General Case (gpt-5.4)

Brief Log of Running

• created_at_utc: “2026-04-29T19:25:46.478905+00:00”,
• updated_at_utc: “2026-04-29T19:25:46.478921+00:00”,
• Model: Rethlas (Generator GPT-5.4 xhigh, Verifier GPT-5.4 xhigh)
• statement_type: prove_or_disprove
• statement: Let $G$ be a (not necessarily reductive) smooth connected affine group over a (not necessarily perfect) field $k$. Let $K$ be an algebraic field extension of $k$. Let $\mathbb{X}_ \ast(G):=\mathrm{Hom}_ {k\text{-groups}}(\mathbb{G}_ m,G)$ $k$-homomorphisms. Let $G_K$ be the base change of $G$ to $K$. Prove (or disprove with a counterexample) that $\mathbb{X}_ \ast(G)/G(k)\rightarrow\mathbb{X}_ \ast(G_ K)/G(K)$ is injective. (You can take for granted that this statement is true if $k$ is a perfect field.)

lemma lem:split_unipotent_h1

statement

Let $k$ be a field and let $U$ be a smooth connected $k$-split unipotent group. Then $$ H^1(k,U)=1 $$ for nonabelian fppf cohomology. In particular, if $$ 1\to U\to E\xrightarrow{\pi} Q\to 1 $$ is an exact sequence of smooth affine $k$-groups, then the map $$ E(k)\to Q(k) $$ is surjective.

proof

Since $U$ is $k$-split, there is a composition series by smooth connected normal $k$-subgroups $$ 1=U_ r\subset U_ {r-1}\subset \cdots \subset U_ 1\subset U_ 0=U $$ such that each quotient $U_ i/U_ {i+1}$ is $k$-isomorphic to $\mathbf G_ a$.

We prove $H^1(k,U)=1$ by induction on $r$. The case $r=0$ is trivial. For the induction step, choose an exact sequence $$ 1\to U_ 1\to U\to \mathbf G_ a\to 1. $$ The standard exact sequence in nonabelian fppf cohomology gives $$ H^1(k,U_ 1)\to H^1(k,U)\to H^1(k,\mathbf G_ a). $$ By induction $H^1(k,U_ 1)=1$, and by additive Hilbert 90 we have $$ H^1(k,\mathbf G_ a)=0. $$ Hence $H^1(k,U)=1$.

For the surjectivity statement, let $q\in Q(k)$. The fiber $\pi^{-1}(q)$ is a left $U$-torsor over $\operatorname{Spec}(k)$. Since $H^1(k,U)=1$, that torsor is trivial, so it has a $k$-point. Therefore $q$ lifts to an element of $E(k)$.

lemma lem:reduction_by_split_unipotent_radical

statement

Let $G$ be a smooth connected affine $k$-group, let $$ U:=R_ {us,k}(G) $$ be the maximal $k$-split smooth connected unipotent normal $k$-subgroup, and write $$ \pi:G\to \overline G:=G/U. $$ Assume that for every algebraic extension $K/k$ the natural map $$ \mathbb X_ \ast(\overline G)/\overline G(k)\to \mathbb X_ \ast(\overline G_ K)/\overline G(K) $$ is injective. Then for every algebraic extension $K/k$ the natural map $$ \mathbb X_ \ast(G)/G(k)\to \mathbb X_ \ast(G_ K)/G(K) $$ is injective.

proof

Let $K/k$ be algebraic, and let $\lambda,\mu\in \mathbb X_ \ast(G)$ become $G(K)$-conjugate. Their images $$ \overline\lambda:=\pi\circ \lambda,\qquad \overline\mu:=\pi\circ \mu $$ become $\overline G(K)$-conjugate. By injectivity for $\overline G$, there exists $\overline g\in \overline G(k)$ such that $$ \overline\mu=\operatorname{Int}(\overline g)\circ \overline\lambda. $$ By Lemma lem:split_unipotent_h1, the map $G(k)\to \overline G(k)$ is surjective, so choose $g\in G(k)$ lifting $\overline g$. Replacing $\lambda$ by $\operatorname{Int}(g)\circ\lambda$, we reduce to the case $$ \overline\lambda=\overline\mu=:\nu. $$

Let $$ E:=\pi^{-1}(\nu(\mathbf G_ m))\subset G. $$ Then $$ 1\to U\to E\xrightarrow{\pi} \nu(\mathbf G_ m)\simeq \mathbf G_ m\to 1 $$ is exact, and both $\lambda$ and $\mu$ are $k$-sections of $\pi$.

