The Stacks project

Lemma 109.71.2. Let $G$ be a smooth commutative group algebraic space over a field $K$. Then $H^1_{fppf}(\mathop{\mathrm{Spec}}(K), G)$ is torsion.

Proof. Every $G$-torsor $P$ over $\mathop{\mathrm{Spec}}(K)$ is smooth over $K$ as a form of $G$. Hence $P$ has a point over a finite separable extension $L/K$. Say $[L : K] = n$. Let $[n](P)$ denote the $G$-torsor whose class is $n$ times the class of $P$ in $H^1_{fppf}(\mathop{\mathrm{Spec}}(K), G)$. There is a canonical morphism

\[ P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P \to [n](P) \]

of algebraic spaces over $K$. This morphism is symmetric as $G$ is abelian. Hence it factors through the quotient

\[ (P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P)/S_ n \]

On the other hand, the morphism $\mathop{\mathrm{Spec}}(L) \to P$ defines a morphism

\[ (\mathop{\mathrm{Spec}}(L) \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(L))/S_ n \longrightarrow (P \times _{\mathop{\mathrm{Spec}}(K)} \ldots \times _{\mathop{\mathrm{Spec}}(K)} P)/S_ n \]

and the reader can verify that the scheme on the left has a $K$-rational point. Thus we see that $[n](P)$ is the trivial torsor. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AFA. Beware of the difference between the letter 'O' and the digit '0'.