All Questions

If $G$ is an algebraic group such that $H^1(S, G) = 0$ for all schemes $S$, must $G$ be the trivial group? My original motivation for the question is the rationale I always give students for studying ...

Let $k$ be a characteristic zero field, $V \subset \mathbb{A}^n_k$ an open subscheme, $G$ a split reductive group over $k$ and $T$ a $G$-torsor over $V$ (in the etale, equivalently fppf topology). ...

Setup: Let $k$ be an algebraically closed field. Let $C$ be a smooth connected projective curve over $k$. Let $J$ be a smooth commutative group scheme over $C$ with connected fibers. Let $j:\eta\to C$ ...

Let $X$ be a projective algebraic variety over an algebraically closed field. Let an algebraic group $G$ act algebraically on $X$. Let $\mathcal{F}$ be a $G$-equivariant vector bundle (or, more ...

Let $G$ be an affine algebraic group over ${\mathbb C}$. We denote the sheaf of morphisms from ${\mathbb A}^1$ to $G$ by $\bf G$. Then $H^1({\mathbb A}^1,\bf G)=0$ (Cech cohomlogy). Is this fact true? ...