Proof. By induction, it is enough to prove the result for $n=2$ . First, note that $O_{K,S}/O_{F,S}$ is finite étale by assumption (since $K/F$ is unramified outside of S). In particular, $O_{K,S}$ is finitely-generated and torsion-free as an $O_{F,S}$ -module, and both rings are Dedekind domains integrally closed in their fields of fractions. By base-change, we get a finite étale extension $O_{K,S} \otimes _{O_{F,S}} O_{K,S}/ O_{K,S}$ , which is still finitely-generated, locally free, and torsion-free as an $O_{K,S}$ -module (this last fact follows from using the injection $O_{K,S} \otimes _{O_{F,S}} O_{K,S} \hookrightarrow \prod _{\Gamma _{K/F}} O_{K,S}$ , under which $O_{K,S}$ maps into the diagonally-embedded copy, which clearly acts on the product without torsion).

We are thus in the setting of [Reference Conrad2, Thm. 1.3], which says that the composition

$$ \begin{align*}O_{K,S} \otimes_{O_{F,S}} O_{K,S} \hookrightarrow K \otimes_{O_{F,S}} O_{K,S} \xrightarrow{\sim} \prod_{i} K_{i}\end{align*} $$

maps $O_{K,S} \otimes _{O_{F,S}} O_{K,S}$ isomorphically onto the product of integral closures of $O_{K,S}$ in each $K_{i}$ , where $K_{i}$ is some finite separable extension of K and the last isomorphism comes from the fact that $K \otimes _{O_{F,S}} O_{K,S}$ is finite étale over K a field. It thus suffices to show that each $K_{i}$ is actually K.

The desired result follows from the series of elementary manipulations

$$ \begin{align*}K \otimes_{O_{F,S}} O_{K,S} \xrightarrow{\sim} K \otimes_{F} F \otimes_{O_{F,S}} O_{K,S} \xrightarrow{\sim} K \otimes_{F} K \xrightarrow{\sim} \prod_{\Gamma_{K/F}} K.\end{align*} $$

We leave it to the reader to check that the isomorphism $O_{K,S} \otimes _{O_{F,S}} O_{K,S} \xrightarrow {\sim } \prod _{\Gamma _{K/F}} O_{K,S}$ constructed above agrees with the map in the statement of the Lemma.

Proof. All that one must check is that the isomorphism of Lemma 2.1 preserves cocycles and coboundaries, which is straightforward.

Proof. Since T splits over $O_{S}$ , it suffices to prove the result for $T = \mathbb {G}_{m}$ . For $i=1$ , we have

$$ \begin{align*} H^{1}_{\text{et}}(O_{S}, T_{O_{S}}) = \varinjlim \limits_{F \subset K \subset F_{S} \text{ finite Galois}} H^{1}_{\text{et}}(O_{K,S}, \mathbb{G}_{m}) = \varinjlim \text{Pic}(\text{Spec}(O_{K,S})) = \varinjlim \text{Cl}(O_{K,S}) = 0, \end{align*} $$

where the second equality comes from [Reference Milne17, Prop. II.2.1]) and the third from the proof of [Reference Neukirch, Schmidt and Wingberg19, Prop. 8.3.6].

For $i =2$ , identifying $H^{2}_{\text {et}}(O_{K,S}, \mathbb {G}_{m}) = \text {Br}(O_{K,S})$ gives (by [Reference Poonen20, 6.9.2]) an exact sequence

$$ \begin{align*}0 \to \text{Br}(O_{K,S}) \to \bigoplus_{v \in S_{K}} \text{Br}(K_{v}) \xrightarrow{\sum \text{inv}_{v}} \mathbb{Q}/\mathbb{Z},\end{align*} $$

where $K_{v}$ denotes the completion of K at v. Taking the direct limit of the first two terms shows that we have an injective map $\text {Br}(O_{S}) \hookrightarrow \bigoplus _{v \in S_{F_{S}}} \text {Br}(F_{S} \cdot F_{v})$ . Note that the field extension $F_{S} \cdot F_{v}$ contains $F^{\text {nr}}_{v}$ , the maximal unramified extension of $F_{v}$ , using the fact that $F_{S}/F$ contains all finite extensions of the constant field of F. Moreover, the valuation ring $O_{F_{S} \cdot F_{v}}$ of this field is Henselian, as it is the direct limit of the Henselian rings $O_{K_{v}}$ ([Reference Hochster7, p. 56]), and the previous sentence implies that it has algebraically closed residue field. We may then deduce from the proof of [Reference Milne17, Prop. I.A.1] that $\text {Br}(F_{S} \cdot F_{v}) = 0$ , giving the desired result.

Finally, for $i \geq 2$ , we have that for any $K/F$ a finite Galois extension, we have $H^{i}_{\text {et}}(O_{K,S}, \mathbb {G}_{m}) = 0$ , by [Reference Milne17], Remark II.2.2. Taking the direct limit gives the desired result.

Proof. Since T is unramified outside S, it is enough to show that

$$ \begin{align*}H^{i}_{\text{fppf}}(O_{S}^{\bigotimes_{O_{F,S}} n}, \mathbb{G}_{m}) = \varinjlim \limits_{F \subset K \subset F_{S} \text{ finite Galois}} H^{i}_{\text{fppf}}(O_{K,S}^{\bigotimes_{O_{F,S}} n}, \mathbb{G}_{m}) = 0. \end{align*} $$

By Lemma 2.1, we have a canonical identification $\text {Spec}(O_{K,S}^{\bigotimes _{O_{F,S}} n}) = \coprod _{\underline {\sigma } \in \Gamma _{K/F}^{n-1}} \text {Spec}(O_{K,S})$ and isomorphism $ H^{i}_{\text {fppf}}(\coprod _{\underline {\sigma }} \text {Spec}(O_{K,S}), \mathbb {G}_{m}) \xrightarrow {\sim } \prod _{\underline {\sigma }} H^{i}_{\text {fppf}}(O_{K,S}, \mathbb {G}_{m})$ . Also, if $K^{\prime }/K$ is finite and contained in $F_{S}$ , then the map $H^{i}_{\text {fppf}}(O_{K,S}^{\bigotimes _{O_{F,S}} n}, \mathbb {G}_{m}) \to H^{i}_{\text {fppf}}(O_{K^{\prime },S}^{\bigotimes _{O_{F,S}} n}, \mathbb {G}_{m})$ corresponds via this isomorphism to diagonally embedding each factor of $\prod _{\underline {\sigma } \in \Gamma _{K/F}^{n-1}} H^{i}_{\text {fppf}} (O_{K,S}, \mathbb {G}_{m})$ into some subset of the factors of $\prod _{\underline {\sigma } \in \Gamma _{K^{\prime }/F}^{n-1}} H^{i}_{\text {fppf}}(O_{K^{\prime },S}, \mathbb {G}_{m})$ (by means of the pullback map $H^{i}_{\text {fppf}}(O_{K,S}, \mathbb {G}_{m}) \to H^{i}_{\text {fppf}}(O_{K^{\prime },S}, \mathbb {G}_{m})$ ).

Hence, for any $\alpha \in H^{i}_{\text {fppf}}(O_{K,S}^{\bigotimes _{O_{F,S}} n}, \mathbb {G}_{m})$ , to show that $\alpha $ vanishes in some $H^{i}_{\text {fppf}}(O_{K^{\prime },S}^{\bigotimes _{O_{F,S}} n}, \mathbb {G}_{m})$ for large $K^{\prime }$ , it is enough to show that $\varinjlim _{K/F} H^{i}_{\text {fppf}}(O_{K,S}, \mathbb {G}_{m}) = 0$ for all i, thus reducing the result to the case $n=1$ , which follows from Lemma 2.3.

Proof. Combine Lemma 2.5 with Proposition 2.6.

Proof. It is evidently enough to prove this in the case when $[F^{\prime }:F] = p$ , which we now assume. We claim that $O_{F^{\prime }} = O_{F}^{(p)}$ , the extension of $O_{F}$ obtained by adjoining all p-power roots. There is an obvious inclusion of $O_{F}$ -algebras $O_{F^{\prime }} \hookrightarrow O_{F}^{(p)}$ because $F^{\prime } = F^{(p)}$ . The morphism of smooth projective curves $X^{\prime } \to X$ corresponding to the inclusion $F \to F^{\prime }$ is purely inseparable of degree p, so by [24, 0CCV], we obtain an isomorphism of $O_{F}$ -algebras $O_{F^{\prime }} \xrightarrow {\sim } O_{F}^{(p)}$ , giving the claim. The claim implies that, at the level of local rings, a uniformizer $\varpi \in O_{F,\mathfrak {p}}$ has a pth root in $O_{F^{\prime }, \mathfrak {p^{\prime }}}$ for any prime $\mathfrak {p}^{\prime }$ above $\mathfrak {p}$ , giving the desired result.

