2
$\begingroup$

Let $S$ be a finite set of primes and $G_{\mathbb{Q},S}$ the Galois group of maximal extension of $\mathbb{Q}$ which is unramified ouside $S$. Since $\mathbb{Q}$ has no unramified extensions, $G_{\mathbb{Q},S}$ is topologically generated by inertia groups at primes in $S$, i.e. the subgroup generated by inertia groups at primes in $S$ is dense in $G_{\mathbb{Q},S}$. It leads to the following question:

Is the union of conjugates of the inertia groups at primes in $S$ dense in $G_{\mathbb{Q},S}$?

Any comments and suggestions will be appreciated.

$\endgroup$
2
  • $\begingroup$ I don't know if it's an issue of English, but I don't understand the meaning of the question. Is the question whether the union of all conjugates of all inertia groups dense? $\endgroup$
    – YCor
    Oct 14, 2022 at 14:51
  • $\begingroup$ @YCor Yes, I ask whether the union of all conjugates of all inertia groups at primes in $S$ in $G_{\mathbb{Q},S}$ is dense. $\endgroup$
    – Nobody
    Oct 14, 2022 at 14:56

2 Answers 2

7
$\begingroup$

Not unless $S$ has one element.

If $S$ contains two primes, $p$ and $q$, the Galois group of the cyclotomic field $\mathbb Q(\mu_{p^\infty q^\infty})$ is a quotient of $G_{\mathbb Q,S}$ and is isomorphic to $\mathbb Z_p^\times \times \mathbb Z_q^\times$. The images of the inertia groups at $p$, $q$, and any other prime are $\mathbb Z_p^\times \times \{1\}, \{1\}\times \mathbb Z_q^\times$, and $\{(1,1)\}$ respectively, and the union of these is not dense.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks so much for your counterexample! But why is it true when $S$ has only one element? $\endgroup$
    – Nobody
    Oct 14, 2022 at 14:19
  • 3
    $\begingroup$ @Nobody I don't think I said that it was! It's not true for $S =\{253381\}$ since the extension generated by the roots of $x^5+ x+3$ is ramified only at that prime (since its discriminant is $253381$) and the inertia group consists of an element of order $2$ in the Galois group and the trivial element, but not every element in the Galois group has order $1$ or $2$. $\endgroup$
    – Will Sawin
    Oct 14, 2022 at 15:19
5
$\begingroup$

Not if $S$ has one element.

If $p$ is an irregular prime and $H$ is the Hilbert class field of $\mathbf{Q}(\zeta_p)$ then $\mathrm{Gal}(H/\mathbf{Q})$ has order divisible by $p$ but no inertial element has order $p$.

If $p$ is a regular prime, then $p$ does not divide the class number of $\mathbf{Q}(\zeta_{p^n})$ for any $n$. A theorem of Washington implies that the $\ell$-part of the class group of this tower is uniformly bounded for every $\ell \ne p$. From the class number formula the order of the class groups also become arbitrarily large as $n$ increases. Taken together this implies that the class number of $\mathbf{Q}(\zeta_{p^n})$ is divisible by at least one prime $\ell$ with $(\ell,p(p-1)) = 1$ for sufficiently larger $n$. If $H$ is the Hilbert class field of $\mathbf{Q}(\zeta_{p^n})$ for such an $n$, then $\mathrm{Gal}(H/\mathbf{Q})$ will have order divisible by $\ell$ but no inertial element can have order $\ell$.

Since the prime above $p$ in $\mathbf{Q}(\zeta_{p^n})$ is principal, it (the prime above $p$) splits completely in the Hilbert class field. So in these cases the inertia group above $p$ coincides with the decomposition group. Hnce these examples also demonstrate the stronger claim that not even the union of the decomposition groups above $p$ are dense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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