2.1 The groups  $\mathbb{U}_{3}(\mathbb{F}_{2})$ and $\mathbb{U}_{5}(\mathbb{F}_{2})$

We shall need special notation for these two groups. First note that we define the dihedral group of order  $8$ to be  $\mathbb{U}_{3}(\mathbb{F}_{2})$ , and we may write  $D_{4}=\mathbb{U}_{3}(\mathbb{F}_{2})$ . An element  $g\in D_{4}$ is a matrix of the form

$$\begin{eqnarray}g=\left(\begin{array}{@{}ccc@{}}1 & s_{1}(g) & t(g)\\ 0 & 1 & s_{2}(g)\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Thus  $D_{4}$ is equipped with maps  $s_{1},s_{2},t:D_{4}\rightarrow \mathbb{F}_{2}$ . (The letter  $t$ is for ‘top’.) Note that the first two are group homomorphisms, but  $t$ is not. Our favourite generators are the involutions  $\unicode[STIX]{x1D70E}_{1}$ and  $\unicode[STIX]{x1D70E}_{2}$ , with  $s_{i}(\unicode[STIX]{x1D70E}_{i})=1$ and  $s_{j}(\unicode[STIX]{x1D70E}_{i})=t(\unicode[STIX]{x1D70E}_{i})=0$ for  $j\neq i$ .

Similarly, an element  $g\in \mathbb{U}_{5}(\mathbb{F}_{2})$ will be written

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & s_{1}(g) & t_{1}(g) & u_{1}(g) & z(g)\\ 0 & 1 & s_{2}(g) & u_{3}(g) & u_{2}(g)\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & s_{4}(g)\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

This endows  $\mathbb{U}_{5}(\mathbb{F}_{2})$ with maps  $s_{1},\ldots ,z:\mathbb{U}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ , and  $s_{i}$ is a group homomorphism for $1\leqslant i\leqslant 4$ .

More generally the group  $\mathbb{U}_{n}(\mathbb{F}_{2})$ has homomorphisms  $s_{i}:\mathbb{U}_{n}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ for  $1\leqslant i\leqslant n-1$ , already mentioned in the Introduction, obtained by looking at the entries on what we call the near-diagonal. If we define elements  $\unicode[STIX]{x1D70E}_{i}$ by requiring  $s_{i}(\unicode[STIX]{x1D70E}_{i})=1$ while all the other entries of  $\unicode[STIX]{x1D70E}_{i}$ above the diagonal are  $0$ , then each  $\unicode[STIX]{x1D70E}_{i}$ is an involution, and these generate  $\mathbb{U}_{n}(\mathbb{F}_{2})$ .

We note that  $\mathbb{U}_{n}(\mathbb{F}_{2})$ has an automorphism which exchanges  $\unicode[STIX]{x1D70E}_{i}$ with  $\unicode[STIX]{x1D70E}_{n-i}$ . Most of our considerations respect this symmetry, and this motivates the notation above for  $\mathbb{U}_{5}(\mathbb{F}_{2})$ . (The automorphism is given by ‘the transpose but along the other diagonal’, followed by  $g\mapsto g^{-1}$ .)

2.2 Around the group  $D_{4}$

We write  $s=(s_{1},s_{2}):D_{4}\rightarrow C_{2}\times C_{2}$ , where we have identified  $\mathbb{F}_{2}$ with the cyclic group of order  $2$ in multiplicative notation, written  $C_{2}$ . There is an exact sequence

$$\begin{eqnarray}1\longrightarrow \mathbb{F}_{2}\longrightarrow D_{4}\stackrel{s}{\longrightarrow }C_{2}\times C_{2}\longrightarrow 1,\end{eqnarray}$$

the kernel of  $s$ being generated by  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ (which is the element  $g$ with  $t(g)=1$ and  $s_{i}(g)=0$ , $i=1,2$ ).

The quotient group  $C_{2}\times C_{2}$ is generated by the images of  $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}$ , written  $\overline{\unicode[STIX]{x1D70E}}_{1},\overline{\unicode[STIX]{x1D70E}}_{2}$ . The cohomology group  $\operatorname{H}^{1}(C_{2}^{2},\mathbb{F}_{2})=\operatorname{Hom}(C_{2}^{2},\mathbb{F}_{2})$ is endowed with the dual basis  $\overline{s}_{1},\overline{s}_{2}$ . The next lemma is very well known.

We introduce the two elementary abelian subgroups  $E_{1},E_{2}$ , where  $E_{1}$ is the kernel of  $s_{2}$ and  $E_{2}$ is the kernel of  $s_{1}$ (the switch is justified by the next lemma). A very useful observation is that  $t$ , when restricted to either of these, is a group homomorphism.

Lemma 2.2. The corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{1}(E_{i},\mathbb{F}_{2})\longrightarrow \operatorname{H}^{1}(D_{4},\mathbb{F}_{2})\end{eqnarray}$$

carries  $t|_{E_{i}}$ to  $s_{i}$ , for  $i=1,2$ . More generally if  $\unicode[STIX]{x1D6E4}$ is any profinite group with a continuous homomorphism  $\unicode[STIX]{x1D711}:\unicode[STIX]{x1D6E4}\rightarrow D_{4}$ , and if  $H=\unicode[STIX]{x1D711}^{-1}(E_{i})$ is assumed to have index  $2$ in  $\unicode[STIX]{x1D6E4}$ , then the corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{1}(H,\mathbb{F}_{2})\longrightarrow \operatorname{H}^{1}(\unicode[STIX]{x1D6E4},\mathbb{F}_{2})\end{eqnarray}$$

carries  $t\circ \unicode[STIX]{x1D711}$ to  $s_{i}\circ \unicode[STIX]{x1D711}$ , for  $i=1,2$ .

Proof. We recall some properties of the corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{i}(N,M)\longrightarrow \operatorname{H}^{i}(G,M),\end{eqnarray}$$

where  $N$ is a subgroup of finite index of the profinite group  $G$ , and  $M$ is a  $G$ -module. In fact, we only need to consider the case when  $M$ has a trivial action, $N$ is closed of index  $2$ (and thus is normal) in  $G$ , and  $i=1$ , so that

$$\begin{eqnarray}\operatorname{cores}:\operatorname{Hom}(N,M)\longrightarrow \operatorname{Hom}(G,M).\end{eqnarray}$$

Here if  $f:N\rightarrow M$ , then the map  $\operatorname{cores}(f):G\rightarrow M$ is characterized as follows. Pick  $\unicode[STIX]{x1D70F}\in G\smallsetminus N$ . Then (i) $\operatorname{cores}(f)(n)=f(n)+f(\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F})$ for $n\in N$ , and (ii) $\operatorname{cores}(f)(\unicode[STIX]{x1D70F})=f(\unicode[STIX]{x1D70F}^{2})$ . This follows from the material in [Reference TateTat76, § 2], for example.

Let us use this for  $N=E_{1}$ and  $G=D_{4}$ . If

$$\begin{eqnarray}n=\left(\begin{array}{@{}ccc@{}}1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right)\quad \text{and}\quad \unicode[STIX]{x1D70F}=\left(\begin{array}{@{}ccc@{}}1 & c & d\\ 0 & 1 & 1\\ 0 & 0 & 1\end{array}\right),\end{eqnarray}$$

then

$$\begin{eqnarray}\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F}=\left(\begin{array}{@{}ccc@{}}1 & a & a+b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Moreover

$$\begin{eqnarray}\unicode[STIX]{x1D70F}^{2}=\left(\begin{array}{@{}ccc@{}}1 & 0 & c\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Thus  $t(n)+t(\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F})=b+a+b=a=s_{1}(n)$ , and $t(\unicode[STIX]{x1D70F}^{2})=c=s_{1}(\unicode[STIX]{x1D70F})$ . Hence the first statement of the lemma for  $i=1$ . The other cases are treated similarly.◻

2.3 $D_{4}$ -extensions of fields

We proceed to apply the above observations in a Galois-theoretic context, but a couple of comments are in order. First, the group  $D_{4}$ has an automorphism exchanging  $\unicode[STIX]{x1D70E}_{1}$ and  $\unicode[STIX]{x1D70E}_{2}$ , but the proposition below is not ‘symmetric’ in this way; it involves the subgroup  $E_{2}$ and not  $E_{1}$ , for example (so that one could get a new proposition by exchanging the roles of various players). Second, when asked for a basis for  $E_{2}$ , the reader would probably offer  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2}$ ; however, later considerations with  $\mathbb{U}_{5}(\mathbb{F}_{2})$ compel us to work with  $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2}$ instead (specifically, we want Lemma 2.5 to have the simple form given below). This is reflected in the proposition below, since the dual basis of  $\operatorname{H}^{1}(E_{2},\mathbb{F}_{2})$ is  $t,s_{2}+t$ (or, in more complete notation, $t|_{E_{2}},s_{2}|_{E_{2}}+t|_{E_{2}}$ ).

