_{4}h\sigma(X_{\alpha_{2}+\alpha_{4}})=t_{1}t_{3}^{-1}t_{2}t_{4}\sigma
t(X_{\alpha_{2}+\alpha_{3}})=t_{1}t_{3}^{-1}t_{2}t_{4}\sigma
t(X_{\varepsilon_{2}-\varepsilon_{4}})=t_{1}t_{3}^{-1}t_{2}t_{4}t_{2}t_{4}^{-1}\sigma(X_{\alpha_{2}+\alpha_{3}})=t_{1}t_{2}^{2}t_{3}^{-1}(X_{\alpha_{1}+\alpha_{2}})$,
and

$h^{3}(X_{\alpha_{1}+\alpha_{2}+\alpha_{3}})=h^{2}\sigma
t(X_{\alpha_{1}+\alpha_{2}+\alpha_{3}})=h^{2}\sigma
t(X_{\varepsilon_{1}-\varepsilon_{4}})=t_{1}t_{4}^{-1}h^{2}(X_{\alpha_{1}+\alpha_{2}+\alpha_{4}})=\\\
t_{1}t_{4}^{-1}h\sigma
t(X_{\varepsilon_{1}+\varepsilon_{4}})=t_{1}t_{4}^{-1}t_{1}t_{4}h\sigma(X_{\alpha_{1}+\alpha_{2}+\alpha_{4}})=t_{1}t_{4}^{-1}t_{1}t_{4}\sigma
t(X_{\alpha_{2}+\alpha_{3}+\alpha_{4}})=t_{1}^{2}\sigma
t(X_{\varepsilon_{2}+\varepsilon_{3}})=t_{1}^{2}t_{2}t_{3}\sigma(X_{\alpha_{1}+\alpha_{2}+\alpha_{4}})=t_{1}^{2}t_{2}t_{3}(X_{\alpha_{1}+\alpha_{2}+\alpha_{3}})$.

It follows that the eigenvalues of $h$ obtained from this are
$(t_{1}t_{2}^{-1}t_{3}^{2})^{1/3}$ $(t_{1}t_{2}^{2}t_{3}^{-1})^{1/3}$,
$(t_{1}^{2}t_{2}t_{3})^{1/3}$, $\zeta(t_{1}t_{2}^{-1}t_{3}^{2})^{1/3}$,
$\zeta(t_{1}t_{2}^{2}t_{3}^{-1})^{1/3}$, $\zeta(t_{1}^{2}t_{2}t_{3})^{1/3}$,
$\zeta^{2}(t_{1}t_{2}^{-1}t_{3}^{2})^{1/3}$,
$\zeta^{2}(t_{1}t_{2}^{2}t_{3}^{-1})^{1/3}$ and
$\zeta^{2}(t_{1}^{2}t_{2}t_{3})^{1/3}$, where $\zeta^{3}=1\neq\zeta$. In
addition, a similar action of $h$ on $X_{-\alpha}$ with $\alpha$ a positive
root above yields the inverses of these eigenvalues.

The basis of the zero weight space $L_{0}$ of $L(G)$ can be chosen to be
$[X_{\alpha_{i}},X_{-\alpha_{i}}]$ for $i=1,\ldots,4$ [2, Ch. VIII, §13.4].
Therefore, $\sigma$ has 3 distinct eigenvalues on it; as the composition
series of $L(G)$ as a $G$-module has two trivial factors and one isomorphic to
$V_{0}$, it follows that the eigenvalues of $\sigma$ on $V_{0}$ are
$\zeta,\zeta^{-1}$.