Because $U$ is unipotent, every $k$-split torus in $E$ maps injectively into $\nu(\mathbf G_ m)$, so every $k$-split torus in $E$ has dimension at most $1$. Hence $\lambda(\mathbf G_ m)$ and $\mu(\mathbf G_ m)$ are maximal split $k$-tori of $E$.

We use the following cited result.

Complete cited statement: Theorem 4.2.9 in Structure and classification of pseudo-reductive groups states: “Any two maximal split $k$-tori in a smooth connected affine $k$-group $G$ are conjugate under $G(k)$.”

paper_id: Structure and classification of pseudo-reductive groups
theorem_id: Theorem 4.2.9
arXiv id:

Applying this to $E$, there exists $e\in E(k)$ such that $$ e\lambda(\mathbf G_ m)e^{-1}=\mu(\mathbf G_ m). $$ Since $\pi(e)\in \nu(\mathbf G_ m)(k)=\mu(\mathbf G_ m)(k)$, choose $t\in \mathbf G_ m(k)$ such that $$ \mu(t)=\pi(e). $$ Then $$ u:=\mu(t)^{-1}e\in U(k), $$ and $u$ still conjugates $\lambda(\mathbf G_ m)$ onto $\mu(\mathbf G_ m)$ because $\mu(t)$ centralizes $\mu(\mathbf G_ m)$.

Finally, $\operatorname{Int}(u)\circ\lambda$ and $\mu$ are both sections of $\pi:E\to \nu(\mathbf G_ m)$ with the same image torus $\mu(\mathbf G_ m)$. Since $\pi|_ {\mu(\mathbf G_ m)}:\mu(\mathbf G_ m)\to \nu(\mathbf G_ m)$ is an isomorphism, there is only one such section. Therefore $$ \mu=\operatorname{Int}(u)\circ\lambda. $$ So $\lambda$ and $\mu$ are $G(k)$-conjugate.

lemma lem:reduction_by_central_split_torus

statement

Let $G$ be a smooth connected affine $k$-group, let $$ 1\to Z\to G\xrightarrow{\pi} \overline G\to 1 $$ be an exact sequence with $Z$ a central split $k$-torus, and assume that for every algebraic extension $K/k$ the natural map $$ \mathbb X_ \ast(\overline G)/\overline G(k)\to \mathbb X_ \ast(\overline G_ K)/\overline G(K) $$ is injective. Then for every algebraic extension $K/k$ the natural map $$ \mathbb X_ \ast(G)/G(k)\to \mathbb X_ \ast(G_ K)/G(K) $$ is injective.

proof

Let $K/k$ be algebraic and let $\lambda,\mu\in \mathbb X_ \ast(G)$ become $G(K)$-conjugate. Their images $\overline\lambda,\overline\mu\in \mathbb X_ \ast(\overline G)$ become $\overline G(K)$-conjugate. By injectivity for $\overline G$, there exists $\overline g\in \overline G(k)$ such that $$ \overline\mu=\operatorname{Int}(\overline g)\circ \overline\lambda. $$

Since $Z\simeq \mathbf G_ m^n$ is split, $H^1(k,Z)=1$ by Hilbert 90, so $G(k)\to \overline G(k)$ is surjective. Choose $g\in G(k)$ lifting $\overline g$. Replacing $\lambda$ by $\operatorname{Int}(g)\circ\lambda$, we reduce to the case $$ \overline\lambda=\overline\mu=:\nu. $$

Let $$ M:=\pi^{-1}(\nu(\mathbf G_ m)). $$ Then $$ 1\to Z\to M\to \nu(\mathbf G_ m)\simeq \mathbf G_ m\to 1 $$ is an exact sequence of tori. Because both kernel and quotient are split, $M$ is itself a split $k$-torus: on character lattices we get a short exact sequence of free abelian groups, hence a split exact sequence. In particular, $M$ is commutative.