Again, this is essentially known, but we need the precise version given here. Note that it is necessary to deal with the cases when either  $a$ or  $b$ is a square, and that the proof below gives more concrete information in some situations. Also note that, as promised, the elements  $t,s_{2}+t$ , related to  $E_{2}$ , make an uncanny appearance.

Proof. We can combine  $\unicode[STIX]{x1D712}_{a}$ and  $\unicode[STIX]{x1D712}_{b}$ into a homomorphism  $G_{F}\rightarrow C_{2}\times C_{2}$ . The obstruction to lifting it to  $D_{4}$ is the cohomology class of the extension, so that Lemma 2.1 gives immediately the equivalence of (1) and (2).

We first conduct the rest of the proof under the following assumption.

Assumption.

Assume for the moment that  $a$ is not a square in  $F$ .

Suppose (2) holds. Note that  $\unicode[STIX]{x1D711}^{-1}(E_{2})=G_{F[\sqrt{a}]}$ . Let  $B^{\prime }=x^{\prime }+y^{\prime }\sqrt{a}\in F[\sqrt{a}]$ be such that $\unicode[STIX]{x1D712}_{B^{\prime }}=t\circ \unicode[STIX]{x1D711}|_{G_{F[\sqrt{a}]}}$ . Since  $a$ is not a square in  $F$ , the subgroup  $G_{F[\sqrt{a}]}$ has index  $2$ in  $G_{F}$ . Lemma 2.2 shows then that  $\operatorname{cores}(\unicode[STIX]{x1D712}_{B^{\prime }})=s_{2}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{b}$ . Now recall that under the identifications of  $\operatorname{H}^{1}(F,\mathbb{F}_{2})$ with  $F^{\times }\!/F^{\times 2}$ and of  $\operatorname{H}^{1}(F[\sqrt{a}],\mathbb{F}_{2})$ with  $F[\sqrt{a}]^{\times }\!/F[\sqrt{a}]^{\times 2}$ , the corestriction becomes the usual norm  $N_{F[\sqrt{a}]/F}$ . It follows that

$$\begin{eqnarray}N_{F[\sqrt{a}]/F}(B^{\prime })=(x^{\prime })^{2}-a(y^{\prime })^{2}=b\text{ mod squares}.\end{eqnarray}$$

This gives (3), clearly, but we need to modify  $B^{\prime }$ to get the consistency statement. And indeed, we put  $B=bB^{\prime }$ , so that  $B^{\prime }=bB$ modulo squares, and the result follows.

Next, we must prove that (3) implies (2), or equivalently (1). We may as well suppose that  $a$ and  $b$ are both not squares, for (1) holds trivially otherwise. The assumption is that  $b$ is a norm from  $F[\sqrt{a}]$ , or in more cohomological terms, that  $\unicode[STIX]{x1D712}_{b}$ is the corestriction of an element from the subgroup  $G_{F[\sqrt{a}]}$ . That the cup product  $\unicode[STIX]{x1D712}_{a}\unicode[STIX]{x1D712}_{b}=0$ then follows from the Arason exact sequence [Reference ArasonAra75].

However, to prove the claimed consistency, a more explicit argument is needed. Assume  $b$ is not a square. The element  $B=x+y\sqrt{a}$ , where  $x,y$ are as in (3), is fixed up to squares by  $\operatorname{Gal}(F[\sqrt{a},\sqrt{b}]/F)$ , as is readily checked. It follows that  $K=F[\sqrt{a},\sqrt{b},\sqrt{B}]$ is Galois over  $F$ (by equivariant Kummer theory, if you will).

We distinguish two cases, and assume first that  $a$ and  $b$ are not equal modulo squares. We now introduce elements  $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}\in \operatorname{Gal}(K/F)$ which are dual to  $\sqrt{a},\sqrt{b}$ in the obvious sense. Direct computation shows that  $\unicode[STIX]{x1D70E}_{1}^{2}=\unicode[STIX]{x1D70E}_{2}^{2}=1$ , and that  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}](\sqrt{B})=-\sqrt{B}$ , so that  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]\neq 1$ . From this one draws readily that  $\operatorname{Gal}(K/F)\cong D_{4}$ and (again!) that (2) holds. Also, one computes that  $\unicode[STIX]{x1D70E}_{2}(\sqrt{B})=\pm \sqrt{B}$ . We want to ensure that  $\unicode[STIX]{x1D70E}_{2}(\sqrt{B})=-\sqrt{B}$ , and to achieve this we replace  $\unicode[STIX]{x1D70E}_{2}$ by  $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ if needed. Having done this, the elements  $\sqrt{B}$ , $\sqrt{bB}$ are dual to  $\unicode[STIX]{x1D70E}_{2}$ , $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ , and one checks that the corresponding map  $\unicode[STIX]{x1D711}:G_{F}\rightarrow D_{4}$ has precisely the required consistency.

If  $a=b$ modulo squares, one sees that  $\operatorname{Gal}(K/F)$ has order 4, so is abelian, and if  $\unicode[STIX]{x1D70E}$ is the non-trivial element of  $\operatorname{Gal}(F[\sqrt{a}]/F)$ extended to  $\operatorname{Gal}(K/F)$ , another direct calculation shows that  $\unicode[STIX]{x1D70E}$ does not have order  $2$ . So  $\operatorname{Gal}(K/F)\cong C_{4}$ can be identified with the subgroup of  $D_{4}$ generated by  $\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{2}$ , and (2) follows. Consistency is automatic.

Finally, if  $b$ is a square in  $F$ , we use the same extension  $K=F[\sqrt{a},\sqrt{B}]$ of  $F$ , but compute that  $\operatorname{Gal}(K/F)\cong C_{2}^{2}$ . We identify this group with  $E_{1}$ appropriately, yielding a consistent  $\unicode[STIX]{x1D711}$ .

The case when  $a$ is a square.

In this situation (1) holds trivially, and thus (2) also holds. As for (3), if  $a=u^{2}$ then put

$$\begin{eqnarray}x_{0}=\frac{b+1}{2}\quad \text{and}\quad y_{0}=\frac{1-b}{2u}\end{eqnarray}$$

and compute that  $x_{0}^{2}-ay_{0}^{2}=b$ .

Let us see how we can adjust  $\unicode[STIX]{x1D711}$ from  $x,y,z$ . Put  $B=x+y\sqrt{a}\in F$ , and consider the characters  $\unicode[STIX]{x1D712}_{bB}$ and  $\unicode[STIX]{x1D712}_{b}$ , together defining a homomorphism  $G_{F}\rightarrow C_{2}\times C_{2}$ . Identifying Klein’s group with  $E_{2}$ sitting in  $D_{4}$ appropriately, we obtain  $\unicode[STIX]{x1D711}:G_{F}\rightarrow D_{4}$ satisfying our requirements.