Proof. For the inclusion of the right-hand side into the left-hand side, note that if $x \in \overline {F}$ is such that $x^{p^{m}} \in O_{E,S}$ for some finite (Galois) $E \subset F_{S}$ , then $x \in E^{\prime } := E \cdot F_{m}$ , which is unramified over $F_{m}$ outside of $S_{m}$ , and so $x \in O_{E^{\prime },S_{m}} \subset O_{S_{m}}$ . For the other inclusion, consider a finite Galois extension $K^{\prime }$ of the finite purely inseparable extension $F^{\prime } := F_{m}/F$ with $S^{\prime }:= S_{m}$ . We may factor $K^{\prime }/F$ as a tower $K^{\prime }/K/F$ , where $K/F$ is the separable (Galois) closure of F in $K^{\prime }$ and $K^{\prime }/K$ is purely inseparable. Note that $K \cdot F^{\prime } = K^{\prime }$ ; one containment is clear, and the other follows from the fact that K and $F^{\prime }$ are linearly disjoint and $[K^{\prime }: F^{\prime }] = [K: F]$ .

We want to show that $K/F$ is unramified outside S; this follows because for any prime $\mathfrak {p}$ of $O_{F}$ , we know from Lemma 2.9 that $\mathfrak {p}$ splits as $(\mathfrak {p}^{\prime })^{[F^{\prime }:F]}$ in $O_{F^{\prime }}$ , and if $\mathfrak {p}^{\prime }$ is a prime of $O_{F^{\prime },S^{\prime }}$ , then it factors in $O_{K^{\prime }}$ as $\mathfrak {P}^{\prime }_{1} \cdot \dots \cdot \mathfrak {P}^{\prime }_{r}$ , which means that $\mathfrak {p}$ splits in $O_{K^{\prime }}$ as $(\mathfrak {P}^{\prime }_{1} \cdot \dots \cdot \mathfrak {P}^{\prime }_{r})^{[F^{\prime }:F]}$ . Since $[F^{\prime } : F] = [K^{\prime } : K]$ , we know that $\mathfrak {p}$ must not ramify in $O_{K}$ , or else the ramification degree would be too large. Now for any element $x \in O_{K^{\prime },S^{\prime }}$ , we have that $x^{p^{m}} \in K$ and is integral over $O_{F,S}$ , and hence lies in $O_{K,S}$ , showing that $O_{K^{\prime },S^{\prime }} \subseteq O_{K,S}^{(p^{m})} \subset O_{S}^{\text {perf}}$ , giving the other inclusion.

Proof. It is enough to prove the result for $A = \mathbb {G}_{m}$ and $A = \mu _{m}$ . We focus on the former first: Since $\mathbb {G}_{m}$ is smooth, we can use [Reference Rosengarten22, Lem. 2.2.9] to replace $(O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n}$ by $[(O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n}]_{\text {red}} = (O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}^{\text {perf}}} n}$ and reduce the claim to showing that each $H^{j}_{\text {fppf}}((O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}^{\text {perf}}} n}, \mathbb {G}_{m})$ vanishes. By Lemma 2.10, we have

$$ \begin{align*}(O_{S}^{\text{perf}})^{\bigotimes_{O_{F,S}^{\text{perf}}} n} = \varinjlim_{m} O_{S_{m}}^{ \bigotimes_{O_{F_{m}, S_{m}}} n},\end{align*} $$

and hence, it is enough to show that $\varinjlim _{m} H^{j}_{\text {fppf}}(O_{S_{m}}^{ \bigotimes _{O_{F_{m}, S_{m}}} n}, \mathbb {G}_{m}) = 0$ for all $j, n \geq 1$ , which folllows from Lemma 2.5.

We now prove the $\mu _{m}$ -case. For $i>1$ , we immediately deduce that $H^{i}_{\text {fppf}}((O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n}, \mu _{m})$ vanishes from the long exact sequence in fppf cohomology (induced by the Kummer sequence) and the $\mathbb {G}_{m}$ -case. For $i=1$ , since $H^{1}_{\text {fppf}}((O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n}, \mathbb {G}_{m}) = 0$ , we have from the long exact sequence in fppf cohomology that $H^{1}_{\text {fppf}}((O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n}, \mu _{m})$ is the quotient

$$ \begin{align*}((O_{S}^{\text{perf}})^{\bigotimes_{O_{F,S}} n})^{*}/(((O_{S}^{\text{perf}})^{\bigotimes_{O_{F,S}} n})^{*})^{m}.\end{align*} $$

Since $O_{S}^{*}$ is $n^{\prime }$ -divisible for $n^{\prime }$ coprime to p by [Reference Neukirch, Schmidt and Wingberg19, Prop. 8.3.4], so is $(O_{S}^{\bigotimes _{O_{F,S}} n})^{*}$ (using Lemma 2.1). Now $O_{S}^{\text {perf},*}$ is $\mathbb {N}$ -divisible since it is obtained from $O_{S}$ by adjoining all p-power roots, and once again, this implies that $((O_{S}^{\text {perf}})^{\bigotimes _{O_{F,S}} n})^{*} = (O_{S}^{\bigotimes _{O_{F,S}} n})^{\text {perf}, \ast }$ is as well.

Proof. We leave it to the reader to check that the $\Sigma $ -condition of Proposition 2.13 is satisfied (since everything in $\Sigma $ is an $O_{E,S}$ -algebra, we may replace T with $\mathbb {G}_{m}$ for this condition and use the fact that $O_{E,S}$ and $O_{E^{\prime },S}$ are principal ideal domains, along with [Reference Rosengarten22, Lem. 2.2.9]). It thus suffices to show that $\check {H}^{2}(O_{E^{\prime },S}/O_{E,S}, \mathbb {G}_{m})$ vanishes. For any n, $\mathbb {G}_{m}(O_{E^{\prime },S}^{\bigotimes _{O_{E,S}}n}) = \mathbb {G}_{m}([O_{E^{\prime },S}^{\bigotimes _{O_{E,S}}n}]_{\text {red}})$ , and now $[O_{E^{\prime },S}^{\bigotimes _{O_{E,S}}n}]_{\text {red}} = O_{E^{\prime },S}^{\bigotimes _{O_{E^{\prime },S}}n} = O_{E,S}$ , so our Čech cohomology computations on this cover reduce to that of the trivial cover $O_{E^{\prime },S}/O_{E^{\prime },S}$ , giving the desired vanishing.

Proof. Suppose that we know the result for $F^{\prime } = F_{1}$ . Then, proceeding by induction, $F_{m-1}$ and $F_{v}$ are linearly disjoint, the valuation v extends uniquely to a valuation $v^{\prime }$ on $F_{m-1}$ , and $F_{m-1} \cdot F_{v}$ is the completion of $F_{m-1}$ with respect to $v^{\prime }$ . Thus, $F_{m}/F_{m-1}$ is of degree p, and we may replace $F_{v}$ by $(F_{m-1})_{v^{\prime }}$ and use the $m=1$ case to deduce that $(F_{m-1})_{v^{\prime }} = F_{m-1} \cdot F_{v}$ and $F_{m}$ are linearly disjoint over $F_{m-1}$ , which implies the desired result.

Setting $F^{\prime } = F_{1}$ , note that the extension $F^{\prime } \cdot F_{v}/F_{v}$ is either degree $1$ or degree p since $[F^{\prime } \cdot F_{v} \colon F_{v}] = [F^{\prime } \colon F_{v} \cap F^{\prime }] \mid p$ , and $F^{\prime }$ and $F_{v}$ are linearly disjoint if and only if this degree equals p. Hence, it is enough to show that $F^{\prime } \cap F_{v} = F$ . Thus, suppose that $x \in F_{v}$ is such that $x^{p} \in F$ . If $F(x) \neq F$ , then $F(x) = F^{\prime }$ , so that $F_{v}$ contains all pth roots of F; in particular, $\varpi ^{1/p} \in F_{v}$ , where $\varpi \in O_{F,v}$ (the localization of $O_{F}$ at v) is a v-adic uniformizer, which is clearly false.

