Let $G$ be a linear algebraic group scheme, and let $R$ be a complete discrete valuation ring, with quotient field $K$ and residue field $k$.


If $T$ is an $R$-torsor, it yields by base change a $k$-torsor $T\_k$.


Apparently, we have the following theorem:


**Thm.** If $G$ is smooth and $T,T'$ are two $R$-torsors, then $T\_k\simeq T'\_k$ implies that $T\simeq T'$.


This is proved in SGA3,p. 401,prop. 8.1. In fact, more is proved in SGA3 (it says that any $k$-torsor can be lifted in a unique way to an $R$-torsor, up to isomorphism), but I only need this particular case.


Unfortunately, I do not have access to SGA3, and I'm not really comfortable with the different topologies, so here is a couple of questions:


1. Is the topology used in the theorem the étale topology ?
2. Would it be true if $G$ is not necessarily smooth but if we use torsors for the fppf topology ?


Typically, I wonder what would happen for $G=\mu\_2$ defined over a field of characteristic $2$ ? 


Thanks a lot!



Torsors over complete local fields
I am rewriting my comment as an answer. That is **false** in characteristic $p$ for torsors for the finite, flat group scheme $\mu\_p=\text{Spec}\ \mathbb{Z}[t]/\langle t^p -1 \rangle$ with the usual group multiplication, $$\mathbb{Z}[t]/\langle t^p-1 \rangle \to \mathbb{Z}[t\_1,t\_2]/\langle t\_1^p-1,t\_2^p-1\rangle, \ \ t\mapsto t\_1t\_2.$$ Let $k$ be an algebraically closed field of characteristic $p$. Let $R$ be $k[[x]]$. Let $T$ be the $R$-scheme $$T=\text{Spec} \ S, \ \ S=R[y]/\langle y^p-(1+x) \rangle.$$ Let $m:\mu\_p\times T \to T$ be the $R$-morphism given by $$m^\*:S\to S[t]/\langle t^p-1 \rangle, \ \ y \mapsto yt.$$ I claim that this is a $\mu\_p$-torsor for the fppf topology. Indeed, the $R$-scheme $T$ is itself fppf. We need to check that the induced map, $$(m,\text{pr}\_2):\mu\_p \times T \to T\times\_{\text{Spec}\ R} T,$$ is an isomorphism. On the level of rings, this is given by the $S$-algebra homomorphism, $$S[y\_1]/\langle y\_1^p-(1+x) \rangle \to S[z]/\langle z^p -1 \rangle, \ \ y\_1 \mapsto yz.$$ The inverse isomorphism sends $z$ to $y\_1 y^{p-1}(1+x)^{-1}$. Thus, $T$ is a $\mu\_p$-torsor over $\text{Spec} \ R$.


The reduction of this torsor modulo $x$ is just $\text{Spec}\ k[y]/\langle y^p -1 \rangle$, which equals $\mu\_{p,k}$. However, there is no trivialization over all of $\text{Spec}\ R$. Indeed, for every $f\in R$, the element $(1+xf)^p$ equals $1+x^pf^p$, which cannot equal $1+x$. Thus, the torsor $T$ is nontrivial over $\text{Spec}\ R$.


When $G$ is smooth over $\text{Spec}\ R$, for torsors $T$ and $T'$ over $\text{Spec}\ R$, also the "inner twist" of $G$ by $T$ is smooth over $\text{Spec}\ R$, and the Isom scheme $\text{Isom}\_R(T,T')$ is a torsor for this smooth group $R$-scheme. For a torsor for a smooth group scheme over $\text{Spec}\ R$, every $k$-point lifts to an $R$-point by Hensel's Lemma.