Our very last step is to see how one can adjust  $x,y,z$ from  $\unicode[STIX]{x1D711}$ . First put  $B_{0}=x_{0}+y_{0}\sqrt{a}\in F$ where  $x_{0}$ and  $y_{0}$ are as above, which is a non-zero element since  $B_{0}(x-y\sqrt{a})=b\neq 0$ . For  $f\in F$ , put  $x=(f/B_{0})x_{0}$ and  $y=(f/B_{0})y_{0}$ , so that  $x^{2}-ay^{2}=bz^{2}$ for some  $z\in F^{\times }$ , while  $B=x+y\sqrt{a}=f$ , an arbitrary element of  $F$ . Of course  $\unicode[STIX]{x1D711}$ lands in the subgroup  $E_{2}$ , so we only need to pick  $f$ so that  $\unicode[STIX]{x1D712}_{f}=(s_{2}+t)\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{B}$ ; we have then (2) and (3) simultaneously and consistently.◻

2.4 The group  $\mathbb{U}_{5}(\mathbb{F}_{2})$ and its subquotients

Let  $S$ (for ‘square’) be the subgroup of matrices of the form

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & y_{1} & y_{2}\\ 0 & 1 & 0 & y_{3} & y_{4}\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Then  $S\cong C_{2}^{4}$ , and our favourite $\mathbb{F}_{2}$ -basis, denoted $e_{1},e_{2},e_{3},e_{4}$ , will be given by

$$\begin{eqnarray}\displaystyle & \displaystyle e_{1}=e=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right),\quad e_{2}=\unicode[STIX]{x1D70E}_{1}e\unicode[STIX]{x1D70E}_{1}^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle e_{3}=\unicode[STIX]{x1D70E}_{4}e\unicode[STIX]{x1D70E}_{4}^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right),\quad e_{4}=(\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4})e(\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4})^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right). & \displaystyle \nonumber\end{eqnarray}$$

The centralizer of  $S$ in  $\mathbb{U}_{5}(\mathbb{F}_{2})$ , which we denote by  $\mathscr{C}(S)$ , is easily seen to be composed of the elements  $g$ for which  $s_{1}(g)=s_{4}(g)=0$ , that is  $\mathscr{C}(S)=\ker s_{1}\cap \ker s_{4}$ . In particular  $\unicode[STIX]{x1D70E}_{2}$ and  $\unicode[STIX]{x1D70E}_{3}$ centralize  $S$ , and from the formulae above we see that  $S$ is normal in  $\mathbb{U}_{5}(\mathbb{F}_{2})$ . We shall write  $G=\mathbb{U}_{5}(\mathbb{F}_{2})/S$ , which we identify with  $D_{4}\times D_{4}$ , as we visibly may. The image of  $\mathscr{C}(S)$ in  $G$ , that is  $\mathscr{C}(S)/S$ , will be denoted by  $N$ , a normal subgroup of  $G$ .

One has  $G/N=\mathbb{U}_{5}(\mathbb{F}_{2})/\mathscr{C}(S)=\langle \unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{4}\rangle \cong C_{2}^{2}$ (we shall often write  $\unicode[STIX]{x1D70E}_{i}$ for the image of this element in various quotients, whenever no confusion can arise). The next observation is now clear, but it is crucial.

The group  $N$ itself also has a simple structure: one has  $N\cong C_{2}^{4}$ , a basis being 

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2},\unicode[STIX]{x1D70E}_{3},\unicode[STIX]{x1D70E}_{3}[\unicode[STIX]{x1D70E}_{4},\unicode[STIX]{x1D70E}_{3}].\end{eqnarray}$$

(A choice which respects the ambient ‘symmetry’ already alluded to.) The corresponding dual basis of $\operatorname{H}^{1}(N,\mathbb{F}_{2})$ will be denoted  $x_{1},x_{2},x_{3},x_{4}$ . To bridge the notation with that of the previous sections, we regard  $N$ as sitting in  $D_{4}\times D_{4}$ , which itself possesses six maps  $s_{1},s_{2},t_{1},s_{3},s_{4},t_{2}$ to  $\mathbb{F}_{2}$ , using names adapted from § 2.1. With this notation, one has  $x_{1}=t_{1}$ , $x_{2}=s_{2}+t_{1}$ , $x_{3}=s_{3}+t_{2}$ , and $x_{4}=t_{2}$ (where restrictions to  $N$ are implicit).

Next we introduce some subgroups of  $S$ , and use them to produce extensions of  $N$ . These extensions turn out to control the entire situation, as will be explained. So we let

$$\begin{eqnarray}S_{1}:=\langle e_{2},e_{3},e_{4}\rangle ,\quad S_{2}:=\langle e_{1},e_{3},e_{4}\rangle ,\quad S_{3}:=\langle e_{1},e_{2},e_{4}\rangle .\end{eqnarray}$$

(The group  $S_{4}$ which could be defined using the same logic will not play any role, as it happens. Also note that among these three, only  $S_{1}$ respects the ambient ‘symmetry’.)

Proof. An element of  $\mathscr{C}(S)$ has the form

$$\begin{eqnarray}g=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & u_{1}(g) & z(g)\\ 0 & 1 & s_{2}(g) & u_{3}(g) & u_{2}(g)\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Let us multiply two of these, say  $g$ and  $g^{\prime }$ , using the shorthand  $s_{2}=s_{2}(g)$ and  $s_{2}^{\prime }=s_{2}(g^{\prime })$ , and so on:

We shall use the set-theoretic section  $\sec :N\rightarrow \mathscr{C}(S)$ given by

$$\begin{eqnarray}\sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & 0 & 0\\ 0 & 1 & s_{2}(g) & 0 & 0\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Let us prove (1). Using the section  $N\rightarrow \mathscr{C}(S)/S_{1}$ induced by  $\sec$ , we end up with the bijection of sets $\unicode[STIX]{x1D6F7}:S/S_{1}\times N\rightarrow \mathscr{C}(S)/S_{1}$ given by

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x,g)=x\sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & x & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right)\times \sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & 0 & 0\\ 0 & 1 & s_{2}(g) & x & 0\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Here we have used (∗) to perform the calculation. A caveat: in these expressions, we have identified  $S/S_{1}$ with  $\mathbb{F}_{2}$ (this can be done uniquely!), so that  $x\in S/S_{1}$ can be seen as an entry ( $0$ or $1$ ) of a matrix. A second caveat is that the matrix displayed is understood modulo  $S_{1}$ only.

From the theory of group extensions, we have $\unicode[STIX]{x1D6F7}(x,g)\unicode[STIX]{x1D6F7}(y,g^{\prime })=\unicode[STIX]{x1D6F7}(x+y+c(g,g^{\prime }),gg^{\prime })$ , where the expression $c(g,g^{\prime })$ , is what we are after, that is, it is a two-cocycle representing the cohomology class of the extension under scrutiny. So we compute, from (∗), that

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x,g)\unicode[STIX]{x1D6F7}(y,g^{\prime })=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}+t_{1}^{\prime } & t_{1}s_{3}^{\prime } & t_{1}t_{2}^{\prime }\\ 0 & 1 & s_{2}+s_{2}^{\prime } & s_{2}s_{3}^{\prime }+x+y & s_{2}t_{2}^{\prime }\\ 0 & 0 & 1 & s_{3}+s_{3}^{\prime } & t_{2}+t_{2}^{\prime }\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Another useful computational remark is that, in $S$ identified with the additive group of  $2\times 2$ -matrices, we have

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}a & b\\ c & d\end{array}\right)=(a+b+c+d)e_{1}+(a+b)e_{2}+(b+d)e_{3}+be_{4}\equiv \left(\begin{array}{@{}cc@{}}0 & 0\\ a+b+c+d & 0\end{array}\right)\text{ mod }S_{1}.\end{eqnarray}$$

Thus the last matrix displayed, viewed in  $\mathscr{C}(S)/S_{1}$ , is also

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}+t_{1}^{\prime } & 0 & 0\\ 0 & 1 & s_{2}+s_{2}^{\prime } & x+y+s_{2}s_{3}^{\prime }+t_{1}s_{3}^{\prime }+s_{2}t_{2}^{\prime }+t_{1}t_{2}^{\prime } & 0\\ 0 & 0 & 1 & s_{3}+s_{3}^{\prime } & t_{2}+t_{2}^{\prime }\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

We conclude that, as predicted, $c(g,g^{\prime })=s_{2}(g)s_{3}(g^{\prime })+t_{1}(g)s_{3}(g^{\prime })+s_{2}(g)t_{2}(g^{\prime })+t_{1}(g)t_{2}(g^{\prime })=x_{2}(g)x_{3}(g^{\prime })$ , and  $c$ is indeed the cup-product of  $x_{2}$ and  $x_{3}$ .

The proofs of (2) and (3) are similar. ◻

3.1 Proving that the four cup-products vanish

Note that the mere existence of  $\widetilde{B}$ and  $\widetilde{C}$ implies also that  $(a,b)_{F}=0$ and  $(c,d)_{F}=0$ , by Proposition 2.3.