Proof. The ring $O_{K_{w}}$ is finite and torsion-free over the principal ideal domain $O_{F_{v}}$ and is thus free as an $O_{F_{v}}$ -module. We may thus pick a basis (which is also an $F_{v}$ -basis for $K_{w}$ ) which allows us to view the map in question as the natural map $(O_{F_{v}}^{\bigoplus m})^{\bigotimes _{O_{F_{v}}}n} \to (F_{v}^{\bigoplus m})^{\bigotimes _{F_{v}}n}$ , which may be rewritten as the obvious inclusion $O_{F_{v}}^{\bigoplus mn} \hookrightarrow F_{v}^{\bigoplus mn}$ , giving the result.

Proof. It is enough to show that $\mathbb {A}_{K}^{\bigotimes _{\mathbb {A}}n} = (K^{\bigotimes _{F}n}) \otimes _{F} \mathbb {A}$ is isomorphic to the restricted product $\prod _{v \in V}^{\prime } (K^{\bigotimes _{F}n} \otimes _{F} F_{v})$ , where the restriction is with respect to the image of $O_{K}^{\bigotimes _{O_{F}}n} \otimes _{O_{F}} O_{F_{v}} \to K^{\bigotimes _{F}n} \otimes _{F} F_{v}$ ; the claimed isomorphism is defined on simple tensors by sending $x \otimes (a_{v})_{v}$ to $(x \otimes a_{v})_{v}$ . To prove that this map gives a well-defined isomorphism, it suffices to show that we have an isomorphism

$$ \begin{align*}O_{K} \otimes_{O_{F}} O_{F_{v}} \xrightarrow{\sim} \prod_{w \mid v} O_{K_{w}}\end{align*} $$

for any $v \in V$ . Letting $K^{\prime }$ be the maximal separable subextension of K, we already know that $O_{K^{\prime }} \otimes _{O_{F}} O_{F_{v}}$ is isomorphic to $\prod _{w^{\prime } \mid v} O_{K^{\prime }_{w^{\prime }}}$ , and so we are left with the ring $O_{K} \otimes _{O_{K^{\prime }}} [\prod _{w^{\prime } \mid v} O_{K^{\prime }_{w^{\prime }}}]$ .

We claim that the natural map $O_{K} \otimes _{O_{K^{\prime }}} O_{K^{\prime }_{w^{\prime }}} \to O_{K_{w}}$ (for w the unique extension of $w^{\prime }$ to K) is an isomorphism. For surjectivity, note that by the proof of Lemma 2.9, we have $O_{K} = O_{K^{\prime }}^{(1/p^{m})}$ , where $p^{m}$ is $[K \colon K^{\prime }]$ . We know that $O_{K^{\prime }_{w^{\prime }}}$ spans $O_{K_{w}}$ over $O_{K^{\prime }}^{(1/p^{m})}$ since the ring $O_{K^{\prime }_{w^{\prime }}} \cdot O_{K^{\prime }}^{(1/p^{m})}$ is finitely-generated over the complete discrete valuation ring $O_{K^{\prime }_{w^{\prime }}}$ , using that $O_{K^{\prime }}^{(1/p^{m})}$ is finite over $O_{K^{\prime }}$ by the finiteness of the relative Frobenius morphism (by [24], OCC6, using that $O_{K^{\prime }}$ is of finite type over $\mathbb {F}_{q}$ , being the coordinate ring of an affine open subscheme of a smooth curve over $\mathbb {F}_{q}$ ), and hence is complete as a topological ring, contains $O_{K}$ , and thus must be the w-adic completion $O_{K_{w}}$ . Injectivity follows from the linear disjointness given by Lemma 2.15.

Proof. (1) is immediate and (2) follows from (1) and [Reference Česnavičius1, Lem. 2.4]. Note that it makes sense to view $\mathcal {G}(O_{K,v}^{\bigotimes _{O_{F_{v}}}n})$ as a subgroup of $G(\mathbb {A}_{K,v}^{\bigotimes _{F_{v}}n})$ by Proposition 2.17.

Proof. It is enough to check that the restricted product identifications are compatible with the three inclusion maps $p_{i} \colon \mathbb {A}_{K}^{\bigotimes _{\mathbb {A}}n} \to \mathbb {A}_{K}^{\bigotimes _{\mathbb {A}}n+1}$ , $p^{v}_{i} \colon \mathbb {A}_{K,v}^{\bigotimes _{F_{v}}n} \to \mathbb {A}_{K,v}^{\bigotimes _{F_{v}}n+1}$ and $p^{v,\circ }_{i} \colon O_{K,v}^{\bigotimes _{O_{F_{v}}}n} \to O_{K,v}^{\bigotimes _{O_{F_{v}}}n+1}$ for $1 \leq i \leq n+1$ , which is straightforward.

Proof. For $E^{\prime }/F$ a finite algebraic extension with maximal separable and purely inseparable subextensions $E,F^{\prime }$ , respectively, note that by Lemma 2.15, we have a sequence of isomorphisms

$$ \begin{align*}(E^{\prime} \otimes_{F} F_{v})^{\bigotimes_{F_{v}}k} \xrightarrow{\sim} [F^{\prime} \otimes_{F} (E \otimes_{F} F_{v})]^{\bigotimes_{F_{v}}k} \xrightarrow{\sim} [\prod_{w \mid v} E^{\prime}_{w^{\prime}}]^{\bigotimes_{F_{v}}k} \xrightarrow{\sim} \prod_{w_{1}, \dots , w_{k} \mid v} \bigotimes_{F_{v}}^{i=1, \dots, k} E^{\prime}_{w_{i}^{\prime}},\end{align*} $$

where $E^{\prime }_{w^{\prime }}$ is the completion of $E^{\prime }$ with respect to the unique extension $w^{\prime }$ of the valuation w on E to the purely inseparable extension $E^{\prime }$ , for all $w \mid v$ in $V_{E}$ , and in the third term above, $F_{v}$ is embedded into the direct product diagonally. and so we obtain an identification

$$ \begin{align*}H^{n}((E^{\prime} \otimes_{F} F_{v})^{\bigotimes_{F_{v}} k}, M) \xrightarrow{\sim} \prod_{w_{1}, \dots, w_{k} \mid v_{F}} H^{n}((E^{\prime}_{w_{i}^{\prime}})^{\bigotimes_{F_{v}} k}, M).\end{align*} $$

Moreover, for $K^{\prime }/E^{\prime }$ two such extensions, the inductive map $(E^{\prime } \otimes _{F} F_{v})^{\bigotimes _{F_{v}}k} \to (K^{\prime } \otimes _{F} F_{v})^{\bigotimes _{F_{v}}k}$ gets translated to the map on the corresponding products defined by the product over all k-tuples $(w_{1}, \dots , w_{k})$ of the maps

$$ \begin{align*}\bigotimes_{F_{v}}^{j=1, \dots, k} E^{\prime}_{w_{j}^{\prime}} \to \prod_{\tilde{w}_{1}, \dots \tilde{w}_{k}; \tilde{w}_{j} \mid w_{j} \forall j} \bigotimes_{F_{v}}^{j=1, \dots, k} K^{\prime}_{\tilde{w}_{j}^{\prime}}\end{align*} $$

given in the obvious way. The upshot is that it suffices to show that each $\varinjlim _{K^{\prime }/F} H^{n}(K^{\prime }_{(w_{K})^{\prime }}/F_{v}, M)$ vanishes, where $\{w_{K}\}$ is a coherent system of places lifting v (equivalent to fixing a place $\dot {v}$ on $F^{\text {sep}}$ lifting v). But each direct limit of this form is isomorphic to $H^{n}(\overline {F_{v}}, M)$ , which we know vanishes.

Proof. For any finite algebraic field extension $E^{\prime }/F$ , the extension of rings $F_{v} \to F_{v} \otimes _{F} E^{\prime }$ is fppf. Thus, we get a natural map

$$ \begin{align*}\check{H}^{k}(\overline{\mathbb{A}}_{v}/F_{v}, M) \xrightarrow{\sim} \varinjlim_{E^{\prime}/F} \check{H}^{k}((E^{\prime} \otimes_{F} F_{v})/F_{v}, M) \to \check{H}^{k}_{\text{fppf}}(F_{v}, M) \xrightarrow{\sim} H^{k}(F_{v}, M)\end{align*} $$

via the natural comparison homomorphism $\check {H}_{\text {fppf}}^{k}(F_{v}, M) \to H^{k}(F_{v}, M)$ (from [24, Lem. 03AX]). By taking the direct limit of the spectral sequence from [24, Lem. 03AZ], we deduce that the above map $\check {H}^{k}(\overline {\mathbb {A}}_{v}/F_{v}, M) \to H^{k}(F_{v}, M)$ is an isomorphism since the cohomology groups $H^{j}(\overline {\mathbb {A}}_{v}^{\bigotimes _{F_{v}} m}, M)$ vanish for all $j, m \geq 1$ by Lemma 2.20. Now the commutative diagram

implies that $S_{v}^{k}$ is an isomorphism.

