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On quadratic subfields of generalized quaternion extensions D. D. Kiselev All-Russian Academy of International Trade
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 1, DOI: https://doi.org/10.4213/im9430 
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   § 1. IntroductionThe embedding problem is $(K/k, G, \varphi)$ for an exact sequence of finite groups where $K/k$ is the Galois extension of fields, calls for finding the Galois algebra $L$ over $k$ with the group $G$, with $K\subset L$, such that the epimorphism of the restriction of automorphisms from $L$ to $K$ coincides with $\varphi$ from (1.1). In this case, $A$ is called the kernel of the embedding problem $(K/k, G, \varphi)$.The search for a solution to the problem (1.1) in the class of Galois algebras (not necessarily fields) makes it possible to involve the apparatus of homological algebra, and, in particular, to give a solution to this problem, in appropriate terms, for the case of an Abelian kernel (see, for example, [1]). In meaningful cases, the solvability to problem (1.1) in the class of Galois algebras implies implies that in the class of fields: for example, if $k$ is a number field and $A$ is a nilpotent group (see [2]); if $k$ is a $p$-local number field of sufficiently high degree over $\mathbb Q_p$, and $G$ is a $p$-group (see [3], Ch. 4, § 2, Theorem 4.2.5). On the other hand, of special interest are conditions under which all solutions of a solvable problem (1.1) are fields. In connection with this question, A. V. Yakovlev posed the following problem. Problem 1. Under which conditions, for an extension of finite groups (1.1) with abelian kernel $A$, there exists a Galois extension of number fields $K/k$ with group $F$ such that the corresponding embedding problem $(K/k, G, \varphi)$ is solvable and has only fields as solutions? In what follows, following [4], extension (1.1) will be called ultra solvable if Problem 1 is solved positively for this extension. In [4]–[6], a systematic study of Problem 1 was carried out for the case of the cyclic kernel $A$. In particular, in [5] a complete solution of Problem 1 was obtained for the cyclic kernel $A$ and the group $G$ of odd order, and in [6], for a fairly wide class of $2$-extensions with cyclic kernel $A$. It turned out that (1.1) is ultrasolvable if and only if, for the extension of (1.1) all the Sylov extensions (see § 1.2 in [5]) do not split. In particular, if (1.1) is an odd-order $p$-extension with cyclic kernel, then it is ultrasolvable if and only if it is not semidirect. For $p=2$, the proof of a similar result faces the following difficulties. Almost all known results on Problem 1 were obtained in two stages: first, the ultrasolvability of extension (1.1) (or $p$-extensions directly related to it) over $p$-local number fields was shown, then the Yakovlev local-global principle was applied (see [5], Proposition 1) to the effect that the ultrasolvability of a group extension over local number fields implies that over number fields as well. However, in Theorem 1 of [6] it was shown, in particular, that, for quaternion extensions, where
$$
\begin{equation}
G_{b, c, n}=\langle b, c\mid b^{2^n}=1, \,c^2=b^{2^{n-1}}, \,b^c=b^{-1}\rangle,
\end{equation}
\tag{1.3}
$$
$\varphi(c)=f$ is an element of order $2$, the Yakovlev local-global principle is not applicable for $n\geqslant 4$. At the same time, extension (1.2) with $n=4$ is ultrasolvable (see Theorem 3 in [7]): the problem $(\mathbb Q(\sqrt{7})/\mathbb Q, G_{b, c, 4}, \varphi)$ is ultrasolvable (that is, solvable, and all its solutions are fields). For the case $n>4$, it was not known whether extension (1.2) is ultrasolvable. In Theorem 1, we show that extension (1.2) for $n>4$ is not ultrasolvable, although it is non-direct. So far, such examples have been constructed only for the case of an Abelian non-cyclic kernel (see Theorem 3 in [8]). Let extension (1.1) with abelian kernel $A$ be given by a class $h\in H^2(F, A)$ and let $B$ be some $F$-submodule of the $F$-module $A$. Consider the homomorphism $\alpha\colon H^2(F, B)\to H^2(F, A)$ induced by the embedding $B\hookrightarrow A$. Any class $h_1\in H^2(F, B)$ such that $h=\alpha(h_1)$ (such a class exists if and only if1 $A\nleqslant\Phi(G)$) defines the adjoint extension to (1.1), and, accordingly, the adjoint problem $(K/k, H, \varphi)$ to the problem $(K/k, G, \varphi)$. If $B$ is a maximal subgroup in $A$, then the corresponding adjoint extension (adjoint problem) is called maximal. It is well known (see Theorem 1 in [9]) that the problem $(K/k, G, \varphi)$ is ultrasolvable if and only if all its maximal adjoint problems are unsolvable, and, in this case, the problem $(K/k, G, \varphi)$ itself is solvable. In the case of the quaternion extension (1.2) with $n\geqslant 3$, the maximal subgroups not containing the kernel $\langle b\rangle$ have the form $G_{b^2, c, n-1}$ and $G_{b^2, cb, n-1}$. It is easy to see that $(cb)^2=c^2$, and therefore, the solvability conditions of the adjoint problems are $(k(\sqrt d)/k, G_{b^2, c, n-1}, \varphi)$ and $(k(\sqrt d)/k, G_{b^2, cb, n-1}, \varphi)$ are the same, because the corresponding group extensions are equivalent. In particular, the problem $(k(\sqrt{d})/k, G_{b, c, n}, \varphi)$ for extention (1.2) with $n\geqslant 3$ is ultrasolvable if and only if the extension $k(\sqrt{d})/k$ is embedded into the Galois algebra with the group $G_{b, c, n}$, but is not embedded into the Galois algebra with the group $G_{b, c, n-1}$. This, together with Theorem 1, allows us to give a solvability criterion both for the problem $(k(\sqrt{d_1})/k, G_{b, c, n}, \varphi)$ with cyclic kernel of order $2^n$ and for the problem $(k(\sqrt{d_2})/k, G_{b, c, n}, \psi)$ with the kernel $G_{b^2, cb, n-1}$. This is done in Theorem 2. The problem of complete classification of quadratic subfields of the generalized quaternion extensions evokes certain interest. This problem can be reduced to the embedding problem of the extensions $k(\sqrt{d_1})/k$ and $k(\sqrt{d_2})/k$ in the same field $L$ with the Galois group $G_{b, c, n}$ over $k$. Unfortunately, Theorem 2 does not allow us to answer this question, but nevertheless, in Theorem 3 sufficient conditions for such an embedding are obtained. In particular, if $k=\mathbb Q$, the numbers $d_1$ and $d_2$ are prime, $d_i\equiv 1\ (\operatorname{mod} 2^n)$, and $d_1$ is a square in the field $\mathbb F_{d_2}$, then the fields $k(\sqrt{d_1})$, $k(\sqrt{d_2})$, $k(\sqrt{d_1d_2})$ are quadratic subfields of some extension of the field $k$ with the group $G_{b, c, n}$. § 2. Some facts2.1. NotationAll the groups considered below are finite, unless otherwise specified. The exponent of a group $A$ is denoted by $\exp A$. By a number field, we will mean the finite extension of the field $\mathbb Q$. The symbol $\varepsilon_m$ will denote some primitive root of unity of degree $m$. At the same time, if ${m_1\mid m_2}$, then the corresponding roots of unity will always be selected with the normalization condition $\varepsilon_{m_1}=\varepsilon_{m_2}^{m_2/m_1}$. For a finite extension of the fields $K/k$, by $(K:k)$ we denote the dimension of $K$ considered as the vector space over $k$. For fields of zero characteristic $k_1, k_2$, their composite is denoted by $k_1\cdot k_2$. Let $k$ be a number field. For some $a, b\in k^*$, by $k[a, b]$ we denote the algebra of generalized quaternions of degree $2$, that is, $k$ is an algebra with generators $\xi, \eta$ satisfying $\xi^2=a$, $\eta^2=b$, $\xi\eta=-\eta\xi$. The tensor product $k[a, b]\otimes_k k[c, d]$ is denoted by $k[a, b][c, d]$. The Brauer group of a field $k$ is denoted by $B(k)$. For central simple finite-dimensional $k$-algebras $A$ and $B$, the equivalence in the group $B(k)$ is denoted by $A\sim B$. In what follows, the group $G_{b, c, n}$ will be understood as the group of generalized quaternions defined by conditions (1.3). The group generated by a set $X$ will be denoted by $\langle X\rangle$; $\langle X\rangle_k$ denotes the minimal $k$-subalgebra of some $k$-algebra containing elements of a set $X$. 2.2.Let us recall the basic constructions in the embedding theory. Let $F=\mathrm{Gal}(K/k)$, and let the Galois extension of the fields $K/k$ be embedded in the field $K_1$ with the Galois group $F_1$ over $k$. Let $\theta\colon F_1\to F$ be an epimorphism of the restriction of $k$-automorphisms of the field $K_1$ to $K$. In this case, extension (1.1) admits an inflation relative to $\theta$ for the extension In this case, the problems $(K/k, G, \varphi)$ and $(K_1/k, G\times_F F_1, \widehat{\varphi})$ are equivalent in the sense of solvability. In the case of an Abelian kernel $A$, the cohomology class of the inflated extension is obtained from the cohomology class of the original extension via the inflation homomorphism $\lambda\colon H^2(F, A)\to H^2(F_1, A)$, because by construction $A^{\operatorname{ker}\theta}=A$.For the embedding problem $(K/k, G, \varphi)$, there is an important necessary condition of solvability known as the Faddeev–Hasse concordance condition. Let us recall some equivalent formulations of this condition. Namely, consider the crossed product $G\times K$, that is, $k$ is the algebra of formal sums $\sum_{g\in G}u_gx_g$, where $x_g \in K$, with the multiplication rules $u_{g_1}u_{g_2}=u_{g_1g_2}$ and $xu_g=u_gx^{\varphi(g)}$ for $x\in K$, $g\in G$. We say that the concordance condition is met for the problem $(K/k, G, \varphi)$ if the algebra $G\times K$ is a matrix algebra of order $(K:k)$ over some its subalgebra. It is well known that the concordance condition in the lifted problem and in the original problem hold or fail to hold simultaneously. Now let the kernel $A$ of the problem $(K/k, G, \varphi)$ with $F=\mathrm{Gal}(K/k)$ be abelian, with $\operatorname{char}k=0$, and $\varepsilon_{\exp A}\in K$. The character group $\widehat A=\mathrm{Hom}(A, K^*)$ can be turned into an $F$-module by
$$
\begin{equation*}
\chi^f(a)=\bigl(\chi(a^{f^{-1}})\bigr)^f, \qquad a\in A, \quad f\in F, \quad \chi\in\widehat A.
\end{equation*}
\notag
$$
Let $\mathbb Z[\widehat A]$ be a free $\mathbb Z$ module with generators $\{c_{\chi}\mid\chi\in\widehat A\}$. The module $\mathbb Z[\widehat A]$ can be turned into an $F$-module by putting $c_{\chi}^f=c_{\chi^f}$. Now let $\overline{F}$ be the Galois group of the algebraic closure $\overline k$ of the field $k$, and $\mu\colon\overline F\to F$ be the natural restriction epimorphism. In this case, $\widehat A$ and $\mathbb Z[\widehat A]$ can be turned into $\overline F$ modules via an offset along $\mu$. We have the following an exact sequence of $\overline{F}$-modules where $\alpha(c_{\chi})=\chi$. Since the group $\overline k^{\,*}$ is complete, sequence (2.1) induces the exact sequence where $\mathrm{Hom}(\widehat A, \overline k^{\,*})$ and $A$ are canonically identified, and therefore, sequence (2.2) induces the diagram with the exact top row. It is well known (see [3], Ch. 3, § 13) that the group $H^1(\overline F, \mathrm{Hom}(\mathbb Z[\widehat A], \overline k^{\,*}))$ is trivial, and therefore, $\delta$ is a mononorphism. Let $h\in H^2(F, A)$ be the class of the group extension of the problem $(K/k, G, \varphi)$. This problem is solvable if and only if $\lambda(h)=1$. In the problem $(K/k, G, \varphi)$, the concordance