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. $\square$

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