Proof. (1) It is clear that the formula $g(x\otimes x'):= g(x)\otimes g(x')$ makes $L^H \otimes _{L^S} L^{H'}$ into a $G$ -algebra. Consider the following commutative square of $F$ -schemes.

After base change to a separable closure of $F$ , this square becomes the cartesian square (2.2), and therefore it is cartesian. Passing to coordinate rings, we deduce that the homomorphism $L^H \otimes _{L^S} L^{H'} \rightarrow L$ is an isomorphism of $G$ -algebras. In particular, since $L$ is a Galois $G$ -algebra, so is $L^H \otimes _{L^S} L^{H'}$ .

(2) We have the following $G$ -equivariant cartesian diagram.

Every $G$ -equivariant morphism between $G/H$ and $G/S$ is isomorphic to the projection map $G/H\to G/S$ . Therefore, the base change of $\operatorname {Spec}(K) \to \operatorname {Spec}(E)$ to $F_s$ is $G$ -equivariantly isomorphic to the projection $G/H \to G/S$ . Similarly for $\operatorname {Spec}(K')\to \operatorname {Spec}(E)$ . Therefore, the base change of $\operatorname {Spec}(L)\to \operatorname {Spec}(F)$ over $F_s$ is $G$ -equivariantly isomorphic to $(G/H)\times _{G/S}(G/H')\simeq G$ , that is, the morphism $\operatorname {Spec}(L)\to \operatorname {Spec}(F)$ is an étale $G$ -torsor.

Proof. See [Reference Serre and GreenbergSer79, Chapter XIV, Proposition 4(iii)].

Proof. See [Reference Pál and QuickPQ22, Theorem 3.8].

Proof. See [Reference DwyerDwy75] for Dwyer’s original proof in the setting of abstract groups, and see [Reference EfratEfr14] or [Reference Harpaz and WittenbergHW23, Proposition 2.2] for the statement in the case of profinite groups.

Proof. This follows from Theorem 2.4 and (2.1).

Proof. (1) The automorphism $\sigma _b\colon F_{a,b}\to F_{a,b}$ extends to an isomorphism of étale algebras $f\colon (F_{a,b})_{ax}\to (F_{a,b})_x$ by sending $(ax)^{1/p}$ to $a^{1/p}x^{1/p}$ . The map $f$ is well defined because $f((ax)^{1/p})^p=(a^{1/p}x^{1/p})^p=ax$ . We now show that $f$ is $U_3$ -equivariant. The restriction of $f$ to $F_{a,b}$ is $U_3$ -equivariant because $\sigma _a\sigma _b=\sigma _b\sigma _a$ on $F_{a,b}$ . We have

\begin{equation*} \sigma _a(f((ax)^{1/p}))=\sigma _a(a^{1/p})\cdot \sigma _a(x^{1/p})=\zeta \cdot a^{1/p}\cdot \frac {b^{1/p}}{\alpha }\cdot x^{1/p}=\frac {\zeta a^{1/p}b^{1/p}x^{1/p}}{\alpha } \end{equation*}

and

\begin{equation*}f(\sigma _a((ax)^{1/p}))=f((b^{1/p}/\alpha )\cdot (ax)^{1/p})=\zeta \cdot \frac {b^{1/p}}{\alpha }\cdot a^{1/p}\cdot x^{1/p}=\frac {\zeta a^{1/p}b^{1/p}x^{1/p}}{\alpha }.\end{equation*}

Thus, $f$ is $\langle {\sigma _a}\rangle$ -equivariant. We also have

\begin{equation*}\sigma _b(f((ax)^{1/p}))= \sigma _b(a^{1/p}) \cdot \sigma _b(x^{1/p}) = a^{1/p}\cdot x^{1/p} \end{equation*}

and

\begin{equation*}f(\sigma _b((ax)^{1/p}))=f((ax)^{1/p})=a^{1/p}\cdot x^{1/p}.\end{equation*}

Thus, $f$ is $\langle {\sigma _b}\rangle$ -equivariant. Since $\sigma _a$ and $\sigma _b$ generate $U_3$ , we conclude that $f$ is $U_3$ -equivariant, as desired.

(2) The proof is similar to that of (1).