Altogether the eigenvalues of $h$ on $V$ are $\zeta,\zeta^{2}$,
$(t_{2}t_{3}^{-1})^{\pm 1}$, $(t_{1}t_{3})^{\pm 1}$, $(t_{1}t_{2})^{\pm 1}$,
$\zeta^{i}(t_{1}t_{2}^{-1}t_{3}^{2})^{\pm 1/3}$,
$\zeta^{i}(t_{1}t_{2}^{2}t_{3}^{-1})^{\pm 1/3}$,
$\zeta^{i}(t_{1}^{2}t_{2}t_{3})^{\pm 1/3}$ for $i=0,1,2$. One observes that
there are $t_{1},t_{2},t_{3}\in F$ such that all these eigenvalues are
distinct. So we have

######  Lemma 6.1.

Let $t\in T$ be such that $\varepsilon_{i}(t)=t_{i}\in F$ and let $\sigma$ be
such that $\sigma(\alpha_{2})=\alpha_{2}$, $\sigma(\alpha_{1})=\alpha_{4}$,
$\sigma(\alpha_{3})=\alpha_{1}$, $\sigma(\alpha_{4})=\alpha_{3}$. Let
$h=\sigma t$. Then the eigenvalues of $h$ on V are

$\zeta,\zeta^{2},(t_{2}t_{3}^{-1})^{\pm 1},\,\,(t_{1}t_{3})^{\pm
1},\,\,(t_{1}t_{2})^{\pm 1},\,\,(t_{1}t_{2}^{-1}t_{3}^{2})^{\pm
1/3}\zeta^{i},\,\,(t_{1}t_{2}^{2}t_{3}^{-1})^{\pm
1/3}\zeta^{i},\,\,(t_{1}^{2}t_{2}t_{3})^{\pm 1/3}\zeta^{i}$ (2)

for $i=0,1,2$, where $\zeta$ is a primitive $3$-root of unity in $F$. These
are of multiplicity $1$ for a suitable choice of $t$.

###  6.1. Group $\langle D_{4}(q),\sigma\rangle$

Observe that $V_{\omega_{2}}|_{G(q)}$ is irreducible. Moreover, all
irreducible $FG(q)$-module of dimension 26 are Galois conjugate to each other.

######  Corollary 6.2.

Let $G(q)=D_{4}(q)$, q even. Suppose that there exist
$t_{1},t_{2},t_{3}\in\mathop{\mathbb{F}}\nolimits_{q}^{\times}$ such that the
elements (of $F$) in (2) of Lemma 6.1 are distinct. Then there is an element
in $H(q):=\langle G(q),\sigma\rangle$ with simple spectrum on the $2$-modular
irreducible representation of dimension $26$.

###### Proof.

It is well known that the elements $t\in T$ with $t^{q-1}=1$ lies in $G(q)$
(when this is properly included into $G$). So the claim follows from Lemma
6.1. ∎

The following lemma gives a sufficient condition for the assumption of
Corollary 6.2 to be satisfied.

######  Lemma 6.3.

Set $M_{1}=\\{(t_{2}t_{3}^{-1})^{\pm 1},(t_{1}t_{3})^{\pm 1},(t_{1}t_{2})^{\pm
1}\\}$ and $M_{2}=\\{(t_{1}t_{2}^{-1}t_{3}^{2})^{\pm
1},(t_{1}t_{2}^{2}t_{3}^{-1})^{\pm 1}$, $(t_{1}^{2}t_{2}t_{3})^{\pm 1}\\}$.
The assumption of Corollary 6.2 is satisfied if there are elements
$t_{1},t_{2},t_{3}\in\mathop{\mathbb{F}}\nolimits_{q}$ such that
$|M_{1}|=|M_{2}|=6$ and either $3|(q-1)$ or $3\not|(q-1)$ and $s^{3}\notin
M_{2}$ for every $s\in M_{1}$.

###### Proof.

Let $x\in\mathop{\mathbb{F}}\nolimits_{q}^{\times}$ with $|x|=q-1$, and let
$\zeta,s\in F$ with $s^{3}=x,\zeta^{3}=1\neq\zeta$. If $3|(q-1)$ then
$s\notin\mathop{\mathbb{F}}\nolimits_{q}$ whenever $s^{3}=x$. As $\zeta^{i}s$
are distinct for $i=0,1,2$, it follows that the elements in Corollary 6.2 are
distinct if and only if and $|M_{1}|=6$ and $|M_{2}|=6.$ If $3\not|(q-1)$ then
there exists $s\in\mathop{\mathbb{F}}\nolimits_{q}$ with $s^{3}=x$. As $\zeta
s,\zeta^{2}s\notin\mathop{\mathbb{F}}\nolimits_{q}$, the elements in Corollary
6.2 are distinct if and only if $|M_{1}|=6$, $|M_{2}|=6$ and $s^{3}\notin
M_{2}$ for every $s\in M_{1}$.∎

######  Corollary 6.4.

The assumptions of Corollary 6.2 can be satisfied for $q>8$.