Choose $h\in G(K)$ such that $$ \mu=\operatorname{Int}(h)\circ\lambda. $$ Since $\pi\circ\lambda=\pi\circ\mu=\nu$, the element $\pi(h)$ centralizes $\nu(\mathbf G_ m)$. Hence $h$ normalizes $M_ K$, and conjugation by $h$ induces an automorphism of the split torus $M_ K$.

This automorphism acts trivially on the quotient $\nu(\mathbf G_ m)_ K$, because $\pi(h)$ centralizes that torus, and it acts trivially on the kernel $Z_ K$, because $Z$ is central in $G$. Therefore the induced automorphism of the split torus $M_ K$ is the identity: on character lattices it acts trivially on the quotient lattice and on the sublattice coming from $Z$, so it is trivial on all of $X^\ast(M_ K)$.

Thus conjugation by $h$ is trivial on $M_ K$, and hence $$ \lambda_ K=\mu_ K. $$ By faithful flat descent for morphisms, $\lambda=\mu$. So $\lambda$ and $\mu$ are already $G(k)$-conjugate.

lemma lem:wound_radical_centralizes_split_tori

statement

Let $H$ be a smooth connected affine $k$-group with $$ R_ {us,k}(H)=1,\qquad R_ {s,k}(H)=1. $$ Let $R:=R_ k(H)$ be the $k$-radical of $H$, so $R$ is a smooth connected solvable normal $k$-subgroup. Then every split $k$-torus $S\subset H$ centralizes $R$.

proof

We first show that the solvable group $R$ has no nontrivial split smooth connected normal $k$-subgroup. Indeed, let $$ N\subset R $$ be such a subgroup. By the split solvable structure theorem, quoted below, $N$ is a semi-direct product of a split torus against a split unipotent normal subgroup $U=R_ {u,k}(N)$. Since $N$ is normal in $R$ and $R$ is normal in $H$, the subgroup $N$ is normal in $H$, so $U$ is a split smooth connected unipotent normal $k$-subgroup of $H$. Hence $$ U\subset R_ {us,k}(H)=1. $$ Thus $N$ is a split torus. But a normal torus in a connected affine group is central, so $$ N\subset R_ {s,k}(H)=1. $$ Therefore $N=1$.

We use the following cited result.

Complete cited statement: Theorem 5.4 in The structure of solvable groups over fields states that for any solvable smooth connected affine $k$-group $G$, the quotient $G/G_ {\mathrm{split}}$ is a central extension of a $k$-wound unipotent group by a $k$-wound torus, where $G_ {\mathrm{split}}$ is the maximal $k$-split smooth connected normal $k$-subgroup of $G$. In particular, if $G_ {\mathrm{split}}=1$ then $G$ itself is a central extension $$ 1\to T_ w\to G\to U_ w\to 1 $$ with $T_ w$ a $k$-wound torus and $U_ w$ a $k$-wound unipotent group.

paper_id: The structure of solvable groups over fields
theorem_id: Theorem 5.4
arXiv id:

Applying this to $R$, the vanishing just proved shows that $R_ {\mathrm{split}}=1$, so $R$ is $k$-wound.

Now let $$ G:=SR\subset H $$ be the smooth connected solvable $k$-subgroup generated by $S$ and $R$. Let $G_ {\mathrm{split}}\subset G$ be its maximal split smooth connected normal $k$-subgroup from Theorem 5.4.

Because the inclusion $$ S\hookrightarrow G $$ is a homomorphism from a split smooth connected affine $k$-group into $G$, the finality assertion in Theorem 5.4 forces this inclusion to factor through $$ G_ {\mathrm{split}}\hookrightarrow G. $$ Hence $$ S\subset G_ {\mathrm{split}}. $$

On the other hand, $$ G_ {\mathrm{split}}\cap R $$ is a split smooth connected normal $k$-subgroup of the $k$-wound solvable group $R$, so it is trivial. Therefore the projection $$ G\to G/R\simeq S $$ restricts to an injective homomorphism $G_ {\mathrm{split}}\to S$. Since the image contains $S$, we obtain $$ G_ {\mathrm{split}}=S. $$

Thus $S$ is a normal torus in the connected group $G$. The conjugation action $$ G\to \underline{\mathrm{Aut}}(S) $$ factors through an étale $k$-group, so it is trivial on the connected group $G$. Hence $S$ is central in $G$, and therefore $S$ centralizes $R$.

