MATHEMATICS

A. V. Yakovlev

THE FIELD EMBEDDING PROBLEM

(Presented by Academician I. M. Vinogradov on 28 XII 1962)

Let

\[ 1 \to \mathfrak A \to \mathfrak G \xrightarrow{\pi} \mathfrak F \to 1 \]

be an exact sequence of groups, where \(\mathfrak A\) is an abelian group of period \(n\), and \(k\) is a finite normal extension of the field \(k_0\) with group \(\mathfrak F\), containing a primitive \(n\)-th root of unity. It is required to construct a field (or Galois algebra) \(K\), containing \(k\) and normal over \(k_0\) with group \(\mathfrak G\), such that for \(g \in \mathfrak G\) we have \(g|_k=\pi(g)\). Arguing analogously to \((^2)\), one can reduce this problem to the following one.

Let \(C\) be the multiplicative group of the field \(k\) (which we shall write additively), \(\overline{\mathfrak A}=\operatorname{Hom}_{\mathbb Z}(\mathfrak A,C)\), and suppose that an operator \(f\) from \(\mathfrak F\) acts on an element \(\overline a\in\overline{\mathfrak A}\) by the formula

\[ (\overline a f)(a)=[\overline a(faf^{-1})]f \]

for every \(a\in\mathfrak A\). The groups \(C\) and \(\overline{\mathfrak A}\), by means of \(\pi\), are \(\mathfrak G\)-modules. It is required to construct a right \(\mathfrak G\)-module \(X\) such that: 1) the sequence of \(\mathfrak G\)-modules

\[ 0\to C \xrightarrow{\psi} X \xrightarrow{\varphi} \overline{\mathfrak A}\to 0 \]

is exact; 2) for any \(a\in\mathfrak A\), \(x\in X\), we have the equality

\[ xa=x+\psi[\varphi(x)(a)]. \]

Condition 1) means that \(X\) is an extension of \(\overline{\mathfrak A}\) by means of \(C\) and may be specified as follows. Let \(\beta:P\to\overline{\mathfrak A}\) be an epimorphism, where \(P\) is a \(\mathfrak G\)-projective module. Denote by \(Z(\overline{\mathfrak A})\) the integral group module for \(\overline{\mathfrak A}\), with the operators from \(\mathfrak F\) permuting the generators of this module in the same way as the corresponding elements of \(\overline{\mathfrak A}\). For the sequel, we choose \(P\) so that \(\beta\) is the composition of epimorphisms

\[ \beta_1:P\to Z(\overline{\mathfrak A}),\qquad \beta_2:Z(\overline{\mathfrak A})\to\overline{\mathfrak A}, \]

where \(\beta_2\) assigns to the generators of the module \(Z(\overline{\mathfrak A})\) the corresponding elements of \(\overline{\mathfrak A}\). Denote \(\operatorname{Ker}\beta\) by \(M\). \(X\) is completely determined by an element

\[ \gamma\in\operatorname{Hom}_{\mathfrak G}(M,C). \]

Condition 2) means that

\[ \gamma[p(a-1)]=(\beta p)(a)\qquad (p\in P,\ a\in\mathfrak A). \]

Let \(M_0\) be the \(\mathfrak G\)-submodule of \(M\) generated by all elements \(p(a-1)\). Thus, the restriction of \(\gamma\) to \(M_0\) must coincide with the homomorphism known to us, which we denote by \(\gamma_1\).

There is a commutative diagram

\[ \begin{array}{cc} 0 & 0\\ \downarrow & \downarrow\\ M_0 & = M_0\\ \downarrow & \downarrow\\ 0\to M & \to P \xrightarrow{\beta} \overline{\mathfrak A}\to 0\\ \downarrow & \downarrow\\ 0\to M/M_0 & \to P/M_0 \xrightarrow{\overline{\beta}} \overline{\mathfrak A}\to 0\\ \downarrow & \downarrow\\ 0 & 0 \end{array} \]

with naturally defined homomorphisms. It induces the following diagram with exact rows and a commutative triangle:

\[ \begin{array}{c} \operatorname{Hom}_{\mathfrak G}(M,C)\\ \downarrow\alpha_1\\ \operatorname{Hom}_{\mathfrak G}(M_0,C)\\ \swarrow\alpha_3\qquad \downarrow\alpha_2\\ \operatorname{Ext}^1_{\mathfrak G}(\overline{\mathfrak A},C) \xrightarrow{\alpha_4} \operatorname{Ext}^1_{\mathfrak G}(P/M_0,C) \xrightarrow{\alpha_5} \operatorname{Ext}^1_{\mathfrak G}(M/M_0,C) \end{array} \]

From the preceding it follows that, for solvability of the embedding problem, it is necessary and sufficient that there exist a

\[ \gamma\in\operatorname{Hom}_{\mathfrak G}(M,C) \]

such that

\[ \alpha_1\gamma=\gamma_1. \]

The last diagram makes it possible to give two more forms of this condition:
1) \(\alpha_2\gamma_1=0\), or, what is the same, \(\alpha_5\gamma_2=0\), where \(\gamma_2=\alpha_3\gamma_1\in \operatorname{Ext}^1_{\mathfrak G}(P/M_0,C)\); 2) there exists \(\gamma_3\in \operatorname{Ext}^1_{\mathfrak G}(\overline{\mathfrak A},C)\) for which \(\alpha_4\gamma_3=\gamma_2\).

