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Let $K=K^{sep}$ be a separably closed field with $K|\mathbb{F}_q$, where $\mathbb{F}_q$ is the field with $q$ elements. Let $\mathbb{G}$ be a connected linear algebraic group over $\mathbb{F}_q$. Lastly, let $B\in\mathbb{G}(K)$ be a $K$-valued point of $\mathbb{G}$. Define the Lang map relative to B $$f_B:\mathbb{G}(K)\rightarrow\mathbb{G}(K), A\mapsto A^{-1}\cdot B\cdot\mathbb{G}(Frob_q)(A),$$ where $Frob_q:K\rightarrow K, x\mapsto x^q$. My question is: Is $f_1$ surjective?

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I guess, I figured out a way to deduce this from the theorem of Lang-Steinberg: Let $K^{alg}|K$ be an algebraic closure. The map $$f_{1,alg}:\mathbb{G}(K^{alg})\rightarrow\mathbb{G}(K^{alg}), A\mapsto A^{-1}\cdot\mathbb{G}(Frob_q)(A)$$ is surjective by [Steinberg, Endomorphisms of Linear algebraic groups Theorem 10.1]. On the other hand, by [https://ivv5hpp.uni-muenster.de/u/pschnei/publ/lectnotes/Theorie-des-Anstiegs.pdf, Satz 2.1 (german lecture notes)] it follows that $$\tilde{f}_1:GL_n(K)\rightarrow GL_n(K), A\mapsto A^{-1}\cdot GL_n(Frob_q)(A)$$ is surjective. Since $\mathbb{G}$ is a linear algebraic group, we can fix an embedding $\mathbb{G}\subset GL_n$. Now let $A_0\in\mathbb{G}(K)$ be arbitrary. Since $f_{1,alg}$ is surjective, we find $B\in\mathbb{G}(K^{alg})$ with $f_1(B)=A_0$. Since $\tilde{f}_1$ is surjective, there also exists $\tilde{B}\in GL_n(K)$, such that $\tilde{f}_1(\tilde{B})=A_0$. Let $$\tilde{f}_{1,alg}:GL_n(K^{alg})\rightarrow GL_n(K^{alg}), A\mapsto A^{-1}\cdot GL_n(Frob_q)(A).$$ One calculates that $$B\in f_{1,alg}^{-1}(\{A_0\})\subset\tilde{f}_{1,alg}^{-1}(\{A_0\})=GL_n(\mathbb{F}_q)\cdot \tilde{B}\subset GL_n(K).$$ So $B\in\mathbb{G}(K^{alg})\cap GL_n(K)=\mathbb{G}(K)$, hence $f_1$ is surjective.

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  • $\begingroup$ I guess with this method you can prove the following for every integral domain $A$, which is a $\mathbb{F}_q$-algebra: the Lang map is surjective on the $A$-valued points for any connected linear algebra group over $\mathbb{F}_q$, if it is surjective on the $A$-valued points for any $GL_n$. $\endgroup$ – Estus Feb 18 at 18:15

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