###### Proof.

Let $\xi$ be a primitive $(q-1)$-root of unity. Set $t_{1}=\xi$,
$t_{2}=\xi^{2}$, $t_{3}=1$. Then $M_{1}=\\{\xi^{\pm 2},\xi^{\pm 1},\xi^{\pm
3},\\}$ and $M_{2}=\\{\xi^{\pm 1},\xi^{\pm 5},\xi^{\pm 4}\\}$. So
$|M_{1}|=6=|M_{2}|$ for $q>4$. If $3|(q-1)$ then $q>8$ and the result follows
by Lemma 6.3.

Let $(3,q-1)=1$. Set $M_{3}=\\{s^{3}:s\in M_{1}\\}$ so $M_{3}=\\{\xi^{\pm
6},\xi^{\pm 3},\xi^{\pm 9}\\}$. Then $M_{3}\cap M_{2}=\emptyset$ for $q\geq
32$ and the result follows by Lemma 6.3.∎

###  6.2. Group $\langle{}^{3}D_{4}(q)\cdot\sigma\rangle$

Let $G$ be a simple algebraic group of type $D_{4}$ in defining characteristic
$p$ and ${\rm Fr}$ a Frobenius endomorphism of $G$. This can be chosen so that
the group $G^{{\rm Fr}}=\\{g\in G:{\rm Fr}(g)=g\\}$ is $D_{4}(q)$ or
${}^{3}D_{4}(q)$ [6, §10]. Let $\sigma$ be a graph automorphism of $G$ of
order 3 which permutes the root subgroups
$X_{\alpha_{1}},X_{\alpha_{3}},X_{\alpha_{4}}$ and stabilizes
$X_{\alpha_{2}}$, see [6, §10]. Then $\sigma\cdot{\rm Fr}={\rm
Fr}\cdot\sigma$, so $\sigma(G(q))=G(q)$. Set $H(q)=\langle
G(q),\sigma\rangle$.

######  Proposition 6.5.

Let $p=2$, let V be an irreducible G-module with highest weight $\omega_{2}$
and $H=\langle G,\sigma\rangle$. Then V extends to H and remains irreducible
on $H(q)$. If $q>8$ then there exists $h\in H(q)$ with simple spectrum on V.

Let $D$ be the group of diagonal matrices in $K:=SL_{2}(F)$ viewed as a
maximal torus of $K$, and let $U_{+}=\begin{pmatrix}0&1\\\ 0&0\end{pmatrix}$
be a root element of the Lie algebra of $SL_{2}(F)$. Let $G_{\alpha_{i}}\cong
SL_{2}(F)$ be the $A_{1}$-type subgroup, $i=1,2,3,4$. Then there exists an
algebraic group homomorphism $\eta_{i}:SL_{2}(F)\rightarrow G_{\alpha_{i}}$
which yields Lie algebra homomorphism $\eta_{i}^{\prime}:L(K)\rightarrow L(G)$
compatible with the action of these groups on their Lie algebras (that is,
$\eta_{i}^{\prime}$ is a $K$-module homomorphism $L(K)\rightarrow
L(G))|_{K}$). In particular, $\eta_{i}^{\prime}(U_{+})=X_{\alpha_{i}}$. Let
$x=\mathop{\rm diag}\nolimits(y,y^{-1})\in K$. Then $xU_{+}x^{-1}=y^{2}U_{+}$
and hence $\eta_{i}(x)(X_{\alpha_{i}})\eta_{i}(x^{-1})=y^{2}X_{\alpha_{i}}$.
Denote by $y_{i}$ the eigenvalue of $x$ in its action on $X_{\alpha_{i}}$.
This allows us to express $y_{1},y_{2},y_{3},y_{4}$ in terms of
$t_{1},t_{2},t_{3},t_{4}$ in (1). We have