condition is met if and only if $\theta(\lambda(h))=1$.Let the concordance condition be met for the problem $(K/k, G, \varphi)$. Then there is a single element $z\in H^1(\overline F, \mathrm{Hom}(V, \overline k^{\,*}))$ such that $\delta(z)=\lambda(h)$, and the problem $(K/k, G, \varphi)$ is solvable if and only if $z=1$. Let $F_0$ be an arbitrary subgroup of the group $\overline F$ acting trivially on $V$ (obviously, $\operatorname{ker}\mu$ is a subgroup in $F_0$). Then the lifting of the group $H^1(\overline F/F_0, \mathrm{Hom}(V, K_0^*))$ to $H^1(\overline F, \mathrm{Hom}(V, \overline k^{\,*}))$ is an isomorphism (by construction $K_0\subset K$, where $K_0=\overline k^{F_0}$). Thus, under the concordance condition in the problem $(K/k, G, \varphi)$, the additional condition for embedding calls for the triviality of an unambiguously defined element $z\in H^1(\overline F/F_0, \mathrm{Hom}(V, K_0^*))$. This is Yakovlev’s theorem (see [1]). 2.3.Now let the embedding problem $(K/k, G, \varphi)$ with abelian kernel $A$ and $F=\mathrm{Gal}(K/k)$ is given over the number field $k$. Consider some prime divisor $\mathfrak p$ of the field $k$. Then $k_{\mathfrak p}$ is a Galois algebra $L_{\mathfrak p}=K\otimes_k k_{\mathfrak p}$ with Galois group $F=\mathrm{Gal}(K/k)$. Let $F_{\mathfrak p}$ be a Galois group of the kernel-field $K_{\mathfrak p}$ of the algebra $L_{\mathfrak p}$. Then $F_{\mathfrak p}$ is a subgroup in $F$ (the decomposition group of the prime divisor $\mathfrak p$ in $K$). The problem $(K_{\mathfrak p}/k_{\mathfrak p}, G_{\mathfrak p}, \varphi)$, where $G_{\mathfrak p}$ is the coimage of $F_{\mathfrak p}$ relative to $\varphi$, will be called the $\mathfrak p$-localization of the problem $(K/k, G, \varphi)$. In the problem $(K/k, G, \varphi)$, the concordance condition is met if and only if the localization problem $(K_{\mathfrak p}/k_{\mathfrak p}, G_{\mathfrak p}, \varphi)$ is solvable for all prime divisors $\mathfrak p$ of the field $k$. Let $T=\mathbb Q/\mathbb Z$ be considered as an $F$-module with trivial action. The $F$-module $X'=\mathrm{Hom}(X, T)$ is called the dual module of the $F$-module $X$. Now let $\varepsilon_{\exp A}\in K$, and in the problem $(K/k, G, \varphi)$ with $F=\mathrm{Gal}(K/k)$ the concordance condition is met. In this case (see § 2.2), there exists a well-defined element $z\in H^1(F, \mathrm{Hom}(V, K^*))$ whose triviality is necessary and sufficient for solvability of the problem $(K/k, G, \varphi)$. According to Yakovlev (see [10], and Ch. IV, Theorem 9.11 in [11]), the group $H^1(F, \mathrm{Hom}(V, K^*))$ is canonically isomorphic to the cokernel of the homomorphism
$$
\begin{equation*}
\theta\colon\biggl(\prod_{\mathfrak p}H^1(F_{\mathfrak p}, \widehat A)\biggr)' \to H^1(F, \widehat A)'
\end{equation*}
\notag
$$
induced by the restrictions $\theta_{\mathfrak p}\colon H^1(F, \widehat A) \to H^1(F_{\mathfrak p}, \widehat A)$. Thus, for number fields, an additional condition for embedding is that some uniquely defined element $x\in H^1(F, \widehat A)'$ should lie in the image of the homomorphism $\theta$. In particular, if the intersection of all kernels of homomorphisms of the restriction $\theta_{\mathfrak p}$ is trivial, then the additional embeddable condition disappears. Yakovlev (see [12]) obtained the following beautiful result: if the Galois extension $K/k$ of number fields is such that $2$ is completely decomposed in $K$, and the kernel of the embedding problem is a cyclic group, then $\bigcap_{\mathfrak p}\operatorname{ker}\theta_{\mathfrak p}=\{1\}$, that is, in this situation, the concordance condition is sufficient for an embedding. § 3. Auxiliary results3.1. Calculation of some cohomologiesLet $k$ be a number field such that the elements $-1$ and $2$ of the $\mathbb F_2$-vector space $k^*/k^{*2}$ are linearly independent. In this case, $\varepsilon_4\notin k$, $\varepsilon_8\notin k(\varepsilon_4)$. The following result is almost obvious. Proof. Denote $K=k(\varepsilon_4)$. Consider the cyclic Galois extension $\mathbb Q(\varepsilon_{16})/\mathbb Q(\varepsilon_4)$ of order $4$. Since $\varepsilon_8\notin K$, we have $K\cap\mathbb Q(\varepsilon_{16})=\mathbb Q(\varepsilon_4)$. But then the algebra $K\otimes_{\mathbb Q(\varepsilon_4)}\mathbb Q(\varepsilon_{16})$ is a field, and therefore, $(K(\varepsilon_{16}):K)= (\mathbb Q(\varepsilon_{16}):\mathbb Q(\varepsilon_4))=4$. Hence $\varepsilon_{16}\notin k(\varepsilon_8)$, the result required. So, the extension $k(\varepsilon_{16})/k$ is a Galois extension of order $8$ with the group $\langle g_1\rangle\times\langle g_2\rangle$, and
$$
\begin{equation*}
\varepsilon_{16}^{g_1}=\varepsilon_{16}^{-1}, \qquad \varepsilon_{16}^{g_2}=\varepsilon_{16}^5.
\end{equation*}
\notag
$$
We fix an element $c\in k^*\setminus k^{*2}$ such that the elements $-1$, $2$, $c$ of the space $k^*/k^{*2}$ are linearly independent over $\mathbb F_2$. The Galois group $G$ of the extension $k(\varepsilon_{16}, \sqrt{c})/k$ is generated by the elements $g_1$, $g_2$, $g_3$, with $\varepsilon_{16}^{g_3}=\varepsilon_{16}$, and $\sqrt{c}^{\,g_3}=-\sqrt{c}$. Let $B$ be a cyclic group of order $16$ with generator $b$. Let us turn $B$ into a $G$-module as follows: Lemma 2. The group $H^1(G, \widehat B)$ is isomorphic to the Klein group $V_4$ and is generated by classes $d_1$, $d_2$ such that Proof. Consider an arbitrary normalized cocycle $d$ of the group $G$ with values in $\widehat B$. We set $d_{g_i}=\varphi_i$, $i\in\{1, 2, 3\}$. Next, let $\varphi_i(b)=\varepsilon_{16}^{k_i}$. We have, for all $i$,
$$
\begin{equation}
\varphi_i^{g_1}=\varphi_i^{-1}, \qquad \varphi_i^{g_2}=\varphi_i^5, \qquad \varphi_i^{g_3}=\varphi_i^{-1}.
