On the exponent of geometric unipotent radicals of pseudo-reductive groups

Abstract

Let $k'/k$ be a finite purely inseparable field extension and let $G'$ be a reductive $k'$-group. We denote by $G=\mathrm {R}_{k'/k}(G')$, the Weil restriction of $G'$ across $k'/k$, a pseudo-reductive group. This article gives bounds for the exponent of the geometric unipotent radical ${\mathscr {R}}_{u}(G_{\bar{k}})$ in terms of invariants of the extension $k'/k$, starting with the case $G'={{\,\mathrm{GL}\,}}_n$ and applying these results to the case where $G'$ is a simple group.

1 Introduction

Let k be a field, which we assume is imperfect of characteristic p. In this case, there exist pseudo-reductive k-groups which are not reductive. Recall that a smooth, connected, affine algebraic k-group G is pseudo-reductive if the largest smooth connected unipotent normal k-subgroup ${\mathscr {R}}_{u,k}(G)$ is trivial.

We are interested in the structure of the geometric unipotent radical $R:={\mathscr {R}}_u(G_{{\bar{k}}})={\mathscr {R}}_{u,{\bar{k}}}(G_{{\bar{k}}})$ of G. Since R is a p-group, it makes sense to study its exponent: the minimal integer s such that the $p^s$-power map on the geometric unipotent radical factors through the trivial group. We denote the exponent of R by e(R).

M. Bate, B. Martin, G. Röhrle, and the third author previously gave some bounds for e(R) in [1]. For example, [1, Lem. 4.1] implies that if $k'/k$ is a simple purely inseparable field extension with $(k')^{p^e}\subseteq k$, then $e(R)\le e$.

The monographs [4] and [5] contain a classification of pseudo-reductive groups. They are all related in some way or other to the Weil restrictions of reductive groups, which are themselves pseudo-reductive. We focus on groups which are Weil restrictions, since general pseudo-reductive groups contain central pseudo-reductive subgroups whose classification is thought to be intractable. Thus we assume unless stated otherwise that $G=\mathrm {R}_{k'/k}(G')$ for some reductive $k'$-group $G'$ where $k'$ is a finite non-zero reduced k-algebra. Since we are interested in e(R), by the remarks in [1, §4], we may as well consider the case that k is separably closed and $k'$ is a purely inseparable field extension of k. In this case, $G'$ is a split reductive group and as such is classified by its root datum.

Recall that the exponent of the extension $k'/k$ is the smallest integer e such that $(k')^{p^e}\subseteq k$. To describe our results, we need more sensitive data about the structure of $k'/k$. Since $k'/k$ is purely inseparable, we can appeal to the results in [10] and [11], which guarantee a normal generating sequence $\alpha _1,\ldots ,\alpha _l$ of $k'$ over k with certain properties—see Definition 2.6 below for more details. In particular, these elements come with certain exponents which are invariants of $k'/k$. These are the integers $e_1\ge \cdots \ge e_l$ such that $e_i$ is minimal subject to

$$\begin{aligned} \alpha _i^{p^{e_i}}\in k(\alpha _1^{p^{e_i}},\ldots ,\alpha _{i-1}^{p^{e_i}}). \end{aligned}$$

It is immediate that $e_1=e$ and it follows from the tower law that $[k':k]=\prod p^{e_i}$. Let $r\in {\mathbb {N}}$. We define integers

$$\begin{aligned} {{\,\mathrm{m}\,}}(k'/k)&:=\sum _{i=1}^l(p^{e_i}-1)+1=\sum _{i=1}^{l}p^{e_i}-l+1,\\ {{\,\mathrm{em}\,}}(k'/k)&:=\lceil \log _p({{\,\mathrm{m}\,}}(k'/k))\rceil ,\\ {{\,\mathrm{mr}\,}}(k'/k,r)&:=\sum _{i=1}^r p^{e_i}\text {,\quad where we take\,} e_i\,=\,0~\text {for}\, i>l,\\ {{\,\mathrm{emr}\,}}(k'/k,r)&:=\lceil \log _p({{\,\mathrm{mr}\,}}(k'/k,r))\rceil ,\\ \text {and finally } {{\,\mathrm{E}\,}}(k'/k,r)&:=\min \{{{\,\mathrm{em}\,}}(k'/k),{{\,\mathrm{emr}\,}}(k'/k,r)\}. \end{aligned}$$

In other words, ${{\,\mathrm{em}\,}}(k'/k)$ is the minimal s such that $p^s\ge {{\,\mathrm{m}\,}}(k'/k)$, and similarly for ${{\,\mathrm{emr}\,}}$.

In answering a question of [1] on the exponent of the intersection of R with a pseudo-Borel subgroup, we discovered that these numbers give lower bounds on e(R) and for large enough rank determine the exponent of R exactly. We first state our results in the case that $G'$ is a general linear group.

Theorem 1.1

Let $k'/k$ be a finite purely inseparable field extension of a separably closed field k of characteristic p. Suppose $k'/k$ has exponent e and a normal generating sequence $\alpha _1,\ldots ,\alpha _l$ with sequence of exponents $e_1,\ldots ,e_l$. For $r\in {\mathbb {N}}$, let $G':={{\,\mathrm{GL}\,}}_r$, $G:=\mathrm {R}_{k'/k}(G')$, and $R:={\mathscr {R}}_u(G_{{\bar{k}}})$. Furthermore let $B'$ denote a Borel subgroup of $G'$ and $B:=\mathrm {R}_{k'/k}(B')\subseteq G$ the corresponding pseudo-Borel subgroup of G. The following hold:

• (i) if s satisfies $p^s\ge p^e r$, then e(R) is at most s; i.e. $e(R)\le \lceil e+\log _p(r)\rceil$;

• (ii) we have $e(R\cap B_{{\bar{k}}})={{\,\mathrm{E}\,}}(k'/k,r)$;

• (iii) in particular, $e(R)\ge {{\,\mathrm{E}\,}}(k'/k,r)$;

• (iv) if $\sum _{i=r+1}^l(p^{e_i}-1)<r-1$, then $e(R)={{\,\mathrm{E}\,}}(k'/k,r)={{\,\mathrm{em}\,}}(k'/k)$.

Remark 1.2

• (i) The bounds in Theorem 1.1 are both arithmetic and Lie-theoretic in character: the integers ${{\,\mathrm{m}\,}}(k'/k)$ and ${{\,\mathrm{em}\,}}(k'/k)$ depend only on the arithmetic structure of $k'/k$ but ${{\,\mathrm{mr}\,}}(k'/k,r)$, ${{\,\mathrm{emr}\,}}(k'/k,r)$, and ${{\,\mathrm{E}\,}}(k'/k,r)$ depend on both $k'/k$ and the rank r of $G'$.

• (ii) Part (i) of Theorem 1.1 improves [1, Lem. 4.4], which required $p^s\ge r^2(p^e-1)$. As a corollary of (i) and (iv), if $G=\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_r)$ for $p\ge r\ge 2$, then the exponent of ${\mathscr {R}}_u(G_{\bar{k}})$ is e if $k'/k$ is a simple extension and is $e+1$ otherwise. If $r=1$, then $e(R)=e$ as $\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_1)$ is sent into the canonical ${{\,\mathrm{GL}\,}}_1$ k-subgroup by the $p^e$-power map, since it also sends $k'$ into k.