Proof. Since the Shapiro homomorphisms are constructed via the direct limit over finite algebraic extensions, it is enough to prove that for any fixed $2$ -cocycle $x \in M((E^{\prime }_{\dot {v}_{E^{\prime }}})^{\bigotimes _{F_{v}}3})$ , $E^{\prime }/F$ a finite extension of fields, there is a 1-cochain $c \in M((E^{\prime } \otimes _{F} F_{v})^{\bigotimes _{F_{v}}2})$ such that $dc = \dot {S}_{v}^{2}(x) \cdot \dot {S}_{v}^{^{\prime } 2}(x)^{-1}$ , and that if we have a inductive system $\{x_{E^{\prime }}\}_{E^{\prime }}$ of such 2-cocycles, as $E^{\prime }/F$ ranges over an exhaustive tower of finite extensions, then the system $\{c_{E^{\prime }}\}_{E^{\prime }}$ is also inductive. We will construct each $c_{E^{\prime }}$ explicitly using x (it will be useful later to have an explicit cochain to work with).

Assume that $E^{\prime }/F$ is of the form on p.12, let $E/F$ (resp. $F^{\prime }/F$ ) denote a maximal separable (resp. purely inseparable) subextension of $E^{\prime }/F$ , set $E_{v} := E_{\dot {v}_{E}}$ , and denote the extension of $\dot {v}_{E}$ to $E^{\prime }$ by $v^{\prime }$ . For $w \mid v$ in $V_{E}$ , denote by $r_{w}, \bar {r}_{w}$ the corresponding isomorphisms $E^{\prime }_{v^{\prime }} \xrightarrow {\sim } E^{\prime }_{w^{\prime }}$ (induced by applying $F^{\prime } \otimes _{F} -$ to the isomorphisms $E_{v} \xrightarrow {\sim } E_{w}$ defined by our sections). We define

$$ \begin{align*}c \in \prod_{w_{i_{1}},w_{i_{2}} \mid v_{F}} M(E^{\prime}_{w^{\prime}_{i_{1}}} \otimes_{F_{v}} E^{\prime}_{w^{\prime}_{i_{2}}})\end{align*} $$

to be given on the $(w_{i_{1}}, w_{i_{2}})$ -factor by

$$ \begin{align*}(r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},3}} \otimes r_{w_{i_{2},2}})(x) \cdot (\bar{r}_{w_{i_{1},2}} \otimes r_{w_{i_{2},1}} \cdot \bar{r}_{w_{i_{2},3}})(x)^{-1},\end{align*} $$

where $r_{w_{i_{j}},k}$ denotes that the source is the kth tensor factor of $(E^{\prime }_{v^{\prime }})^{\bigotimes _{F_{v}}3}$ , $1 \leq k \leq 3$ . It is clear that such a system of 1-cochains $\{c_{E^{\prime }}\}$ is inductive if the system $\{x_{E^{\prime }}\}$ is. Recall that $\dot {S}_{v}^{2}$ , $\dot {S}_{v}^{^{\prime } 2}$ are group homomorphisms

$$ \begin{align*}M((E^{\prime}_{v^{\prime}})^{\bigotimes_{F_{v}}3}) \to \prod_{w_{i_{1}},w_{i_{2}}, w_{i_{3}} \mid v_{F}} M(E^{\prime}_{w^{\prime}_{i_{1}}} \otimes_{F_{v}} E^{\prime}_{w^{\prime}_{i_{2}}} \otimes_{F_{v}} E^{\prime}_{w^{\prime}_{i_{3}}}).\end{align*} $$

To show that $dc = \dot {S}_{v}^{2}(x) \cdot \dot {S}_{v}^{^{\prime } 2}(x)^{-1}$ , we may focus on a fixed $(w_{i_{1}}, w_{i_{2}}, w_{i_{3}})$ -factor of the right-hand side. In this factor, the differential of c is given by the six-term product

$$ \begin{align*} &(1 \otimes r_{w_{i_{2},1}} \cdot \bar{r}_{w_{i_{2},3}} \otimes r_{w_{i_{3},2}})(x) \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},3}} \otimes 1 \otimes r_{w_{i_{3},2}})(x)^{-1} \cdot ( r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},3}} \otimes \ r_{w_{i_{2},2}} \otimes 1)(x)\\ & \cdot (1 \otimes \bar{r}_{w_{i_{2},2}} \otimes r_{w_{i_{3},1}} \cdot \bar{r}_{w_{i_{3},3}})(x)^{-1} \cdot ( \bar{r}_{w_{i_{1},2}} \otimes 1 \otimes r_{w_{i_{3},1}} \cdot \bar{r}_{w_{i_{3},3}})(x) \cdot ( \bar{r}_{w_{i_{1},2}} \otimes r_{w_{i_{2},1}} \cdot \bar{r}_{w_{i_{2},3}} \otimes 1)(x)^{-1}. \end{align*} $$

Since x is a 2-cocycle, we claim that the term $(1 \otimes r_{w_{i_{2},1}} \cdot \bar {r}_{w_{i_{2},3}}\otimes r_{w_{i_{3},2}})(x)$ equals

(2) $$ \begin{align} (\bar{r}_{w_{i_{1},1}}\otimes r_{w_{i_{2},3}} \otimes r_{w_{i_{3},2}})(x) \cdot ( \bar{r}_{w_{i_{1},2}} \otimes r_{w_{i_{2},1}} \cdot \bar{r}_{w_{i_{2},3}} \otimes 1)(x) \cdot (\bar{r}_{w_{i_{1},2}} \otimes \bar{r}_{w_{i_{2},1}} \otimes r_{w_{i_{3},3}})(x)^{-1}. \end{align} $$

To see this, note that

$$ \begin{align*}(1 \otimes \text{id}_{1} \otimes \text{id}_{2} \otimes \text{id}_{3})(x) \cdot (\text{id}_{1} \otimes 1 \otimes \text{id}_{2} \otimes \text{id}_{3})(x)^{-1} \cdot (\text{id}_{1} \otimes \text{id}_{2} \otimes 1 \otimes \text{id}_{3})(x) \cdot (\text{id}_{1} \otimes \text{id}_{2} \otimes \text{id}_{3} \otimes 1)(x)^{-1} = 1\end{align*} $$

(inside the group $M((E^{\prime }_{v^{\prime }})^{\bigotimes _{F_{v}} 4})$ ), and now applying $(\text {id}_{2} \otimes \text {id}_{1} \cdot \text {id}_{4} \otimes \text {id}_{3}) \circ (r_{w_{i_{2}}} \otimes \bar {r}_{w_{i_{1}}} \otimes r_{w_{i_{3}}} \otimes \bar {r}_{w_{i_{2}}})$ to the above expression gives the desired equality. We will leave the checking of similar equalities to the reader throughout the proof. The second factor in (2) cancels with the last factor in the main six-term equation. Next, we may rewrite the first term of (2) as

(3) $$ \begin{align} ( 1 \otimes \bar{r}_{w_{i_{2},2}} \otimes r_{w_{i_{3},1}} \cdot \bar{r}_{w_{i_{3},3}})(x) \cdot (\bar{r}_{w_{i_{1},1}} \otimes \bar{r}_{w_{i_{2},2}} \otimes \bar{r}_{w_{i_{3},3}})(x)^{-1} \cdot (\bar{r}_{w_{i_{1},1}} \otimes 1 \otimes r_{w_{i_{3},2}} \cdot \bar{r}_{w_{i_{3},3}})(x). \end{align} $$

We may also replace $(r_{w_{i_{1},1}} \cdot \bar {r}_{w_{i_{1},3}} \otimes r_{w_{i_{2},2}} \otimes 1)(x)$ from the main equation by the expression

$$ \begin{align*}(\bar{r}_{w_{i_{1},2}} \otimes r_{w_{i_{2},1}} \otimes r_{w_{i_{3},3}})(x) \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},2}} \otimes 1 \otimes r_{w_{i_{3},3}})(x)^{-1} \cdot (r_{w_{i_{1},1}} \otimes r_{w_{i_{2},2}} \otimes r_{w_{i_{3},3}})(x),\end{align*} $$

reducing us to showing the equality

(4) $$ \begin{align} \begin{aligned} &(\bar{r}_{w_{i_{1},1}} \otimes 1 \otimes r_{w_{i_{3},2}} \cdot \bar{r}_{w_{i_{3},3}})(x) \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},2}} \otimes 1 \otimes r_{w_{i_{3},3}})(x)^{-1} \\&\cdot (\bar{r}_{w_{i_{1},2}} \otimes 1 \otimes r_{w_{i_{3},1}} \cdot \bar{r}_{w_{i_{3},3}})(x) \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},3}} \otimes 1 \otimes r_{w_{i_{3},2}})(x)^{-1} =1. \end{aligned} \end{align} $$

