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 $S_L$?

  • 6
    $\begingroup$ 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 $\mathrm{Gal}(K'/K)$. $\endgroup$ – GH from MO Oct 18 '12 at 23:56
  • $\begingroup$ @GH. You're right. I forgot to mention why I fix $S$. I'll edit the question. My apologies. $\endgroup$ – Harry Oct 19 '12 at 6:34
  • 4
    $\begingroup$ 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. $\endgroup$ – David E Speyer Oct 19 '12 at 14:21
  • $\begingroup$ I suspect you have left out a condition again. Perhaps you wanted to impose a relation between $S$ and $L$? $\endgroup$ – David E Speyer Oct 19 '12 at 14:22
  • $\begingroup$ @David Speyer. Thank you for the example. $\endgroup$ – Harry Oct 22 '12 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.

  • $\begingroup$ 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. $\endgroup$ – Harry Oct 20 '12 at 12:32
  • $\begingroup$ I think the issue is whether T is allowed to depend on L. $\endgroup$ – Kevin Ventullo Oct 20 '12 at 17:34
  • $\begingroup$ @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 $\mathbb{Q}(\sqrt{-5})$. $\endgroup$ – Filippo Alberto Edoardo Oct 21 '12 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.