• (iii) In [1, Rem. 4.5(iv)], it was asked if the exponent e(R) always coincides with the exponent $e(R\cap B_{{\bar{k}}})$. Clearly Theorem 1.1(ii)-(iv) answers this question in the affirmative when $G'={{\,\mathrm{GL}\,}}_r$ of large enough rank. We give an example to show that in general the answer is ‘no’ (see Proposition 4.3), coming from ${{\,\mathrm{SL}\,}}_2$ in characteristic 2. In fact, in Section 4, we give a complete description of e(R) when $G'$ is ${{\,\mathrm{SL}\,}}_2$, ${{\,\mathrm{PGL}\,}}_2$, or ${{\,\mathrm{GL}\,}}_2$.

Together with the rank 1 results in Section 4, the following (proved in Section 5) confirms the large rank behaviour in all simple groups, with mild conditions on the characteristic.

Corollary 1.3

Let $k'/k$ be a finite, purely inseparable field extension of a separably closed field k of characteristic p. Suppose $k'/k$ has exponent e and a normal generating sequence $\alpha _1,\ldots ,\alpha _l$ with sequence of exponents $e_1,\ldots ,e_l$. Let $G'$ be a (split) simple algebraic k-group of rank $r\ge 2$, $G:=R_{k'/k}(G')$, and let $R:={\mathscr {R}}_{u}(G_{{\bar{k}}})$. If $G'$ is type B, C, D, F, then assume $p\ne 2$ or $G'$ is ${{\,\mathrm{SO}\,}}_{n}$ or ${{\,\mathrm{Sp}\,}}_{2n}$, and if G is of type $E_6$, then assume $p\ne 3$. Then the exponent $e(R)\ge {{\,\mathrm{E}\,}}(k'/k,r)$; if $\sum _{i=r+1}^l(p^{e_i}-1)<r-1$, then in fact, $e(R)={{\,\mathrm{E}\,}}(k'/k,r)={{\,\mathrm{em}\,}}(k'/k)$.

2 Preliminaries and notation

Let k be a field of characteristic p and G be a k-group, by which we mean an affine algebraic group scheme of finite type over k. The k-unipotent radical is the maximal smooth connected normal unipotent k-subgroup of G and we denote this by ${\mathscr {R}}_{u,k}(G)$. The geometric unipotent radical is the $\bar{k}$-unipotent radical of the base change $G_{\bar{k}}$ of G to ${\bar{k}}$. This is denoted by ${\mathscr {R}}_{u}(G_{\bar{k}})$. We call a smooth connected k-group G reductive if ${\mathscr {R}}_{u}(G_{\bar{k}})=1$ and pseudo-reductive if ${\mathscr {R}}_{u,k}(G)=1$. From now on, G will always be pseudo-reductive. We recall the definition and key properties of Weil restriction from [4, A.5].

Definition 2.1

Let $B\rightarrow B'$ be a finite flat map of Noetherian rings, and $X'$ a quasi-projective $B'$-scheme. The Weil restriction is a finite type B-scheme satisfying the universal property

$$\begin{aligned} \mathrm {R}_{B'/B}(X')(A)=X'(B'\otimes _B A) \end{aligned}$$

for all B-algebras A.

The following proposition makes use of the natural map

$$\begin{aligned} q_{G'}:\mathrm {R}_{k'/k}(G')_{k'}\rightarrow G', \end{aligned}$$

(2.2)

which is induced by $k'\otimes _k A \rightarrow A, a\otimes b \mapsto ab$ for any $k'$-algebra A.

Proposition 2.3

([4, Prop. A.5.11, Thm. 1.6.2], [1, §2], [1, Lem. 3.4]). Let $k'/k$ be a finite and purely inseparable field extension, $G'$ a non-trivial smooth connected $k'$-group, and $G=\mathrm {R}_{k'/k}(G')$.

• (i) The kernel of $q_{G'}$ is a smooth connected unipotent $k'$-subgroup and thus is contained in ${\mathscr {R}}_{u,k'}(G_{k'})$.

• (ii) If $G'$ is reductive over $k'$, then the kernel of $q_{G'}$ has field of definition over k equal to $k'\subseteq \bar{k}$. Thus the kernel $\ker (q_{G'})$ coincides with ${\mathscr {R}}_{u,k'}(G_{k'})$, which is a $k'$-descent of ${\mathscr {R}}_u(G_{\bar{k}})$ (i.e. ${\mathscr {R}}_{u,k'}(G_{k'})_{{\bar{k}}}\cong {\mathscr {R}}_u(G_{\bar{k}})$).

• (iii) If $H'$ is a reductive subgroup of $G'$, then the geometric unipotent radical of $\mathrm {R}_{k'/k}(H')$ is a subgroup of the geometric unipotent radical of $\mathrm {R}_{k'/k}(G')$.

• (iv) If $f:G_1\rightarrow G_2$ is an étale isogeny of reductive $k'$-groups, then $\mathrm {R}_{k'/k}(f)_{{\bar{k}}}$ induces an isomorphism of geometric unipotent radicals of the respective Weil restrictions.

We are interested in the following invariant of a unipotent k-group:

Definition 2.4

Let k be a field of characteristic p and U a unipotent k-group. The exponent $\exp (U)$ of U is the minimal s such that the $p^s$-power map on U factors through the trivial group.

Clearly, this definition is insensitive to base change. As in [1, §4], we may therefore assume k is separably closed for the remainder of the article.

When calculating with matrices, we make use of the fact that if $U(k)\subset U$ is dense (e.g. if k is perfect [6, Thm. 34.4]), then a map factors through the trivial group if and only if it maps to the identity on U(k). Since $R={\mathscr {R}}_u(G_{{\bar{k}}})$ descends to $R':={\mathscr {R}}_{u,k'}(G_{k'})$ whose $k'$-points are dense (by the fact that $k'$ is separably closed), we have that e(R) is the smallest s such that $p^s$ kills $R'(k')$.