Replacing the third factor of (4) by the expression

$$ \begin{align*}(r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},3}} \otimes 1 \otimes \bar{r}_{w_{i_{3},2}})(x) \cdot (r_{w_{i_{1},1}} \otimes 1 \otimes r_{w_{i_{3},2}} \cdot \bar{r}_{w_{i_{3},3}})(x)^{-1} \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},2}} \otimes 1 \otimes \bar{r}_{w_{i_{3},3}})(x)\end{align*} $$

reduces (4) to the equality

$$ \begin{align*} &(\bar{r}_{w_{i_{1},1}} \otimes 1 \otimes r_{w_{i_{3},2}} \cdot \bar{r}_{w_{i_{3},3}})(x) \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},2}} \otimes 1 \otimes r_{w_{i_{3},3}})(x)^{-1}\\ &\cdot (r_{w_{i_{1},1}} \otimes 1 \otimes r_{w_{i_{3},2}} \cdot \bar{r}_{w_{i_{3},3}})(x)^{-1} \cdot (r_{w_{i_{1},1}} \cdot \bar{r}_{w_{i_{1},2}}\otimes 1 \otimes \bar{r}_{w_{i_{3},3}})(x) = 1,\end{align*} $$

which follows easily from the fact that x is a 2-cocycle.

Construction 2.24 ([Reference Dillery6], Construction 2.38).

Let $A \xrightarrow {f} B$ be an F-morphism of commutative group schemes and $a, b \in A(S^{\bigotimes _{R}3}), B(S^{\bigotimes _{R}3})$ Čech 2-cocycles such that $[f(a)]= [b]$ in $\check {H}^{2}(S/R, B)$ . Any $x \in B(S \otimes _{R} S)$ satisfying $d(x)\cdot b = f(a)$ defines a morphism $\mathcal {E}_{a} \xrightarrow {\phi _{a,b,x}} \mathcal {E}_{b}$ of fibered categories over R as follows.

For any a-twisted torsor $(T,\psi )$ over an R-scheme V, we define the b-twisted torsor $(T^{\prime },\psi ^{\prime }) =: \phi _{a,b,x}(T, \psi )$ over V as follows. Define the $B_{V \times _{R} S}$ torsor $T^{\prime }$ to be $T \times ^{A_{V \times S},f} B_{V \times S}$ , and take the gluing map to be $\psi ^{\prime } := \overline {m_{x^{-1}} \circ \psi }$ , where $\overline {m_{x^{-1}} \circ \psi }$ denotes the isomorphism of contracted products

$$ \begin{align*} &p_{2}^{*}(T \times^{A_{V \times S},f} B_{V \times S}) =\\ (p_{2}^{*}T) \times^{A_{V \times (S \otimes_{R}S)},f} B_{V \times (S \otimes_{R}S)} \to &(p_{1}^{*}T) \times^{A_{V \times (S \otimes_{R}S)},f} B_{V \times (S \otimes_{R}S)} = p_{1}^{*}(T \times^{A_{V \times S},f} B_{V \times S}) \end{align*} $$

induced by $(m_{x^{-1}} \circ \psi ) \times \text {id}_{B}$ (we are implicitly identifying x with its image in $B(V \times _{F} (S \otimes _{R}S))$ ).

A morphism $(T_{1},\psi _{1}) \xrightarrow {\varphi } (T_{2},\psi _{2})$ of a-twisted torsors induces a morphism of the corresponding b-twisted torsors via the map on contracted products induced by $\varphi \times \text {id}$ , giving the desired functor.

Convention 2.25. When discussing an abelian R-group scheme A and cover $S/R$ , we will always assume that $\check {H}^{1}_{\text {fppf}}(S^{\bigotimes _{R}n}, A) = 0$ for all $n \geq 0$ .

Proof. This is an easy calculation using Convention 2.25.

Proof. We denote the differential $A_{k}(S^{\bigotimes _{R}i}) \to A_{k}(S^{\bigotimes _{R}(i+1)})$ by $d^{(k)}$ . First, note that since $\varprojlim _{k}^{(1)} \check {H}^{i-1}(S/R, A_{k}) = 0$ , the natural map

$$ \begin{align*}\varprojlim_{k} [A_{k}(S^{\bigotimes_{R}i})/B^{i-1}(k)] \to \varprojlim_{k}[(A_{k}(S^{\bigotimes_{R}i})/B^{i-1}(k))/(\check{H}^{i-1}(S/R, A_{k}))]\end{align*} $$

is surjective. Moreover, the natural map $A(S^{\bigotimes _{R}i}) = \varprojlim _{k} A_{k}(S^{\bigotimes _{R}i}) \to \varprojlim _{k} [A_{k}(S^{\bigotimes _{R}i})/B^{i-1}(k)]$ is surjective since we assume that $\varprojlim _{n}^{(1)} B^{i-1}(n) = 0$ .

Now by left-exactness of the inverse-limit functor, we have the exact sequence

In particular, if $x \in A(S^{\bigotimes _{R}(i+1)})$ is such that its image in $\varprojlim _{k} \frac {A_{k}(S^{\bigotimes _{R}(i+1)})}{d^{(k)}(A_{k}(S^{\bigotimes _{R}i}))}$ is zero (which is the hypothesis of the Proposition), then it lies in the image of $\bar {d}:= \varprojlim d^{(k)}$ . But now the diagram

commutes, and since the vertical composition is surjective and such an x lies in the image of the lower-diagonal map, it lies in the image of the upper-diagonal map, giving the desired result.

Conditions 3.2. For each $w \in V_{E}$ , there exists $w^{\prime } \in S_{E}$ such that $\text {Stab}(w, \Gamma _{E/F}) = \text {Stab}(w^{\prime }, \Gamma _{E/F})$ .

Proof. The right-hand isomorphisms on the first two lines follow from the fact that all T-torsors over $O_{F,S}$ are trivial over $O_{E,S}$ . The ‘connecting homomorphism’ $\delta $ is the standard connecting homomorphism in Čech cohomology (see. for example, [Reference Dillery6, §2.2]), and the right-hand column is exact because, applying the isomorphisms $\check {H}^{i}(O_{S}^{\text {perf}}/O_{F,S}, M) \xrightarrow {\sim } H^{i}(O_{F,S}, M)$ for $i=1,2$ and $M=T, \bar {T}, Z$ , the resulting two-column diagram commutes, by functoriality of the Čech-to-derived comparison maps ([Reference Rosengarten22, Prop. E.2.1]). From here, the identical argument as in [Reference Kaletha9] gives the result, using basic properties of the unbalanced cup product in fppf cohomology discussed in [Reference Dillery6, §4.2].

Proof. As in the proof of [Reference Kaletha9, Prop. 3.2.4], we may choose $\bar {Y}$ to be a free $\mathbb {Z}[\Gamma _{E/F}]$ -module, so that the connecting homomorphism of the left-hand column is injective, and $\Theta _{E,S}$ is the restriction of ‘ $-\text {TN}$ ’, which is an isomorphism independent of the choices of $c_{E,S}$ , $k_{i}$ or $(-)^{(1/p^{m_{i}})}$ .

Proof. The proof of [Reference Kaletha9, Lem. 3.2.7] works verbatim here, replacing $H^{i}(\Gamma _{S}, M(O_{S}))$ with $\check {H}^{i}(O_{S}^{\text {perf}}/O_{F,S}, M)$ for $M = T, \bar {T}, Z$ and $i=1,2$ .

Proof. This proof closely follows the proof of [Reference Kaletha9, Corollary 3.2.9]. It is enough to prove the result with $H^{2}(F, Z)$ replaced by $\check {H}^{2}(\overline {F}/F, Z)$ . By Corollary 3.4, Lemma 3.5 and [Reference Kaletha9, Lem. 3.2.8], we obtain a functorial injective homomorphism $\Theta $ as claimed, which is independent of the choices of (appropriately chosen) $S^{(E)}$ , so all that remains to prove is surjectivity.