Proof. (1) Since $Q_3=\langle {\tau }\rangle \simeq \mathbb Z/p\mathbb Z$ and $K^{Q_3}\simeq F_{a,b}$ as $(\mathbb Z/p\mathbb Z)^2$ -algebras, we have an isomorphism of étale $F_{a,b}$ -algebras $K\simeq (F_{a,b})_z$ for some $z\in F_{a,b}^\times$ such that $(\tau -1)z^{1/p}=\zeta ^{-1}$ . We may suppose that $K=(F_{a,b})_z$ . As $\tau$ commutes with $\sigma _b$ ,

\begin{equation*} (\tau -1)(\sigma _b-1)z^{1/p}=(\sigma _b-1)(\tau -1)z^{1/p}=(\sigma _b-1)\zeta ^{-1}=1, \end{equation*}

and hence $(\sigma _b-1)z^{1/p}\in F_{a,b}^\times$ . By Hilbert’s Theorem 90 for the extension $F_{a,b}/F_a$ , there is $t\in F_{a,b}^\times$ such that $(\sigma _b-1)z^{1/p}=(\sigma _b-1)t$ . Replacing $z$ by $zt^{-p}$ , we may thus assume that $(\sigma _b-1)z^{1/p}=1$ . In particular, $z\in F_a^\times$ . Since $(\tau -1)z^{1/p}=\zeta ^{-1}$ , we have $\sigma _b\sigma _a(z^{1/p})=\zeta \sigma _a\sigma _b(z^{1/p})$ . Thus,

\begin{equation*}(\sigma _b-1) (\sigma _a-1)z^{1/p} =(\sigma _b\sigma _a-\sigma _a\sigma _b+ (\sigma _a-1)(\sigma _b-1))z^{1/p}=\zeta (\sigma _a-1) (\sigma _b-1)z^{1/p} = \zeta ,\end{equation*}

and hence $(\sigma _a-1)z^{1/p} = b^{1/p}/\alpha '$ for some $\alpha '\in F_a^\times$ . Moreover, $N_a(\alpha '/\alpha )=b/b=1$ , and so by Hilbert’s Theorem 90 there exists $\theta \in F_a^{\times }$ such that $\alpha '/\alpha =(\sigma _a-1)\theta$ . We define $x:= z\theta ^p\in F_a^\times$ , and we set $x^{1/p}:= z^{1/p}\theta \in (F_{a,b})_z^\times$ . Then $K=(F_{a,b})_x$ , where

\begin{equation*}(\sigma _a-1)x^{1/p}=(\sigma _a-1)z^{1/p}\cdot (\sigma _a-1)\theta =\frac {b^{1/p}}{\alpha '}\cdot \frac {\alpha '}{\alpha }=\frac {b^{1/p}}{\alpha },\end{equation*}

and $(\sigma _b-1)x^{1/p}=1$ , as desired.

(2) The proof is analogous to that of (1).

(3) Suppose we are given an isomorphism of Galois $U_3$ -algebras between $(F_{a,b})_x$ and $(F_{a,b})_y$ . Let $t\in (F_{a,b})_x$ be the image of $y^{1/p}$ under the isomorphism and set

\begin{equation*}w':= x^{1/p} t \in (F_{a,b})_x.\end{equation*}

Set $y':= t^p$ . We have $(\tau -1)w'=\zeta ^{-1}\cdot \zeta =1$ , and hence $w'\in F_{a,b}^\times$ . We have $(w')^p=xy'$ . Since $F_b$ coincides with the $\langle {\sigma _a,\tau }\rangle$ -invariant subalgebra of $(F_{a,b})_x$ and $(F_{a,b})_y$ , the isomorphism $(F_{a,b})_y\to (F_{a,b})_x$ restricts to an isomorphism of Galois $\mathbb Z/p\mathbb Z$ -algebras $F_b \to F_b$ . Since the automorphism group of $F_b$ as a Galois $(\mathbb Z/p\mathbb Z)$ -algebra is $\mathbb Z/p\mathbb Z$ , generated by $\sigma _b$ , this isomorphism $F_b \to F_b$ is equal to $\sigma _b^i$ for some integer $i\geq 0$ . Thus, $y'=\sigma _b^i(y)$ . Define

\begin{equation*}w:= w' a^{-i/p}\mathop {{\prod }}\limits _{j=0}^i\sigma _b^{{\kern1.1pt}j}(\beta )\in F_{a,b}^\times .\end{equation*}

We have $(1-\sigma _b)y=\beta ^p/a$ , and hence

\begin{align*}(1-\sigma _b^i)y= \biggl(\mathop {{\sum }}\limits _{j=0}^{i-1}\sigma _b^{{\kern1.1pt}j}(1-\sigma _b) \biggr)y= \biggl(\mathop {{\prod }}\limits _{j=0}^i\sigma _b^{{\kern1.1pt}j}(\beta ^p) \biggr)/a^i=w^p/(w')^p.\end{align*}

Therefore,

(3.5) \begin{align} w^p=(w')^p(1-\sigma _b^i)y=x\sigma _b^i(y)(1-\sigma _b^i)y=xy. \end{align}