lemma lem:pseudo_reductive_fixed_torus

statement

Let $Q$ be a pseudo-reductive $k$-group, let $$ S\subset Q $$ be a maximal split $k$-torus, and let $K/k$ be an algebraic extension. If $$ \lambda,\mu\in X_ \ast(S) $$ become $Q(K)$-conjugate, then they have the same unique representative in the closed dominant chamber for the relative root system $\Phi(Q,S)$. In particular, they are $N_ Q(S)(k)$-conjugate.

proof

Choose a minimal pseudo-parabolic $k$-subgroup $$ P_ 0\subset Q $$ containing $S$. We use the following cited results.

Complete cited statement: Theorem 5.3.2(i) in Structure and classification of pseudo-reductive groups states that for a smooth connected affine $k$-group $G$, a maximal split $k$-torus $S$, and a minimal pseudo-parabolic $k$-subgroup $P$ containing $S$, the set $\Phi(G/R_ {u,k}(G),S)$ is a root system in $X(S)_ \mathbf Q$, the roots coming from $P$ form a positive system, and the natural map $$ N_ G(S)(k)/Z_ G(S)(k)\to W(\Phi(G/R_ {u,k}(G),S)) $$ is an isomorphism.

paper_id: Structure and classification of pseudo-reductive groups
theorem_id: Theorem 5.3.2(i)
arXiv id:

Since $Q$ is pseudo-reductive, $R_ {u,k}(Q)=1$, so Theorem 5.3.2(i) gives a root system $$ {}^k\Phi:=\Phi(Q,S) $$ in $X(S)_ \mathbf Q$, with positive system ${}^k\Phi^+$ determined by $P_ 0$, and identifies $$ N_ Q(S)(k)/Z_ Q(S)(k) $$ with the Weyl group $W({}^k\Phi)$.

Let $$ C=\{\nu\in X_ \ast(S)\mid \langle \beta,\nu\rangle\ge 0\text{ for all }\beta\in {}^k\Phi^+\} $$ be the closed dominant chamber. By root-system combinatorics, each $W({}^k\Phi)$-orbit in $X_ \ast(S)$ meets $C$ in exactly one point.

Hence every $\nu\in X_ \ast(S)$ is $N_ Q(S)(k)$-conjugate to a unique element of $C$. It therefore remains to show that $\lambda$ and $\mu$ have the same representative in $C$.

Let $k_ s$ be a separable closure of $k$. Then $Q_ {k_ s}$ is pseudo-split pseudo-reductive. Choose a split maximal $k_ s$-torus $$ T\subset (P_ 0)_ {k_ s} $$ containing $S_ {k_ s}$, and choose a minimal pseudo-parabolic $k_ s$-subgroup $$ B\subset (P_ 0)_ {k_ s} $$ containing $T$. Let $$ \Phi_ {\mathrm{abs}}:=\Phi(Q_ {k_ s},T) $$ be the absolute root system, with positive system $\Phi_ {\mathrm{abs}}^+$ determined by $B$.

We use one further cited result.

Complete cited statement: Theorem 5.3.2(iv) in Structure and classification of pseudo-reductive groups states that if $G$ is a smooth connected affine $k$-group with $R_ {u,k}(G)$ $k$-wound, then the root system $\Phi(G,S)$ consists of the nontrivial $S$-weights on $\operatorname{Lie}(G)$.

paper_id: Structure and classification of pseudo-reductive groups
theorem_id: Theorem 5.3.2(iv)
arXiv id:

For $\nu\in X_ \ast(S)\subset X_ \ast(T)$ and $\alpha\in \Phi_ {\mathrm{abs}}$, one has $$ \langle \alpha,\nu\rangle=\langle \alpha|_ S,\nu\rangle. $$ Since $Q$ is pseudo-reductive, $R_ {u,k}(Q)=1$ is certainly $k$-wound, so Theorem 5.3.2(iv) identifies the relative roots with the nontrivial $S$-weights on $\operatorname{Lie}(Q)$. The positive relative roots are the $S$-weights occurring in $\operatorname{Lie}(P_ 0)$. On the other hand, the absolute positive roots are exactly the nontrivial $T$-weights occurring in $\operatorname{Lie}(B)\subset \operatorname{Lie}((P_ 0)_ {k_ s})$. Therefore every root in $\Phi_ {\mathrm{abs}}^+$ restricts either to $0$ or to an element of ${}^k\Phi^+$, and every element of ${}^k\Phi^+$ arises as the nonzero restriction of some root in $\Phi_ {\mathrm{abs}}^+$. Hence a cocharacter $\nu\in X_ \ast(S)$ lies in $C$ if and only if it is dominant for $\Phi_ {\mathrm{abs}}^+$.

