For each $n \geq 1$, we denote by $\mathbb{Q}_n$ the unique subfield of $\mathbb{Q}(\zeta_{p^{n+1}})$ for which $[\mathbb{Q}_n:\mathbb{Q}]=p^n$. Let now $K$ be a finite Galois extension of $\mathbb{Q}$ with the following properties:

  1. K contains $\mathbb{Q}_m$ for some $m \geq 1$;
  2. There exists an element $g \in Gal(K/\mathbb{Q})$ with the property that the order of $g$ is $p^n$ with $n>m$ and $\left. g \right|_{\mathbb{Q}_m}$ is the generator of the Galois group $Gal(\mathbb{Q}_m/\mathbb{Q})$.

Is it then true that $K$ must contain $\mathbb{Q}_t$ for some $t>m$?

I have tried to use quite a lot of theory around wild ramification and local considerations (including this post wildly ramified extensions and $p$-power roots of unity ), without success in producing a proof/counterexample.

  • 1
    $\begingroup$ Do you impose extra conditions on K/Q, especially about ramification ? Your title mentions wild ramification. $\endgroup$ – nguyen quang do Jan 7 '17 at 8:21
  • $\begingroup$ No, the conditions are just those mentioned. The "wild ramification" bit is there due to the local behaviour (in my notations $\mathbb{Q}_m,p/\mathbb{Q}_p$ is wildly ramified and the hope is that at every prime above p, the condition on $g$ will give a larger wildly ramified extension at every such prime). Also, if it turns out that there is a counterexample, I would be very grateful to know some further conditions which would make it work. $\endgroup$ – Jake Haider Jan 7 '17 at 11:29

Since the prime $p$ ($\neq 2$ for simplification) plays a special role, let us concentrate on the case when $K/Q$ is a $p$-extension, i.e. a Galois extension whose group $G$ is a finite $p$-group. The case of a general $G$ can be derived from this provided we have sufficient information on the structure of $G$. Suppose also that $K/Q$ is $S$-ramified, i.e. unramified outside a finite set of primes $S$={$p$} $\cup$ $T$, with $T$ = {$l_1,...,l_r$} or $T$ empty. Note that the presence of $p$ is necessary because we'll deal with the so-called cyclotomic $Z_p$-extension of $Q$, namely $G_S (k) = \cup Q_n$.

For a number field $k$ and the set $S_k$ of places of $k$ above $S$ (we'll write $S$ for $S_k$, no confusion is possible), denote by $k_S$ the maximal pro-$p$-extension of $k$ which is $S$-ramified, and $G_S (k) = Gal(k_S/k)$. The description of $G_S (k)$ by generators and relations can be approached using CFT and Galois cohomology, as for example in H. Koch's book "Galois Theory of p-extensions". In particular, the minimal number $d$ of generators (resp. the number $r$ of relations) of $G_S (k)$ is equal to the $Z/pZ$-dimension of $H^1(G_S (k), Z/pZ)$ (resp. of $H^2(.)$). Actually, $d$ has been computed by Shafarevitch (op. cit.,§11, thm. 11.8), and $r$ is given by a so-called Euler-Poincaré characteristic formula : $1-d+r = -c$, where $c$ is the number of complex places of $k$.

After these preliminaries, we can attack your problem over $k = Q$. Since no ramification condition is specified, we must consider two separate cases:

1) If $T$ is empty, i.e. only wild ramification is allowed, Shafarevitch's formula over $Q$ gives $d=1$ and $r=0$, in other words, $G_S (Q)$ is the free pro-$p$-group on a single generator, which is $Z_p$. Equivalently, $Q_S = Q_{\infty}$ , which implies that $K$ = a certain $Q_n$ , and the answer to your question is trivially YES

2) If $T$ is not empty, i.e. tame ramification is allowed, let $ Q_m = K \cap Q_{\infty}$ , $G = Gal(K/Q)$ and $H = Gal(K/Q_m)$. Suppose, as you do, that there exists $g \in G$, of order $n>m$, such that its image $ g$ mod $H$ in $G/H$ generates $G/H = Gal(Q_m/Q)$. Replace $K$ by the fixed field $L$ of $(H \cap<g>)$, so that the image $g$ mod $(H\cap<g>)$ in $Gal(L/Q)$ generates a subgroup isomorphic to $Gal(Q_m/Q)$. This means that that the exact sequence $1 \to Gal(L/Q_m) \to Gal(L/Q) \to Gal(Q_m/Q) \to 1$ splits, i.e. describes a semi-direct product. So in this case the answer to your question will be NO if we can construct the adequate extension $L/Q_m$ . If $T$ = {$l_1,...,l_r$}, the number of relations of $G_S (Q)$ will be $r$, more precisely the relations will come from $r$ local tame relations of the type (op. cit., §10, thm. 10.2) $t^{l-1}[t^{-1}, s^{-1}]=1$ (necessarily, $l\equiv 1$ mod $p$ to allow ramification). This means that the maximal abelian quotient of $G_S (Q)$, considered as a $Z_p$-module, will have rank 1 and torsion isomorphic to a product of cyclic groups $Z/(l -1)Z$. By adjusting our $l$, we can immediately construct the desired extension $L/Q_m$ ./.

