[Chapter II, Section 3, Proposition 5, SerreGaloisCohomology]

Proposition 59.67.4. Let $K$ be a field with separable algebraic closure $K^{sep}$. Assume that for any finite extension $K'$ of $K$ we have $\text{Br}(K') = 0$. Then

1. $H^ q(\text{Gal}(K^{sep}/K), (K^{sep})^*) = 0$ for all $q \geq 1$, and

2. $H^ q(\text{Gal}(K^{sep}/K), M) = 0$ for any torsion $\text{Gal}(K^{sep}/K)$-module $M$ and any $q \geq 2$,

Proof. Set $p = \text{char}(K)$. By Lemma 59.59.2, Theorem 59.61.6, and Example 59.59.3 the proposition is equivalent to showing that if $H^2(\mathop{\mathrm{Spec}}(K'),\mathbf{G}_ m|_{\mathop{\mathrm{Spec}}(K')_{\acute{e}tale}}) = 0$ for all finite extensions $K'/K$ then:

• $H^ q(\mathop{\mathrm{Spec}}(K),\mathbf{G}_ m|_{\mathop{\mathrm{Spec}}(K)_{\acute{e}tale}}) = 0$ for all $q \geq 1$, and

• $H^ q(\mathop{\mathrm{Spec}}(K),\mathcal{F}) = 0$ for any torsion sheaf $\mathcal{F}$ and any $q \geq 2$.

We prove the second part first. Since $\mathcal{F}$ is a torsion sheaf, we may use the $\ell$-primary decomposition as well as the compatibility of cohomology with colimits (i.e, direct sums, see Theorem 59.51.3) to reduce to showing $H^ q(\mathop{\mathrm{Spec}}(K),\mathcal{F}) = 0$, $q \geq 2$ for all $\ell$-power torsion sheaves for every prime $\ell$. This allows us to analyze each prime individually.

Suppose that $\ell \neq p$. For any extension $K'/K$ consider the Kummer sequence (Lemma 59.28.1)

$0 \to \mu _{\ell , \mathop{\mathrm{Spec}}{K'}} \to \mathbf{G}_{m, \mathop{\mathrm{Spec}}{K'}} \xrightarrow {(\cdot )^{\ell }} \mathbf{G}_{m, \mathop{\mathrm{Spec}}{K'}} \to 0$

Since $H^ q(\mathop{\mathrm{Spec}}{K'},\mathbf{G}_ m|_{\mathop{\mathrm{Spec}}(K')_{\acute{e}tale}}) = 0$ for $q = 2$ by assumption and for $q = 1$ by Theorem 59.24.1 combined with $\mathop{\mathrm{Pic}}\nolimits (K) = (0)$. Thus, by the long-exact cohomology sequence we may conclude that $H^2(\mathop{\mathrm{Spec}}{K'}, \mu _\ell ) = 0$ for any separable $K'/K$. Now let $H$ be a maximal pro-$\ell$ subgroup of the absolute Galois group of $K$ and let $L$ be the corresponding extension. We can write $L$ as the colimit of finite extensions, applying Theorem 59.51.3 to this colimit we see that $H^2(\mathop{\mathrm{Spec}}(L), \mu _\ell ) = 0$. Now $\mu _\ell$ must be the constant sheaf. If it weren't, that would imply there exists a Galois extension of degree relatively prime to $\ell$ of $L$ which is not true by definition of $L$ (namely, the extension one gets by adjoining the $\ell$th roots of unity to $L$). Hence, via Lemma 59.67.3, we conclude the result for $\ell \neq p$.

Now suppose that $\ell = p$. We consider the Artin-Schrier exact sequence (Section 59.63)

$0 \longrightarrow \underline{\mathbf{Z}/p\mathbf{Z}}_{\mathop{\mathrm{Spec}}{K}} \longrightarrow \mathbf{G}_{a, \mathop{\mathrm{Spec}}{K}} \xrightarrow {F-1} \mathbf{G}_{a, \mathop{\mathrm{Spec}}{K}} \longrightarrow 0$

where $F - 1$ is the map $x \mapsto x^ p - x$. Then note that the higher Cohomology of $\mathbf{G}_{a, \mathop{\mathrm{Spec}}{K}}$ vanishes, by Remark 59.23.4 and the vanishing of the higher cohomology of the structure sheaf of an affine scheme (Cohomology of Schemes, Lemma 30.2.2). Note this can be applied to any field of characteristic $p$. In particular, we can apply it to the field extension $L$ defined by a maximal pro-$p$ subgroup $H$. This allows us to conclude $H^ n(\mathop{\mathrm{Spec}}{L}, \underline{\mathbf{Z}/p\mathbf{Z}}_{\mathop{\mathrm{Spec}}{L}}) = 0$ for $n \geq 2$, from which the result follows for $\ell = p$, by Lemma 59.67.3.

To finish the proof we still have to show that $H^ q(\text{Gal}(K^{sep}/K), (K^{sep})^*) = 0$ for all $q \geq 1$. Set $G = \text{Gal}(K^{sep}/K)$ and set $M = (K^{sep})^*$ viewed as a $G$-module. We have already shown (above) that $H^1(G, M) = 0$ and $H^2(G, M) = 0$. Consider the exact sequence

$0 \to A \to M \to M \otimes \mathbf{Q} \to B \to 0$

of $G$-modules. By the above we have $H^ i(G, A) = 0$ and $H^ i(G, B) = 0$ for $i > 1$ since $A$ and $B$ are torsion $G$-modules. By Lemma 59.57.6 we have $H^ i(G, M \otimes \mathbf{Q}) = 0$ for $i > 0$. It is a pleasant exercise to see that this implies that $H^ i(G, M) = 0$ also for $i \geq 3$. $\square$

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