Proof. First we consider the composition

$$\begin{eqnarray}\overline{\overline{\unicode[STIX]{x1D711}}}:G_{F}\longrightarrow \overline{U}_{5}(\mathbb{F}_{2})\longrightarrow \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})/S=G=D_{4}\times D_{4}.\end{eqnarray}$$

Projecting further onto the left factor, we make a first use of Proposition 2.3. We draw the existence of  $x,y,z\in F$ satisfying $x^{2}-ay^{2}=bz^{2}$ , that is  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=bz^{2}$ , with  $\widetilde{B}=x+yX\in F_{a}$ . If  $B=x+y\sqrt{a}$ is the corresponding element, then the proposition says that we can arrange to have  $\unicode[STIX]{x1D712}_{bB}=t_{1}\circ \overline{\unicode[STIX]{x1D711}}=x_{1}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{B}=(s_{2}+t_{1})\circ \overline{\unicode[STIX]{x1D711}}=x_{2}\circ \overline{\unicode[STIX]{x1D711}}$ . (We fully use the notation from § 2.4.)

Using the right factor, we find elements  $\widetilde{C}$ and  $C$ similarly, such that  $\unicode[STIX]{x1D712}_{C}=x_{3}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{cC}=x_{4}\circ \overline{\unicode[STIX]{x1D711}}$ . This also proves the first assertion.

We prove that the cup-products vanish as announced in the second assertion, starting with $(b,c)_{F}=0$ (which of course does not depend on the choices for  $B$ and  $C$ ). For this, consider the map

$$\begin{eqnarray}\mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow \mathbb{U}_{3}(\mathbb{F}_{2})=D_{4}\end{eqnarray}$$

which discards the top row and the rightmost column of an element of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ ; this factors through  $\overline{U}_{5}(\mathbb{F}_{2})$ . Postcomposing  $\overline{\unicode[STIX]{x1D711}}$ with this, we draw from Proposition 2.3 that  $(b,c)_{F}=0$ , as required.

Next we turn to the proof of  $(B,c)_{F[\sqrt{a}]}=0$ , the cup-product  $(b,C)_{F[\sqrt{d}]}$ being treated in a similar way. Define a group homomorphism  $\unicode[STIX]{x1D70B}:\ker (s_{1})\subset \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{U}_{3}(\mathbb{F}_{2})=D_{4}$ by

$$\begin{eqnarray}g\mapsto \left(\begin{array}{@{}ccc@{}}1 & t_{1}(g) & u_{1}(g)\\ 0 & 1 & s_{3}(g)\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

One checks that  $\unicode[STIX]{x1D70B}$ is well defined. Note that  $\unicode[STIX]{x1D70B}\circ \overline{\unicode[STIX]{x1D711}}:G_{F[\sqrt{a}]}\rightarrow D_{4}$ is a lift for  $(t_{1}\circ \overline{\unicode[STIX]{x1D711}},s_{3}\circ \overline{\unicode[STIX]{x1D711}}):G_{F[\sqrt{a}]}\rightarrow \mathbb{F}_{2}\times \mathbb{F}_{2}$ (using that  $G_{F[\sqrt{a}]}=\overline{\unicode[STIX]{x1D711}}^{-1}(\ker (s_{1}))$ ). Therefore, we see from Proposition 2.3 that the cup product of  $t_{1}\circ \overline{\unicode[STIX]{x1D711}}$ and  $s_{3}\circ \overline{\unicode[STIX]{x1D711}}$ is zero. This means, in alternative notation, that  $(Bb,c)_{F[\sqrt{a}]}=0$ , so  $(B,c)_{F[\sqrt{a}]}=0$ .

We now turn to the third assertion. Let  $\unicode[STIX]{x1D6FC}$ be the cohomology class of the extension

$$\begin{eqnarray}0\longrightarrow \mathbb{F}_{2}\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})\longrightarrow 1.\end{eqnarray}$$

Let us write down an explicit two-cocycle  $\unicode[STIX]{x1D6FE}$ representing  $\unicode[STIX]{x1D6FC}$ . First, recall the functions $s_{1},t_{1},\ldots :\mathbb{U}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ introduced in § 2.1. Multiplying two matrices  $g,h\in \mathbb{U}_{5}(\mathbb{F}_{2})$ , the top-right coefficient must be  $z(g)+z(h)+\unicode[STIX]{x1D6FE}(g,h)$ , and we deduce

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}(g,h)=s_{1}(g)u_{2}(h)+t_{1}(g)t_{2}(h)+u_{1}(g)s_{4}(h).\end{eqnarray}$$

To obtain a two-cocycle representing the pull-back  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ , we only need compose with  $\overline{\unicode[STIX]{x1D711}}$ . Restricting to the subgroup  $G_{E}$ where  $s_{1}\circ \overline{\unicode[STIX]{x1D711}}$ and  $s_{4}\circ \overline{\unicode[STIX]{x1D711}}$ both vanish, we obtain that  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}$ is represented by the two-cocyle

$$\begin{eqnarray}\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\mapsto (t_{1}\circ \overline{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D70E}))(t_{2}\circ \overline{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D70F})).\end{eqnarray}$$

It follows that  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}=t_{1}\circ \overline{\unicode[STIX]{x1D711}}\cup t_{2}\circ \overline{\unicode[STIX]{x1D711}}$ , the cup-product of the classes  $t_{i}\circ \overline{\unicode[STIX]{x1D711}}\in \operatorname{H}^{1}(E,\mathbb{F}_{2})$ . Or in other words $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}=(bB,cC)_{E}$ .

Given that  $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , and  $(b,c)_{F}=0$ , it follows that one has $u=\unicode[STIX]{x1D711}^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ satisfies  $u_{E}=(B,C)_{E}$ .

Finally, suppose that  $\unicode[STIX]{x1D711}$ exists as in the fourth assertion. We note that  $G_{E}=\unicode[STIX]{x1D711}^{-1}(\mathscr{C}(S))$ (again the notation $\mathscr{C}(S)$ for the centralizer of  $S$ is from § 2.4, and we had noted  $\mathscr{C}(S)=\ker (s_{1})\,\cap \,\ker (s_{4})$ ). The composition

$$\begin{eqnarray}f:G_{E}\longrightarrow \mathscr{C}(S)\longrightarrow N\end{eqnarray}$$

factors via  $\mathscr{C}(S)/S_{i}$ (for  $i=1,2,3$ ), so from the Lemma 2.5, we must have  $f^{\ast }(x_{2}x_{3})=0$ , $f^{\ast }(x_{1}x_{3})=0$ and  $f^{\ast }(x_{2}x_{4})=0$ . But these translate as  $(B,C)_{E}=0$ , which we were after, and $(bB,C)_{E}=0$ ,   $(B,cC)_{E}=0$ , consistently with the above.◻

3.2 Shapiro’s lemma and the converse

Let  $G$ be a finite group, let  $N$ be a subgroup, and let  $k$ be a field. For any  $kN$ -module  $A$ , we let  $\operatorname{Coind}_{N}^{G}(A)$ denote  $\operatorname{Hom}_{N}(kG,A)$ , which is a (left) $kG$ -module with action  $(\unicode[STIX]{x1D70E}\cdot f)(x)=f(x\unicode[STIX]{x1D70E})$ . (We are thinking of  $G$ and  $N$ as being the groups bearing those names in the discussion above, with  $k=\mathbb{F}_{2}$ , and  $A$ having trivial action.) The well-known Shapiro’s lemma states the existence of an isomorphism

$$\begin{eqnarray}sh:\operatorname{H}^{2}(G,\operatorname{Coind}_{N}^{G}(A))\longrightarrow \operatorname{H}^{2}(N,A).\end{eqnarray}$$

More precisely, the map is obtained using  $\operatorname{ev}:\operatorname{Coind}_{N}^{G}(A)\rightarrow A$ which evaluates at  $1\in G$ , followed by restriction (in cohomology) to  $N$ . Note, if the class  $\unicode[STIX]{x1D6FC}\in \operatorname{H}^{2}(G,\operatorname{Coind}_{N}^{G}(A))$ describes the extension

$$\begin{eqnarray}0\longrightarrow \operatorname{Coind}_{N}^{G}(A)\longrightarrow \unicode[STIX]{x1D6E4}\stackrel{p}{\longrightarrow }G\longrightarrow 1,\end{eqnarray}$$

then  $sh(\unicode[STIX]{x1D6FC})$ corresponds to

$$\begin{eqnarray}0\longrightarrow A\longrightarrow \frac{p^{-1}(N)}{\ker (\operatorname{ev})}\longrightarrow N\longrightarrow 1,\end{eqnarray}$$

as is easily verified.

