TITLE: Torsors trivializing over a fixed finite etale cover
QUESTION [9 upvotes]: Let $S$ be an integral regular scheme and let $T\to S$ be a finite etale morphism. Let $G$ be a smooth affine finite type group scheme over $S$.
Is the set of $S$-isomorphism classes of $G$-torsors over $S$ which are trivial over $T$ finite?
I guess the set-up is ridiculously general. So what if 
1) $S$ is of finite type over $\mathbb C$, or
2) $S$ is of finite type $\mathbb F_p$, or
4) $S$ is of finite type over $\mathbb Z_p$, or
5) $S$ is of finite type over $\mathbb Q_p$, or
6) $S$ is of finite type over $\mathbb Z$?

REPLY [6 votes]: The answer to the general question is NO.
We take $S=\mathrm{Spec}\, K$, where $K$ is a number field.
Let $L/K$ be a finite Galois extension. We set $T=\mathrm{Spec}\, L$.
We consider the norm homomorphism
$$N_{L/K}\colon R_{L/K}\mathbb{G}_{m,L}\to \mathbb{G}_{m,K}\,,$$
where $R_{L/K}$ denotes the Weil restriction of scalars.
Set $G=\mathrm{ker}\, N_{L/K}$.
Then our torus $G$ splits over $L$, hence $H^1(L,G)=1$ by Theorem 90.
From the short exact sequence 
$$1\to G\to R_{L/K}\mathbb{G}_{m,L}\to \mathbb{G}_{m,K}\to 1$$
and the induced Galois cohomology exact sequence
$$ L^*\to K^* \to H^1(K,G)\to H^1(K,R_{L/K}\mathbb{G}_{m,L})=1$$
we obtain that $H^1(K,G)= K^*/ N_{L/K}(L^*)$, and this group is infinite.
Thus the kernel
$$\mathrm{ker}[H^1(K,G)\to H^1(L,G)]=H^1(K,G)= K^*/ N_{L/K}(L^*)$$
is infinite, as required.