\end{equation}
\tag{3.2}
$$
From (3.2) we have $d_{g_1^2}=d_{g_3^2}=1$ for any $k_i$. Let us indicate the non-trivial relations. Using (3.2), we have
$$
\begin{equation}
\begin{alignedat}{3} d_{g_2^4} &=1 &\quad &\Longrightarrow &\quad &\varphi_2^{5^3+5^2+5+1}=1, \\ d_{g_1g_2} &=d_{g_2g_1} &\quad &\Longrightarrow &\quad &\varphi_1^5\varphi_2 =\varphi_2^{-1}\varphi_1, \\ d_{g_1g_2^2} &=d_{g_2^2g_1} &\quad &\Longrightarrow &\quad &\varphi_1^{5^2}\varphi_2^{5+1} =\varphi_2^{-5-1}\varphi_1, \\ d_{g_1g_3} &=d_{g_3g_1} &\quad &\Longrightarrow &\quad &\varphi_1^{-1}\varphi_3 =\varphi_3^{-1}\varphi_1, \\ d_{g_2g_3} &=d_{g_3g_2} &\quad &\Longrightarrow &\quad &\varphi_2^{-1}\varphi_3 =\varphi_3^5\varphi_2, \\ d_{g_2^2g_3} &=d_{g_3g_2^2} &\quad &\Longrightarrow &\quad &\varphi_2^{-5-1}\varphi_3 =\varphi_3^{5^2}\varphi_2^{5+1}. \end{alignedat}
\end{equation}
\tag{3.3}
$$
In terms of $k_i$, relations (3.3) can be written as
$$
\begin{equation}
\begin{alignedat}{3} k_2&\equiv 0\pmod 4, &\qquad 2k_1+k_2&\equiv 0 \pmod 8, &\qquad 2k_1+3k_2&\equiv 0\pmod 4, \\ k_1&\equiv k_3\pmod 8, &\qquad k_2{+}\,2k_3&\equiv 0 \pmod 8, &\qquad 2k_3{+}\,3k_2&\equiv 0\pmod 4. \end{alignedat}
\end{equation}
\tag{3.4}
$$
Relations (3.4) are equivalent to the system
$$
\begin{equation}
\begin{cases} k_2=4m_1, \\ k_1=2m_2, \\ k_3=2m_3, \\ k_1=k_3+8m_4, \\ 2k_1+k_2=8m_5, \\ k_2=-2k_3+8m_6 \end{cases}
\end{equation}
\tag{3.5}
$$
for integer parameters $m_i$, which are related by
$$
\begin{equation}
\begin{cases} m_2=m_3+4m_4, \\ m_1+m_2=2m_5, \\ m_1=-m_3+2m_6 \end{cases} \Longleftrightarrow\quad \begin{cases} m_1=-m_3+2m_6, \\ m_2=m_3+4m_4, \\ m_5=2m_4+m_6, \end{cases}
\end{equation}
\tag{3.6}
$$
where the right-hand system is free from linear relations for the integer parameters $m_3$, $m_4$, $m_6$. In view of (3.5) and (3.6),
$$
\begin{equation}
\varphi_1(b)=\varepsilon_8^{m_3}\cdot(-1)^{m_4}, \qquad \varphi_2(b) =\varepsilon_4^{-m_3}\cdot(-1)^{m_6}, \qquad \varepsilon_3(b)=\varepsilon_8^{m_3}.
\end{equation}
\tag{3.7}
$$
Now let $\varkappa\in\widehat B$ be a character such that $\varkappa(b)=\varepsilon_{16}^m$ for some $m$. We have
$$
\begin{equation*}
\varkappa^{g_1-1}(b)=\varepsilon_8^{-m}, \qquad \varkappa^{g_2-1}(b)=\varepsilon_4^m, \qquad \varkappa^{g_3-1}(b)=\varepsilon_8^{-m},
\end{equation*}
\notag
$$
and hence the cocycle $d$ is homologous to either the cocycle $d_1$ or the cocycle $d_2$ of (3.1). It is not difficult to see that the cocycles $d_1$ and $d_2$ are non-trivial, not homologous to each other, and, in the group $H^1(G, \widehat B)$ are of order $2$. Lemma 2 is proved. 3.2. The additional embeddability condition and the adjoint problemsLet $K/k$ be a Galois extension of fields with finite group $F$. Let $A$, $B$ be finite $F$-modules, and let there exist an $F$-modular monomorphism $\alpha\colon A\to B$. Next, let $\operatorname{char}k=0$, and $\varepsilon_{\exp B}\in K$. By $\overline F$ we denote the group $\mathrm{Gal}(\overline k/k)$. Let us transform $A$ and $B$ into $\overline F$-modules using a deviation along the epimorphism of the constraint $\mu\colon\overline F\to F$. Consider the diagram where the rows are exact and constructed as in § 2.2, $\alpha^*$ is induced by the embedding $\alpha$, $\psi_1(c_{\chi})=c_{\alpha^*(\chi)}$, $\psi_2=\left.\psi_1\right|_{\operatorname{ker}\varphi_B}$. It is easily checked that $\psi_1$, $\psi_2$ are correctly defined $\overline F$-modular homomorphisms, and diagram (3.8) is commutative. By completeness of the group $\overline k^{\,*}$, diagram (3.8) induces the commutative diagram with exact rows where, as in § 2.2, $\mathrm{Hom}(\widehat A, \overline k^{\,*})$ is canonically identified with $A$, and $\mathrm{Hom}(\widehat B, \overline k^{\,*})$, with $B$. Diagram (3.9) induces the commutative diagram where as in § 2.2, $\delta_A$ and $\delta_B$ are monomorphisms.Let $h_A\in H^2(F, A)$ and $h_B\in H^2(F, B)$ be classes of group extensions, where $h_B=\gamma(h_A)$, and $\gamma\colon H^2(F, A)\to H^2(F, B)$ is a homomorphism induced by the embedding $\alpha$. Consider the embedding problems of the extension $K/k$ defined by the classes $h_A$ and $h_B$. Let us assume that in such problems the concordance condition is met (it is enough to require the concordance condition for a problem with the class $h_A$, because this problem is adjoined to a problem with the class $h_B$, and therefore, the concordance condition for a problem with the class $h_B$ is met automatically – this follows, for example, from diagram (3.10)). Let $\lambda_A\colon H^2(F, A)\to H^2(\overline F, A)$, $\lambda_B\colon H^2(F, B)\to H^2(\overline F, B)$ be the lifting homomorphisms induced by the epimorphism $\mu$. In this case, $\lambda_A(h_A)=\delta_A(x_A)$, and $\lambda_B(h_B)=\delta_B(x_B)$ for some uniquely defined elements $x_A\in H^1(\overline F, \mathrm{Hom}(V_A, \overline k^{\,*}))$ and $x_B\in H^1(\overline F, \mathrm{Hom}(V_B, \overline k^{\,*}))$. However,
$$
\begin{equation*}
\begin{gathered} \, \delta_B(x_B)=\lambda_B(h_B)=\lambda_B(\gamma(h_A))= (\alpha_*\lambda_A)(h_A)=\alpha_*(\delta_A(x_A)), \\ (\alpha_*\delta_A)(x_A)=(\delta_B)((\psi_2^*)_*)(x_A). \end{gathered}
\end{equation*}
\notag
$$
Here, $x_B=(\psi_2^*)_*(x_A)$, since $\delta_B$ is a monomorphism. Applying, as in § 2.2, the exact Hochschild–Serre sequence, we get the following result. Lemma 3. Let $K/k$ be a Galois extension of fields with finite group $F$. Let $A$, $B$ be finite $F$-modules, and let there exist an $F$-modular monomorphism $\alpha\colon A\to B$. Next, let $\operatorname{char}k=0$, and $\varepsilon_{\exp B} \in K$. Let $h_A \in H^2(F, A)$ and $h_B\in H^2(F, B)$ be classes of group extensions such that $h_B=\gamma(h_A)$, and $\gamma\colon H^2(F, A)\to H^2(F, B)$ be a homomorphism induced by the embedding $\alpha$. Consider the embedding problems of the extension $K/k$ defined by the classes $h_A$ and $h_B$. Suppose that the concordance condition for these problems is met. Let $y_A\in H^1(F, \mathrm{Hom}(V_A, K^*))$, $y_B\in H^1(F, \mathrm{Hom}(V_B, K^*))$ be the obstructions to the problems under consideration. Then $y_B=\varkappa(y_A)$, where $\varkappa\colon H^1(F, \mathrm{Hom}(V_A, K^*)) \to H^1(F, \mathrm{Hom}(V_B, K^*))$ is the homomorphism induced by the homomorphism $\psi_2$ from diagram (3.8) and the lifting isomorphisms of the cohomology from § 2.2. Corollary 1. Under the conditions of Lemma 3, let, in addition, $k$ be a number field. Consider the commutative diagram where, as in § 2.3, $\theta_A$ and $\theta_B$ are homomorphisms dual to the restriction homomorphism, and $\alpha_1$ and $\alpha_2$ are homomorphisms induced by the embedding $\alpha$. Let $h_A\in H^2(F, A)$ and $h_B\in H^2(F, B)$ be classes of group extensions with the condition $h_B=\gamma(h_A)$, where $\gamma\colon H^2(F, A)\to H^2(F, B)$ is a homomorphism induced by the embedding $\alpha$. Consider the embedding problems of the extension $K/k$ defined by the classes $h_A$ and $h_B$. Suppose that the concordance condition for these problems is met. Let $z_A\in\operatorname{coker}\theta_A$, $z_B\in\operatorname{coker}\theta_B$ be obstructions for the problems under consideration. Then $z_B=\alpha_2(z_A)$. Proof. By Lemma 3, Theorem 9.11 in [11], and Theorem D.3.5 in [3], which imply that there exist canonical isomorphisms we have the required result.