$y_{1}^{2}=t_{1}t_{2}^{-1},\,\,y_{2}^{2}=t_{2}t_{3}^{-1},\,\,y_{3}^{2}=t_{3}t_{4}^{-1},\,\,y_{4}^{2}=t_{3}t_{4},$
(3)

whence

$t_{1}=y_{1}^{2}y_{2}^{2}y_{3}y_{4},\,\,t_{2}=y_{2}^{2}y_{3}^{-1}y_{4}^{-1},\,\,t_{3}=y_{3}y_{4},\,\,t_{4}=y_{3}^{-1}y_{4}.$
(4)

######  Lemma 6.6.

Let $G(q)={}^{3}D_{4}(q)$, $q\geq 16$ even, and $H(q)=\langle
G(q),\sigma\rangle$, where $\sigma$ is as above. Let $V$ be an irreducible
$FH(q)$-module such that $V|_{G(q)}$ is irreducible with highest weight
$\omega_{2}$. Then $h$ has simple spectrum on V for some $h\in(H(q)\setminus
G(q))$.

###### Proof.

By (4) we have

$t_{3}^{2}=y_{3}^{2}y_{4}^{2},\,\,\,t_{4}^{2}=y_{3}^{-2}y_{4}^{2},\,\,\,t_{1}t_{3}=y_{1}^{2}y_{2}^{2}y_{3}^{2}y_{4}^{2},\,\,\,t_{2}^{2}=y_{2}^{4}y_{3}^{2}y_{4}^{2},\,\,\,t_{1}t_{2}=y_{1}^{2}y_{2}^{4}y_{3}^{2}y_{4}^{2},$

so

$t_{1}t_{2}^{-1}t_{3}^{2}=y_{1}^{2}y_{3}^{2}y_{4}^{2},\,\,\,t_{1}t_{2}^{2}t_{3}^{-1}=y_{1}^{2}y_{2}^{6}y_{3}^{2}y_{4}^{2},\,\,\,t_{1}^{2}t_{2}t_{3}=y_{1}^{4}y_{2}^{6}y_{3}^{4}y_{4}^{4}.$

Suppose that $y_{3}=y_{1}^{q}$, $y_{4}=y_{1}^{q^{2}}$, where
$|y_{1}|=q^{3}-1$. Then we have

$t_{2}t_{3}^{-1}=y_{2}^{2},\,\,t_{1}t_{3}=y_{1}^{2(1+q+q^{2})}y_{2}^{2},\,\,t_{1}t_{2}=y_{1}^{2(1+q+q^{2})}y_{2}^{4},$
$t_{1}t_{2}^{-1}t_{3}^{2}=y_{1}^{2(1+q+q^{2})},\,\,t_{1}t_{2}^{2}t_{3}^{-1}=y_{1}^{2(1+q+q^{2})}y_{2}^{6},\,\,t_{1}^{2}t_{2}t_{3}=y_{1}^{4(1+q+q^{2})}y_{2}^{6}.$

Set $u=y_{1}^{2(1+q+q^{2})}$. Then $|u|=q-1$,
$u\in\mathop{\mathbb{F}}\nolimits_{q}$ as $(q^{3}-1)/(q-1)=q^{2}+q+1$. Then we
have

$t_{2}t_{3}^{-1}=y_{2}^{2},\,\,\,t_{1}t_{3}=uy_{2}^{2},\,\,\,t_{1}t_{2}=uy_{2}^{4},\,t_{1}t_{2}^{-1}t_{3}^{2}=u,\,\,\,t_{1}t_{2}^{2}t_{3}^{-1}=uy_{2}^{6},\,\,\,t_{1}^{2}t_{2}t_{3}=u^{2}y_{2}^{6}.$

Therefore, the eigenvalues of $h=t\sigma$ on $V$ are

$\zeta^{\pm 1}\,\,$, $y_{2}^{\pm 2}\,\,$, $(uy_{2}^{2})^{\pm 1}\,\,$,
$(uy_{2}^{4})^{\pm 1}\,\,$, $u^{\pm 1/3}\zeta^{i}\,\,$, $(uy_{2}^{6})^{\pm
1/3}\zeta^{i}\,\,$, $(u^{2}y_{2}^{6})^{\pm 1/3}\zeta^{i}\,\,$ for $i=0,1,2$.