In order to recognize that a given $G$ -module, say  $S$ , is isomorphic to  $\operatorname{Coind}_{N}^{G}(A)$ for some  $A$ , let us merely consider the case where  $A=k^{r}$ with trivial  $N$ -action, and assume that  $N$ is normal in  $G$ to boot. Then  $\operatorname{Coind}_{N}^{G}(A)=(k[G/N]^{\ast })^{r}$ . The dual module  $(k[G/N])^{\ast }$ is free of rank one, that is, it is isomorphic to  $k[G/N]$ . (Consider the map taking  $1\in G$ to $\unicode[STIX]{x1D6FF}_{1}$ , the Dirac delta function at  $1$ . Thus, $\unicode[STIX]{x1D6FF}_{1}(N)=1$ and $\unicode[STIX]{x1D6FF}_{1}(g\cdot N)=0$ if $g\notin N$ .) We conclude that  $S$ is isomorphic to  $\operatorname{Coind}_{N}^{G}(k^{r})$ if and only if  $N$ acts trivially and we can find a basis for  $S$ as a free  $k[G/N]$ of rank  $r$ . If this basis is  $\unicode[STIX]{x1D700}_{1},\ldots ,\unicode[STIX]{x1D700}_{r}$ , then  $\ker (ev)$ is the  $k$ -vector space spanned by  $g\unicode[STIX]{x1D700}_{i}$ , for  $g\in G$ , $g\neq 1$ , $i=1,\ldots ,r$ .

For example, let  $G,N,S$ recover their concrete meanings as in § 2.4 (all the accompanying notation will be used, too). Then Lemma 2.4 asserts that  $S\cong \operatorname{Coind}_{N}^{G}(\mathbb{F}_{2})$ , in such a way that  $\ker (ev)$ is identified with  $S_{1}$ . As a result, the cohomology class of

corresponds via  $sh$ to the cohomology class of the first extension treated in Lemma 2.5, that is, $x_{2}x_{3}$ .

But  $S$ can be regarded in another way. Consider the subgroup  $G^{\prime }\subset G$ of elements mapping into  $\langle \unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4}\rangle$ under  $G\rightarrow G/N$ (equivalently, $g\in G^{\prime }$ if  $s_{1}(g)=s_{4}(g)$ ), and view  $S$ as a  $G^{\prime }$ -module. Lemma 2.4 shows that  $S$ is a free  $\mathbb{F}_{2}[\langle \unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4}\rangle ]=\mathbb{F}_{2}[G^{\prime }/N]$ -module, with basis  $e_{1},e_{2}$ (or alternatively  $e_{1},e_{3}$ ). Thus we also have  $S\cong \operatorname{Coind}_{N}^{G^{\prime }}(\mathbb{F}_{2}\oplus \mathbb{F}_{2})$ , and  $\ker (ev)$ is spanned by  $e_{3},e_{4}$ (or  $e_{2},e_{4}$ in the alternative). Now the cohomology class of

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})^{\prime }\longrightarrow G^{\prime }\longrightarrow 1,\end{eqnarray}$$

where  $\mathbb{U}_{5}(\mathbb{F}_{2})^{\prime }$ is the preimage of  $G^{\prime }$ , is taken by Shapiro to that of the extension

$$\begin{eqnarray}0\longrightarrow \mathbb{F}_{2}e_{1}\oplus \mathbb{F}_{2}e_{2}\longrightarrow \frac{\mathscr{C}(S)}{\langle e_{3},e_{4}\rangle }\longrightarrow N\longrightarrow 1.\end{eqnarray}$$

This extension is described by two classes in  $\operatorname{H}^{2}(N,\mathbb{F}_{2})$ , corresponding to the exact sequences obtained by factoring out  $e_{1}$ and  $e_{2}$ respectively. From Lemma 2.5, these are  $x_{1}x_{3}$ and  $x_{2}x_{3}$ respectively. With the alternative choice of basis for  $S$ , this discussion ends with  $x_{2}x_{4}$ and  $x_{2}x_{3}$ .

There is a well-known version of Shapiro’s lemma for profinite groups (cf. [Reference Neukirch, Schmidt and WingbergNSW08, ch. 1, § 6]), which can be deduced from the version mentioned above using a straightforward limit argument. We record this version below in the context of Galois cohomology, since we will use it later on.

Lemma 3.2. Let  $E/F$ be a finite Galois extension, let  $A$ be a trivial, discrete  $G_{E}$ -module, and consider  $\operatorname{Coind}_{G_{E}}^{G_{F}}(A)$ , a discrete  $G_{F}$ -module. Then Shapiro’s map

$$\begin{eqnarray}\operatorname{H}^{2}(F,\operatorname{Coind}_{G_{E}}^{G_{F}}(A))\longrightarrow \operatorname{H}^{2}(E,A),\end{eqnarray}$$

defined as above, is an isomorphism.

With the preparations above, we can now prove our primary converse to Theorem 3.1.

Proof. We start by assuming that neither  $a$ nor  $d$ is a square in  $F$ (we identify  $F_{a}$ and  $F_{d}$ with  $F[\sqrt{a}]$ and  $F[\sqrt{d}]$ respectively). From Proposition 2.3 (applied twice), we obtain a homomorphism

$$\begin{eqnarray}f:G_{F}\longrightarrow D_{4}\times D_{4}=G\end{eqnarray}$$

such that  $s_{i}\circ f=\unicode[STIX]{x1D712}_{a},\unicode[STIX]{x1D712}_{b},\unicode[STIX]{x1D712}_{c},\unicode[STIX]{x1D712}_{d}$ according as  $i=1,2,3,4$ , and also  $f^{\ast }(x_{1})=\unicode[STIX]{x1D712}_{bB}$ , $f^{\ast }(x_{2})=\unicode[STIX]{x1D712}_{B}$ , $f^{\ast }(x_{3})=\unicode[STIX]{x1D712}_{C}$ , $f^{\ast }(x_{4})=\unicode[STIX]{x1D712}_{cC}$ . If  $\unicode[STIX]{x1D6FC}\in \operatorname{H}^{2}(D_{4}\times D_{4},S)$ is the class of the extension

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow D_{4}\times D_{4}\longrightarrow 1,\end{eqnarray}$$

then its pull-back  $f^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,S)$ is represented by the fibred-product of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ with  $G_{F}$ over  $D_{4}\times D_{4}$ (via $f$ ). To conclude the proof of the theorem, we will show that one has  $f^{\ast }(\unicode[STIX]{x1D6FC})=0$ , so that this fibred-product is a split extension of $G_{F}$ by $S$ . The composition of such a splitting with the projection to $\mathbb{U}_{5}(\mathbb{F}_{2})$ provides the necessary homomorphism $\unicode[STIX]{x1D711}$ .

Note that  $G_{E}=f^{-1}(N)$ . Suppose first that  $[E:F]=4$ . The  $G_{F}$ -module  $S$ is isomorphic to  $\operatorname{Coind}_{G_{E}}^{G_{F}}(\mathbb{F}_{2})$ , the trivial module of  $G_{E}$ (co)induced to  $G_{F}$ . Shapiro’s lemma applies, yielding the isomorphism

$$\begin{eqnarray}\operatorname{H}^{2}(F,S)\longrightarrow \operatorname{H}^{2}(E,\mathbb{F}_{2}).\end{eqnarray}$$

From the comments above and the naturality of Shapiro’s isomorphism, we see that  $f^{\ast }(\unicode[STIX]{x1D6FC})$ is taken to  $f^{\ast }(x_{2}x_{3})=f^{\ast }(x_{2})f^{\ast }(x_{3})=(B,C)_{E}=0$ , so we are done in this case.

We turn to the case where $a=d$ mod squares and $[E:F]=2$ ; another way to phrase this is by saying that the image of  $f$ lies within  $G^{\prime }$ . Now the  $G_{F}$ -module  $S$ is isomorphic, although not canonically, to  $\operatorname{Coind}_{G_{E}}^{G_{F}}(\mathbb{F}_{2}\oplus \mathbb{F}_{2})$ : we have at least the two possibilities given in the discussion preceding the proof, and for definiteness say we pick the basis  $e_{1},e_{2}$ .