Following [9], let ${\mathfrak {B}}:=k'\otimes _k k'$. Then ${\mathfrak {B}}$ is a local ring with maximal ideal

$$\begin{aligned} {\mathfrak {m}}=\langle 1\otimes x - x \otimes 1 \mid x\in k' \rangle =\ker (\phi ), \end{aligned}$$

where $\phi :k'\otimes _k k' \rightarrow k', a\otimes b \mapsto ab$. Moreover, $\mathrm {R}_{k'/k}(G')_{k'}$ can be identified with $\mathrm {R}_{{\mathfrak {B}}/k'}(G'_{\mathfrak {B}})$ and the $k'$-points of the $k'$-unipotent radical of the former is the kernel $G'({\mathfrak {m}})$ of the map $G'({\mathfrak {B}})\rightarrow G'(k)$ induced by $\phi$. Suppose now $G'={{\,\mathrm{GL}\,}}_r$ and $G=\mathrm {R}_{k'/k}(G')$. As the entries of any $M\in {{\,\mathrm{Mat}\,}}(r,{\mathfrak {m}})$ are nilpotent elements of ${\mathfrak {B}}$, it follows that $I_r+M$ has invertible determinant, hence is an element of $G'({\mathfrak {m}})$. Therefore

$$\begin{aligned} {\mathscr {R}}_{u,k}(\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_r)_{k'})(k')=\{I_r+M\mid M\in {{\,\mathrm{Mat}\,}}(r,{\mathfrak {m}})\}. \end{aligned}$$

As $I_r$ and M commute, we have $A^{p^s}=I_r+M^{p^s}=I_r$ is equivalent to $M^{p^s}=0$. Hence:

Lemma 2.5

The exponent of the geometric unipotent radical of $\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_r)$ is the smallest s such that $M^{p^s}=0$ for all $M\in {{\,\mathrm{Mat}\,}}(r,{\mathfrak {m}})$.

It was observed in [1, Lem. 4.1] that as ${\mathfrak {m}}^n=0$ for some $n\in {\mathbb {N}}$, one knows that s in the lemma above is bounded above by the minimal t such that $p^t\ge n$: the entries of $M^{p^t}$ are homogeneous polynomials of degree $p^t$ in elements of ${\mathfrak {m}}$ and so must vanish.

Definition 2.6

Let $k'/k''/k$ be a tower of finite purely inseparable field extensions.

• (i) If $x\in k'$, then the minimal $s\in {\mathbb {N}}$ such $x^{p^s}\in k''$ is called the exponent of x with respect to $k''$ and denoted by $e_{k''}(x)$.

• (ii) We say that $x\in k'$ is normal in $k'/k''$ if $e_{k''}(x)\ge e_{k''}(y)$ for all $y\in k'$.

• (iii) A sequence $\alpha _1,\ldots , \alpha _l$ in $k'$ is called a normal sequence if for every $1\le i \le l$ and $k_i=k[\alpha _1,\ldots , \alpha _i]$, we have that $\alpha _i$ is normal in $k'/k_{i-1}$, and $\alpha _i \notin k_{i-1}$.

• (iv) A normal sequence $\alpha _1,\ldots , \alpha _l$ is called a normal generating sequence of $k'/k$ if $k[\alpha _1,\ldots , \alpha _l]=k'$.

• (v) For a normal generating sequence $\alpha _1,\ldots , \alpha _l$ of $k'/k$, we set $e_i=e_{k_{i-1}}(\alpha _i)$. We call $e_1,\ldots ,e_l$ a sequence of exponents for $k'/k$

Note that if $e_1,\ldots ,e_l$ is a sequence of exponents for $k'/k$, then we have $e_1\ge e_2 \ge \cdots \ge e_l$ due to [11, Prop. 5]. Crucially, by loc. cit., we also have

$$\begin{aligned} \alpha _i^{p^{e_i}}\in k(\alpha _{1}^{p^{e_i}},\ldots ,\alpha _{i-1}^{p^{e_i}}) \text { for all }1\le i\le l.\end{aligned}$$

(2.7)

In particular, as the $e_i$ are decreasing,

$$\begin{aligned} (k')^{p^{e_i}}\subseteq k(\alpha _1^{p^{e_i}},\ldots ,\alpha _{i-1}^{p^{e_i}}). \end{aligned}$$

(2.8)

Combining this with the main result of [10], we get that the sequence of exponents is an invariant of the field extension $k'/k$. More precisely:

Lemma 2.9

Let $k'/k$ be a finite purely inseparable field extension. Suppose that $\alpha _{1},\ldots ,\alpha _l$ and $\beta _1,\ldots ,\beta _{l'}$ are normal generating sequences of $k'/k$ with sequences of exponents $e_1,\ldots , e_l$ and $e'_1,\ldots , e'_{l'}$. Then $l=l'$ and $e_i=e'_i$ for $1\le i< l$.

By the lemma, the numbers ${{\,\mathrm{m}\,}}(k'/k)$, ${{\,\mathrm{mr}\,}}(k'/k,r)$ defined in the introduction are both invariants of $k'/k$ and $(k'/k,r)$ respectively. To relate these with ${\mathfrak {m}}$, we first prove:

Lemma 2.10

For a normal generating sequence $\alpha _1,\ldots , \alpha _l$ of $k'/k$, define $a_i:=1\otimes \alpha _i-\alpha _i\otimes 1\in {\mathfrak {m}}$. Then:

• (i) the ideal ${\mathfrak {m}}$ is generated by the $a_i$ over $k'$;

• (ii) the $p^{e_i}$-power map takes ${\mathfrak {m}}$ into the $k'$-subspace $k'[a_1^{p^{e_i}},\ldots ,a_{i-1}^{p^{e_i}}]\cap {\mathfrak {m}}$.

Proof

(i) Since ${\mathfrak {m}}$ is a codimension 1 (left) $k'$-subspace of ${\mathfrak {B}}$, it is a codimension $[k':k]$ k-subspace of ${\mathfrak {B}}$. Define $J=\{ (b_1,b_2,\ldots ,b_l)\in {\mathbb {Z}}^l \mid 0\le b_i\le p^{e_i}-1 \}$. For $b=(b_1,...,b_l)\in J$, we set $x^b=\prod _{i=1}^{l}\alpha _{i}^{b_i}$. Then $\{x^b\}_{b\in J}$ is a k-basis of $k'$ and $\{1\otimes x^b\}$ is a $k'$-basis of ${\mathfrak {B}}$. For $b\ne (0,\ldots , 0)$, the elements $a_b:=1\otimes x^b-x^b\otimes 1=1\otimes x^b-x^b (1\otimes 1)$ are therefore linearly independent, generating a codimension 1 subspace of ${\mathfrak {B}}$, which is contained in the codimension 1 kernel ${\mathfrak {m}}$ of the multiplication map $x\otimes y\mapsto xy$; hence they form a $k'$-basis of ${\mathfrak {m}}$. We wish to show $a_b\in {\mathfrak {m}}':={\mathfrak {m}}\cap k'[a_1,\ldots ,a_l]$ by induction on $\sum b_i$, the case $\sum b_i=1$ being trivial. So let $\sum b_i>1$. Without loss of generality, suppose $b_1>0$ and let $b'=b-(1,0,\ldots ,0)$. Then inductively $a_{b'}=1\otimes x^{b'}-x^{b'}\otimes 1\in {\mathfrak {m}}'$. Now one checks directly that $a_b=x^{b'}a_1+\alpha _1x^{b'}-a_{b'}\cdot a_1\in {\mathfrak {m}}'$ as required. (And it follows that ${\mathfrak {m}}={\mathfrak {m}}'$.)

