Cyclic extensions of a number field of full local degree in a given set $S$

Question

Let $K$ be a number field, and let $S=S_f\cup S_{\mathbb R}\cup S_{\mathbb C}$ be a finite set of places of $K$, where $S_f$ denotes the set of finite places in $S$, $S_{\mathbb R}$ denotes the set of real places in $S$, and $S_{\mathbb C}$ denotes the set of complex places in $S$. Let $m=m_1 m_2 \ge 4$ be a natural number, where the natural numbers $m_1$ and $m_2$ are both even. We say that a Galois extension $L/K$ of degree $m$ is of full local degree in $S$ if for any place $v\in S_f$ and for any place $w$ of $L$ over $v$ we have $[L_w:K_v]=m=[L:K]$, and for any place $v\in S_{\mathbb R}$ all places $w$ of $L$ over $v$ are complex.

By Theorem X.6 in the second edition of the book "Class field theory" by Artin and Tate, there exists a cyclic extension $L/K$ of degree $m$ of full local degree in $S$. Consider the unique subextension $L/E/K$ of degree $[E:K]=m_1$. Clearly, $E/K$ is a cyclic extension of degree $m_1$ of full local degree in $S_f$.

Question. For given $K$, $S$, and $m=m_1 m_2$, does there exist a cyclic extension $L/K$ of degree $m$ of full local degree in $S$ such that the subextension $L/E/K$ of degree $m_1$ over $K$ is of full local degree in $S_{\mathbb R}$ (that is, all places of $E$ over $S_{\mathbb R}$ are complex)?

Here by the definition of "of full local degree in $S$" we know that all places of $L$ over $S_{\mathbb R}$ are complex, but I want this property already for $E$, not only for $L$.

Answer 1

This is not posssible if $S_{\mathbf{R}}$ is non-empty.

If $v$ is a real place in $K$, then complex conjugation $c_v \in \mathrm{Gal}(L/K)$ will have order dividing $2$. Since $m_2 = [L:E]$ is assumed to be even and $\mathrm{Gal}(L/K)$ is cyclic, it follows that $c_v \in \mathrm{Gal}(L/E)$. But now the image of $c_v$ is therefore trivial in $\mathrm{Gal}(E/K)$, and hence all infinite places of $E$ will be real.