Let $K_1/k_1, \ldots, K_r/k_r$ be "separable cyclic" extensions of degree $n$ where each $k_i$ is a local field of characteristic 0 (archimedean or not). By separable cyclic of degree $n$, I mean the localization of some global cyclic extension of degree $n$, so a given $K_i/k_i$ may be split or not. (E.g., if $n=2$, either $K_i/k_i$ is a quadratic field extension or $K_i = k_i \oplus k_i$.)

Does there exist an extension of number fields $E/F$ globalizing all $K_i/k_i$ and unramified everywhere else? That is, is there a global $E/F$ with a set $S = \{v_1, \ldots v_r \}$ of places of $k$ such that $E_{v_i} = K_i$, $F_{v_i} = k_i$ and $E_v/F_v$ is unramified for $v \not \in S$.

Results are known about globalizing local field extensions when one fixed a global field $F$, e.g. Grunwald-Wang, but I don't want to do that. Indeed, a positive answer is not always possible if one, say, requires $F=\mathbb Q$, even if each $F_{v_i}$ specified is a distinct local completion of $\mathbb Q$. (For instance, take $n=2$, specify that $E$ should be $\mathbb C$ at infinity, ramified at 3 and unramified everywhere else---there is a unique choice globally $E=\mathbb Q(\sqrt{-3})$, but locally there are 2 ramified extensions at 3 one can choose.)

Rather I want to prescribe stronger local conditions than what one gets from Grunwald-Wang. A positive answer (even with some conditions) should allow one to use global methods to prove results about local representation theory via trace formulas.

Note this is related to finding number fields $F$ such that a ray class field has prescribed local behaviour.

I imagine such questions are hard, so partial results are welcome. (I'm not even sure whether the answer is yes when $n=2$ and $r=1$, though I think it is doable.)

**Edit:** Based on the answer and comments, I want to clarify that when $r=1$, the question is asking for a global extension $E/F$ which is unramified at all but 1 place, where it is $K_1/k_1$. The current comments and answer seem to give an extension $E/F$ that at all ramified places $v \in R$ looks like $K_1/k_1$, but at least a priori we do not have $|R| = 1$.