Lifting Lang-Steinberg to DVR's in Characteristic 0

Let $A$ be a compact DVR in characteristic $0$, uniformizer $\pi$ and residue field $k$. Let $A\subset B$ be a complete DVR with the same uniformizer $\pi$ and algebraicly closed residue field $F$. Let $\varphi:B\rightarrow B$ be a surjective morphism of $A$-Algebras, such that the induced morphism of $k$-Algebras $\overline{\varphi}: F\rightarrow F$ has a finite subfield of fixed points $\{x\in F~|~\overline{\varphi}(x)=x\}$.

Now let $\mathbb{G}$ be a linear algebraic group over $A$, i.e. $\mathbb{G}\subset GL_{n,A}$ is a closed subgroup-scheme, such that the basechange $\mathbb{G}_F:=\mathbb{G}\underset{B}{\otimes}F$ is integral (and maybe $\mathbb{G}$ needs to satisfy other properties too). By theorem of Lang-Steinberg, the map of varieties $$\overline{\Psi}:\mathbb{G}(F)\rightarrow\mathbb{G}(F),x\mapsto x^{-1}\mathbb{G}(\overline{\varphi})(x)$$ is surjective. Furthermore, it is $\mathbb{G}(B)\cong\underset{\leftarrow}{lim}\mathbb{G}(B/\pi^mB)$.

My question now is, if these information can be used to lift this surjectivity in the sense that I want to show, that $$\Psi:\mathbb{G}(B)\rightarrow\mathbb{G}(B),x\mapsto x^{-1}\mathbb{G}(\varphi)(x)$$ is surjective.

1 Answer

Yes, as long as you are careful about the hypotheses, this is a result of M. J. Greenberg [Schemata over local rings II, Section 3, Proposition 3]. You need to assume that $\mathbb{G}$ is smooth over $A$ (Greenberg says 'simple') and of course that its fibre over $F$ is connected (otherwise the Lang--Steinberg theorem doesn't hold). On the other hand, there is no need to assume that $A$ has characteristic $0$.

The proof is based on the existence and basic properties of the Greenberg functor (as well as the classical Lang--Steinberg theorem). Your $\varphi$ is slightly more general than the one in Greenberg's result mentioned above, but the proof is easily adapted to this case.

• Only smoothness and connectedness at the special fiber is really great, since I need this at another point anyways, thank you very much! – Estus Aug 6 '18 at 12:21