Let $K$ be a number field, $d$ a positive integer and $S$ a finite set of places of $K$.


By Cebotarev, there exists a finite set of finite places $T$ disjoint from $S$ such that the conjugacy classes of geometric Frobeni $F\_v$ ($v\in T$) fill up $\mathrm{Gal}(K^\prime/K)$ for any $K^\prime/K$ Galois of degree at most $d$ (**Edit**) and unramified outside $S$.


For a finite field extension $L/K$, let $T\_L$ be the set of places of $L$ lying over $T$. We use similar notation for $S\_L$.


**Question.** Let $L/K$ be a finite extension, not necessarily Galois. Then $T\_L$ is disjoint from $S\_L$. Do the conjugacy classes of geometric Frobeni $F\_w$ ($w\in T\_v$) fill up $\mathrm{Gal}(L^\prime/L)$ for any $L^\prime/L$ Galois of degree at most $d$ (**Edit**) and unramified outside $S\_L$?



A question on Cebotarev's density theorem
**Edit:** I understand that you are happy with a $T$ which depends upon $L$ and works simultaneously for all $L'/L$ of degree bounded by $d$ and unramified outside of $S\_L$. If you are looking for *a uniform* $T$ which does the job for *all possible $L$ at once*, then the answer is no, as explained in David's comment.


The answer is yes. Let $[L:K]=n$ and let $R$ be the finite set of ramified primes in $L/K$. By Hermite-Mikowsky, there are only finitely many extensions of $K$ of degree $nd$ unramified outside of $S\cup R$. Let $K'$ be the compositum of all of these, which is a finite (compositum of finitely many extensions of finite degree), Galois (by maximality) extension of $K$. If now you apply your first part to the extension $K'/K$ you find a set $T$ disjoint from $R\cup S$ fulfilling your condition.


