Perfect Field Case (gpt-5.5-pro-web)

Statement

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

Proof

Let $$ U:=R_ u(G) $$ be the unipotent radical of $G$. Since $k$ is perfect and $G$ is smooth connected affine, $U$ is a smooth connected unipotent normal $k$-subgroup, and $$ H:=G/U $$ is a connected reductive $k$-group. Let $$ q:G\to H $$ be the quotient map.

We will prove that the natural map $$ q_ \ast:\mathbb X_ \ast(G)/G(k)\longrightarrow \mathbb X_ \ast(H)/H(k) $$ is injective. Then the result follows from the reductive case applied to $H$.

First recall two standard facts about smooth connected unipotent groups over perfect fields.

  1. If $U$ is smooth connected unipotent over $k$, then $U$ is $k$-split. Hence $$ H^1(k,U)=1. $$ Indeed, $U$ has a normal $k$-composition series with quotients $\mathbb G_ a$, and $H^1(k,\mathbb G_ a)=0$.

  2. If $T=\mathbb G_ m$ acts $k$-rationally on such a $U$, then every algebraic $1$-cocycle $$ c:T\to U $$ is a coboundary. Equivalently, any two $k$-sections of $$ U\rtimes T\to T $$ are conjugate by an element of $U(k)$.

Here is the proof of the second fact. Because $U$ is split and $T$ is split, there is a $T$-stable normal series $$ 1=U_ 0\subset U_ 1\subset\cdots\subset U_ n=U $$ with $U_ i/U_ {i-1}\cong \mathbb G_ a$. So it suffices by induction to treat $U=\mathbb G_ a$. Then the $T$-action is given by a character $t\mapsto t^m$, and a cocycle is a regular function $f:\mathbb G_ m\to\mathbb G_ a$, hence a Laurent polynomial $f(t)\in k[t,t^{-1}]$, satisfying $$ f(tt’)=f(t)+t^m f(t’). $$

If $m=0$, this forces $f=0$. If $m\neq 0$, comparing Laurent monomials gives $$ f(t)=a(1-t^m) $$ for some $a\in k$. In either case $$ f(t)=a-t\cdot a, $$ so $f$ is a coboundary. The induction then proves the claim for $U$.

Now we prove that $q_ \ast$ is injective. Let $$ \lambda,\mu\in \mathbb X_ \ast(G) $$ and suppose that $q\circ\lambda$ and $q\circ\mu$ are $H(k)$-conjugate. Thus there is $\bar g\in H(k)$ such that $$ q\circ\mu=\operatorname{Int}(\bar g)\circ q\circ\lambda. $$

Since $H^1(k,U)=1$, the map $$ G(k)\to H(k) $$ is surjective. Choose $g\in G(k)$ lifting $\bar g$. Replacing $\lambda$ by $\operatorname{Int}(g)\circ\lambda$, we may assume $$ q\circ\lambda=q\circ\mu. $$

Set $$ \nu:=q\circ\lambda=q\circ\mu:\mathbb G_ m\to H. $$

Form the pullback group $$ E:=G\times_ {H,\nu}\mathbb G_ m. $$ Thus, for a $k$-algebra $R$, $$ E(R)=\{(x,t)\in G(R)\times R^\times : q(x)=\nu(t)\}. $$

There is an exact sequence $$ 1\to U\to E\to \mathbb G_ m\to 1. $$

The cocharacters $\lambda$ and $\mu$ give two $k$-sections of $E\to\mathbb G_ m$: $$ t\mapsto (\lambda(t),t),\qquad t\mapsto (\mu(t),t). $$

Using the first section, we identify $$ E\cong U\rtimes \mathbb G_ m, $$ where $\mathbb G_ m$ acts on $U$ by conjugation through $\lambda$: $$ t\cdot u=\lambda(t)u\lambda(t)^{-1}. $$

Under this identification, the second section is given by a cocycle $$ c(t)=\mu(t)\lambda(t)^{-1}\in U. $$

By the unipotent cocycle fact above, there is $u\in U(k)$ such that $$ c(t)=u\lambda(t)u^{-1}\lambda(t)^{-1}. $$

Therefore $$ \mu(t)=c(t)\lambda(t)=u\lambda(t)u^{-1} $$ for every $t$. Hence $$ \mu=\operatorname{Int}(u)\circ\lambda. $$

Thus $\lambda$ and $\mu$ are $G(k)$-conjugate. So $$ q_ \ast:\mathbb X_ \ast(G)/G(k)\to \mathbb X_ \ast(H)/H(k) $$ is injective.

Now suppose $\lambda,\mu\in \mathbb X_ \ast(G)$ become $G(K)$-conjugate after base change to $K$. Applying $q_ K:G_ K\to H_ K$, the cocharacters $$ (q\circ\lambda)_ K,\qquad (q\circ\mu)_ K $$ are $H(K)$-conjugate. Since $H$ is connected reductive, the reductive case gives that $$ q\circ\lambda $$ and $$ q\circ\mu $$ are already $H(k)$-conjugate. By the injectivity of $q_ \ast$, $\lambda$ and $\mu$ are $G(k)$-conjugate.

Therefore $$ \mathbb X_ \ast(G)/G(k)\longrightarrow \mathbb X_ \ast(G_ K)/G(K) $$ is injective.