  • $\begingroup$ This is a very nice and beautifully presented answer, thank you very much for it! Before I can fully unfold the counter-example in the second part of your answer, I would like to understand 2 basic things: I might be missing something obvious, but why is $L$ in your notation necessarily Galois over $\mathbb{Q}$? Also, once the adequate $L$ is constructed, how does one recover the element $g$ in the larger extension $K/\mathbb{Q}$? $\endgroup$ – Jake Haider Jan 10 '17 at 16:09
  • $\begingroup$ I supposed implicitly that <g> is normal (hence <g^n/m> obviously is) but on re-reading your post, I see that you do not assume that. It does not matter anyway, since we are only looking for a counter-example, and we may as well take K/Q abelian.Then the same reasoning as before shows that the abelian Gal(L/Q) is a direct product, but perhaps it would be clearer to construct directly an abelian K/Q (K, not L) with the required required g. $\endgroup$ – nguyen quang do Jan 11 '17 at 7:14

On re-reading my second comment, I do not find it very enlightening, so I propose to replace it by this more detailed answer (mainly in order to be able to type math symbols (*)). I also edited two unfortunate misprints in my first answer (just before the exact sequence of Galois groups).

So let us construct our abelian extension $K/Q$ with the required $g$ of order $p^n$, with $n>m$, and such that $Gal(K/Q)\cong Gal(K/Q_m) \times Gal(Q_m/Q)$. We have seen that he Galois group of the maximal abelian $S$-ramified $S$-extension of $Q$ is a direct product ($Z_p, +$)$\times \mathcal T$, where the torsion group $\mathcal T$ is a finite product of cyclic groups of the form $Z/(l-1)Z$. Because a subgroup of $Z_p$ is isomorphic to $Z_p$ hence is pro-$p$-free, $Gal(K/Q)$ will be viewed as a quotient of $Gal(Q_m/Q) \times\mathcal T$ and $Gal(K/Q_m)$ as a quotient of $\mathcal T$. Then choose $l \in T$, such that $p^n$ divides exactly $ 1- p^n$, so that $Gal(K/Q)$ will contain an element $g$ of order $p^n$ coming from $\mathcal T$. Taking further $l' \in T, l' \neq l $, $T=${$l, l'$}, and $Gal(K/Q_m) \cong$ the factor $Z/(l'-1)Z$ of $\mathcal T$, we get our extension $K$ with the desired properties. Things could be seen much more clearly by drawing Galois diagrams.

  • $\begingroup$ Thank you very much for this update, it did clarify the questions I had in the previous comments. The second part of the previous answer also contains much more information and analyzing it lead me to the following question: Can one exhibit a necessary and sufficient condition (or at least sufficient and non-trivial) for $K$ so that if $K \cap \mathbb{Q}_{\infty}=\mathbb{Q}_m$, then we have $Gal(K/\mathbb{Q})$ is the semidirect product between $Gal(K/\mathbb{Q}_m)$ and $Gal(\mathbb{Q}_m/\mathbb{Q})$? $\endgroup$ – Jake Haider Jan 11 '17 at 10:44
  • $\begingroup$ I think I have a proof using local considerations that this is always the case (as in, for any $K$ one has a semi-direct product), I would be happy to know if you think it shoudln't be this way. $\endgroup$ – Jake Haider Jan 11 '17 at 15:44
  • $\begingroup$ Perhaps we should discuss this in the "chat" space ? $\endgroup$ – nguyen quang do Jan 12 '17 at 9:45
  • $\begingroup$ sure, any particular room? $\endgroup$ – Jake Haider Jan 12 '17 at 10:07
  • $\begingroup$ What do you mean by a "room" ? I know nothing of the procedure. I suggest that you choose your own "room" and then "invite" me. $\endgroup$ – nguyen quang do Jan 12 '17 at 15:51

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