We have $(\sigma _b-1)x^{1/p}=1$ and

\begin{equation*}(\sigma _a-1)(\sigma _b-1)t=(\sigma _a-1)(\sigma _b-1)y^{1/p}=(\sigma _a-1)(a^{1/p}/\beta )=\zeta .\end{equation*}

Therefore,

\begin{equation*}(\sigma _a-1)(\sigma _b-1)w'=(\sigma _a-1)(\sigma _b-1)t=\zeta .\end{equation*}

Since $(\sigma _a-1)(\sigma _b-1)a^{1/p}=1$ and $(\sigma _a-1)(\sigma _b-1)\beta =1$ , we conclude that

(3.6) \begin{align} (\sigma _a-1)(\sigma _b-1)w=(\sigma _a-1)(\sigma _b-1)w'=\zeta . \end{align}

Putting (3.5) and (3.6) together, we see that $w$ satisfies the conditions of (3). Conversely, suppose we are given $w'\in F_{a,b}^\times$ such that

\begin{equation*}xy=(w')^p,\quad (\sigma _a-1)(\sigma _b-1)w'=\zeta .\end{equation*}

Claim 3.3. There exists $w\in F_{a,b}^\times$ such that

\begin{equation*}xy=w^p,\quad (\sigma _a-1)w=\zeta ^{-i}\frac {b^{1/p}}{\alpha },\quad (\sigma _b-1)w=\zeta ^{-j}\frac {a^{1/p}}{\beta }, \end{equation*}

for some integers $i$ and $j$ .

Proof of Claim 3.3 . First, we find $\eta _a\in F_a^\times$ such that

(3.7) \begin{align} \eta _a^p=1, \quad (\sigma _a-1)(w'/\eta _a)=\zeta ^{-i}\frac {b^{1/p}}{\alpha }. \end{align}

We have

\begin{equation*} (\sigma _a - 1) (w')^p=(\sigma _a - 1) x=\frac {b}{\alpha ^p}. \end{equation*}

Let

\begin{equation*}\zeta _a:= (\sigma _a-1)w'\cdot \alpha \cdot b^{-1/p}\in F_{a,b}^\times .\end{equation*}

We have $\zeta _a^p=1$ . Moreover, $(\sigma _b-1)\zeta _a=\zeta \cdot 1\cdot \zeta ^{-1}=1$ , that is, $\zeta _a$ belongs to $F_a^\times$ . If $F_a$ is a field, this implies that $\zeta _a=\zeta ^{-i}$ for some integer $i$ , and (3.7) holds for $\eta _a=1$ .

Suppose that $F_a$ is not a field. Then $F_a\simeq F^p$ , where $\sigma _a$ acts on F^p by cyclically permuting the coordinates, that is,

\begin{equation*}\sigma _a(x_1,x_2,\ldots ,x_p)=(x_2,\ldots ,x_p, x_1).\end{equation*}

We have $\zeta _a=(\zeta _1,\ldots ,\zeta _p)$ in $F_a=F^p$ , where $\zeta _i\in F^\times$ is a $p$ th root of unity for all $i$ . We have $N_a(\zeta _a)=N_a(\alpha )/b=1$ , and so $\zeta _1\cdots \zeta _p=1$ . Inductively, define $\eta _1:= 1$ and $\eta _{i+1}:= \zeta _i\eta _i$ for all $i=1,\ldots ,p-1$ . Then

\begin{equation*}\eta _1/\eta _p=(\eta _1/\eta _2)\cdot (\eta _2/\eta _3)\cdots (\eta _{p-1}/\eta _p)=\zeta _1^{-1}\zeta _2^{-1}\cdots \zeta _{p-1}^{-1}=\zeta _p.\end{equation*}

Therefore, the element $\eta _a:= (\eta _1,\ldots ,\eta _p)\in F^p=F_a$ satisfies $\eta _a^p=1$ and

\begin{equation*}(\sigma _a-1)\eta _a=(\eta _2/\eta _1,\ldots ,\eta _p/\eta _{p-1},\eta _1/\eta _p)=(\zeta _1,\ldots ,\zeta _{p-1},\zeta _p)=\zeta _a.\end{equation*}

Thus,

\begin{equation*}\eta _a^p=1,\quad (\sigma _a-1)(w'/\eta _a)=(\sigma _a-1)w'\cdot \zeta _a^{-1}=\frac {b^{1/p}}{\alpha },\end{equation*}

All in all, independent of whether $F_a$ is a field or not, we have found $\eta _a$ satisfying (3.7).