Note that the element $(y_{1},y_{2},y_{1}^{q},y_{1}^{q^{2}})\in T$ lies in
$G(q)$ as ${\rm Fr}(y_{1},y_{2},y_{1}^{q},y_{1}^{q^{2}})=\\\ \sigma\cdot{\rm
Fr}_{q}(y_{1},y_{2},y_{1}^{q},y_{1}^{q^{2}})=\sigma(y_{1}^{q},y_{2},y_{1}^{q^{2}},y_{1}^{q^{3}})=\sigma(y_{1}^{q},y_{2},y_{1}^{q^{2}},y_{1})=(y_{1},y_{2},y_{1}^{q},y_{1}^{q^{2}})$.
Here ${\rm Fr}_{q}$ is an endomorphism of $G$ arising from the mapping
$x\rightarrow x^{q}$ for $x\in F$.

If $(3,q-1)=1$ then there exists $y_{2}\in\mathop{\mathbb{F}}\nolimits_{q}$
such that $u=y_{2}^{3}$. Then the eigenvalues are as follows, where we use $y$
in place of $y_{2}$:

$\zeta^{\pm 1},\,\,y^{\pm 2},\,\,y^{\pm 5},\,\,y^{\pm 7},\,\,y^{\pm
1}\zeta^{i},\,\,y^{\pm 3}\zeta^{i},\,\,y^{\pm 4}\zeta^{i}\,\,\,{\rm
for}\,\,\,i=0,1,2.$

If $q\geq 32$ and $(3,q-1)=1$ then $y^{\pm 2}$, $y^{\pm 5}$, $y^{\pm 7}$,
$y^{\pm 1}$, $y^{\pm 3}$, $y^{\pm 4}$ are distinct, and hence the above
eigenvalues are distinct too.

If $3|(q-1)$ then we choose $y_{2}$ with $y_{2}^{2}=u$, and the eigenvalues
are

$\zeta^{\pm 1},\,\,y^{\pm 2},\,\,y^{\pm 4},\,\,y^{\pm 6},\,\,y^{\pm
8/3}\zeta^{i},\,\,y^{\pm 10/3}\zeta^{i},\,\,y^{\pm 2/3}\zeta^{i}\,\,\,{\rm
for}\,\,\,i=0,1,2.$

If $q\geq 16$ and $3|(q-1)$ then $\zeta^{\pm 1}$, $y^{\pm 2}$, $y^{\pm 4}$,
$y^{\pm 6}$ are distinct and lie in $\mathop{\mathbb{F}}\nolimits_{q}$. In
addition, $y^{\pm 8/3}\zeta^{i}$, $y^{\pm 10/3}\zeta^{i}$, $y^{\pm 2/3}$ are
distinct and do not lie in $\mathop{\mathbb{F}}\nolimits_{q}$. It follows that
the eigenvalues above are distinct.

It is clear that that there exist elements $t=(t_{1},t_{2},t_{3},t_{4})\in
G(q)$ satisfying (4). Then
$u=y_{1}^{2(1+q+q^{2})}=(t_{1}t_{2}^{-1})^{1+q+q^{2}}$ and the conditions
$y_{3}=y_{1}^{q},y_{4}=y_{1}^{q^{2}}$ gives
$t_{3}t_{4}^{-1}=(t_{1}t_{2}^{-1})^{q}$,
$t_{3}t_{4}=(t_{1}t_{2}^{-1})^{q^{2}}$. To satisfy this, one can take
$t_{3}^{2}=t_{1}^{q+q^{2}}t_{2}^{-q-q^{2}}$,
$t_{4}^{2}=t_{1}^{-q+q^{2}}t_{2}^{q-q^{2}}$. So the lemma follows. ∎

##  7\. The representations of $H$ reducible on $G$

###### Proof of Proposition 1.2.

Let $V_{1}$ be an irreducible constituent of $V|_{G}$. For a weight $\mu$ of
$V_{1}$ let $V_{\mu}$ be the $\mu$-weight space of $V_{1}$.