For any $h \in \check {H}^{2}(\overline {F}/F, Z)$ , we may find $E^{\prime }/F$ finite such that $h \in \check {H}^{2}(E^{\prime }/F, Z)$ ; denote the separable closure of F in $E^{\prime }$ by E, so that $E^{\prime } = E \cdot F_{m}$ for some unique $m \in \mathbb {N}$ . Moreover, since $E^{\prime \bigotimes _{F} 3} = \varinjlim _{S^{(E)}} O_{E^{\prime },S^{(E)}}^{\bigotimes _{O_{F,S^{(E)}}} 3}$ , where the direct limit is over all finite $S^{(E)} \subset V$ satisfying the required conditions with respect to $E/F$ , there is some finite $S^{(E)}$ satisfying the required conditions with respect to $E/F$ such that we can find $h_{E^{\prime },S^{(E)}} \in Z(O_{E^{\prime },S^{(E)}}^{\bigotimes _{O_{F,S^{(E)}}} 3})$ with image in $Z(E^{\prime \bigotimes _{F} 3}) \to \check {H}^{2}(E^{\prime }/F, Z)$ equal to h. We may enlarge $S^{(E)}$ even further to assume that $h_{E^{\prime },S^{(E)}} \in Z^{2}(O_{E^{\prime },S^{(E)}}/O_{F,S^{(E)}}, Z)$ since the Čech differential on $Z(O_{E^{\prime },S^{(E)}}^{\bigotimes _{O_{F,S^{(E)}}} 3})$ is the same as that of $Z(E^{\prime \bigotimes _{F} 3})$ , and we may use finitely many elements of F and $E^{\prime }$ to encode the fact that $dh_{E^{\prime },S^{(E)}} = 1$ in $Z(E^{\prime \bigotimes _{F} 4})$ . Denote by $\bar {h}_{E^{\prime },S^{(E)}}$ the image of $h_{E^{\prime },S^{(E)}}$ in $\check {H}^{2}(O_{E^{\prime },S^{(E)}}/O_{F,S^{(E)}}, Z)$ .

Once we have such an $\bar {h}_{E^{\prime },S^{(E)}}$ , choose an $O_{F,S}$ -torus $Z \hookrightarrow T$ with $\bar {T} :=T/Z$ such that $\bar {Y} = X_{*}(\bar {T})$ is free over $\Gamma _{E/F}$ , and denote the image of $\bar {h}_{E^{\prime },S^{(E)}}$ in $\check {H}^{2}(O_{E^{\prime },S^{(E)}}/O_{F,S^{(E)}}, T)$ by $\bar {h}_{E^{\prime },S^{(E)},T}$ . Note that we have a commutative diagram of isomorphisms from Corollary 2.14:

and so we may pick a (unique) preimage, denoted by $\bar {h}_{E, S^{(E)},T}$ , of $\bar {h}_{E^{\prime },S^{(E)},T}$ in $\check {H}^{2}(O_{E,S}/O_{F,S}, T)$ , and by the commutativity of the diagram, the image of $\bar {h}_{E,S^{(E)},T}$ in $\check {H}^{2}(O_{E,S^{(E)}}/O_{F,S^{(E)}}, \bar {T})$ is zero. We may thus lift $\text {-TN}^{-1}(\bar {h}_{E,S^{(E)},T}) \in \widehat {H}^{0}(\Gamma _{E/F}, Y[S^{(E)}_{E}]_{0})$ to some $g \in \widehat {H}^{-1}(\Gamma _{E/F}, A^{\vee }[S_{E}^{(E)}]_{0})$ , and then the same argument as in [Reference Kaletha9, Corollary 3.2.9] shows that $\Theta _{E,S^{(E)}}(g) \in \check {H}^{2}(O_{S^{(E)}}^{\text {perf}}/O_{F,S^{(E)}}, Z)$ has image in $\check {H}^{2}(\overline {F}/F, Z)$ equal to h, as desired (even though we need to take the image of $\bar {h}_{E^{\prime },S^{(E)},T}$ in $\check {H}^{2}(O_{E,S}/O_{F,S}, T)$ , the argument of [Reference Kaletha9] uses that the image of their $\Theta _{E,S^{(E)}}(g)$ in $\check {H}^{2}(\overline {F}/F, T)$ is the same as that of h, which is still true for our g obtained via the above adjustment for non-separability).

Conditions 3.7.

1. (1) S contains all places that ramify in E.

2. (2) Every ideal class of E contains an ideal with support in $S_{E}$ (i.e., $\text {Cl}(O_{E,S}) = 0$ ).

3. (3) For every $w \in V_{E}$ , there exists $w^{\prime } \in S_{E}$ with $\text {Stab}(w, \Gamma _{E/F}) = \text {Stab}(w^{\prime }, \Gamma _{E/F})$ .

4. (4) For every $\sigma \in \Gamma _{E/F}$ , there exists $\dot {v} \in \dot {S}_{E}$ such that $\sigma \dot {v} = \dot {v}$ .

Proof. We claim first that the natural map $\check {H}^{2}(\overline {F_{v}}/F_{v}, (P_{\dot {V}})_{F_{v}}) \to \varprojlim _{i} \check {H}^{2}(\overline {F_{v}}/F_{v}, P_{E_{i}, \dot {S}_{i}, n_{i}})$ is an isomorphism. To simplify notation, set $P_{i} := (P_{E_{i}, \dot {S}_{i}, n_{i}})_{F_{v}}$ . By Proposition 2.29, it suffices to show that $\varprojlim _{i}^{(1)} \check {H}^{1}(\overline {F_{v}}/F_{v}, P_{i}) = 0$ and $\varprojlim _{i} d(P_{i}(\overline {F_{v}})) = 0$ ; the latter trivially follows because it is a system of finite groups. We now explain the vanishing of $\varprojlim _{i}^{(1)} \check {H}^{1}(\overline {F_{v}}/F_{v}, P_{i})$ (all group schemes are of finite type so we replace the notation ‘ $\check {H}^{1}(\overline {F_{v}}/F_{v}, -)$ ’ with ‘ $H^{1}(F_{v}, -)$ ’ for notational convenience):

As in [Reference Kaletha9, §3.4], we may fit any $M_{E, \dot {S}_{E}, n}$ into a sequence (where the middle $0$ -subscript denotes the kernel of the $\Gamma _{E/F}$ -augmentation map)

(10) $$ \begin{align} 0 \to M_{E, \dot{S}_{E}, n} \to \frac{1}{n}\mathbb{Z}/\mathbb{Z}[\Gamma_{E/F} \times S]_{0} \to \frac{1}{n}\mathbb{Z}/\mathbb{Z}[S_{E}]_{0} \to 0. \end{align} $$

These identifications are compatible with the transition maps, which on the right-most terms are

$$ \begin{align*}\sum_{(\sigma, w) \in \Gamma_{E/F} \times S_{E}} a_{\sigma, w}[(\sigma, w)] \mapsto \sum_{(\gamma, u) \in \Gamma_{K/F} \times S_{K}^{\prime}} (\# \Gamma_{K/E}^{u})a_{\bar{\gamma}, u_{E}}[(\gamma, u)].\end{align*} $$

We have the exact sequence

$$ \begin{align*} H^{1}(F_{v}, A_{i}) \to H^{1}(F_{v}, P_{i}) \to C_{i}, \end{align*} $$

where $A_{i}$ denotes the multiplicative group scheme Cartier dual to $\frac {1}{n_{i}}\mathbb {Z}/\mathbb {Z}[\Gamma _{E_{i}/F} \times S_{i}]_{0} $ and $C_{i}$ is the image of $H^{1}(F_{v}, P_{i})$ in $H^{2}(F_{v}, B_{i})$ , where $B_{i}$ is the kernel of $A_{i} \to P_{i}$ . Note that $\varprojlim ^{(1)} C_{i} = 0$ since these are all finite groups (because $H^{2}(F_{v}, B_{i})$ is, using local Poitou-Tate duality), so it suffices to prove that $\varprojlim ^{(1)} H^{1}(F_{v}, A_{i}) = 0$ .

Denoting by $\tilde {A}_{i}$ the multiplicative group scheme Cartier dual to $\frac {1}{n_{i}}\mathbb {Z}/\mathbb {Z}[\Gamma _{E_{i}/F} \times S_{i}]$ , we have an exact sequence

$$ \begin{align*} H^{1}(F_{v}, \tilde{A}_{i}) \to H^{1}(F_{v}, A_{i}) \to \tilde{C}_{i}, \end{align*} $$

where $\tilde {C}_{i}$ is the image of $H^{1}(F_{v}, A_{i}) \to H^{2}(F, \tilde {B}_{i})$ and $\tilde {B}_{i}$ is the kernel of $\tilde {A}_{i} \to A_{i}$ (it is Cartier dual to $\frac {1}{n_{i}}\mathbb {Z}/\mathbb {Z}[\Gamma _{E_{i}/F}]$ ). Since $H^{2}(F_{v}, B_{i})$ is finite, we have $\varprojlim ^{(1)} \tilde {C}_{i} =0$ , and we thus reduce further to showing that $\varprojlim ^{(1)} H^{1}(F_{v}, \tilde {A}_{i}) = 0$ .