Similarly, we construct $\eta _b\in F_b^\times$ such that

(3.8) \begin{align} \eta _b^p=1,\quad (\sigma _b-1)(w'/\eta _b)=\zeta ^{-j}\frac {a^{1/p}}{\beta }, \end{align}

for some integer $j$ . Set $w:= w'/(\eta _a\eta _b)\in F_{a,b}^\times$ . Putting together (3.7) and (3.8), we deduce that $w$ satisfies the conclusion of Claim 3.3.

Let $w\in F_{a,b}^\times$ be as in Claim 3.3. By Lemma 3.1(1), applied $i$ times, the Galois $U_3$ -algebra $(F_{a,b})_x$ is isomorphic to $(F_{a,b})_{a^ix}$ , where

\begin{equation*}(\sigma _a-1)(a^ix)^{1/p}=\frac {b^{1/p}}{\alpha },\quad (\sigma _b-1)(a^ix)^{1/p}=1.\end{equation*}

By Lemma 3.1(2), applied $j$ times, the Galois $U_3$ -algebra $(F_{a,b})_y$ is isomorphic to $(F_{a,b})_{b^{{\kern1.1pt}j}y}$ , where

\begin{equation*}(\sigma _a-1)(b^{{\kern1.1pt}j}y)^{1/p}=1,\quad (\sigma _b-1)(b^{{\kern1.1pt}j}y)^{1/p}=\frac {a^{1/p}}{\beta }.\end{equation*}

Thus, it suffices to construct an isomorphism of $U_3$ -algebras $(F_{a,b})_{a^ix}\simeq (F_{a,b})_{b^{{\kern1.1pt}j}y}$ . Let

\begin{equation*}\tilde {w}:= wa^{i/p}b^{{\kern1.1pt}j/p}\in F_{a,b}^\times ,\end{equation*}

so that

\begin{equation*}(\sigma _a-1)\tilde {w}=\frac {b^{1/p}}{\alpha },\quad (\sigma _b-1)\tilde {w}=\frac {a^{1/p}}{\beta }.\end{equation*}

Let $f\colon (F_{a,b})_{a^ix}\to (F_{a,b})_{b^{{\kern1.1pt}j}y}$ be the isomorphism of étale algebras which is the identity on $F_{a,b}$ and sends $(a^ix)^{1/p}$ to $\tilde {w}/(b^{{\kern1.1pt}j}y)^{1/p}$ . Note that $f$ is well defined because

\begin{equation*}(\tilde {w})^p=wa^ib^{{\kern1.1pt}j}=(a^ix)(b^{{\kern1.1pt}j}y).\end{equation*}

Moreover,

\begin{align*}(\sigma _a-1) (\tilde {w}/(b^{{\kern1.1pt}j}y)^{1/p})=\frac {b^{1/p}}{\alpha }=(\sigma _a-1)(a^ix)^{1/p}, \\[-24pt] \end{align*}

\begin{equation*}(\sigma _b-1) (\tilde {w}/(b^{{\kern1.1pt}j}y)^{1/p})=\frac {a^{1/p}}{\beta }\cdot \frac {\beta }{a^{1/p}}=1=(\sigma _b-1)(a^ix)^{1/p},\end{equation*}

and hence $f$ is $U_3$ -equivariant.

Proof. Let $H$ (respectively, $H'$ ) be the subgroup of $\overline {U}_4$ generated by $\sigma _c$ and $\tau _{bc}$ (respectively, $\sigma _a$ and $\tau _{ab}$ ), and let $S$ be the subgroup of $\overline {U}_4$ generated by $H$ and $H'$ . Note that $K^H$ is a Galois $U_3$ -algebra over $F$ such that $(K^H)^{Q_3}\simeq F_{a,b}$ as $(\mathbb Z/p\mathbb Z)^2$ -algebras and $K^S\simeq F_b$ as $(\mathbb Z/p\mathbb Z)$ -algebras. Thus, by Proposition 3.2(1), there exists $x\in F_a^\times$ such that $K^H\simeq (F_{a,b})_x$ as Galois $U_3$ -algebras. Similarly, by Proposition 3.2(2), there exists $x'\in F_c^\times$ such that $K^{H'}\simeq (F_{b,c})_{x'}$ as Galois $U_3$ -algebras. Therefore, $x$ satisfies (3.10) and $x'$ satisfies (3.12). We apply Lemma 2.1(2) to (2.14). We obtain the isomorphisms of $\overline {U}_4$ -algebras

\begin{equation*}K\simeq K^H\otimes _{K^S}K^{H'}\simeq (F_{a,b,c})_{x,x'},\end{equation*}

where $(F_{a,b,c})_{x,x'}$ is the $\overline {U}_4$ -algebra determined by (3.10) and (3.12).