The ”only if” part. Since $k$ is a prime, $V|_{G}=V_{1}\oplus\cdots\oplus
V_{k}$, where $V_{1},\ldots,V_{k}$ are irreducible $G$-modules, and $H$
permutes $V_{1},\ldots,V_{k}$ transitively (Clifford’s theorem). As
$h^{k}V_{1}=V_{1}$ and $h^{k}\in G$, we can apply Lemma 2.2 to conclude that
$h$ has an eigenvalue of multiplicity at least $d=\dim V_{\mu}$. So $d=1$.

The ”if” part. Set $V^{+}_{\mu}=\sum h^{i}V_{\mu}$. Then the multiplicity of
$h^{k}$ on $V^{+}_{\mu}$ equals $k$. By Lemma 2.2, the multiplicity of the
eigenvalues of $h$ on $V^{+}_{\mu}$ equals 1. Clearly, $V^{+}_{\mu}\cap
V^{+}_{\mu^{\prime}}=0$ whenever $\mu^{\prime}\neq\mu$ is a weight of $V_{1}$.
As the eigenvalues of $h^{k}$ on $V_{\mu}$ and $V_{\mu^{\prime}}$ are
distinct, the result follows. ∎

Suppose that all weights of $V_{1}$ are of multiplicity 1. A natural question
is whether there always exists $h\in H$ with $h^{k}$ with simple spectrum on
$V_{1}$. The answer is negative.

Indeed, let $G=SL_{4}(F)$, where $F$ is an algebraically closed field of
characteristic not 2. Let $V_{1}=V_{2\omega_{1}}$ be an irreducible $G$-module
with highest weight $2\omega_{1}$ and $V=V_{1}\uparrow H$, the induced module.
Then $V|_{G}=V_{1}+V_{2}$, where $V_{2}$ is irreducible of highest weight
$2\omega_{3}$. We show that no element $h\in H$ has simple spectrum on $V$.

As in Section 4 we can express $h$ as $\sigma n_{w}t$, where $T$ is the group
of diagonal matrices in $G$, $t\in T$ and $n_{w}=\mathop{\rm
diag}\nolimits(\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},1,1)$ or $\mathop{\rm
diag}\nolimits(\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},\begin{pmatrix}0&1\\\
-1&0\end{pmatrix})$. Then $h^{2}=\mathop{\rm diag}\nolimits(x,x^{-1},1,1)$ or
$\mathop{\rm diag}\nolimits(x,x^{-1},y,y^{-1})$ for some $x,y\in{\rm
F}^{\times}$, see Section 4. Note that $V_{2\omega_{1}}$ has weights
$\pm\omega_{2}$. As
$\omega_{2}=\frac{1}{2}(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3}-\varepsilon_{4})$
[1, Planche 1], we have $\omega_{2}(h^{2})=\pm 1$, and hence
$\omega_{2}(h^{2})=(-\omega_{2})(h^{2})$. So the spectrum of $h^{2}$ on
$V_{2\omega_{1}}$ is not simple, and hence, by Lemma 1.2, the spectrum of $h$
on $V$ is not simple.

Acknowledgement. I am indebted to R. Guralnick for rising the problem
discussed in this paper and his encouraging comments.