Let $\lambda^+,\mu^+\in C$ be the unique representatives of $\lambda$ and $\mu$ in $C$. Since the relative Weyl action is realized by $N_ Q(S)(k)\subset Q(k)$, the pairs $(\lambda,\lambda^+)$ and $(\mu,\mu^+)$ are already $Q(k)$-conjugate. Thus $\lambda^+$ and $\mu^+$ remain $Q(K)$-conjugate, and hence $Q(\overline k)$-conjugate. We claim that they are in the same orbit under the absolute Weyl group $W(\Phi_ {\mathrm{abs}})$. Let $g\in Q(\overline k)$ satisfy $$ \mu^+=\operatorname{Int}(g)\circ\lambda^+. $$ Let $$ H:=Z_ {Q_ {\overline k}}(\mu^+)^\circ. $$ Since both $T_ {\overline k}$ and $gT_ {\overline k}g^{-1}$ contain the image of $\mu^+$, they lie in $H$. They are maximal tori of $H$, because any larger torus in $H$ would be a torus in $Q_ {\overline k}$ properly containing $T_ {\overline k}$. By Theorem 4.2.9 applied over the algebraically closed field $\overline k$, any two maximal tori in the smooth connected affine group $H$ are conjugate. So there exists $$ c\in H(\overline k) $$ such that $$ c g T_ {\overline k} g^{-1} c^{-1}=T_ {\overline k}. $$ Thus $$ n:=cg\in N_ Q(T)(\overline k) $$ and still $$ \mu^+=\operatorname{Int}(n)\circ\lambda^+. $$ Applying Theorem 5.3.2(i) over the separably closed field $k_ s$ to the maximal split torus $T\subset Q_ {k_ s}$, the quotient $N_ Q(T)/Z_ Q(T)$ is the constant Weyl group $W(\Phi_ {\mathrm{abs}})$. Hence $\lambda$ and $\mu$ lie in the same $W(\Phi_ {\mathrm{abs}})$-orbit. Hence $\lambda^+$ and $\mu^+$ lie in the same $W(\Phi_ {\mathrm{abs}})$-orbit.

Every absolute Weyl-group orbit in $X_ \ast(T)$ has a unique $\Phi_ {\mathrm{abs}}^+$-dominant element. Since $\lambda^+,\mu^+\in C$, the equivalence between relative and absolute dominance on $X_ \ast(S)$ shows that both $\lambda^+$ and $\mu^+$ are $\Phi_ {\mathrm{abs}}^+$-dominant. Therefore $$ \lambda^+=\mu^+. $$ So $\lambda$ and $\mu$ have the same unique representative in $C$, and in particular they are $N_ Q(S)(k)$-conjugate.

theorem thm:main

statement

Let $G$ be a (not necessarily reductive) smooth connected affine group over a (not necessarily perfect) field $k$. Let $K$ be an algebraic field extension of $k$. Let $\mathbb{X}_ \ast(G):=\mathrm{Hom}_ {k\text{-groups}}(\mathbb{G}_ m,G)$ $k$-homomorphisms. Let $G_ K$ be the base change of $G$ to $K$. Prove (or disprove with a counterexample) that $\mathbb{X}_ \ast(G)/G(k)\rightarrow\mathbb{X}_ \ast(G_ K)/G(K)$ is injective. (You can take for granted that this statement is true if $k$ is a perfect field.)

proof

We prove that the map is injective for every smooth connected affine $k$-group $G$, so there is no counterexample.

Let $$ \lambda,\mu\in \mathbb X_ \ast(G) $$ and assume that they become $G(K)$-conjugate.