Proof. Suppose that $K=(F_{a,b,c})_{x,x'}$ , with $\overline {U}_4$ -action determined by (3.10)–(3.13). Let $L$ be a Galois $U_4$ -algebra over $F$ such that $L^{Z_4}=K$ , and let $y\in K^\times$ be such that $L=K_y$ .

We have $\operatorname {Gal}(L/F_{a,b,c})=Q_4=\langle {\tau _{ab}, \tau _{bc},\rho }\rangle \simeq (\mathbb Z/p\mathbb Z)^3$ , and hence one may choose $y$ in $F_{a,b,c}^\times$ and such that

\begin{equation*} (\tau _{ab}-1)y^{1/p}=1,\quad (\tau _{bc}-1)y^{1/p}=1,\quad (\rho -1)y^{1/p}=\zeta ^{-1}. \end{equation*}

The element $\sigma _b$ commutes with $\tau _{ab}, \tau _{bc}$ and $\rho$ . Hence,

\begin{equation*} \tau _{ab}(\sigma _b-1)(y^{1/p})=(\sigma _b-1)\tau _{ab}(y^{1/p})=(\sigma _b-1)(y^{1/p}). \end{equation*}

Similarly,

\begin{equation*} \tau _{bc}(\sigma _b-1)(y^{1/p})=(\sigma _b-1)(y^{1/p}) \end{equation*}

and

\begin{equation*}\rho (\sigma _b-1)(y^{1/p})=(\sigma _b-1)(\zeta \cdot y^{1/p})=(\sigma _b-1)(y^{1/p}).\end{equation*}

It follows that $(\sigma _b-1)(y^{1/p})\in F_{a,b,c}^\times$ . By Hilbert’s Theorem 90, applied to $F_{a,b,c}/F_{a,c}$ , there is $q\in F_{a,b,c}^\times$ such that $(\sigma _b-1)(y^{1/p})=(\sigma _b-1)q$ . Replacing $y$ by $y/ q^p$ , we may assume that $\sigma _b(y^{1/p})=y^{1/p}$ . In particular, $y\in F_{a,c}^\times$ . We have

\begin{align*} \rho (\sigma _a-1) (y^{1/p})= &\ (\sigma _a -1)\rho (y^{1/p})=(\sigma _a -1)(\zeta ^{-1}\cdot y^{1/p})=(\sigma _a -1)(y^{1/p}), \\ \sigma _b(\sigma _a-1) (y^{1/p})= &\ (\sigma _a\sigma _b{\tau _{ab}}^{-1} -\sigma _b)(y^{1/p})=(\sigma _a-1) (y^{1/p}),\\ \tau _{ab}(\sigma _a-1) (y^{1/p})= &\ (\sigma _a -1)\tau _{ab}(y^{1/p})=(\sigma _a -1)(y^{1/p}),\\ \tau _{bc}(\sigma _a-1) (y^{1/p})= &\ (\rho ^{-1} \sigma _a -1)\tau _{bc}(y^{1/p})=( \sigma _a \rho ^{-1} -1)(y^{1/p})=\zeta \cdot (\sigma _a-1) (y^{1/p}). \end{align*}

By (3.12)–(3.13), analogous identities are satisfied by $(x')^{1/p}$ , that is,

\begin{equation*}(\rho -1)(x')^{1/p}=(\sigma _b-1)(x')^{1/p}=(\tau _{ab}-1)(x')^{1/p}=1,\quad (\tau _{bc}-1)(x')^{1/p}=\zeta .\end{equation*}

Therefore,

\begin{equation*} (\sigma _a-1) (y^{1/p})=\frac {(x')^{1/p}}{u'}, \end{equation*}

for some $u'\in F_{a,c}^\times$ . In particular, $x'= N_a(u')$ . A similar computation shows that

\begin{equation*} (\sigma _c-1) (y^{1/p})=\frac {x^{1/p}}{u}, \end{equation*}

for some $u\in F_{a,c}^\times$ . In particular, $x= N_c(u)$ . In addition,

\begin{equation*} \frac {b^{1/p}}{\alpha }=(\sigma _a-1) (x^{1/p})=(\sigma _a-1) [u\cdot (\sigma _c-1)(y^{1/p})], \end{equation*}

\begin{equation*} \frac {b^{1/p}}{\gamma }=(\sigma _c-1) ((x')^{1/p})=(\sigma _c-1) [u'\cdot (\sigma _a-1)(y^{1/p})]. \end{equation*}

Therefore,

\begin{equation*}\alpha \cdot (\sigma _a-1)u=\gamma \cdot (\sigma _c-1)u'.\end{equation*}