Now Shapiro’s lemma gives an isomorphism

$$\begin{eqnarray}\operatorname{H}^{2}(F,S)\longrightarrow \operatorname{H}^{2}(E,\mathbb{F}_{2}\oplus \mathbb{F}_{2})=\operatorname{H}^{2}(E,\mathbb{F}_{2})\oplus \operatorname{H}^{2}(E,\mathbb{F}_{2}).\end{eqnarray}$$

This takes  $f^{\ast }(\unicode[STIX]{x1D6FC})$ to a pair of cohomology classes, and again from naturality, they are  $f^{\ast }(x_{2}x_{3})=(B,C)_{E}$ and  $f^{\ast }(x_{1}x_{3})=(bB,C)_{E}$ . These are both zero by assumption, and  $f^{\ast }(\unicode[STIX]{x1D6FC})=0$ also in this case.

Finally, suppose that  $a$ is a square in  $F$ (a symmetric argument deals with the case when  $d$ is a square). A neat way to handle this is to use the Massey vanishing conjecture for  $n=3$ , (which is now a theorem, see [Reference MatzriMat11, Reference Efrat and MatzriEM17, Reference Mináč and TânMT17c, Reference Mináč and TânMT16]). First we claim that  $(b,c)_{F}=0$ . Indeed, $(b,c)_{E}=0$ by assumption; if  $d$ is also a square in  $F$ then  $E=F$ , and otherwise  $E=F[\sqrt{d}]$ and  $\operatorname{N}_{E/F}(C)=c$ mod squares, so  $(b,C)_{E}=0$ implies  $(b,c)_{F}=0$ by the projection formula in group cohomology. We also have  $(c,d)_{F}=0$ by Proposition 2.3. We deduce that the Massey product  $\langle b,c,d\rangle$ vanishes, and so that there is a continuous  $\unicode[STIX]{x1D713}:G_{F}\rightarrow \mathbb{U}_{4}(\mathbb{F}_{2})$ compatible with  $b,c,d$ . However, if we see  $\mathbb{U}_{4}(\mathbb{F}_{2})$ as the subgroup of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ consisting of those matrices whose first row is that of the identity matrix, we see that we have built  $G_{F}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ compatible with  $1,b,c,d$ , as required.◻

Remark 3.4. It is possible to avoid relying on the Massey vanishing conjecture for  $n=3$ if one wishes to do so. Here is a sketch. When  $d$ is not a square, use another argument based on Shapiro’s lemma, this time replacing  $G^{\prime }$ by  $G^{\prime \prime }$ , the subgroup of  $G$ of elements mapping to  $\langle \unicode[STIX]{x1D70E}_{d}\rangle \subset G/N$ . When  $d$ is a square, define  $f$ as in the first part of the proof; this time  $f$ takes values in  $N$ . The exact sequence

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathscr{C}(S)\longrightarrow N\longrightarrow 1\end{eqnarray}$$

is central, and controlled by four cohomology classes in  $\operatorname{H}^{2}(N,\mathbb{F}_{2})$ , pulling back to  $(B,C)$ , $(b,C)$ , $(B,c)$ and  $(b,c)$ in  $\operatorname{H}^{2}(F,\mathbb{F}_{2})$ under  $f^{\ast }$ .

The proof of Theorem A is now almost complete. In fact, what we have is the following.

The only difference with Theorem A is the presence of the elements  $\widetilde{B}$ and  $\widetilde{C}$ instead of just  $B,C$ . Clearly, if neither  $a$ nor  $d$ is a square in  $F$ , then the two results are the same. On the other hand, when one of these two elements is a square, in fact when one of  $a,b,c,d$ is a square, things become very easy, as we proceed to show in the following subsection.

3.3 Some trivial cases

We proceed to show how we can improve the statement of Theorem 3.5 to that of Theorem A from the Introduction, so that the two are in fact equivalent. We have already mentioned that this is obvious when neither  $a$ nor  $d$ is a square in  $F$ .

Let us call (1A), (2A), (3A) the conditions of Theorem A, and keep (1), (2), (3) for those of Theorem 3.5, which we know are equivalent. Note (1) $=$ (1A).

Suppose  $a$ is a square in  $F$ , but not  $d$ . Assume condition (1A). We must show that condition (2A) holds. Indeed, from condition (2), we have the element  $C$ with  $\operatorname{N}_{F[\sqrt{d}]/F}(C)=c$ mod squares, and satisfying  $(b,C)_{F[\sqrt{d}]}=0$ , and moreover  $(b,c)_{F}=0$ . Now put  $B=b\in F=F[\sqrt{a}]$ , so that  $\operatorname{N}_{F[\sqrt{a}]/F}(B)=B=b$ . Then  $(B,C)_{E}=(b,C)_{E}=0$ , while  $(B,c)_{F[\sqrt{a}]}=(b,c)_{F}=0$ . We do have condition (2A), and so also (3A).

Conversely, condition (3A) is enough to ensure that  $(a,b)_{F}=(1,b)_{F}=0$ , $(b,c)_{F}=0$ (apply the projection formula to  $(b,C)_{E}=0$ ) and  $(c,d)_{F}=0$ (merely because  $C$ exists, cf. Proposition 2.3). So Lemma 3.6 applies and shows that condition (1A) holds.

The situation when  $a$ is not a square, but  $d$ is, is clearly similar.

Now suppose  $a$ and  $d$ are both squares. Suppose condition (1A) holds, and so also (2), and we have  $(a,b)=(c,d)=0$ (because  $\widetilde{B}$ and  $\widetilde{C}$ exist, cf. remark after Theorem 3.1) and  $(b,c)=0$ . Thus condition (2A) holds with  $B=b$ and  $C=c$ , and (2A) $=$ (3A) here. Conversely, condition (3A) contains the statement $(b,c)=0$ , while $(a,b)=(1,b)=0$ and  $(c,d)=(c,1)=0$ , and we see by the last lemma that condition (1A) holds. We have therefore just proven Theorem A.

3.4 Maps from profinite groups into  $\mathbb{U}_{5}(\mathbb{F}_{2})$

A routine modification of the arguments given above produces the next result.

We shall have no use for this theorem in the sequel, so we leave the proof to the reader.

5.1 The fundamental equation

Proposition 2.3 gives a necessary and sufficient condition for a cup-product to vanish in Galois cohomology, in terms of a simple polynomial equation. We shall see that the four cup-products of Theorem 3.5 are likewise controlled by a single polynomial equation, taking place in a certain finite étale $F$ -algebra, namely

$$\begin{eqnarray}\mathscr{E}=F[X,Y]/(X^{2}-a,Y^{2}-d)\cong F_{a}\otimes _{F}F_{d}.\end{eqnarray}$$

More precisely, we establish the following.

Proposition 5.1. Let  $a,b,c,d\in F^{\times }$ . Let  $\widetilde{B}\in F_{a}\subset \mathscr{E}$ satisfy  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=bz_{1}^{2}$ , and let  $\widetilde{C}\in F_{d}\subset \mathscr{E}$ satisfy  $\operatorname{N}_{F_{d}/F}(\widetilde{C})=cz_{2}^{2}$ , for some  $z_{1},z_{2}\in F^{\times }$ . Let  $B\in F[\sqrt{a}]$ and  $C\in F[\sqrt{d}]$ be the corresponding elements. Then the equation

$$\begin{eqnarray}u^{2}-\widetilde{B}v^{2}=\widetilde{C}\end{eqnarray}$$

has some solution with  $u,v\in \mathscr{E}$ if and only if we have simultaneously  $(B,C)_{F[\sqrt{a},\sqrt{d}]}=0$ , $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , $(b,c)_{F}=0$ . Alternatively, the same holds with the equation

$$\begin{eqnarray}u^{2}\widetilde{B}+v^{2}\widetilde{C}=1.\end{eqnarray}$$

The proof will be done in a case-by-case manner (each time getting a slightly more precise statement than that in the proposition). A quick remark about the equivalence of the two equations, though: it is not a completely general fact, as for example the equation $u^{2}+2v^{2}=1$ has four solutions over  $\mathbb{Z}/4\mathbb{Z}$ while  $u^{2}-v^{2}=2$ has no solution over the same ring. When working over a field of characteristic different from  $2$ however, the two problems are equivalent, as elementary computations reveal; since  $\mathscr{E}$ is a direct sum of such fields, we may indeed use either equation. The second is symmetric in  $\widetilde{B}$ and  $\widetilde{C}$ , and will be used in the sequel, but the proofs in the remainder of this section will be dealing with the first.