Step 1: remove the split unipotent radical

Let $$ U:=R_ {us,k}(G). $$ By Lemma lem:reduction_by_split_unipotent_radical, it is enough to prove the theorem for $$ G_ 1:=G/U. $$ So from now on we replace $G$ by $G_ 1$ and assume $$ R_ {us,k}(G)=1. $$

Step 2: remove the central split-torus part of the solvable radical

We use the following cited result.

Complete cited statement: Corollary 5.12 in The structure of solvable groups over fields states: for any smooth connected affine $k$-group $G$, if $R_ {us,k}(G)=1$ then $R_ {s,k}(G)$ is the maximal central $k$-split torus in $G$.

paper_id: The structure of solvable groups over fields
theorem_id: Corollary 5.12
arXiv id:

So $$ Z:=R_ {s,k}(G) $$ is a central split $k$-torus. By Lemma lem:reduction_by_central_split_torus, it is enough to prove the theorem for $$ G_ 2:=G/Z. $$ Thus we may replace $G$ by $G_ 2$ and assume $$ R_ {us,k}(G)=R_ {s,k}(G)=1. $$

Step 3: pass from the wound solvable radical to the pseudo-reductive quotient

Let $$ R:=R_ k(G) $$ be the $k$-radical. Because $$ R_ {us,k}(G)=R_ {s,k}(G)=1, $$ the argument in Lemma lem:wound_radical_centralizes_split_tori shows that $R$ has no nontrivial split smooth connected normal $k$-subgroup. Hence Theorem 5.4 implies that $R$ is $k$-wound. Let $$ Q:=G/R. $$ Then $Q$ is pseudo-reductive.

Fix a maximal split $k$-torus $$ S\subset G. $$ By Lemma lem:wound_radical_centralizes_split_tori, the torus $S$ centralizes $R$. Hence the quotient map $G\to Q$ identifies $S$ with a maximal split $k$-torus of $Q$.

We now use the following cited result.

Complete cited statement: Theorem 4.2.9 in Structure and classification of pseudo-reductive groups states: “Any two maximal split $k$-tori in a smooth connected affine $k$-group $G$ are conjugate under $G(k)$.”

paper_id: Structure and classification of pseudo-reductive groups
theorem_id: Theorem 4.2.9
arXiv id:

Applying this to $G$, after conjugating $\lambda$ and $\mu$ by suitable elements of $G(k)$, we may assume $$ \lambda,\mu\in X_ \ast(S). $$ Since $R$ centralizes $S$, the images $$ \overline\lambda,\overline\mu\in X_ \ast(S)\subset X_ \ast(Q) $$ are still $Q(K)$-conjugate.

By Lemma lem:pseudo_reductive_fixed_torus, $\overline\lambda$ and $\overline\mu$ are $N_ Q(S)(k)$-conjugate. Equivalently, they lie in the same orbit under the Weyl group $$ N_ Q(S)(k)/Z_ Q(S)(k). $$

We next use the following cited result.

Complete cited statement: Proposition 5.3.1 in Structure and classification of pseudo-reductive groups states that for a maximal split $k$-torus $S$ in a smooth connected affine $k$-group $G$, the quotient $N_ G(S)/Z_ G(S)$ is constant and the natural map $$ N_ G(S)(k)/Z_ G(S)(k)\to (N_ G(S)/Z_ G(S))(k) $$ is an equality.

paper_id: Structure and classification of pseudo-reductive groups
theorem_id: Proposition 5.3.1
arXiv id:

Because $R$ centralizes $S$, the natural map $$ N_ G(S)/Z_ G(S)\to N_ Q(S)/Z_ Q(S) $$ is an isomorphism. So the Weyl element carrying $\overline\lambda$ to $\overline\mu$ is represented by some $$ n\in N_ G(S)(k). $$ Again because $R$ centralizes $S$, the action of $n$ on $S$ is the same before and after passage to $Q$. Hence $$ \mu=\operatorname{Int}(n)\circ \lambda. $$ So $\lambda$ and $\mu$ are $G(k)$-conjugate.

We have proved that whenever two $k$-defined cocharacters of $G$ become conjugate under $G(K)$, they were already conjugate under $G(k)$. Equivalently, $$ \mathbb X_ \ast(G)/G(k)\longrightarrow \mathbb X_ \ast(G_ K)/G(K) $$ is injective.