Conversely, suppose we are given $u,u'\in F_{a,c}^\times$ such that

\begin{equation*}\alpha \cdot (\sigma _a-1)u=\gamma \cdot (\sigma _c-1)u',\quad x=N_c(u),\quad x'=N_a(u').\end{equation*}

Then

\begin{equation*} (\sigma _a-1)x=(\sigma _a-1)N_c(u)=N_c(\sigma _a-1)u=N_c\Big(\frac {\gamma }{\alpha }\Big)=\frac {b}{\alpha ^p}, \end{equation*}

\begin{equation*} (\sigma _c-1)x'=(\sigma _c-1)N_a(u')=N_a(\sigma _c-1)u'=N_a\Big(\frac {\alpha }{\gamma }\Big)=\frac {b}{\gamma ^p}. \end{equation*}

We have

\begin{equation*} N_c \Big(\frac {x}{u^p}\Big)=\frac {N_c(x)}{N_c(u^p)}=\frac {x^p}{x^p}=1, \end{equation*}

\begin{equation*} N_a \Big(\frac {x'}{(u')^p}\Big)=\frac {N_a(x')}{N_a((u')^p)}=\frac {(x')^p}{(x')^p}=1, \end{equation*}

\begin{equation*} (\sigma _a-1)\Big(\frac {x}{u^p}\Big)=\frac {b}{\alpha ^p\cdot (\sigma _a-1)u^p}= \frac {b}{\gamma ^p\cdot (\sigma _c-1)(u')^p}=(\sigma _c-1)\Big(\frac {x'}{(u')^p}\Big). \end{equation*}

By Hilbert’s Theorem 90 applied to $F_{a,c}/F$ , there is $y\in F_{a,c}^\times$ such that

\begin{equation*} (\sigma _a-1)y=\frac {x'}{(u')^p} \quad \text {and}\quad (\sigma _c-1)y=\frac {x}{u^p}. \end{equation*}

We consider the étale $F$ -algebra $L:= K_y$ and make it into a Galois $U_4$ -algebra such that $L^{Z_4}=K$ . It suffices to describe the $U_4$ -action on $y^{1/p}$ . We set

\begin{equation*} (\sigma _a-1)(y^{1/p})=\frac {(x')^{1/p}}{u'},\quad (\sigma _b-1)(y^{1/p})=1,\quad (\sigma _c-1)(y^{1/p})=\frac {x^{1/p}}{u}. \end{equation*}

One can check that this definition is compatible with relations (2.4)–(2.7), and hence that it makes $L$ into a Galois $U_4$ -algebra such that $L^{Z_4}=K$ .

Proof. It is clear that (3) implies (2) and that (2) implies (1). We now prove that (1) implies (3). The first part of the proof is a reduction to the case when $\operatorname {char}(F)\neq p$ and $F$ contains a primitive $p$ th root of unity, and it follows [Reference Mináč and TânMT16, Proposition 4.14].

Consider the short exact sequence

(3.14) \begin{align} 1\to Q_4\to U_4\to (\mathbb Z/p\mathbb Z)^3\to 1, \end{align}

where the map $U_4\to (\mathbb Z/p\mathbb Z)^3$ comes from (2.13). Recall from the paragraph preceding Proposition 3.5 that the group $Q_4$ is abelian. Therefore, the group homomorphism $\chi :=(\chi _1,\chi _2,\chi _3)\colon \Gamma _F \to (\mathbb Z/p\mathbb Z)^3$ induces a pullback map

\begin{equation*}H^2((\mathbb Z/p\mathbb Z)^3, Q_4)\to H^2(F, Q_4).\end{equation*}

We let $A\in H^2(F, Q_4)$ be the image of the class of (3.14) under this map. By Theorem 2.4, for every finite extension $F'/F$ the Massey product $\langle {\chi _1,\chi _2,\chi _3}\rangle$ vanishes over $F'$ if and only if the restriction of $\chi$ to $\Gamma _{F'}$ lifts to $U_4$ , and this happens if and only if $A$ restricts to zero in $H^2(F', Q_4)$ .

When $\operatorname {char}(F)=p$ , we have $\operatorname {cd}_p(F)\leq 1$ by [Reference Serre and IonSer97, § 2.2, Proposition 3]. Therefore, $H^2(F,Q_4)=0$ and hence $A=0$ . Thus, (1) implies (3) when $\operatorname {char}(F)=p$ .

Suppose that $\operatorname {char}(F)\neq p$ . There exists an extension $F'/F$ of prime-to- $p$ degree such that $F'$ contains a primitive $p$ th root of $1$ . If (1) implies (3) for $F'$ , then $A$ restricts to zero in $H^2(F', Q_4)$ . By a restriction-corestriction argument, we deduce that $A$ vanishes, that is, (1) implies (3) for $F$ . Thus, we may assume that $F$ contains a primitive $p$ th root of unity $\zeta$ .