(ii) Define ${\mathfrak {m}}':=k'[a_1^{p^{e_i}},\ldots ,a_{i-1}^{p^{e_i}}]\cap {\mathfrak {m}}$. From (i), we can express any element of ${\mathfrak {m}}$ as a polynomial in $a_1,\ldots , a_l$ over $k'$. Since $k'$ has characteristic p, the $p^{e_i}$-power map distributes over addition and so it suffices to check that $(a_1^{r_1}\cdots a_l^{r_l})^{p^{e_i}}\in {\mathfrak {m}}'$ for $r_1,\ldots , r_l\in {\mathbb {Z}}_{\ge 0}$. For this, it suffices to check that $a_j^{p^{e_i}}\in {\mathfrak {m}}'$ for $j\ge i$. Now $a_j^{p^{e_i}}=(1\otimes \alpha _j -\alpha _j\otimes 1)^{p^{e_i}}=1\otimes \alpha _j^{p^{e_i}}-\alpha _j^{p^{e_i}}\otimes 1$ and by (2.8), we can write $\alpha _j^{p^{e_i}}$ as a k-polynomial P in $k[\alpha _1^{p^{e_i}},\ldots ,\alpha _{i-1}^{p^{e_i}}]$ with no constant term. But applying part (i) with $l=i-1$, $a_j^{p^{e_i}}$ in place of $a_j$ for $1\le j\le i-1$, and $k[a_1^{p^{e_i}},\dots , a_{i-1}^{p^{e_i}}]$ in place of $k'$, we deduce that $1\otimes P-P\otimes 1$ is indeed a polynomial in the $a_j^{p^{e_i}}$ with $1\le j\le i-1$ as required.$\square$

We may now identify the minimal integer n such that ${\mathfrak {m}}^n=0$ with the integer ${{\,\mathrm{m}\,}}(k'/k)$. In fact, we also prove something stronger:

Lemma 2.11

With notation as above, let $d\le l$ and suppose $m_1,\ldots ,m_d\in {\mathfrak {m}}$ and $f_1,\ldots , f_d\in {\mathbb {Z}}_{\ge 0}$.

If $\sum f_i\ge {{\,\mathrm{mr}\,}}(k'/k,d)-d+1=\sum p^{e^i}-d+1$, then $m_1^{f_1}\cdots m_d^{f_d}=0$.

In particular, the minimum n such that ${\mathfrak {m}}^{n}=0$ is ${{\,\mathrm{m}\,}}(k'/k)$.

Proof

Assume for a contradiction that $m:=m_1^{f_1}\cdots m_d^{f_d}\ne 0$. We proceed by induction on d, observing that the product of any factors of m is also non-zero. It does no harm to assume that $f_1\ge \cdots \ge f_d$. If $d=1$, then ${{\,\mathrm{mr}\,}}(k'/k,1)=p^{e_1}=p^e$ and indeed $m=0$, a contradiction. Otherwise, let c be maximal such that $f_c\ge p^{e_c}$. This implies $\sum _{i=1}^c f_i\ge {{\,\mathrm{mr}\,}}(k'/k,c)-c+1$ so if $c<d$, then $m_1^{f_1}\cdots m_c^{f_c}$ is zero by induction, a contradiction. Therefore $c=d$. Let $(q_i,r_i)$ be the quotient and remainder when $p^{e_d}$ is divided into $f_i$. As $f_i\ge f_d\ge p^{e_d}$, we have each $q_i\ge 1$. Now

$$\begin{aligned} f_1+\cdots +f_d=p^{e_d}(q_1+\cdots +q_d)+(r_1+\cdots +r_d)\ge p^{e_1}+\cdots +p^{e_d}-d+1 \end{aligned}$$

and so

$$\begin{aligned} p^{e_d}(q_1+\cdots +q_d)&\ge p^{e_d}(q_1+\cdots +q_{d-1}+1)\\&\ge p^{e_1}+\cdots +p^{e_d}-d+1-d(p^{e_d}-1)\\&=p^{e_d}(p^{e_1-e_d}+\cdots +p^{e_{d-1}-e_d}-d+1)+1. \end{aligned}$$

Since the second expression above is a multiple of $p^{e_d}$, we have

$$\begin{aligned}q_1+\cdots +q_{d}\ge p^{e_1-e_d}+\cdots +p^{e_{d-1}-e_d}-(d-2). \end{aligned}$$

Now as $q_ip^{e_d}\le f_i$ for each i, we have $m'=m_1^{q_1p^{e_d}}\cdots m_d^{q_dp^{e_d}}\ne 0$. By Lemma 2.10(ii), we may write each $m_i^{p^{e_d}}$ as a polynomial in $k'[a_1^{p^{e_d}},\ldots ,a_{d-1}^{p^{e_d}}]$. After expanding $m'$, any constituent monomial will be of the form $a_1^{\nu _1}\cdots a_{d-1}^{\nu _{d-1}}$ where $\sum _{i=1}^{d-1}\nu _i=p^{e_d}\sum _{i=1}^d q_i\ge p^{e_1}+\cdots +p^{e_{d-1}}-p^{e_d}(d-2)$ and each $\nu _i$ is divisible by $p^{e_d}$. Furthermore, as $a_{d-1}^{p^{e_{d-1}}}$ can be re-expressed in terms of the $a_i$ with $i<d-1$, we may assume $\nu _{d-1}\le p^{e_{d-1}}-p^{e_d}$. It follows that

$$\begin{aligned} \sum _{i=1}^{d-2}\nu _i\ge p^{e_1}+\cdots +p^{e_{d-2}}-p^{e_d}(d-3). \end{aligned}$$

Continuing inductively, $\nu _1\ge p^{e_1}-p^{e_d}+1$. Thus $\nu _1\ge p^{e_1}$ and hence all monomials in $m'$ are zero, a contradiction.

We now tackle the second statement. To see that $n\ge {{\,\mathrm{m}\,}}(k'/k)$, observe that $a:=\prod _{i=1}^{l}a_i^{p^{e_i}-1}$ is a product of ${{\,\mathrm{m}\,}}(k'/k)-1$ elements from $\{a_1,\ldots , a_l\}$. Expanding this product in terms of the k-basis $\{x^b\otimes x^c\mid b,c\in J\}$ of ${\mathfrak {B}}$, the coefficient of $1\otimes x^b$ where $b=(p^{e_1}-1,\ldots ,p^{e_l}-1)$ is easily seen to be 1. It follows that $a\ne 0$.

For the upper bound, take $d=l$ in the first part of the lemma, noting that ${{\,\mathrm{m}\,}}(k'/k)={{\,\mathrm{mr}\,}}(k'/k,l)-l+1$. $\square$

3 Bounds in the case of the general linear group

In this section, $G'={{\,\mathrm{GL}\,}}_r$ and $G=\mathrm {R}_{k'/k}(G')$, where $k'/k$ is purely inseparable of exponent e. In [1, Lem. 4.4], it was shown that the exponent e(R) of $R:={\mathscr {R}}_u(G_{{\bar{k}}})$ is at most s, where s is chosen such that $p^s\ge r^2(p^e-1)$. Using the Cayley–Hamilton theorem in its full generality (see [2, III.8.11]) yields an improvement.