§ 4. Formulations and proofs of the theorems4.1.Let $k$ be a number field, $d\in k^*\setminus k^{*2}$. We set $K=k(\sqrt{d})$ and consider the embedding problem of the extension $K/k$ into a Galois algebra with the group $G_{b, c, n}$ with $n\geqslant 2$. Namely, we consider the problem $(K/k, G_{b, c, n}, \varphi)$, where $\varphi(c)$ is the generator of the group $\mathrm{Gal}(K/k)$, and $\varphi(b)=1$. Proposition 1. For $n>3$, the embedding problem $(K/k, G_{b, c, n}, \varphi)$ is solvable if and only if the condition $K\cdot\mathbb R=\mathbb R$ is met for $k\cdot\mathbb R=\mathbb R$. Proof. Note that if $K\cdot\mathbb R=\mathbb C$ under $k\cdot\mathbb R=\mathbb R$, then the problem is unsolvable: in fact, since $\mathbb C$ is algebraically closed, the kernel-field of any Galois algebra solving the problem coincides with $\mathbb C$, but then the problem is semidirect according to [3], Ch. 1, § 9, Theorem 1.9, a contradiction. So, the condition specified in the proposition is necessary. Let us prove its sufficiency. Note first that by Theorem 1 in [7], under the conditions of the proposition, the concordance condition is met for the problem $(K/k, G_{b, c, n}, \varphi)$ for $n\geqslant 3$. It remains thus to show that the additional condition for the embedding disappears for all $n\geqslant 4$. Note also that the problem $(K/k, G_{b, c, 4}, \varphi)$ is adjoined for all problems $(K/k, G_{b, c, n}, \varphi)$ with $n>4$. Therefore, it is sufficient to show that the additional condition of embedding disappears in the problem $(K/k, G_{b, c, 4}, \varphi)$. Next, let $B=\langle b\rangle$.
Consider the problems $(K/k, G_{b, c, 3}, \varphi)$ and $(K/k, G_{b, c, 4}, \varphi)$ and attach to the field $K$ an element $\varepsilon_{16}$ by lifting. Let $L=K(\varepsilon_{16})$, and $G=\mathrm{Gal}(L/k)$. There are several cases to consider. Case 1. The elements $d$, $2$, $-1$ of the space $k^*/k^{*2}$ are linearly independent over $\mathbb F_2$. In this case, Lemma 1 implies that $G=\langle g_1\rangle\times\langle g_2\rangle\times\langle g_3\rangle$, where
$$
\begin{equation*}
\varepsilon_{16}^{g_1}=\varepsilon_{16}^{-1}, \qquad \varepsilon_{16}^{g_2}=\varepsilon_{16}^5, \qquad \varepsilon_{16}^{g_3}=\varepsilon_{16}, \qquad\sqrt{d}^{\,g_3}=-\sqrt{d}.
\end{equation*}
\notag
$$
At the same time Therefore, by Lemma 2 the group $H^1(G, \widehat B)$ is generated by the classes $d_1$ and $d_2$ from (3.1). Let $A=\langle b^2\rangle$. Consider the cohomology homomorphism $\alpha_2$ defined in (3.11). We show that $\alpha_2$ is trivial. Let $\delta$ be a fixed element of $H^1(G, \widehat A)'$. Let also $\overline d_i=\alpha^*(d_i)$, where $\alpha^*\colon H^1(G, \widehat B)\to H^1(G, \widehat A)$ is the homomorphism induced by the embedding $\alpha\colon A\to B$. Next, let the character of $\psi\in\widehat B$ be such that $\psi(b)=-1$. We have
$$
\begin{equation*}
\begin{gathered} \, \alpha_2(\delta)(d_1)=\delta(\overline d_1), \qquad (\overline d_1)_{g_1}=\psi|_{A}, \qquad (\overline d_1)_{g_2}=(\overline d_1)_{g_3}=1, \\ \alpha_2(\delta)(d_2)=\delta(\overline d_2), \qquad (\overline d_2)_{g_2}=\psi|_{A}, \qquad (\overline d_1)_{g_1}=(\overline d_1)_{g_3}=1. \end{gathered}
\end{equation*}
\notag
$$
Hence $\alpha_2$ is trivial since $\psi|_A=1$. Hence by Corollary 1 the problem $(K/k, G_{b, c, 4}, \varphi)$ is solvable.
Case 2. Let the elements $d$, $2$, $-1$ of the space $k^*/k^{*2}$ be linearly dependent over $\mathbb F_2$. We set $L=K(\varepsilon_8)$. If the group $F=\mathrm{Gal}(L/k)$ is cyclic, then, for some prime divisor $\mathfrak p$ of the field $k$, the group $F$ coincides with the decomposition group $F_{\mathfrak p}$ of the point $\mathfrak p$ in $L$. In this case, the map $\theta_A$ from (3.11) is an epimorphism, and therefore, the problem $(K/k, G_{b, c, 3}, \varphi)$ is solvable, and hence so is the problem $(K/k, G_{b, c, 4}, \varphi)$. Suppose now that $F$ is not cyclic. In this case, the elements $d$, $2$ and $-1$ generate a two-dimensional subspace in $k^*/k^{*2}$, and $d\notin k^{*2}$. We also have $F\cong V_4$. If $|\mathrm{Gal}(L\cdot \mathbb Q_2/k\cdot \mathbb Q_2)|=4$, then the decomposition group of the prime divisor $\mathfrak p$ of the field $k$ over $2$ coincides with $F$; in this case, the homomorphism $\theta_A$ from (3.11) is again an epimorphism, which implies the solvability of the problem $(K/k, G_{b, c, 3}, \varphi)$, and, therefore, of the problem $(K/k, G_{b, c, 4}, \varphi)$. So, we can assume that the group $F_2=\mathrm{Gal}(L\cdot\mathbb Q_2/k\cdot\mathbb Q_2)$ is cyclic. We claim that in this case the problem is $(K/k, G_{b, c, 3}, \varphi)$ is solvable. According to Proposition 2 in [7], the solvability conditions of the problem have the form
$$
\begin{equation}
\exists\,x\in k^*\colon k[-d, x]\sim 1, \quad k(\sqrt{-d})[2, x]\sim k(\sqrt{-d})[-1, -1].