In the ‘generic case’ first, that is when  $a$ and  $d$ are linearly independent in  $F^{\times }\!/F^{\times 2}$ , we can and we do identify  $\mathscr{E}$ with  $E=F[\sqrt{a},\sqrt{d}]$ . Moreover, we have the following simple situation.

Lemma 5.2. Let  $\widetilde{B},B,\widetilde{C},C$ be as in the proposition. Suppose that $(B,C)_{E}=0$ .

1. (i) If neither  $d$ nor  $ad$ is a square in  $F$ , then we have  $(B,c)_{F[\sqrt{a}]}=0$ . If neither  $a$ nor  $ad$ is a square in  $F$ , we have  $(b,C)_{F[\sqrt{d}]}=0$ .

2. (ii) If $(B,c)_{F[\sqrt{a}]}=0$ and  $a$ is not a square in  $F$ , we have  $(b,c)_{F}=0$ . Likewise, if $(b,C)_{F[\sqrt{d}]}=0$ and  $d$ is not a square in  $F$ , we also conclude that  $(b,c)_{F}=0$ .

In particular, when  $[E:F]=4$ , the four cup-products from Theorem 3.5 vanish precisely when there are  $u,v\in E$ satisfying

$$\begin{eqnarray}u^{2}-v^{2}B=C.\end{eqnarray}$$

Proof. If we suppose that  $a$ is not a square in  $F$ , then $\operatorname{N}_{F[\sqrt{a}]/F}(B)=bz_{1}^{2}$ ; if  $ad$ is not a square either, than we can also write  $\operatorname{N}_{E/F[\sqrt{d}]}(B)=bz_{1}^{2}$ . Thus we can use the projection formula from group cohomology, asserting that

$$\begin{eqnarray}\operatorname{cores}((B,C)_{E})=(b,C)_{F[\sqrt{d}]},\end{eqnarray}$$

where  $\operatorname{cores}:\operatorname{H}^{2}(E,\mathbb{F}_{2})\rightarrow \operatorname{H}^{2}(F[\sqrt{d}],\mathbb{F}_{2})$ is the corestriction. Thus $(B,C)_{E}=0$ does imply $(b,C)_{F[\sqrt{d}]}=0$ . The other case is treated similarly.

One also proves (ii) using the projection formula.

For the last statement, we invoke Proposition 2.3 which states that the proposed equation has a solution if and only if  $(B,C)_{E}=0$ . By the first part, this implies that the other three cup-products also vanish.◻

Now suppose that  $a$ and  $d$ are both squares in  $F$ . Then we have four  $F$ -homomorphisms $p_{i}:\mathscr{E}\rightarrow F$ , for  $i=1,2,3,4$ , mapping  $X$ to  $\pm \sqrt{a}$ and  $Y$ to  $\pm \sqrt{d}$ . Together they induce an isomorphism  $\mathscr{E}\cong F\times F\times F\times F$ , by the Chinese remainder theorem. (Note that the map  $F\rightarrow \mathscr{E}$ which turns  $\mathscr{E}$ naturally into an  $F$ -algebra is the diagonal embedding of  $F$ in  $F^{4}$ , under this isomorphism.)

Lemma 5.3. Suppose  $a$ and  $d$ are both squares in  $F$ . Let  $\widetilde{B},B,\widetilde{C},C$ be as in the proposition. Then

$$\begin{eqnarray}u^{2}-\widetilde{B}v^{2}=\widetilde{C}\end{eqnarray}$$

has a solution with  $u,v\in \mathscr{E}$ if and only if we have simultaneously

$$\begin{eqnarray}(B,C)_{F}=(B,c)_{F}=(b,C)_{F}=(b,c)_{F}=0.\end{eqnarray}$$

Proof. Applying  $p_{i}$ with  $i=1,2,3,4$ , the equation becomes equivalent to the four equations

$$\begin{eqnarray}u_{i}^{2}-p_{i}(\widetilde{B})v_{i}^{2}=p_{i}(\widetilde{C})\end{eqnarray}$$

with unknowns  $u_{i},v_{i}\in F$ . This is possible if and only if  $(p_{i}(\widetilde{B}),p_{i}(\widetilde{C}))=0$ for  $i=1,2,3,4$ . (Note that the calculations to follow will establish that  $p_{i}(B)\neq 0$ , $p_{i}(C)\neq 0$ for all  $i$ , so that the cup-products make sense.)

We need some notation. Let  $B=x+y\sqrt{a}$ and  $C=x^{\prime }+y^{\prime }\sqrt{d}$ , and introduce  $B^{\prime }=x-y\sqrt{a}$ and  $C^{\prime }=x^{\prime }-y^{\prime }\sqrt{d}$ , so that  $BB^{\prime }=bz_{1}^{2}\neq 0$ , and likewise  $CC^{\prime }=cz_{2}^{2}\neq 0$ . To fix our ideas, we assume that the numbering of the homomorphisms  $p_{i}$ has been made such that

$$\begin{eqnarray}\displaystyle & \displaystyle p_{1}(\widetilde{B})=p_{3}(\widetilde{B})=B,\quad p_{2}(\widetilde{B})=p_{4}(\widetilde{B})=B^{\prime }, & \displaystyle \nonumber\\ \displaystyle & \displaystyle p_{1}(\widetilde{C})=p_{2}(\widetilde{C})=C,\quad p_{3}(\widetilde{C})=p_{4}(\widetilde{C})=C^{\prime }. & \displaystyle \nonumber\end{eqnarray}$$

The four cup-products we consider are then  $(B,C)$ , $(B^{\prime },C)$ , $(B,C^{\prime })$ and  $(B^{\prime },C^{\prime })$ for  $i=1,2,3,4$ respectively, all in the cohomology of  $F$ . However

$$\begin{eqnarray}(b,C)=(BB^{\prime },C)=(B,C)+(B^{\prime },C)=0,\end{eqnarray}$$

and similarly we draw  $(B,c)=0$ and  $(b,c)=0$ . One can also work backwards, clearly.◻

When  $a$ is a square in  $F$ , but  $d$ is not, we view  $\mathscr{E}$ as  $F[\sqrt{d}][X]/(X^{2}-a)\cong F[\sqrt{d}]\times F[\sqrt{d}]$ with its two  $F[\sqrt{d}]$ -homomorphisms  $p_{1},p_{2}:\mathscr{E}\rightarrow F[\sqrt{d}]$ . The elements  $Y$ and  $\sqrt{d}$ are identified. A reasoning similar to the above yields the following.

Lemma 5.4. Suppose that  $a$ is a square in  $F$ , and that  $d$ is not. Let  $\widetilde{B},B,\widetilde{C},C$ be as in the Proposition. Then the equation

$$\begin{eqnarray}u^{2}-\widetilde{B}v^{2}=\widetilde{C}\end{eqnarray}$$

has a solution with  $u,v\in \mathscr{E}$ if and only if we have simultaneously  $(B,C)_{F[\sqrt{d}]}=(b,C)_{F[\sqrt{d}]}=0$ . When this is the case, we have automatically  $(B,c)_{F}=(b,c)_{F}=0$ from Lemma 5.2.

Of course a similar result holds with the roles of  $a$ and  $d$ exchanged.

Finally we turn to the case when neither  $a$ nor  $d$ is a square in  $F$ , but  $ad$ is. This is in fact similar to the previous case, except that we now have a choice. Namely, we can produce two  $F[\sqrt{d}]$ -homomorphisms  $p_{1},p_{2}:\mathscr{E}\rightarrow F[\sqrt{d}]$ giving an isomorphism  $\mathscr{E}\cong F[\sqrt{d}]\times F[\sqrt{d}]$ , identifying  $\sqrt{d}$ with  $Y$ all along; or, we can alternatively find two  $F[\sqrt{a}]$ -homomorphisms $p_{1}^{\prime },p_{2}^{\prime }:\mathscr{E}\rightarrow F[\sqrt{a}]$ , giving an isomorphism  $\mathscr{E}\cong F[\sqrt{a}]\times F[\sqrt{a}]$ , identifying  $X$ with  $\sqrt{a}$ all along. These two isomorphisms are distinct, even if  $a=d$ . Using one and then the other, we get the following.