Note that $\tilde {A}_{i} = \prod _{S_{i}} [\mathrm {Res}_{E_{i}/F}(\mu _{n_{i}})]_{F_{v}}$ . We first compute $H^{1}(F_{v}, \mathrm {Res}_{E_{i}/F}(\mu _{n_{i}}))$ ; the Mackey formula and Shapiro’s lemma tell us that

(11) $$ \begin{align} H^{1}(\Gamma_{v}, \frac{1}{n_{i}}\mathbb{Z}/\mathbb{Z}[\Gamma_{E_{i}/F}]) = \bigoplus_{w \in V_{E_{i}}, w \mid v} H^{1}(\Gamma_{E_{i,w}}, \frac{1}{n_{i}}\mathbb{Z}/\mathbb{Z}), \end{align} $$

and so by local Poitou-Tate duality, we have $H^{1}(F_{v}, \mathrm {Res}_{E_{i}/F}(\mu _{n_{i}})) = \bigoplus _{w \in V_{E_{i}}, w \mid v} H^{1} (E_{i,w}, \mu _{n_{i}})$ . It follows that

$$ \begin{align*} H^{1}(F_{v}, \tilde{A}_{i}) = \bigoplus_{S_{i}} \bigoplus_{w \in V_{E_{i}}, w \mid v} H^{1}(E_{i,w}, \mu_{n_{i}}), \end{align*} $$

and we want to show that $\varprojlim ^{(1)}$ of this sequence of groups vanishes. From here, one can use the verbatim argument in the proof of [Reference Dillery6, Proposition 3.1] (which uses a computation involving the explicit resolution computing $\varprojlim ^{(1)}$ as in [Reference Weibel31, §3.5]) to obtain the desired result, finally proving the claimed vanishing.

The isomorphism we just proved implies that the map $\check {H}^{2}(\overline {F}/F, P_{\dot {V}}) \to \check {H}^{2}(\overline {F_{v}}/F_{v}, (P_{\dot {V}})_{F_{v}})$ factors as the composition

$$ \begin{align*}\check{H}^{2}(\overline{F}/F, P_{\dot{V}}) \to \varprojlim_{i} \check{H}^{2}(\overline{F}/F, P_{E_{i}, \dot{S}_{i}, n_{i}}) \to \varprojlim_{i} \check{H}^{2}(\overline{F_{v}}/F_{v}, (P_{E_{i}, \dot{S}_{i}, n_{i}})_{F_{v}}),\end{align*} $$

where the second map is the inverse limit of the localizations for each i. It is thus enough to show that, for each i, the map $\text {loc}_{v}^{P_{E_{i}, \dot {S}_{i}, n_{i}}}$ sends $\xi _{(E_{i})_{v}, n_{i}} \in \check {H}^{2}(\overline {F_{v}}/F_{v}, u_{(E_{i})_{v}/F_{v}, n_{i}})$ to the image of $\xi _{E_{i}, \dot {S}_{i}, n_{i}}$ under the map $\check {H}^{2}(\overline {F}/F, P_{E_{i}, \dot {S}_{i}, n_{i}}) \to \check {H}^{2}(\overline {F_{v}}/F_{v}, (P_{E_{i}, \dot {S}_{i}, n_{i}})_{F_{v}})$ . Once we have reached this step, we get the result from the proof of [Reference Kaletha9, Corollary 3.8].

Proof. The first vanishing follows from the proof of [Reference Kaletha9, Lem. 3.4.4]. For the second vanishing, we first use the identity (11); identifying each $H^{1}(\Gamma _{E_{w}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ with $\text {Hom}(\Gamma _{E_{w}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ , the transition map

$$ \begin{align*}\bigoplus_{w \in V_{E}, w \mid v} \text{Hom}(\Gamma_{E_{w}}, \frac{1}{n}\mathbb{Z}/\mathbb{Z}) \to \bigoplus_{u \in V_{K}, u \mid v} \text{Hom}(\Gamma_{K_{u}}, \frac{1}{m}\mathbb{Z}/\mathbb{Z})\end{align*} $$

is given by the maps $\text {Hom}(\Gamma _{E_{w}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z}) \to \bigoplus _{u \mid w} \text {Hom}(\Gamma _{K_{u}}, \frac {1}{m}\mathbb {Z}/\mathbb {Z})$ induced by the inclusions $\Gamma _{K_{u}} \hookrightarrow \Gamma _{E_{w}}$ . For a fixed $f_{w} \in \text {Hom}(\Gamma _{E_{w}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ , the kernel $H_{f_{w}}$ of $f_{w}$ is an open normal subgroup of $\Gamma _{E_{w}}$ , and so if $K/E$ is a large enough finite Galois extension, we have $\Gamma _{K_{u}} \subseteq H_{f_{w}}$ for all $u \mid w$ places of K. Note that, given such a K, this property also holds for any $K^{\prime }/K/F$ finite Galois and $\tilde {u} \mid w$ a place of $K^{\prime }$ . Now for any $(f_{w}) \in \bigoplus _{w \mid v} \text {Hom}(\Gamma _{E_{w}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ , we can apply the previous sentence to each w and take a compositum to find a finite Galois $K/F$ such that for any w and $u \in V_{K}$ with $u \mid w$ , we have $\Gamma _{K_{u}} \subseteq H_{f_{w}}$ . Thus, the image of $(f_{w})_{w}$ in $\bigoplus _{u \mid v} \text {Hom}(\Gamma _{K_{u}}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ is trivial, proving (2).

Proof. Note that $H^{1}(F_{v}, (P_{\dot {V}})_{F_{v}}) = \varprojlim _{i} H^{1}(F_{v}, (P_{E_{i}, \dot {S}_{i}, n_{i}})_{F_{v}})$ since $\varprojlim _{i}^{(1)} H^{0}(F_{v}, (P_{E_{i}, \dot {S}_{i}, n_{i}})_{F_{v}})$ is trivial (the same argument works for F instead of $F_{v}$ ). Thus, local Poitou-Tate duality gives

$$ \begin{align*}H^{1}(F_{v}, (P_{\dot{V}})_{F_{v}}) = \varprojlim_{i} (H^{1}(\Gamma_{v}, M_{E_{i}, \dot{S}_{i}, n_{i}})^{*}) = (\varinjlim_{i} H^{1}(\Gamma_{v}, M_{E_{i}, \dot{S}_{i}, n_{i}}))^{*}.\end{align*} $$

Now we have (using (10)) the exact sequence

$$ \begin{align*}0 \to C_{i} \to H^{1}(\Gamma_{v}, M_{E_{i}, \dot{S}_{i}, n_{i}}) \to H^{1}(\Gamma_{v}, \frac{1}{n_{i}}\mathbb{Z}/\mathbb{Z}[\Gamma_{E_{i}/F} \times S_{i}]_{0}),\end{align*} $$

where $C_{i}$ is a subquotient of $\frac {1}{n_{i}}\mathbb {Z}/\mathbb {Z}[(S_{i})_{E_{i}}]_{0}$ , and the colimits of the outer two terms are zero, by Lemma 3.13 and the remark preceding it, giving the result.

Proof. For any finite Galois $K/F$ , we have the exact sequence

$$ \begin{align*}0 \to N_{K/F}(C_{K}) \to C_{F} \xrightarrow{(-,K/F)} \Gamma_{K/F}^{\text{ab}} \to 0.\end{align*} $$

Since the group $N_{K/F}(C_{K})$ is open of finite index in $C_{F}$ , the inverse limit over all open subgroups of $C_{F}$ of finite index may be taken over all open subgroups of finite index which lie in $N_{K/F}(C_{K})$ , and for any such subgroup U, we get the exact sequence

$$ \begin{align*}0 \to \frac{N_{K/F}(C_{K})}{U} \to \frac{C_{F}}{U} \to \Gamma_{K/F}^{\text{ab}} \to 0,\end{align*} $$

which after applying the (left-exact) functor $\varprojlim (-)$ yields the exact sequence

$$ \begin{align*}0 \to N_{K/F}(C_{K})^{\wedge} \to \overline{C_{F}} \to \Gamma_{K/F}^{\text{ab}} \to 0;\end{align*} $$

note that surjectivity is preserved because the kernels are all finite groups. Now since $C_{K}$ is dense in $\overline {C_{K}}$ , we have that $N_{K/F}(\overline {C_{K}}) = N_{K/F}(C_{K})^{\wedge }$ inside $\overline {C_{F}}$ , by continuity of the norm map. Applying the inverse limit over all finite Galois $K/F$ then yields the exact sequence

$$ \begin{align*}0 \to \overline{N} \to \overline{C_{F}} \to \Gamma^{\text{ab}},\end{align*} $$

so it is enough to show that the completed universal residue map $\overline {C_{F}} \to \Gamma ^{\text {ab}}$ is injective, which is a basic fact of global class field theory (see, for example, [Reference Neukirch, Schmidt and Wingberg19, Prop. 8.1.26]).