\end{equation}
\tag{4.1}
$$
By the condition, $k(\sqrt{-d})\cdot\mathbb R=\mathbb C$, and hence the algebra $k(\sqrt{-d})[-1, -1]$ may be different from $1$ in the group $B(k(\sqrt{-d}))$ only if $(k(\sqrt{-d})\cdot\mathbb Q_2:\mathbb Q_2) \not\equiv 0\,\,(\operatorname{mod}2)$. In this case, $d\equiv -1\ (\operatorname{mod}(k\cdot\mathbb Q_2)^{*2})$. Since the group $F_2$ is cyclic, some of the numbers $2$, $-1$, $-2$ is a square in the field $k\cdot\mathbb Q_2$. But then the degree of $(k\,\cdot\,\mathbb Q_2:\mathbb Q_2)$ is even. Therefore, conditions (4.1) in this case are met, for example, for $x=1$.
So, in the case under consideration, the problem is $(K/k, G_{b, c, 3}, \varphi)$ is solvable, and hence so is the problem $(K/k, G_{b, c, 4}, \varphi)$. Proposition 1 is proved. Theorem 1. For $n>4$, extension (1.2) is not ultrasolvable. Proof. Assume that, for some $n>4$ and some number field $k$, the problem $(k(\sqrt{d})/k, G_{b, c, n}, \varphi)$ is solvable, and the adjoint embedding problem $\bigl(k(\sqrt{d})/k, G_{b, c, n-1}, \varphi\bigr)$ is unsolvable. In this case, for $k\cdot\mathbb R=\mathbb R$, the condition $k(\sqrt{d})\cdot\mathbb R=\mathbb R$ is met. Now the required result is secured by Proposition 1. 4.2.Let again $k$ be a number field, $x\in k^*\setminus k^{*2}$. Consider, for $n>4$, the embedding problem $(k(\sqrt{x})/k, G_{b, c, n}, \psi)$, where $\operatorname{ker}\psi=G_{b^2, c, n-1}$, and $\psi(b)$ generates the group $\mathrm{Gal}(k(\sqrt{x})/k)$. Proposition 2. For $n>4$, the problem $(k(\sqrt{x})/k, G_{b, c, n}, \psi)$ is solvable if and only if for $k\cdot\mathbb R=\mathbb R$, the condition $k(\sqrt{x})\cdot\mathbb R=\mathbb R$ is met. Proof. The above embedding problem is not semidirect, and therefore, the condition of the proposition is necessary for solvability: if $k\cdot\mathbb R=\mathbb R$ and $k(\sqrt{x})\cdot \mathbb R= \mathbb C$, then by [3], Ch. 1, § 9, Theorem 1.9, even $\mathbb R$-localization of the problem $(k(\sqrt{x})/k, G_{b, c, n}, \psi)$ is unsolvable.
Let us show the sufficiency of the conditions of the proposition. It is easy to see that the problem $(k(\sqrt{x})/k, G_{b^2, cb, n-1}, \psi)$ is adjoined to the original one, and its kernel is a cyclic group with generator $b^2$. By the condition, the order of this group is not smaller than $2^4$, and, therefore, Proposition 1 applies. The next result follows from Propositions 1 and 2. Theorem 2. The embedding problem $(k(\sqrt{d_1})/k, G_{b, c, n}, \varphi)$, where $\varphi(b)=1$, $\varphi(c)$ is an element of order $2$, is solvable if and only if 1) if $n>3$, then, for $k\cdot\mathbb R=\mathbb R$, the condition $k(\sqrt{d_1})\cdot\mathbb R=\mathbb R$ is met; 2) if $n=3$, then $\exists\,x\in k^*\colon k[-d_1, x]\sim 1, k(\sqrt{-d_1})[2, x][-1, -1]\sim 1$; 3) if $n=2$, then $k(\sqrt{-d_1})[-1, -1]\sim 1$; 4) if $n=1$, then $k[-1, -1]\sim 1$. The embedding problem $(k(\sqrt{d_2})/k, G_{b, c, n}, \psi)$, where $\psi(c)=1$, $\psi(b)$ is an element of order $2$, is solvable if and only if 1) if $n>2$, then, for $k\cdot\mathbb R=\mathbb R$, the condition $k(\sqrt{d_2})\cdot\mathbb R=\mathbb R$ is met; 2) if $n=2$, then $k(\sqrt{-d_2})[-1, -1]\sim 1$; 3) if $n=1$, then $k[-1, -1]\sim 1$. Proof. For the problem $(k(\sqrt{d_1})/k, G_{b, c, n}, \varphi)$ for $n>3$, the result follows from Proposition 1; for $n=3$, from Proposition 2 in [7]; for $n\in\{1, 2\}$, the result is well known (the case $n=1$ is described, for example, in the introduction to [3]; for the case $n=2$, see [7], Theorem 1).