Let $a,b,c\in F^\times$ be such that $\chi _a=\chi _1$ , $\chi _b=\chi _2$ and $\chi _c=\chi _3$ in $H^1(F,\mathbb Z/p\mathbb Z)$ . Since $(a,b)=(b,c)=0$ in $\operatorname {Br}(F)$ , there exists $\alpha \in F_a^\times$ and $\gamma \in F_c^\times$ such that $N_a(\alpha )=N_c(\gamma )=b$ . Since $N_{ac}(\gamma /\alpha )=N_c(\gamma )/N_a(\alpha )=1$ in $F_{ac}^\times$ , by Hilbert’s Theorem 90 there exists $t\in F_{a,c}^\times$ such that $\gamma /\alpha = (\sigma _a\sigma _c - 1)t$ . Define $u,u'\in F_{a,c}^\times$ by $u:= \sigma _c(t)$ and $u':= t^{-1}$ . Then

\begin{equation*}\alpha \cdot (\sigma _a-1)u=\alpha \cdot (\sigma _a\sigma _c-\sigma _c)t= \alpha \cdot (\sigma _a\sigma _c-1)t\cdot (\sigma _c-1)t^{-1}=\gamma \cdot (\sigma _c-1)u'.\end{equation*}

Let $x:= N_c(u)\in F_a^\times$ and $x':= N_a(u')\in F_c^\times$ . Since $\sigma _a\sigma _c=\sigma _c\sigma _a$ on $F_{a,c}^\times$ ,

\begin{equation*}(\sigma _a-1)x=N_c((\sigma _a-1)u)=N_c((\sigma _c-1)u'\cdot (\gamma /\alpha ))=N_c(\gamma )/N_c(\alpha )=b/\alpha ^p.\end{equation*}

Similarly, $(\sigma _c-1)x'=b/\gamma ^p$ . Therefore, $x,x'$ satisfy (3.9). Let $K:= (F_{a,b,c})_{x,x'}$ be the Galois $\overline {U}_4$ -algebra over $F$ , with the $\overline {U}_4$ -action given by (3.10)–(3.13). By Proposition 3.5, there exists a Galois $U_4$ -algebra $L$ over $F$ such that $L^{Z_4}\simeq (F_{a,b,c})_{x,x'}$ as $\overline {U}_4$ -algebras. In particular, $L^{Q_4}\simeq F_{a,b,c}$ as $(\mathbb Z/p\mathbb Z)^3$ -algebras. By Corollary 2.5, we conclude that $\langle {a,b,c}\rangle$ vanishes.

Proof. Denote by $U_4^+$ and $U_4^-$ the top-left and bottom-right $4\times 4$ corners of $U_5$ , respectively, and let $S:= U_4^+\cap U_4^-$ be the middle subgroup $U_3$ . Let $Q_4^+$ and $Q_4^-$ be the kernels of the maps $U_4^+\to (\mathbb Z/p\mathbb Z)^3$ and $U_4^-\to (\mathbb Z/p\mathbb Z)^3$ , respectively, and let $P_4^+$ and $P_4^-$ be the kernels of the maps $U_4^+\to U_3$ and $U_4^-\to U_3$ , respectively.

Suppose $\langle a,b,c,d \rangle$ is defined. By Corollary 2.5, there exists a $\overline {U}_5$ -algebra $L$ such that $L^{\overline {Q}_5}\simeq F_{a,b,c,d}$ as $(\mathbb Z/p\mathbb Z)^4$ -algebras. Using Lemma 2.2, we fix $\alpha \in F_a^\times$ and $\gamma \in F_c^\times$ such that $N_a(\alpha )=b$ and $N_c(\gamma )=b$ . By Proposition 3.5, there exist $u,u'\in F_{a,c}^\times$ such that, letting $x':= N_c(u')$ and $x:= N_a(u)$ , the $\overline {U}_4^+$ -algebra $K_1$ induced by $L$ is isomorphic to the $\overline {U}_4^+$ -algebra $(F_{a,b,c})_{x',x}$ , where $\overline {U}_4^+$ acts via (3.10)–(3.13), and where the roles of $x$ and $x'$ have been switched.

