Given a Galois extension of number fields $L/K$ and an exact sequence of groups $$1\to \ker \varphi\to G \overset{\varphi}{\to} \text{Gal}(L/K)\to 1$$ where $G$ is a finite group, $\ker \varphi$ is solvable and $(|\ker \varphi|, |G/\ker\varphi|)=1$, then there exists a Galois extension $N/K$ that contains $L$ and such that $G=\text{Gal}(N/K)$ and $\varphi(\sigma)=\sigma|_L$ (see V.V. Ishkhanov, The Embedding Problem in Galois Theory).

My question: is it true that in this case the embedding problem has infinitely many solutions? i.e. there exist infinitely many such $N/K$ ? I'm particularly interested in the case $\ker \varphi$ is a $2-$group. And what happens when $\ker \varphi$ is abelian?

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    $\begingroup$ This might be a better fit on Math.SE $\endgroup$ – David White Aug 15 '16 at 1:29
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    $\begingroup$ I would like to strongly disagree that this belongs on math.SE . This involves digging into the details of a proof (Shafarevich's theorem that all solvable groups are realizable over $\mathbb{Q}$) that not many people have worked through. I imagine I could work out the abelian case if I put in the time, but it is still work, and the solvable case should involve some real digging in the literature. $\endgroup$ – David E Speyer Aug 15 '16 at 13:12

The answer is positive, and the abelian case can be handled even over 'Hilbertian fields'.

By the Schur-Zassenhaus Theorem the exact sequence splits, so we get a so called split embedding problem. If the kernel is abelian, the embedding problem is regularly solvable. That is, it is (properly) solvable over the function field $K(T)$. Since $K$ is Hilbertian (i.e. satisfies Hilbert's irreducibility theorem), the solution over $K(T)$ can be specialized to a solution over $K$ in infinitely many distinct ways. In fact, the infinitely many specializations can be taken linearly disjoint over the fixed field of $\mathrm{Ker}(\varphi)$ (i.e. $L$).

For more details see the book 'Field Arithmetic' by Fried and Jarden.

More generally, consider the embedding problem where $G$ is replaced by its $n$-th fibered power over $\varphi$. The solvability of this embedding problem (which follows from the result stated by the OP) implies that the original embedding problem has $n$ distinct solutions for each positive integer $n$, as required.

The $n$-th fibered power $G^n_{\varphi}$ of $G$ over $\varphi$ is the subgroup of all $(g_1, \dots,g_n) \in G^n$ such that $\varphi(g_i) = \varphi(g_1)$ for all $1 \leq i \leq n$. It fits into the exact sequence

$$1\to \ker(\varphi|_G)^n \to G^n_{\varphi} \overset{\varphi}{\to} \text{Gal}(L/K)\to 1.$$

Even more generally, it was conjectured that the answer to this question is positive also in the case that the kernel is not solvable. Specifically, Debes and Deschamps conjectured that every finite split embedding problem is solvable over a number field (so in particular, the inverse Galois problem has an affirmative answer).

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Let $K=\mathbf{Q}$, and $L$ a $p$-th cyclotomic extension for $p$ a prime number.

Then by Kummer we can construct infinitely many extensions $N$ of degree $p$ over $L$ by adjoining $p$ th roots of elements of $L/{L^*}^p$. By linear disjointness this will be Galois over $\mathbf{Q}$. This will work if we replace $L$ by any Galois extension of rationals containing $L$ which could be a non-abelian extension of $\mathbf{Q}$ whose degree is coprime to $p$. So we have a large collection of cases where it is true even if the group is non-solvable.

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  • $\begingroup$ When $p\neq2$, not every degree-$p$ cyclic extension $N$ of $L=\mathbf{Q}(\zeta_p)$ is galoisian over $\mathbf{Q}$ ! It is necessary and sufficient for $N$ to correspond to a $G$-stable $\mathbf{F}_p$-line in $L^\times\!/L^{\ast p}$, where $G=\mathrm{Gal}(L| \mathbf{Q})$. $\endgroup$ – Chandan Singh Dalawat Aug 16 '16 at 8:40
  • $\begingroup$ A degree $p$ extension over $\mathbf{Q}$, got by adjoining a $p$th root of a *suitable* element, say a rational prime number, and $p$th cyclotomic extension of $\mathbf{Q}$, their compositum can be the $N$. Is not this Galois over the rationals? $\endgroup$ – P Vanchinathan Aug 16 '16 at 8:47
  • $\begingroup$ For every $a\in\mathbf{Q}^\ast $, of course $\mathbf{Q}(\root p\of{a},\root p\of1)$ is galoisian over $\mathbf{Q}$. $\endgroup$ – Chandan Singh Dalawat Aug 16 '16 at 8:51

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