For the problem $(k(\sqrt{d_2})/k, G_{b, c, n}, \psi)$, for $n>4$, the result follows from Proposition 2; for the case $n=4$, the result is known and proved in [13]; the case $n=3$ follows from [3], Ch. 5, § 1, Theorem 5.1.4: in fact, the proof hold verbatim in the case of $p=2$ if the localization of the problem by real divisors is solvable (see also [13]) – this is indeed so by the assumption. In the cases of $n\in\{1, 2\}$ it turns out a problem equivalent to $(k(\sqrt{d_1})/k, G_{b, c, n}, \varphi)$ with $d_1$ replaced by $d_2$. 4.3.Consider the problem $(K/k, G_{b, c, n}, \gamma)$, where $K=k(\sqrt{d_1}, \sqrt{d_2})$ is the extension of the field $k$ with the Klein group, $b^2$ generates the kernel of the epimorphism $\gamma$, and
$$
\begin{equation}
\begin{alignedat}{2} \sqrt{d_1}^{\,\gamma(c)} &=-\sqrt{d_1}, &\qquad \sqrt{d_1}^{\,\gamma(b)} &=\sqrt{d_1}, \\ \sqrt{d_2}^{\,\gamma(c)} &=\sqrt{d_2}, &\qquad \sqrt{d_2}^{\,\gamma(b)} &=-\sqrt{d_2}. \end{alignedat}
\end{equation}
\tag{4.2}
$$
Let $n\geqslant 4$. Let us find the concordance condition for the problem $(K/k, G_{b, c, n}, \gamma)$. The crossed product $G_{b, c, n}\times K$ has the central pairwise orthogonal idempotents
$$
\begin{equation}
\begin{gathered} \, E_0 =\frac{1+b^{2^{n-1}}}{2}\cdot\frac{1+b^{2^{n-2}}}{2}\cdots \frac{1+b^4}{2}\cdot\frac{1+b^2}{2}, \\ E_1 =\frac{1+b^{2^{n-1}}}{2}\cdot\frac{1+b^{2^{n-2}}}{2}\cdots \frac{1+b^4}{2}\cdot\frac{1-b^2}{2}, \\ \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \\ E_j =\frac{1+b^{2^{n-1}}}{2}\cdot\frac{1+b^{2^{n-2}}}{2}\cdots \frac{1+b^{2^{j+1}}}{2} \cdot\frac{1-b^{2^j}}{2}, \\ \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \\ E_{n-1} =\frac{1-b^{2^{n-1}}}{2}. \end{gathered}
\end{equation}
\tag{4.3}
$$
It is easy to see that $E_0+\dots+E_{n-1}=1$. We set $\Lambda_j=(G_{b, c, n}\times K)E_j$. The above algebra $\Lambda_0$ is a crossed product $V_4\times K$, which corresponds to a trivial embedding problem, so $\Lambda_0$ is a matrix algebra of order $4$ over some its subalgebra. The algebra $\Lambda_1$ contains the subalgebras $\langle bE_1, \sqrt{d_2}\,E_1\rangle_k$ and $\langle c\sqrt{d_2}\,E_1, \sqrt{d_1}\,E_1\rangle_k$, which commute and are isomorphic, respectively, to $k[-1, d_2]$ and $k[d_2, d_1]$. Accordingly, $\Lambda_1$ is a matrix algebra of order $4$ over some its subalgebra if and only if $k[-d_1, d_2]\sim 1$. Consider the sequence
$$
\begin{equation*}
\theta_0=2, \quad \theta_1=\sqrt{2}, \quad \theta_2=\sqrt{2+\sqrt{2}}, \quad \dots, \quad \theta_k=\sqrt{2+\theta_{k-1}}.
\end{equation*}
\notag
$$
The centre of the algebra $\Lambda_j$ for $1<j<n-1$ is generated over $k$ by the elements $(b^2+b^{-2})E_j$ and $b^{2^{j-1}}\sqrt{d_1}\,E_j$. Accordingly, the centre of the algebra $\Lambda_j$ over $k$ is isomorphic to the field $k_j=k(\sqrt{-d_1}, \theta_{j-2})$. Over the field $k_j$, the algebra $\Lambda_j$ contains the commuting subalgebras $\langle(b+b^{-1})E_j, \sqrt{d_2}\,E_j\rangle_{k_j}$ and $\langle cE_j, \sqrt{d_1}\,E_j\rangle_{k_j}$, which are isomorphic, respectively, to the algebras $k_j[2+\theta_{j-2}, d_2]$ and $k_j[1, d_1]$. So, the algebra $\Lambda_j$ is a matrix algebra of order $4$ over some its subalgebra if and only if $k_j[2+\theta_{j-2}, d_2]\sim 1$. Now let us consider the algebra $\Lambda_{n-1}$. The centre of $\Lambda_{n-1}$ is generated over $k$ by the elements $(b^2+b^{-2})E_{n-1}$ and $b^{2^{n-2}}E_{n-1}$. That is, the centre the algebra $\Lambda_{n-1}$ is isomorphic over $k$ to the field $k_{n-1}=k(\sqrt{-d_1}, \theta_{n-3})$. Over the field $k_{n-1}$, the algebra contains the commuting subalgebras $\langle(b+b^{-1})E_{n-1}, \sqrt{d_2}\,E_{n-1}\rangle_{k_{n-1}}$ and $\langle cE_{n-1}, \sqrt{d_1}\,E_{n-1}\rangle_{k_{n-1}}$, which are isomorphic, respectively, to the algebras $k_{n-1}[2+ \theta_{n-3}, d_2]$ and $k_{n-1}[-1, d_1]$. But $k_{n-1}[-1, d_1]\sim k_{n-1}[-1, -1]$, and, therefore, $\Lambda_{n-1}$ is a matrix algebra of order $4$ over some of its subalgebra if and only if $k_{n-1}[2+\theta_{n-3}, d_2][-1, -1]\sim 1$. So, in the problem $(K/k, G_{b, c, n}, \gamma)$ with $n\geqslant 4$ the concordance condition is met if and only if
$$
\begin{equation}
\begin{gathered} \, k[-d_1, d_2]\sim 1, \\ k_j[2+\theta_{j-2}, d_2]\sim 1\quad\forall\,j\in[2, n-2]\cap\mathbb N, \\ k_{n-1}[2+\theta_{n-3}, d_2][-1, -1]\sim 1. \end{gathered}
\end{equation}
\tag{4.4}
$$
Theorem 3. Consider the problem $(K/k, G_{b, c, n}, \gamma)$, where $K=k(\sqrt{d_1}, \sqrt{d_2})$ is the extension of the field $k$ with the Klein group, $b^2$ generates the kernel of the epimorphism $\gamma$, and $\gamma(c)$ acts on $K$ according to (4.2). Suppose that $n\geqslant 4$, $k$ is a number field, and the numbers $\sqrt 2$, $\sqrt{-1}$, $\sqrt{-2}$ do not belong to $K\cdot\mathbb Q_2$. Suppose that the concordance conditions (4.4) are met. Then the problem $(K/k, G_{b, c, n}, \gamma)$ is solvable. Proof. Note that the kernel $B=\langle b^2\rangle $ of the problem $(K/k, G_{b, c, n}, \gamma)$ is a cyclic group of order $2^{n-1}$. Therefore, if the concordance conditions (4.4) are met, we only need to verify that the additional embedding condition disappears. To do this, we apply the idea from Lemma 2 in [12].
