Torsors over complete local fields 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:


*

*Is the topology used in the theorem the étale topology ?

*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!
 A: The isomorphism classes of torsors for $\mu_{n}$ over $R$ and $k$ are
classified by the fppf cohomology groups $H^{1}(R,\mu_{n})$ and $H^{1}%
(k,\mu_{n})$. From the Kummer sequence, one sees that $H^{1}(R,\mu_{n})\simeq
R^{\times}/R^{\times n}$ and $H^{1}(k,\mu_{n})\simeq k^{\times}/k^{\times n}$.
If $n$ is prime to the residue characteristic (so $\mu_{n}$ is etale),
then $R^{\times}/R^{\times n}\rightarrow k^{\times}/k^{\times n}$ is a
bijective, but if $n$ equals the residue characteristic then it is not injective;
this means that there are nonisomorphic torsors over $R$ that become
isomorphic over $k$. (Yes, a key point is that for smooth groups, the torsors are classified by the etale cohomology groups.)
A: 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.