Similarly, there exist $v,v'\in F_{b,d}^\times$ such that, letting $z:= N_d(v)$ and $z':= N_b(v')$ , the $\overline {U}^-_4$ -algebra $K_2$ induced by $L$ is isomorphic to $(F_{b,c,d})_{z,z'}$ . Since the $U_3$ -algebras $(K_1)^{P_4^+}$ and $(K_2)^{P_4^-}$ are equal, by Proposition 3.2(3) there exists $w\in F_{b,c}^\times$ such that

\begin{equation*} N_a(u)\cdot N_d(v) = xz = w^p,\quad (\sigma _b -1)(\sigma _c -1)w = \zeta . \end{equation*}

Conversely, let $u\in F_{a,c}^\times$ , $v\in F_{b,d}^\times$ , and $w\in F_{b,c}^\times$ be such that

\begin{equation*} N_a(u)\cdot N_d(v) = w^p,\quad (\sigma _b -1)(\sigma _c -1)w = \zeta .\end{equation*}

By Lemma 2.2, there exist $\alpha \in F_a^\times$ and $\delta \in F_d^\times$ such that $N_a(\alpha )=b$ and $N_d(\delta )=c$ . We may write

\begin{equation*} (\sigma _b - 1)w = \frac {c^{1/p}}{\beta },\quad (\sigma _c - 1)w = \frac {b^{1/p}}{\gamma }, \end{equation*}

for some $\beta \in F_b^\times$ and $\gamma \in F_c^\times$ . We have

\begin{equation*} N_a((\sigma _c - 1)u\cdot (\gamma /\alpha )) = (\sigma _c - 1)N_a(u)\cdot N_a(\gamma /\alpha )= (\sigma _c - 1)w^p \cdot (\gamma ^p/b) = 1. \end{equation*}

By Hilbert’s Theorem 90, there is $u'\in F_{a,c}^\times$ such that

\begin{equation*} \alpha \cdot (\sigma _a - 1)u' = \gamma \cdot (\sigma _c - 1)u. \end{equation*}

By Proposition 3.5, we obtain a Galois $U_4^+$ -algebra $K_1$ over $F$ with the property that $(K_1)^{Q_4^+}\simeq F_{a,b,c}$ as $(\mathbb Z/p\mathbb Z)^3$ -algebras. Similarly, we get a Galois $U_4^-$ -algebra over $F$ such that $(K_2)^{Q_4^-}\simeq F_{b,c,d}$ as $(\mathbb Z/p\mathbb Z)^3$ -algebras. Since $N_a(u)\cdot N_d(v) = w^p$ and $(\sigma_b-1)(\sigma_c-1)w=\zeta$ , by Proposition 3.2(3) the $U_3$ -algebras $(K_1)^{P_4^+}$ and $(K_2)^{P_4^-}$ are isomorphic. Now Lemma 2.1 applied to the cartesian square (2.14) for $n=4$ yields a $\overline {U}_5$ -Galois algebra $L$ such that $L^{Q_5}\simeq F_{a,b,c,d}$ as $(\mathbb Z/p\mathbb Z)^4$ -algebras. By Corollary 2.5, this implies that $\langle {a,b,c,d}\rangle$ is defined.

Proof. We have $(\sigma _b-1)(\sigma _c-1)(w_bw_c)=(\sigma _b-1)w_c=1$ for all $w_b\in F_b^\times$ and $w_c\in F_c^\times$ . Conversely, if $w\in F_{b,c}^\times$ satisfies $(\sigma _b-1)(\sigma _c-1)w=1$ , then $(\sigma _c-1)w\in F_c^\times$ and $N_c((\sigma _c-1)w)=1$ , and hence by Hilbert’s Theorem 90 there exists $w_c\in F_c^\times$ such that $(\sigma _c-1)w_c=(\sigma _c-1)w$ . Letting $w_b:= w/w_c\in F_{b,c}^\times$ , we have

\begin{equation*}(\sigma _c-1)w_b=(\sigma _c-1)(w/w_c)=1,\end{equation*}

that is, $w_b\in F_b^{\times }$ .

Proof. By Proposition 3.7, there exist $u_0\in F_{a,c}^\times$ , $v_0\in F_{b,d}^\times$ and $w_0\in F_{b,c}^\times$ such that

\begin{equation*}N_a(u_0) \cdot N_d(v_0) =w_0^p,\quad (\sigma _b-1)(\sigma _c-1)w_0=\zeta .\end{equation*}

We have $(\sigma _b-1)(\sigma _c-1)(w_0/w)=1$ . By Lemma 3.8, this implies that $w_0=w w_b w_c$ , where $w_b\in F_b^\times$ and $w_c\in F_c^\times$ . If we define $u=u_0 w_c$ and $v=v_0 w_b$ , then

\begin{equation}N_a(u)N_d(v)=N_a(u_0)N_a(w_c)N_d(v_0)N_d(w_b)=w_0^pw_c^pw_b^p=w^p.\end{equation}