The homomorphism \(\overline\beta:P/M_0\to \overline{\mathfrak A}\) can be represented as a composite homomorphism
\[ P/M_0\to Z(\overline{\mathfrak A})\to \overline{\mathfrak A}, \]
since \(M_0\subset \operatorname{Ker}\beta\). Therefore \(\alpha_4\) is represented as a composite homomorphism
\[ \operatorname{Ext}^1_{\mathfrak G}(\overline{\mathfrak A},C) \xrightarrow{\alpha_6} \operatorname{Ext}^1_{\mathfrak G}(Z(\overline{\mathfrak A}),C) \xrightarrow{\alpha_7} \operatorname{Ext}^1_{\mathfrak G}(P/M_0,C). \]
The existence of \(\gamma_4\in \operatorname{Ext}^1_{\mathfrak G}(Z(\overline{\mathfrak A}),C)\) such that \(\alpha_7\gamma_4=\gamma_2\) (one can see that \(\gamma_4\) is then determined uniquely) is equivalent to the compatibility condition \((1^2)\).

The additional condition for embedding is that there exists
\[ \gamma_3\in \operatorname{Ext}^1_{\mathfrak G}(\overline{\mathfrak A},C) \]
for which \(\alpha_6\gamma_3=\gamma_4\). Denote \(\operatorname{Ker}\beta_2\) by \(D\). We have the exact sequence
\[ \operatorname{Ext}^1_{\mathfrak G}(\overline{\mathfrak A},C) \xrightarrow{\alpha_6} \operatorname{Ext}^1_{\mathfrak G}(Z(\overline{\mathfrak A}),C) \xrightarrow{\alpha_8} \operatorname{Ext}^1_{\mathfrak G}(D,C). \]
Therefore the additional condition takes the form
\[ \gamma_5=\alpha_8\gamma_4=0. \]
But
\[ \operatorname{Ext}^1_{\mathfrak G}(D,C)=H^1(\mathfrak G,\operatorname{Hom}_Z(D,C)) \]
(see (4), Theorem XI.9.2). Let \(\mathfrak G_0\supset \mathfrak A\) be a normal subgroup of \(\mathfrak G\) consisting of all elements acting trivially on \(\overline{\mathfrak A}\); \(\overline{\mathfrak F}=\mathfrak G/\mathfrak G_0\); \(C_0\) the subfield of \(C\) belonging to \(\pi(\mathfrak G_0)\). The exact sequence
\[ 0\to H^1(\overline{\mathfrak F},\operatorname{Hom}_Z(D,C_0)) \xrightarrow{\lambda} H^1(\mathfrak G,\operatorname{Hom}_Z(D,C)) \xrightarrow{i} H^1(\mathfrak G_0,\operatorname{Hom}_Z(D,C)), \]
where \(\lambda\) is the lift homomorphism and \(i\) the restriction homomorphism. \(i\gamma_5=0\), since this equality is equivalent to the solvability of the embedding problem corresponding to the sequence
\[ 1\to \mathfrak A\to \mathfrak G_0\to \pi(\mathfrak G_0)\to 1, \]
and this problem is Brauerian. Therefore
\[ \gamma_5=\lambda\delta,\qquad \delta\in H^1(\overline{\mathfrak F},\operatorname{Hom}_Z(D,C)) =\operatorname{Ext}^1_{\overline{\mathfrak F}}(D,C_0). \]

Thus we have proved

Theorem 1. If the compatibility condition is fulfilled, then for embeddability it is necessary and sufficient in addition that the element \(\delta\) of \(\operatorname{Ext}^1_{\overline{\mathfrak F}}(D,C_0)\) be equal to \(0\).

Theorem 2. If \(\overline{\mathfrak F}=\langle f\rangle\) is a cyclic group, then compatibility is sufficient for embeddability.

The proof is based on a chain of lemmas, most of which are only formulated here. By the theorems of Kochendörffer \((^3)\) it is enough to consider \(\overline{\mathfrak F}\) and \(\overline{\mathfrak A}\) as \(p\)-groups. The group \(H^0(\overline{\mathfrak F},C_0)\) is Abelian and \((\overline{\mathfrak F}:1)\) is its period. By Prüfer’s theorem (see, for example, \((^6)\), p. 156), it is a direct sum of cyclic groups \(\{a_i\}\), \(i\in I\). By \(a_i\) denote the invariant element of \(C_0\) corresponding to \(\overline a_i\). Let \(n_i\) be the order of \(\overline a_i\), \(n_i m_i=(\overline{\mathfrak F}:1)\).

