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## A question on Cebotarev’s density theorem

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 $S_L$?

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Your initial statement "By Cebotarev etc." is false. For example, if $d=2$, $K=\mathbb{Q}$, and $T$ is any finite set of primes, then there is a quadratic extension $K'/K$ which splits at all the primes in $T$, hence the Frobenii $F_v$ $(v\in T)$ are all trivial in $\mathrm{Gal}(K'/K)$. – GH Oct 18 at 23:56
@GH. You're right. I forgot to mention why I fix $S$. I'll edit the question. My apologies. – Harry Oct 19 at 6:34
Take $K = \mathbb{Q}$, $d=2$ and $S = \emptyset$. Since the only unramified extension of $\mathbb{Q}$ is $\mathbb{Q}$ itself, any nonempty $T$ obeys the hypothesis -- for concreteness, say $p=(41)$. Now let $L$ be a number field which does have unramified quadratic extensions. For example, $L=\mathbb{Q}(\sqrt{-5})$ has the unramified extension $L'=\mathbb{Q}(i, \sqrt{-5})$. $41$ splits as $4 + \sqrt{-5}$, $4-\sqrt{-5}$ in $L$, both of which split further in $L'$, so $T_L$ does not give us any nonidentity Frobeniuses. – David Speyer Oct 19 at 14:21
I suspect you have left out a condition again. Perhaps you wanted to impose a relation between $S$ and $L$? – David Speyer Oct 19 at 14:22
@David Speyer. Thank you for the example. – Harry Oct 22 at 19:41

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.
 This shows that there exists some $T_L$ with the right property. It doesn't show that I can take $T_L$ to be the places lying over $T$, though. And as David Speyer's example shows, this is not possible in general. – Harry Oct 20 at 12:32 I think the issue is whether T is allowed to depend on L. – Kevin Ventullo Oct 20 at 17:34 @Harry: Indeed, Kevin is right. If you fix $T$ beforehand, you will certainly produce an $L$ for which that $T$ is the bad choice. But if you start with $L$ you can always find a $T$ which does the job for your extension. In David's example, $T$ is a bad choice for $L$ because $(41)$ splits in $K/mathbb{Q}$. Therefore, its Frobenius do not generate the Galois grp. of the compositum of extensions of $\mathbb{Q}$ unramified outside of $2\cdot 5$ of degree $\leq 4$ - due precisely to the existence of the Hilbert class field of $\mathbb{Q}(\sqrt{-5})$. – Filippo Alberto Edoardo Oct 21 at 4:10