Proof of Theorem 1.1(i)

Let $M\in {{\,\mathrm{Mat}\,}}(r,{\mathfrak {m}})$, as in Lemma 2.5. By the Cayley–Hamilton theorem, we have $\chi _M(M)=0$, where $\chi _M(\lambda )=\lambda ^r+\sum _{t=0}^{r-1}f_t \lambda ^t$ is the characteristic polynomial of M with degree r. Observe that the coefficient $f_t$ is either zero or a homogeneous polynomial in the entries of ${\mathfrak {m}}$ with degree $r-t$. Since the matrices $M^{r-1},\ldots ,M,I_r$ all commute and $k'\otimes _{k} k'$ has characteristic p, we get

$$\begin{aligned} M^{rp^e}=-(f_{r-1}^{p^e}M^{(r-1)p^e}+\cdots +f_1^{p^e}M^{p^e}+f_0^{p^e}I_r)=0, \end{aligned}$$

since $f_t^{p^e}=0$ for all $f_t\in {\mathfrak {m}}$. Choosing s such that $p^s\ge p^er$, we get $M^{p^s}=M^{p^er}M^{p^s-p^er}=0$. $\square$

Let $B'\subseteq G'$ be the upper Borel subgroup, whose points are upper triangular matrices, and $B=\mathrm {R}_{k'/k}(B')\subseteq G$ its Weil restriction—a pseudo-Borel subgroup in the terminology of [5, C.2]. We now specify certain elements $M\in {{\,\mathrm{Mat}\,}}(r,{\mathfrak {m}})$ with $M\in B(k')=B'({\mathfrak {B}})$ to construct lower bounds on e(R) which depend on r and the exponents $e_i$ of a normal generating sequence. Moreover, we calculate exactly the exponent of the intersection $R\cap B_{{\bar{k}}}$.

Before we give the proof, we make some elementary remarks about powers of matrices. If A is a commutative ring and $E_{i,j}\in {{\,\mathrm{Mat}\,}}(r,A)$ is the $r\times r$ elementry matrix with $e_{i,j}=1$ and zeros everywhere else, then we have $E_{i,j}E_{l,s}=E_{i,s}$ if and only if $j=l$ and 0 else. Write $M=\sum a_{ij}E_{i,j}$ for a general element of ${{\,\mathrm{Mat}\,}}(r,A)$ . Then after expansion, any non-zero term in the power $M^n$ is of the form $a_w\cdot E_{i_1,i_{n+1}}$ where $a_w=a_{i_1,i_2}\cdot a_{i_2,i_3}\cdots a_{i_l,i_{n+1}}$, and $w\in (i_1,\ldots ,i_{n+1})\in \{1,\ldots ,r\}^{n+1}$. Furthermore, the (i, j)-th entry of $M^n$ is the sum over all $a_w$ such that $i_1=i$ and $i_{n+1}=j$. For $w\in \{1,\ldots ,r\}^{n+1}$, say n is the length of w. Now, if M is an upper triangular matrix, then $a_w$ will be zero unless $i_j\le i_{j+1}$ for each $1\le j\le n+1$. Hence the only w yielding non-zero $a_w$ are non-decreasing sequences of integers from $\{1,\ldots ,r\}$, whose length is bounded above by ${{\,\mathrm{m}\,}}(k'/k)$ (Lemma 2.11).

Proof of Theorem 1.1

(ii)-(iv). We work inside $B'({\mathfrak {m}})$. There are two cases. Suppose we have $\sum _{i=r+1}^l(p^{e_i}-1)<r-1$. Then there is an integer $q<r$ with q maximal such that $(p^{e_{q+1}}-1)+\cdots +(p^{e_l}-1)\ge q-1$. Define the matrix $M\in B'({\mathfrak {m}})$ as follows.

• (i) Start with $M=0$;

• (ii) fill the first q spaces on the diagonal with $a_1,\ldots , a_q$ respectively;

• (iii) Fill up the first $q-1$ elements of the superdiagonal with $p^{e_l}-1$ entries of $a_l$, then $p^{e_{l-1}}-1$ entries of $a_{l-1}$ and so on.

• (iv) If one has now written down $p^{e_{q+1}}-1$ entries $a_q$, then stop. Otherwise put $a_q$ in the q-th position of the superdiagonal. If, after this, there are only $p^{e_{q+1}}-1-\tau$ entries $a_q$ with $\tau \ge 1$ on the superdiagonal, we set also the $(q+1)$st diagonal element equal to $a_q$.

We claim $M^{{{\,\mathrm{m}\,}}(k'/k)-1}\ne 0$. Take the sequence

$$\begin{aligned}w=(\underbrace{1,\ldots ,1}_{p^{e_1}\text { times}},\underbrace{2,\ldots ,2}_{p^{e_2}\text { times}},\ldots , \underbrace{q,\ldots ,q}_{p^{e_q}\text { times}},\underbrace{q+1,\ldots ,q+1}_{\tau +1\text { times}})\end{aligned}$$

and observe that $a_w\ne 0$. A little thought shows that this is the unique sequence w of this length with $a_w\ne 0$ starting at 1 and ending at $q+1$. This proves the claim. Hence we conclude $e(R\cap B_{{\bar{k}}})\ge {{\,\mathrm{em}\,}}(k'/k)={{\,\mathrm{E}\,}}(k'/k,r)$. Since $M^{{{\,\mathrm{m}\,}}(k'/k)}=0$ by Lemmas 2.11 and 2.5, we have $e(R\cap B_{{\bar{k}}})\le {{\,\mathrm{em}\,}}(k'/k)$ and so this case is done.

Now suppose $\sum _{i=r+1}^l(p^{e_i}-1)\ge r-1$. One constructs a similar M. Start as before with $M=0$. One fills the diagonal with $a_1,\ldots , a_r$ and the superdiagonal with any $r-1$ elements from the list $(\underbrace{a_{r+1},\ldots ,a_{r+1}}_{p^{e_{r+1}}-1\text { times}},\ldots ,\underbrace{a_l,\ldots ,a_l}_{p^{e_l}-1 \text { times}})$. This gives a matrix M such that $M^m\ne 0$ where $m=\sum _{i=1}^r(p^{e_i}-1)+r-1={{\,\mathrm{mr}\,}}(k'/k,r)-1$. Hence $e(R\cap B_{{\bar{k}}})\ge {{\,\mathrm{emr}\,}}(k'/k,r)$.

We need to demonstrate the upper bound for this case. Take a matrix $M\in B'({\mathfrak {m}})$ and any sequence w of length ${{\,\mathrm{mr}\,}}(k'/k)$ such that $a_{w}\ne 0$ in a product of ${{\,\mathrm{mr}\,}}(k'/k,r)$ copies of M. Suppose the entries on the diagonal of M are $m_1,\ldots ,m_r\in {\mathfrak {m}}$. Then $a_w=m_1^{f_1}\cdots m_r^{f_r}\cdot \mu$ for non-negative integers $f_1,\ldots ,f_r$ and $\mu$ a product of s entries from above the diagonal; we have $s\le r-1$. Since ${{\,\mathrm{mr}\,}}(k'/k)=\sum f_i+s$, we have $\sum f_i\ge {{\,\mathrm{mr}\,}}(k'/k,r)-r+1$. We are now done by Lemma 2.11.