Proof. As in the proof of Lemma 3.13, it is enough to show that the colimit $\varinjlim H^{2}(\Gamma _{E}, \frac {1}{n}\mathbb {Z}/\mathbb {Z})$ vanishes, with the transition maps given by the restriction homomorphism. For $(E,n)$ fixed, by [Reference Neukirch, Schmidt and Wingberg19, Thm. 8.4.4] (with $S=V_{E}$ ), we have an isomorphism

$$ \begin{align*}H^{2}(\Gamma_{E}, \frac{1}{n}\mathbb{Z}/\mathbb{Z}) \xrightarrow{\sim} (\widehat{H}^{0}(\Gamma_{E}, \overline{C}[n]))^{\vee},\end{align*} $$

where $\widehat {H}^{0}(\Gamma _{E}, \overline {C}[n]) := \varprojlim _{K/E} \widehat {H}^{0} (\Gamma _{K/E}, \overline {C}[n]^{\Gamma _{K}})$ , with transition maps given by the projections

$$ \begin{align*}\frac{\overline{C_{E}}[n]}{N_{K^{\prime}/E}( \overline{C_{K^{\prime}}}[n])} \to \frac{\overline{C_{E}}[n]}{N_{K/E}( \overline{C_{K}}[n])};\end{align*} $$

recall that for M a locally-compact Hausdorff topological group, $M^{\vee }$ denotes $\text {Hom}_{\text {cts}}(M, \mathbb {R}/\mathbb {Z})$ .

We claim that the natural map $\overline {C_{E}}[n] \to \widehat {H}^{0} (\Gamma _{E}, \overline {C}[n])$ is an isomorphism, which follows from the fact that both $\varprojlim _{K/E} N_{K/E}( \overline {C_{K}}[n])$ and $\varprojlim ^{(1)} N_{K/E}( \overline {C_{K}}[n])$ vanish. For the former, there is an inclusion $N_{K/E}( \overline {C_{K}}[n]) \hookrightarrow N_{K/E}(\overline {C_{K}})[n]$ , and hence an inclusion

$$ \begin{align*}\varprojlim N_{K/E}( \overline{C_{K}}[n]) \hookrightarrow (\varprojlim N_{K/E}(\overline{C_{K}})))[n] = 0,\end{align*} $$

where the last term equals zero by Lemma 3.15. Moreover, one checks using the explicit resolution computing $\varprojlim ^{(1)}$ along with Lemma 3.15 and the compactness of $\overline {C_{E}}$ that $\varprojlim ^{(1)} N_{K/E}( \overline {C_{K}}[n])= 0$ . In conclusion, we obtain an isomorphism $H^{2}(\Gamma _{E}, \frac {1}{n}\mathbb {Z}/\mathbb {Z}) \xrightarrow {\sim } \overline {C_{E}}[n]^{\vee }$ .

It thus suffices to show that $\varinjlim \overline {C_{E}}[n]^{\vee }$ vanishes, where the transition maps are induced by the maps $\overline {C_{K}}[m] \to \overline {C_{E}}[n]$ given by $N_{K/E}$ composed with the $m/n$ -power map. Given $f \in \overline {C_{E}}[n]^{\vee }$ , since it is continuous and $\frac {1}{n}\mathbb {Z}/\mathbb {Z}$ is finite, $\text {ker}(f)$ is an open subgroup of $\overline {C_{E}}[n]$ . The norm groups $N_{K/E}(\overline {C_{K}}) \subseteq \overline {C_{E}}$ shrink to the identity, so there is some $(K,m)> (E, n)$ with $N_{K/E}(\overline {C_{K}}) \subseteq \text {ker}(f)$ (using the finite intersection property), and hence, the image of f in $\overline {C_{K}}[m]^{\vee }$ is zero.

Proof. The first equality is from the proof of Proposition 3.14. Localization $\check {H}^{1}(\overline {F}/F, P_{\dot {V}}) \to \check {H}^{1}(\overline {F_{v}}/F_{v}, P_{\dot {V}})$ gives (via Lemma 3.17) a localization map $H^{1}(F, P_{\dot {V}}) \to H^{1}(F_{v}, P_{\dot {V}})$ for all $v \in V$ . Setting $\text {ker}^{1}(F, P_{\dot {V}}) := \text {ker}[H^{1}(F, P_{\dot {V}}) \to \prod _{\dot {v} \in \dot {V}} H^{1}(F_{v}, P_{\dot {V}})]$ , it suffices by Lemma 3.13 to show $\text {ker}^{1}(F, P_{\dot {V}}) \xrightarrow {\sim } \varprojlim _{i} \text {ker}^{1}(F, P_{E_{i}, \dot {S}_{i}, n_{i}}) = 0$ .

For i fixed, [Reference Česnavičius1, Lem. 4.4] tells us that we have a perfect pairing of finite abelian groups

$$ \begin{align*}\text{ker}^{1}(F, P_{E_{i}, \dot{S}_{i}, n_{i}}) \times \text{ker}^{2}(F, \underline{M_{E_{i}, \dot{S}_{i}, n_{i}}}) \to \mathbb{Q}/\mathbb{Z},\end{align*} $$

where $\underline {M_{E_{i}, \dot {S}_{i}, n_{i}}}$ is the étale F-group scheme associated to the $\Gamma $ -module $M_{E_{i}, \dot {S}_{i}, n_{i}}$ (and is Cartier dual to the finite flat F-group scheme $P_{E_{i}, \dot {S}_{i}, n_{i}}$ ). Thus, it is enough to show that

$$ \begin{align*}\varprojlim_{i} (\text{ker}^{2}(\Gamma, M_{E_{i}, \dot{S}_{i}, n_{i}}))^{*} = (\varinjlim_{i} \text{ker}^{2}(\Gamma, M_{E_{i}, \dot{S}_{i}, n_{i}}))^{*} = 0,\end{align*} $$

which we will do by showing that the direct limit $\varinjlim _{i} \text {ker}^{2}(\Gamma , M_{E_{i}, \dot {S}_{i}, n_{i}})$ vanishes, which follows from an easier version of the analogous argument in the proof of [Reference Kaletha9, Prop. 3.4.6] that we leave to the reader (using our Lemmas 3.13, 3.16 in place of Lemma 3.4.4 loc. cit.).

Proof. The case $k=0$ is trivial, so we only need to focus on $k=1,2$ . By Lemma 2.29, it is enough to show that $\varprojlim _{i}^{(1)} \check {H}^{k}(\overline {\mathbb {A}}_{v}/F_{v}, P_{i}) = 0$ for $k=0,1$ and that $\varprojlim _{i}^{(1)} B^{1}(i) = 0$ (notation as in Lemma 2.29). The vanishing of $\varprojlim _{i}^{(1)} \check {H}^{0}(\overline {\mathbb {A}}_{v}/F_{v}, P_{i})$ follows from the fact that $\check {H}^{0}(\overline {\mathbb {A}}_{v}/F_{v}, P_{i}) = P_{i}(F_{v})$ , and the system $\{P_{i}(F_{v})\}$ consists of finite groups. The vanishing of $\varprojlim _{i}^{(1)} B^{1}(i)$ comes from the fact that the system $\{ B^{1}(i)\}$ has surjective transition maps: On Čech 0-cochains, the transition maps $P_{i+1}(\overline {\mathbb {A}}_{v}) \to P_{i}(\overline {\mathbb {A}}_{v})$ are all surjective by Lemma 2.20, and since the Čech differentials are compatible with F-homomorphisms (in our case, the transition maps $P_{i+1} \to P_{i}$ ), this surjectivity carries over to the group of $1$ -coboundaries.

The proof of Corollary 2.21 shows that the groups $\check {H}^{1}(\overline {\mathbb {A}}_{v}/F_{v}, P_{i})$ are (compatibly) isomorphic to $H^{1}(F_{v}, P_{i})$ , and, as argued in the proof in Corollary 3.12, $\varprojlim _{i}^{(1)} H^{1}(F_{v}, P_{i}) = 0$ .