Lemma 1. For every \(i\in I\) there exists \(c_i\in C_0\), \(c_i(f^{m_i}-1)=0\),
\[ a_i=c_i(1+f+\cdots+f^{m_i-1}). \]

Next, consider the \(\overline{\mathfrak F}\)-module
\[ B=C\oplus \sum \oplus Z(\overline{\mathfrak F})e_i, \]
where \(Z(\overline{\mathfrak F})\) is the integral group module of the group \(\overline{\mathfrak F}\), and its \(\overline{\mathfrak F}\)-submodule \(B_0\), generated by the elements
\[ c_i=e_i(1+f^{m_i}+\cdots+f^{m_i(n_i-1)}). \]
Obviously,
\[ B_0\simeq \sum \oplus Z(\overline{\mathfrak F}/\langle f^{m_i}\rangle). \]

Lemma 2. \(H^1(\overline{\mathfrak F}_1,B/B_0)=0\) for every subgroup \(\overline{\mathfrak F}_1\) of the group \(\overline{\mathfrak F}\);
\[ H^2(\overline{\mathfrak F},B/B_0)=0. \]

The proof of this lemma is quite cumbersome.

Lemma 3. Let \(Z_p\) be a cyclic group of order \(p\), on which the operators from \(\overline{\mathfrak F}\) act trivially. Then
\[ \operatorname{Ext}^2_{\overline{\mathfrak F}}(Z_p,B/B_0)=0. \]

Proof. The exact sequence
\[ \operatorname{Ext}^1_{\overline{\mathfrak F}}(pZ,B/B_0) \to \operatorname{Ext}^2_{\overline{\mathfrak F}}(\mathfrak A,B/B_0) \to \operatorname{Ext}^2_{\overline{\mathfrak F}}(Z,B/B_0). \]
By Lemma 2 the end terms are equal to \(0\).

Lemma 4.
\[ \operatorname{Ext}^2_{\overline{\mathfrak F}}(\overline{\mathfrak A},B/B_0)=0. \]

It is proved by induction on \((\overline{\mathfrak A}:1)\) with the aid of Lemma 3.

Lemma 5.
\[ \operatorname{Ext}^1_{\overline{\mathfrak F}}(D,B/B_0)=0. \]

Proof. The sequence

\[ \operatorname{Ext}^{1}_{\mathfrak F}(Z(\overline{\mathfrak A}), B/B_0) \longrightarrow \operatorname{Ext}^{1}_{\mathfrak F}(D, B/B_0) \longrightarrow \operatorname{Ext}^{2}_{\mathfrak F}(\overline{\mathfrak A}, B/B_0) \]

is exact. Its first term is equal to \(0\) by the first assertion of Lemma 2, and the last one by Lemma 4.

Lemma 6. \(\operatorname{Ext}^{1}_{\mathfrak F}(D, B_0)=0\).

Lemma 7. \(\operatorname{Ext}^{1}_{\mathfrak F}(D, C)=\operatorname{Ext}^{1}_{\mathfrak F}(D, B)=0\).

The assertion of Theorem 2 follows from Lemma 7.

It is proved similarly that, if \(k\) is a local field, then consistency is sufficient for embeddability (5).

The same methods make it possible to describe the set of solutions of the embedding problem:

Theorem 3. If \(\mathfrak G\) is a semidirect extension of \(\mathfrak F\) by means of \(\mathfrak A\), then there exists a one-to-one correspondence between the solutions of the embedding problem and the elements of the group \(\operatorname{Ext}^{1}_{\mathfrak F}(\overline{\mathfrak A}, C)\).

Indeed, let the \(\mathfrak F\)-module \(X\) be an extension of \(\overline{\mathfrak A}\) by means of \(C\). Put

\[ xa=x+\psi[\varphi(x)(a)],\qquad x\in X,\quad a\in\mathfrak A. \]

Then, as is easy to see, \(X\) is a \(\mathfrak G\)-module satisfying conditions 1) and 2), which were formulated at the beginning.

Received
24 XII 1962

CITED LITERATURE

\(^{1}\) B. N. Delone, D. K. Faddeev, Matem. sborn., 15 (57), 2, 243 (1944).
\(^{2}\) H. Hasse, Math. Nachr. 1, 40 (1948).
\(^{3}\) R. Kochendörffer, Math. Nachr. 10, 75 (1953).
\(^{4}\) A. Cartan, S. Eilenberg, Homological Algebra, IL, 1960.
\(^{5}\) S. P. Demushkin, I. R. Shafarevich, Izv. AN SSSR, ser. matem., 23, 823 (1959).
\(^{6}\) A. G. Kurosh, Group Theory, Moscow, 1953.