For part (iii) of the theorem, we note that any lower bound for $e(R\cap B_{{\bar{k}}})$ is automatically one for e(R). For part (iv), we just observe that ${{\,\mathrm{E}\,}}(k'/k,r)={{\,\mathrm{em}\,}}(k'/k)$ is also an upper bound for e(R) by Lemmas 2.11 and 2.5.$\square$

4 Extended example of

${{\,\mathrm{SL}\,}}_2$ and ${{\,\mathrm{PGL}\,}}_2$. In this section, we give a complete description of e(R) in case $G'={{\,\mathrm{SL}\,}}_2,{{\,\mathrm{PGL}\,}}_2$, or ${{\,\mathrm{GL}\,}}_2$.

Proposition 4.1

Let $k'/k$ be purely inseparable of characteristic p and exponent e. Let G be $\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_2)$, $\mathrm {R}_{k'/k}({{\,\mathrm{PGL}\,}}_2)$, or $\mathrm {R}_{k'/k}({{\,\mathrm{SL}\,}}_2)$, the last of these only if $p\ne 2$. Then e(R) is e if $k'/k$ is simple and $e+1$ otherwise.

Proof

If $G'$ is ${{\,\mathrm{GL}\,}}_2$, then Theorem 1.1(iv) gives the exponent $e(R)=e$ for $k'/k$ simple of exponent e, and at least $e+1$ when $k'/k$ is non-simple. However, by Theorem 1.1(i), we have e(R) is at most $e+1$. As the map ${{\,\mathrm{GL}\,}}_2\rightarrow {{\,\mathrm{PGL}\,}}_2$ is smooth, and ${{\,\mathrm{SL}\,}}_2\subset {{\,\mathrm{GL}\,}}_2$, we immediately obtain e or $e+1$ respectively as upper bounds for both the cases $G'={{\,\mathrm{SL}\,}}_2$ and $G'={{\,\mathrm{PGL}\,}}_2$. In fact, the matrices constructed in the proof of Theorem 1.1(ii)-(iv) and their relevant powers are clearly non-zero in their image in ${{\,\mathrm{PGL}\,}}_2({\mathfrak {B}})$ and so the exponents of the radicals in case $G'={{\,\mathrm{GL}\,}}_2$ and $G'={{\,\mathrm{PGL}\,}}_2$ coincide. Now by Proposition 2.3(iv), these also coincide for ${{\,\mathrm{SL}\,}}_2$ outside of characteristic 2, since the isogeny $f:{{\,\mathrm{SL}\,}}_2\rightarrow {{\,\mathrm{PGL}\,}}_2$ is smooth. $\square$

We see some different behaviour for ${{\,\mathrm{SL}\,}}_2$ in characteristic 2; in particular, the exponents of $R_u(G_{\bar{k}})\cap R_u(B_{\bar{k}})$ and $R_u(G_{\bar{k}})$ do not coincide.

Recall that a unital ring R has characteristic 2 if the equation $1+1=0$ holds in R. We prove a general formula for $2^s$-th powers of matrices over rings of characteristic 2.

Lemma 4.2

Let R be a commutative ring of characteristic 2 and $M=\begin{pmatrix} a &{}\quad b \\ c &{}\quad d \end{pmatrix}$ with $a,b,c,d\in R$. Then for all $s\in {\mathbb {N}}_0$, we have

$$\begin{aligned} M^{2^s}=\begin{pmatrix} a^{2^s}+\sum _{i=0}^{s-1} b^{2^i}c^{2^i}t^{2^s-2^{i+1}} &{} bt^{2^s-1} \\ ct^{2^s-1} &{} d^{2^s}+\sum _{i=0}^{s-1} b^{2^i}c^{2^i}t^{2^s-2^{i+1}} \end{pmatrix}, \end{aligned}$$

where $t=a+d$.

Proof

We argue by induction on s. For $s=0$, we just get M on both sides. So assume the result for s and set $l_s=\sum _{i=0}^{s-1} b^{2^i}c^{2^i}t^{2^s-2^{i+1}}$. We note that

$$\begin{aligned} l_s^2+bct^{2^{s+1}-2}=\sum _{i=1}^{s} b^{2^{i}}c^{2^{i}}t^{2^{s+1}-2^{i+1}}+bct^{2^{s+1}-2}=l_{s+1} \end{aligned}$$

holds. And so we get

$$\begin{aligned} M^{2^{s+1}} =(M^{2^s})^2&=\begin{pmatrix} a^{2^{s+1}}+l_s^2+bct^{2^{s+1}-2} &{} bt^{2^s-1}(a^{2^s}+d^{2^s}) \\ ct^{2^s-1}(a^{2^s}+d^{2^s}) &{} d^{2^{s+1}}+l_s^2+bct^{2^{s+1}-2} \end{pmatrix}\\&=\begin{pmatrix} a^{2^{s+1}}+l_{s+1} &{} bt^{2^{s+1}-1} \\ ct^{2^{s+1}-1} &{} d^{2^{s+1}}+l_{s+1} \end{pmatrix}. \end{aligned}$$

$\square$

Proposition 4.3

Let $G'={{\,\mathrm{SL}\,}}_2$, $p=2$, and suppose $k'/k$ has normal generating sequence $\alpha _1,\ldots ,\alpha _l$ with exponents $e_1\ge e_2\ge \cdots \ge e_l$. Let $B'$ be an upper Borel subgroup of $G'$ and $B=\mathrm {R}_{k'/k}(B')$ a pseudo-Borel subgroup.

• (i) We have $e(R\cap B_{{\bar{k}}})=e$.

• (ii) We have $e(R)={\left\{ \begin{array}{ll}e+1\text { if }e_1=e_2,\\ e\text { otherwise.}\end{array}\right. }$

Proof

We first calculate the exponent of $B'({\mathfrak {m}})=\ker B'({\mathfrak {B}})\rightarrow B'(k')$. A general element is $I+M$ with $M=\begin{pmatrix} a &{}\quad b\\ 0 &{}\quad d\\ \end{pmatrix}$, where $a,b,d\in {\mathfrak {m}}=\langle a_1,\ldots ,a_l\rangle$, satisfying $a+d+ad=0$ from the condition $\det (I+M)=1$. As $a\in {\mathfrak {m}}$, $1+a$ is invertible in ${\mathfrak {B}}$, and so it follows that $d=a(1+a)^{-1}$ and $a+d=ad=a^2(1+a)^{-1}$. By Lemma 4.2, the $2^s$-th power of this matrix is

$$\begin{aligned}\begin{pmatrix} a^{2^s} &{}b(a^2(1+a)^{-1})^{2^s-1}\\ 0 &{} d^{2^s}\\ \end{pmatrix}.\end{aligned}$$

Choosing for example $a=a_1=1\otimes \alpha _1-\alpha _1\otimes 1$, one sees the minimal s for which $M^{2^s}=0$ is at least $e_1=e$ and also that if $s\ge e$, then the diagonal entries vanish. For $s=e$, the off-diagonal entry is a multiple of $(a^2)^{2^e-1}$, which has at least $2^e$ factors of a, hence is also 0. So $e(R\cap B_{{\bar{k}}})=e$, proving (i).

