If $G$ is an algebraic group over a field $K$, we can consider the Galois (or flat) cohomology $H^1(K, G)$. If $G = \mathbb{G}_a$ or $\mathbb{G}_m$, it is well known that $H^1(K, G) = 0$ (the latter is the famous Hilbert Theorem 90). What about more general algebraic tori? For example, if $L/K$ is a finite separable extension (we can start with quadratic extensions), and $$ G = \operatorname{Res}_{L/K} \mathbb{G}_m $$ or $$ G = \ker\left(\operatorname{Nm} : \operatorname{Res}_{L/K} \mathbb{G}_m \to \mathbb{G}_m \right), $$ does it follow that $H^1(K,G) = 0$?
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7$\begingroup$ no, for example if you take $L/K = \mathbb{C}/\mathbb{R}$ and $T: x^2 + y^2 =1$ to be the non-split torus (so with real points the group $S^1$). Then $H^1(\mathbb{R},T) = \mathbb{Z}/2$, with the non-trivial class given by the non-trivial torsor $x^2 + y^2 = -1$. $\endgroup$– S. carmeliNov 3, 2022 at 21:17
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2$\begingroup$ Another interesting example of a non-trivial Galois cohomology of a torus is given in this handout by Brian Conrad. $\endgroup$– Gro-TsenNov 3, 2022 at 21:21
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5$\begingroup$ However, tori as described in your first point (Weil restrictions of the multiplicative group or products thereof), known as “quasi-trivial” tori do have trivial $H^1$, because their character module is a permutation module (and in general, cohomology of tori is best computed through their character module with Galois action). See for example §2 (“preliminaries on tori”) of “Tori and Essential Dimension” by Giordano Favi and Mathieu Florence for more details. $\endgroup$– Gro-TsenNov 3, 2022 at 21:31
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1$\begingroup$ Just to add one more comment: Steinberg's proof of Serre's "Conjecture I" (the cohomology $H^1$ vanishes for all connected algebraic groups $G$ over a field of cohomological dimension $1$) uses Steinberg's result $H^1$ of every group $G$ is induced from $H^1$ of some torus (not split, not quasi-trivial) inside $G$. So that explains why we cannot have vanishing of $H^1$ of all tori; it would also imply vanishing of $H^1$ of all $G$, e.g., vanishing of $H^1$ of $\textbf{PGL}_n$ so that also the Brauer group vanishes. $\endgroup$– Jason StarrNov 4, 2022 at 9:46
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1$\begingroup$ We have $$H^n({\Bbb R},T)\cong H^{n-2}(\Gamma, X_*(T))\cong H^{n}(\Gamma, X_*(T))$$ for any $\Bbb R$-torus $T$ and any $n\in\Bbb Z$. Here $X_*(T)$ is the cocharacter group of $T$, and $\Gamma={\rm Gal}({\Bbb C}/{\Bbb R})$. This follows from the Tate-Nakayama theorem. For details, see Theorem 3.6 in this preprint. $\endgroup$– Mikhail BorovoiNov 7, 2022 at 19:33
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