Namely, we attach the root $\varepsilon_{2^{n-1}}$ to the field $K$ by lifting and consider the lifted problem $(K_1/k, \overline G_{b, c, n}, \gamma)$, where $K_1=K(\varepsilon_{2^{n-1}})$. Let $F_1$ be the group $\mathrm{Gal}(K_1/k)$. In the lifted problem, the concordance condition is also fulfilled, and therefore, it is sufficient to show that, under the conditions of the theorem, the intersection of all kernels of the restriction homomorphisms $\theta_{\mathfrak p}\colon H^1(F_1, \widehat B)\to H^1(F_{1\mathfrak p}, \widehat B)$ on the decomposition groups $F_{\mathfrak p}$ of prime divisors $\mathfrak p$ of the field $k$ in the extension $K_1/k$ is trivial. Let us prove this claim. Let $c$ be an arbitrary cocycle of the group $F_1$ with values in $\widehat B$. Let $F_0$ be a subgroup in $F_1$ acting trivially on $\widehat B$. We claim that $c_{f_0}=1$ for any element $f_0\in F_0$. Indeed, $\langle f_0\rangle=F_{1\mathfrak p}$ for some prime divisor $\mathfrak p$ of the field $k$, and therefore, the restriction of the cocycle $c$ to the group $\langle f_0\rangle$ decays, that is, there is a character $\chi\in\widehat B$ such that $c_{f_0}=\chi^{f_0-1}$. Now the required result follows since $\chi^{f_0}=\chi$. Note that $\widehat B$ is a cyclic group of order $2^{n-1}$. Hence, for every $f\in F_1$, there exists an integer $s$ such that $\chi^f=\chi^s$ for all $\chi\in\widehat B$. By the condition of the theorem, the numbers $\sqrt 2$, $\sqrt{-1}$, $\sqrt{-2}$ do not belong to $K\cdot\mathbb Q_2$. But then the fields $K\cdot \mathbb Q_2$ and $k\cdot\mathbb Q_2(\varepsilon_{2^{n-1}})$ are linearly separated over $k\cdot\mathbb Q_2$. In addition, the fields $k\cdot \mathbb Q_2$ and $\mathbb Q_2(\varepsilon_{2^{n-1}})$ are linearly separated over $\mathbb Q_2$. Therefore, for any prime divisor $\mathfrak p$ of the field $k$ lying over $2$, the group $F_{1\mathfrak p}$ contains a subgroup isomorphic to $\mathrm{Gal}(\mathbb Q(\varepsilon_{2^{n-1}})/\mathbb Q)$ and acting trivially on $B$. Let $\widehat F_{1\mathfrak p}$ be the specified subgroup of the above group $F_{1\mathfrak p}$. For the element $f_1\in\widehat F_{1\mathfrak p}$, the character $\chi\in\widehat B$ and the element $x\in B$, we have
$$
\begin{equation*}
\chi^{f_1}(x)=\chi\bigl(x^{f_1^{-1}}\bigr)^{f_1}=\chi(x)^{f_1},
\end{equation*}
\notag
$$
since $\widehat F_{1\mathfrak p}$ acts on $B$ trivially. Therefore, by construction, for each integer $s$, there exists $f_1\in\widehat F_{1\mathfrak p}$ such that $\chi^{f_1}=\chi^s$ for all $\chi\in\widehat B$.
So, any element $f \in F_1$ is represented as the product $f_1f_0$ with $f_1 \in \widehat F_{1\mathfrak p}$ and $f_0\in F_0$. When restricted to the group $F_{1\mathfrak p}$, the cocycle $c$ decays, which means that there exists $\chi\in\widehat B$ with the property
$$
\begin{equation*}
c_{f_1}=\chi^{f_1-1}\quad \forall\,f_1\in F_{1\mathfrak p}.
\end{equation*}
\notag
$$
If $f=f_1f_0$ with $f_1\in\widehat F_{1\mathfrak p}\leqslant F_{1\mathfrak p}$ and $f_0\in F_0$, then
$$
\begin{equation*}
c_f=c_{f_1f_0}=c_{f_1}^{f_0}c_{f_0}=\chi^{(f_1-1)f_0}=\chi^{f-1},
\end{equation*}
\notag
$$
that is, the cocycle $c$ decays. Theorem 3 is proved. Corollary 2. Let, under the conditions of Theorem 3, $k=\mathbb Q$, $d_1$ and $d_2$ be primes comparable to $1$ modulo $2^n$, and such that $d_1$ is a square in the field $\mathbb F_{d_2}$. Then the problem $(K/k, G_{b, c, n}, \gamma)$ is solvable. Proof. Since $d_i\equiv 1\ (\operatorname{mod}2^n)$, it follows that $2$ is completely decomposed in $K$. So, it suffices to verify the concordance condition (4.4). Since $d_1$ is a square in the field $\mathbb Q_{d_2}$, we have $\mathbb Q_{d_2}[d_1, d_2]\sim 1$. By the Artin reciprocity law, $\mathbb Q[d_1, d_2]\sim 1$. Therefore, $\mathbb Q[-d_1, d_2]\sim 1$. So, the first condition in (4.4) is met. It can be directly verified that the other conditions in (4.4) are also met. However, we will proceed in a simpler way.
Let $p$ be an arbitrary prime number. Consider the $p$-localization $(K\cdot\mathbb Q_p/\mathbb Q_p, G_{b, c, n}, \gamma)$ of the original problem. If $p\notin\{d_1, d_2, 2\}$, then the extension $K\cdot\mathbb Q_p/\mathbb Q_p$ is unramified, and therefore, $p$-localization is solvable by Ch. 3, § 14, Lemma 3.14 in [3]. If $p=2$, then $p$-localization is also solvable, because $2$ is completely decomposed in $K$. Let $p=d_1$. Since $\mathbb Q[d_1, d_2]\sim 1$, we have $K\cdot\mathbb Q_p/\mathbb Q_p$ is a completely branched extension of degree $2$. Since $d_1\equiv 1\ (\operatorname{mod} 2^n)$, we have $\varepsilon_{2^n}\in\mathbb Q_p$. Let $m\in\mathbb N$ be such that $\varepsilon_{2^m}\in\mathbb Q_p$, and $\varepsilon_{2^{m+1}}\notin\mathbb Q_p$. In this case, $m\geqslant n$. Note now that the kernel of the problem $(K\cdot\mathbb Q_p/\mathbb Q_p, G_{b, c, n}, \gamma)$ is cyclic of order $2^{n-1}$. Since $m-(n-1)\geqslant 1$, the considered $p$-local problem is solvable by Theorem 6 in [14]. Let $p=d_2$. As before, $\mathbb Q[d_1, d_2]\sim 1$, and now the same argument applies as for $p=d_1$. Finally, if we consider the $\mathbb R$-localization of the problem under consideration, then this problem is solvable because $K\cdot\mathbb R=\mathbb R$. So, all the localizations of the problem $(K/k, G_{b, c, n}, \gamma)$ are solvable, and therefore, condition (4.4) is met. Remark 1. Theorem 3 and Corollary 2 give conditions for the parameters $d_1$ and $d_2$, under which the extensions $k(\sqrt{d_1})/k$, $k(\sqrt{d_2})/k$, $k(\sqrt{d_1d_2})/k$ are embedded into an extension of fields $L/k$ with generalized quaternion group $G_{b, c, n}$ for $n\geqslant 4$. For the case $n=3$, the concordance conditions (4.4) are sufficient for the solvability of the problem $(K/k, G_{b, c, n}, \gamma)$ because the kernel of this problem is cyclic of order $4$. 
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