For (ii), a general element of $G'({\mathfrak {m}})$ is a matrix $I+M$ where $M=\begin{pmatrix} a &{}\quad b\\ c &{}\quad d\\ \end{pmatrix}$, satisfying $a+d+ad+bc=0$, since $\det (I+M)=1$. If $l=1$ (i.e. $k'/k$ is simple of exponent $e=e_1$), then it is clear that $e(R)=e$. Suppose $e_2\le e_1-1=e-1$. We may write $d=(1+a)^{-1}(a+bc)$, hence $t=a+d=(1+a)^{-1}(a^2+bc)$. Therefore $t^{2^{e-1}}=(1+a)^{-2^{e-1}}(a^{2^e}+b^{2^{e-1}}c^{2^{e-1}})=(1+a)^{-2^{e-1}}(b^{2^{e-1}}c^{2^{e-1}})$. Now by Lemma 2.10, we have $m^{2^{e-1}}\in \langle a_1^{2^{e-1}}\rangle \subseteq {\mathfrak {m}}$ for any $m\in {\mathfrak {m}}$. It follows that $b^{2^{e-1}}c^{2^{e-1}}$ has a factor of $a_1^{2^{e-1}}\cdot a_1^{2^{e-1}}=a_1^{2^e}=0$. By the same token, $t^{2^{e-1}}=0$ and hence, using the lemma, $M^{2^e}=0$. Therefore $e(R)=e$ as required.

Now suppose $e_1=e_2$. Since $G\subseteq \mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_2)$, we have $e(R)\le e+1$. Set $b=a_1$, $c=a_2$, $d=0$, and $a=a_1a_2$. Then

$$\begin{aligned}M^{2^e}=\begin{pmatrix} (a_1a_2)^{2^e}+\sum _{i=1}^{e-1} (a_1a_2)^{2^i+2^e-2^{i+1}} &{} a_1(a_1a_2)^{2^e-1}\\ a_2(a_1a_2)^{2^e-1} &{} \sum _{i=1}^{e-1} (a_1a_2)^{2^i+2^e-2^{i+1}}\\ \end{pmatrix}.\end{aligned}$$

The off-diagonal entries are zero, but $\sum _{i=1}^{e-1} (a_1a_2)^{2^i+2^e-2^{i+1}}=\sum _{i=1}^e (a_1a_2)^{2^e-2^{i}}$ and the $i=e-1$ entry is $(a_1a_2)^{2^{e-1}}$. Provided $e_2=e_1$, this element is non-zero by the same argument as in the proof of Lemma 2.11—in a k-basis of ${\mathfrak {B}}$, the coefficient of $1\otimes \alpha _1^{2^{e-1}}\alpha _2^{2^{e-1}}$ in $(a_1a_2)^{2^{e-1}}$ is evidently 1. It follows that $M^{2^e}\ne 0$ and $e(R)=e+1$. $\square$

5 Simple groups of rank at least 2

For a positive integer r, recall that the algebraic k-group $\mu _r$ is the functor returning the r-th roots of unity in any k-algebra A; it is the kernel of the r-power map $\phi _r:{{\,\mathrm{{\mathbb {G}}_m}\,}}\rightarrow {{\,\mathrm{{\mathbb {G}}_m}\,}},x\mapsto x^r$. (We have $\mu _r$ étale if and only if $p\not \mid r$.) Say a $k'$-group L is a skewed ${{\,\mathrm{GL}\,}}_n$ if $L\cong {{\,\mathrm{GL}\,}}_n/K$, where $K\cong \mu _r$, with $r\not \mid n$; the quotient by $\mu _r$ induces the r-power map on the central torus of ${{\,\mathrm{GL}\,}}_n$. Since $K\cap {{\,\mathrm{SL}\,}}_n=1$, the derived subgroup ${\mathscr {D}}(L)$ of L is still isomorphic to ${{\,\mathrm{SL}\,}}_n$ and notably there is still a central torus $C\subset L$ which contains $Z({{\,\mathrm{SL}\,}}_n)$. In [12, Ex. 5.17], it is explained that if $\mu _n\subset T$ is a finite subgroup of a split k-torus of rank at least 2, then any automorphism of $\mu _n$ (which is some r-power map on $\mu _n$ with $(r,n)=1$) lifts to an automorphism of T. In our case, the quotient map by $K\times 1$ on ${{\,\mathrm{GL}\,}}_n\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ lifts to an automorphism of ${{\,\mathrm{GL}\,}}_n\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ defined on the central torus ${{\,\mathrm{{\mathbb {G}}_m}\,}}\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ by $(x,y)\mapsto (x^ry^{-n},x^by^a)$ where a and b are integers such that $ar+bn=1$. It follows that if L is a skewed ${{\,\mathrm{GL}\,}}_n$-group, then $L\times {{\,\mathrm{{\mathbb {G}}_m}\,}}\cong {{\,\mathrm{GL}\,}}_n\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$.

The group ${{\,\mathrm{SL}\,}}_{n+1}$ contains a (Levi) subgroup $L:={{\,\mathrm{GL}\,}}_n$ embedded block diagonally via $g\mapsto {{\,\mathrm{diag}\,}}(g,\det {g}^{-1})$. If $G'$ is a split simple k-group of type $A_n$, then $G'\cong {{\,\mathrm{SL}\,}}_{n+1}/\mu _r$ with $\mu _r\subseteq Z({{\,\mathrm{SL}\,}}_{n+1})\cong \mu _{n+1}$, so $r\mid n+1$. But then $r\not \mid n$, hence L maps under the quotient by $\mu _r$ to a skewed ${{\,\mathrm{GL}\,}}_n$-subgroup of $G'$. This observation establishes the type A case of the following lemma.

Lemma 5.1

Let $G'$ be a split simple $k'$-group of rank r. Then one of the following holds:

• (i) $G'$ contains a skewed ${{\,\mathrm{GL}\,}}_r$ subgroup L.

• (ii) There are étale isogenies $\pi :L\rightarrow L/Z$, and $\phi :{{\,\mathrm{GL}\,}}_r\rightarrow L/Z$ or $\phi :L\rightarrow {{\,\mathrm{GL}\,}}_r$.

• (iii) $p=2$ and either $G'$ is of type $F_4$ or $G'$ is of type B, C, or D, but not isomorphic to ${{\,\mathrm{SO}\,}}_{2r}$, ${{\,\mathrm{SO}\,}}_{2r+1}$, or ${{\,\mathrm{Sp}\,}}_{2r}$.

• (iv) $p=3$ and G is of type $E_6$.

In cases (i) and (ii), if we let $R={\mathscr {R}}_u(\mathrm {R}_{k'/k}(L)_{{\bar{k}}})$, then we have $e(R)=e({\mathscr {R}}_u(\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_r)_{{\bar{k}}}))$.