## References

  * [1] N. Bourbaki, Groupes et algebres de Lie, ch. IV-VI, Masson, Paris, 1981.
  * [2] N. Bourbaki, Groupes et algebres de Lie, ch. VII-VIII, Springer, Berlin, 2006.
  * [3] D. Gorenstein, R. Lyons and R. Solomon, The Classification of the Finite Simple Groups, vol.3, Amer. Math. Soc. Rhode Island, 1998.
  * [4] G.M. Seitz, ‘The Maximal Subgroups of Classical Algebraic Groups’, Memoirs Amer. Math. Soc., no.365, Amer. Math. Soc., Providence, $1987$.
  * [5] R. Steinberg, Endomorphisms of linear algebraic groups, Memoirs of Amer. Math. Soc., vol. 80, AMS, 1968.
  * [6] R. Steinberg, Lectures on Chevalley groups, Amer. Math. Soc. Univ. Lect. Series, vol. 66, Providence, Rhode Island, 2016.
  * [7] I.D. Suprunenko and A.E. Zalesski, Irreducible representations of finite Chevalley groups containing a matrix with a simple spectrum. In: ”Groups, Combinatorics and Geometry”, London Math. Soc. Lecture notes, vol. 165, Cambridge Univ. Press, Cambridge, 1992, $327-332$.
  * [8] I.D. Suprunenko and A.E. Zalesski, Irreducible representations of finite classical groups containing matrices with simple spectra, Comm. Algebra 26(1998), 863 - 888.
  * [9] I.D. Suprunenko and A.E. Zalesski, Irreducible representations of finite groups of exceptional Lie type containing matrices with simple spectra, Comm. Algebra 28(2000), no.4, 1789–1833.
  * [10] D. Testerman and A. Zalesski, Subgroups of simple algebraic groups containing maximal tori and representations with multiplicity $1$ non-zero weights, Transformation Groups 20(2015), $831-861$.
  * [11] A.E. Zalesskiĭ and I.D. Suprunenko, Representations of dimensions $(p^{n}\pm 1)/2$ of the symplectic group of degree $2n$ over a field of characteristic $p$ (in Russian), Vesti AN BSSR, ser. fiz.-mat. navuk 1987, no. 6, $9-15$. (English translation: http://arxiv.org/abs/2108.10650)

$\begin{array}[]{|l|c|c|}\hline\cr\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ G&{\rm highest\leavevmode\nobreak\
weight\leavevmode\nobreak\ of}\leavevmode\nobreak\ V&\mbox{weight }0\mbox{
{multiplicity }}\\\ \hline\cr A_{n},\,n>1,\leavevmode\nobreak\
p\,\not|\leavevmode\nobreak\ (n+1)&\omega_{1}+\omega_{n}&n\\\
(n,p)\neq(2,3),\,p\,|\,(n+1)&\omega_{1}+\omega_{n}&n-1\\\
A_{3},\leavevmode\nobreak\ p>3&2\omega_{2}&2\\\ \hline\cr B_{n},n>2,p\neq
2&\omega_{2}&n\\\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
p\,\,\,|\,(2n+1)&2\omega_{1}&n\\\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
p\not|\,(2n+1)&2\omega_{1}&n+1\\\ \hline\cr C_{2},\,\,p\neq 2&2\omega_{1}&2\\\
C_{2},\,\,p\neq 2,5&2\omega_{2}&2\\\ C_{3},\leavevmode\nobreak\
p=3&2\omega_{1}&3\\\ C_{n},\leavevmode\nobreak\ n>2,\,\,p\neq
2&2\omega_{1}&n\\\
\,\,\,\,\,\,\,\,\,\,(n,p)\neq(3,3),\,\,p\not|n&\omega_{2}&n-1\\\
\,\,\,\,\,\,\,\,\,\,(n,p)\neq(3,3),\,\,p\,\,\,|n&\omega_{2}&n-2\\\ \hline\cr
D_{n},\,\,n>3,\,p>2,\,\,p\,\,\,|\,\,n&2\omega_{1}&n-2\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,n>3,\,p\neq 2,\,\,p\not|\,n&2\omega_{1}&n-1\\\
&\omega_{2}&n\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,n>3,\,\,p=2&\omega_{2}&n-(2,n)\\\
\hline\cr E_{6}\,\,\,\,\,p\neq 3&\omega_{2}&6\\\
\,\,\,\,\,\,\,\,\,\,\,\,p=3&\omega_{2}&5\\\ \hline\cr E_{7},\,\,p\neq
2&\omega_{1}&7\\\ \,\,\,\,\,\,\,\,\,\,\,p=2&\omega_{1}&6\\\ \hline\cr
E_{8}&\omega_{8}&8\\\ \hline\cr F_{4},\,\,p\neq 2&\omega_{1}&4\\\
\,\,\,\,\,\,\,\,\,\,\,p=2&\omega_{1}&2\\\ \,\,\,\,\,\,\,\,\,\,\,p\neq
3&\omega_{4}&2\\\ \hline\cr G_{2},\,p\neq 3&\omega_{2}&2\\\
\hline\cr\end{array}$

Table 1: Irreducible $p$-restricted $G$-modules $V$ whose non-zero weight
spaces are $1$-dimensional and whose zero weight has multiplicity greater than
$1$.