Lemma 5.5. Suppose that neither  $a$ nor  $d$ is a square in  $F$ , but that  $ad$ is. Let  $\widetilde{B},B,\widetilde{C},C$ be as in the proposition. Then the equation

$$\begin{eqnarray}u^{2}-\widetilde{B}v^{2}=\widetilde{C}\end{eqnarray}$$

has a solution with  $u,v\in \mathscr{E}$ if and only if we have simultaneously  $(B,C)_{F[\sqrt{d}]}=(b,C)_{F[\sqrt{d}]}=0$ , if and only if we have simultaneously  $(B,C)_{F[\sqrt{a}]}=(B,c)_{F[\sqrt{a}]}=0$ . When this is the case, we have automatically  $(b,c)_{F}=0$ from Lemma 5.2.

This completes the proof of the proposition. Given that the existence of  $\widetilde{B}$ and  $\widetilde{C}$ is itself controlled by a simple polynomial equation, as in Proposition 2.3, the situation is now entirely rewritten in terms of the existence of a rational point on an algebraic variety.

5.2 Splitting varieties

We proceed to translate our results into the language of algebraic geometry.

As ever, let  $F$ be a field of characteristic not  $2$ , let  $a,b,c,d\in F^{\times }$ , and put

$$\begin{eqnarray}\mathscr{E}=F[X,Y]/(X^{2}-a,Y^{2}-d).\end{eqnarray}$$

Together, Theorem 3.5, Proposition 2.3 and Proposition 5.1 show that the vanishing of the Massey product $\langle a,b,c,d\rangle$ is equivalent to the existence of a solution to the following system of equations, with unknowns  $x_{1},y_{1},z_{1},x_{2},y_{2},y_{3}\in F$ , $u,v\in \mathscr{E}$ :

1. (1) $x_{1}^{2}-ay_{1}^{2}=bz_{1}^{2}$ ;

2. (2) $x_{2}^{2}-dy_{2}^{2}=cz_{2}^{2}$ ;

3. (3) $u^{2}\widetilde{B}+v^{2}\widetilde{C}=1$ , where  $\widetilde{B}=x_{1}+y_{1}X\in \mathscr{E}$ and  $\widetilde{C}=x_{2}+y_{2}Y\in \mathscr{E}$ .

There is also the condition  $z_{1}\neq 0$ , $z_{2}\neq 0$ . Here  $u=u_{1}+u_{2}X+u_{3}Y+u_{4}XY$ , and likewise for  $v$ , so that equation (3) can be written as four equations over  $F$ (carrying out the expansion in practice does not seem to clarify things).

For technical reasons, related to Proposition 4.1, we change coordinates and work with the equations:

1. (i) $x_{1}^{2}-ay_{1}^{2}=b$ ;

2. (ii) $x_{2}^{2}-dy_{2}^{2}=c$ ;

3. (iii) $u^{2}\unicode[STIX]{x1D6FD}\widetilde{B}+v^{2}\unicode[STIX]{x1D6FE}\widetilde{C}=1$ , where  $\widetilde{B}=x_{1}+y_{1}X\in \mathscr{E}$ and  $\widetilde{C}=x_{2}+y_{2}Y\in \mathscr{E}$ .

Here  $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F^{\times }$ are two new unknowns, and  $z_{1},z_{2}$ have disappeared.

Equations (i), (ii), (iii) define an affine subvariety of  $\mathbf{A}_{F}^{14}$ , the affine space of dimension 14 over  $F$ ; we consider its intersection with the open subset defined by  $\unicode[STIX]{x1D6FD}\neq 0$ , $\unicode[STIX]{x1D6FE}\neq 0$ , and call it  $\mathscr{X}_{F}$ . Also, note that equation (i) implies that $\widetilde{B}$ is a unit of $F_{a}:=F[X]/(X^{2}-a)$ , and similarly equation (ii) implies that $\widetilde{C}$ is a unit of $F_{d}:=F[Y]/(Y^{2}-d)$ . In particular, we can describe $\mathscr{X}_{F}$ in a more conceptual way using the Weil-restriction functors $R_{\mathscr{E}|F}$ , $R_{F_{a}|F}$ and $R_{F_{d}|F}$ . Namely, consider the $F$ -variety:

$$\begin{eqnarray}\mathscr{Y}:=\mathbf{G}_{m}\times \mathbf{G}_{m}\times R_{\mathscr{E}/F}\mathbf{A}_{\mathscr{E}}^{2}\times R_{F_{a}/F}\mathbf{G}_{m}\times R_{F_{d}/F}\mathbf{G}_{m}.\end{eqnarray}$$

We view an $F$ -point of $\mathscr{Y}$ as a tuple $(\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE},u,v,\widetilde{B},\widetilde{C})$ , with $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F^{\times }$ , $u,v\in \mathscr{E}$ , $\widetilde{B}\in F_{a}^{\times }$ and $\widetilde{C}\in F_{d}^{\times }$ . Then $\mathscr{X}_{F}$ is the closed subvariety of $\mathscr{Y}$ defined by the three equations:

$$\begin{eqnarray}\operatorname{N}_{F_{a}/F}(\widetilde{B})=b,\quad \operatorname{N}_{F_{d}/F}(\widetilde{C})=c,\quad u^{2}\unicode[STIX]{x1D6FD}\widetilde{B}+v^{2}\unicode[STIX]{x1D6FE}\widetilde{C}=1.\end{eqnarray}$$

We have the following trivial, yet crucial property: for an extension $L/F$ , we have

$$\begin{eqnarray}\mathscr{X}_{F}\times _{\operatorname{Spec}(F)}\operatorname{Spec}(L)=\mathscr{X}_{L}.\end{eqnarray}$$

In particular, we have  $\mathscr{X}_{F}(L)=\mathscr{X}_{L}(L)$ (using the standard notation for the set of rational points). As already noted, the set  $\mathscr{X}_{F}(F)$ is non-empty if and only if the Massey product  $\langle a,b,c,d\rangle$ is defined and vanishes in the cohomology of  $F$ . Now, it is obviously also true, but nicer, that for any field extension  $L/F$ , the set  $\mathscr{X}_{F}(L)$ is non-empty if and only if the Massey product  $\langle a,b,c,d\rangle$ is defined and vanishes in the cohomology of  $L$ . This is a property which is expected of a ‘splitting variety’ for the problem of the vanishing of $\langle a,b,c,d\rangle$ .

As it turns out, we can introduce a second splitting variety $X_{F}$ . It will depend on choices, and so is not canonically associated with the problem alone; on the other hand, the local-global principle is established in the appendix for  $X_{F}$ rather than  $\mathscr{X}_{F}$ . The construction will echo Proposition 4.1 rather precisely. Let  $Z_{F}$ be the subvariety of $R_{F_{a}/F}\mathbf{G}_{m}\times R_{F_{d}/F}\mathbf{G}_{m}$ , defined over  $F$ be the equations

$$\begin{eqnarray}\operatorname{N}_{F_{a}/F}(\widetilde{B})=b,\quad \operatorname{N}_{F_{d}/F}(\widetilde{C})=c.\end{eqnarray}$$

There is an obvious morphism  $\unicode[STIX]{x1D70B}:\mathscr{X}_{F}\rightarrow Z_{F}$ (forgetting  $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE},u,v$ ), and we will define  $X_{F}$ to be a fibre of  $\unicode[STIX]{x1D70B}$ above an  $F$ -rational point of  $Z_{F}$ . That is, we will suppose from now on that $(a,b)_{F}=(c,d)_{F}=0$ , and using Proposition 2.3 twice, we select  $\widetilde{B}_{0}\in F_{a}$ and  $\widetilde{C}_{0}\in F_{d}$ whose norms are  $b$ and  $c$ respectively; then we put  $X_{F}=\unicode[STIX]{x1D70B}^{-1}(\widetilde{B}_{0},\widetilde{C}_{0})$ . The variety  $X_{F}$ is thus defined by the equation

$$\begin{eqnarray}u^{2}\unicode[STIX]{x1D6FD}\widetilde{B}_{0}+v^{2}\unicode[STIX]{x1D6FE}\widetilde{C}_{0}=1.\end{eqnarray}$$

Note that our construction of $X_{F}$ depends on the choice of $\widetilde{B}_{0}$ and $\widetilde{C}_{0}$ as above. In the sequel, this choice will always be clear from context, so we omit the $\widetilde{B}_{0}$ , $\widetilde{C}_{0}$ from the notation.

It is clear that  $X_{F}$ is also compatible with base-change, just like  $\mathscr{X}_{F}$ is. That it is also a splitting variety is part of the next theorem.