Proof

If $G'={{\,\mathrm{SO}\,}}_{2r}$, ${{\,\mathrm{SO}\,}}_{2r+1}$, or ${{\,\mathrm{Sp}\,}}_{2r}$, then [7, 4.5] promises a Levi subgroup $L\cong {{\,\mathrm{GL}\,}}_r$ as required so (i) holds in all these cases. The central isogenies ${{\,\mathrm{Sp}\,}}_{2r}\rightarrow {{\,\mathrm{PSp}\,}}_{2r}$, ${{\,\mathrm{Spin}\,}}_n\rightarrow {{\,\mathrm{SO}\,}}_n$, ${{\,\mathrm{SO}\,}}_n\rightarrow {{\,\mathrm{PSO}\,}}_n$, ${{\,\mathrm{Spin}\,}}_{2r}\rightarrow {{\,\mathrm{HSpin}\,}}_{2r}$, or ${{\,\mathrm{HSpin}\,}}_{2r}\rightarrow {{\,\mathrm{PSO}\,}}_{2r}$ (where ${{\,\mathrm{HSpin}\,}}_{2r}$ occurs only if n even) are all quotients by central subgroups isomorphic to $\mu _2$, hence are étale maps when $p>2$ and hence (ii) holds for all classical groups when $p>2$. Otherwise (iii) holds.

Let now $G'$ be exceptional. Note that groups of type $E_7$ and $E_8$ contain simple $A_7$ and $A_8$ maximal rank subgroups, evident from the Borel–de-Siebenthal algorithm. Hence each contains a skewed ${{\,\mathrm{GL}\,}}_7$ and ${{\,\mathrm{GL}\,}}_8$ subgroup L respectively, satisfying (i) as required. By [8, Thm. A], any Levi subgroup L of $G_2$ is isomorphic to ${{\,\mathrm{GL}\,}}_2$ (rather than ${{\,\mathrm{SL}\,}}_2\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ for example), hence satisfies (i). If $G'$ is of type $F_4$, then we take L to be an $A_3$ Levi subgroup of the ${{\,\mathrm{Spin}\,}}_9$ subgroup of $G'$. Now L admits an étale isogeny with the ${{\,\mathrm{GL}\,}}_4$ Levi subgroup of ${{\,\mathrm{SO}\,}}_9\cong {{\,\mathrm{Spin}\,}}_9/Z({{\,\mathrm{Spin}\,}}_9)$ if $p\ne 2$; so L satisfies (ii) in that case, and if $p=2$, then L is listed in (iii). It remains to deal with the case $G'=E_6$. Using Bourbaki notation [3], we take L to be the $A_5$ Levi subgroup corresponding to the nodes $\{1,3,4,5,6\}$ of the Dynkin diagram explicitly. Assume first $G'$ is simply connected. Then ${\mathscr {D}}(L)$ is simply connected. Since L centralises the ${{\,\mathrm{SL}\,}}_2$ subgroup generated by the $\pm {{\tilde{\alpha }}}$ root groups, where ${{\tilde{\alpha }}}$ is the longest root, the central torus C of L is the image of the cocharacter $h_{\tilde{\alpha }}:{{\,\mathrm{{\mathbb {G}}_m}\,}}\rightarrow G', t\mapsto h_{\tilde{\alpha }}(t)$. One calculates $h_{\tilde{\alpha }}(t)=h_{\alpha _1}(t)h_{\alpha _2}(t^2)h_{\alpha _3}(t^2)h_{\alpha _4}(t^3)h_{\alpha _5}(t^2)h_{\alpha _6}(t)$. This is inside ${\mathscr {D}}(L)\cong {{\,\mathrm{SL}\,}}_6$ iff $t^2=1$; therefore ${\mathscr {D}}(L)\cap C\cong \mu _2$—so L is not a skewed ${{\,\mathrm{GL}\,}}_r$. However, if $p\ne 3$, then there is an étale isogeny from L to $L/Z\cong {{\,\mathrm{GL}\,}}_6$ where Z is a diagonal copy of $\mu _3$ in both factors ${\mathscr {D}}(L)$ and C of L, so that (ii) holds. Besides which, since $Z(G')\cong \mu _3$, then if $p\ne 3$, the quotient by $Z(G')$ is étale and $L/Z(G')$ satisfies (ii) for the adjoint group $G/Z(G')$. Otherwise $p=3$ and (iv) holds.

We prove the last statement. Suppose (ii) holds. Then we note that if $\pi :A\rightarrow B$ is an étale central isogeny of reductive groups, then $\mathrm {R}_{k'/k}(A)\rightarrow \mathrm {R}_{k'/k}(B)$ is an isogeny whose kernel has trivial intersection with the unipotent radical. As in Proposition 2.3(iv), it follows that the induced map ${\mathscr {R}}_u(\mathrm {R}_{k'/k}(A)_{{\bar{k}}})\rightarrow {\mathscr {R}}_u(\mathrm {R}_{k'/k}(B)_{{\bar{k}}})$ is an isomorphism as required. If (i) holds, then L is a skewed ${{\,\mathrm{GL}\,}}_r$-subgroup and by the remarks preceding the lemma, $L\times {{\,\mathrm{{\mathbb {G}}_m}\,}}\cong {{\,\mathrm{GL}\,}}_r\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$. Observe ${\mathscr {R}}_u(\mathrm {R}_{k'/k}(L\times {{\,\mathrm{{\mathbb {G}}_m}\,}})_{{\bar{k}}})\cong {\mathscr {R}}_u(\mathrm {R}_{k'/k}(L)_{{\bar{k}}})\times {\mathscr {R}}_u(\mathrm {R}_{k'/k}({{\,\mathrm{{\mathbb {G}}_m}\,}})_{{\bar{k}}})$; since L already contains $C\cong {{\,\mathrm{{\mathbb {G}}_m}\,}}$, the exponent of the product is equal to that of the first factor, e(R). The same argument with ${{\,\mathrm{GL}\,}}_r\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ in place of $L\times {{\,\mathrm{{\mathbb {G}}_m}\,}}$ implies the claim. $\square$

Proof of Corollary 1.3

With the hypotheses on p, by Lemma 5.1, we get a reductive subgroup L of $G'$, with

$$\begin{aligned} e({\mathscr {R}}_u(\mathrm {R}_{k'/k}(L))_{{\bar{k}}})=e({\mathscr {R}}_u(\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_r))_{{\bar{k}}})\ge {{\,\mathrm{E}\,}}(k'/k,r). \end{aligned}$$

By Proposition 2.3(iii), it follows that $e(R)\ge {{\,\mathrm{E}\,}}(k'/k,r)$. If $\sum _{i=r+1}^l(p^{e_i}-1)<r-1$, then Weil restricting a faithful representation $G'\rightarrow {{\,\mathrm{GL}\,}}_n$ from $k'$ to k gives an embedding of G in ${{\,\mathrm{GL}\,}}_m$ with $m=[k':k]\cdot n\ge r$. Now by Proposition 2.3(iii) again, $e(R)\le e({\mathscr {R}}_u(\mathrm {R}_{k'/k}({{\,\mathrm{GL}\,}}_m))_{{\bar{k}}})$ which is equal to ${{\,\mathrm{E}\,}}(k'/k,m)={{\,\mathrm{em}\,}}(k'/k)$ by Theorem 1.1.$\square$