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Let $K$ be a number field and $S$ a finite set of places of $K$. Then there is a maximal algebraic extension $K^S / K$ unramified outside $S$.

How large do we have to take $S$ to be in order that $K^S / K$ is infinite?

For $K = \mathbf{Q}$ it's necessary and sufficient that $S$ contain at least one finite place. This is not a necessary condition for every $K$, since the Golod--Shafarevich theorem shows that there are fields such that $K^S$ is infinite when $S$ is the empty set; but is it sufficient? That is:

Does there exist a field $K$ and a finite place $v$ such that $K^{\{v\}} / K$ is finite?

There certainly exist examples such that $K^{\{v\}} / K$ contains no solvable extension of $K$, e.g. $K = \mathbf{Q}(\sqrt{3})$, $v$ the prime $1 + 2\sqrt{3}$ of norm 11.

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  • $\begingroup$ Interesting question, and rather out of my experience. Do you have a $K$ that requires more than one place? $\endgroup$ – Lubin Mar 16 '18 at 18:31
  • $\begingroup$ Is there an example of a $K$ and finite place $\mathfrak{p}$ such that the ray class fields of modulus $\mathfrak{p}^k$ stabilize eventually? In particular, can $K^S/K$ be infinite but $(K^S)^{ab}/K$ finite when $S$ contains at least one finite prime? $\endgroup$ – Brandon Carter Mar 16 '18 at 19:21
  • $\begingroup$ @BrandonCarter Sure, this can happen: take $K$ real quadratic and any prime $p = \mathfrak{p} \mathfrak{p}^\sigma$ split in $K$. Then the maximal abelian extension unramified outside $\mathfrak{p}$ is finite. $\endgroup$ – David Loeffler Mar 16 '18 at 19:55
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    $\begingroup$ This question is surely open. But one heuristic is that you should only expect a (general) positive answer under conditions which guarantee that the largest abelian quotient is infinite. Otherwise, it can (and sometimes does) happen that the largest abelian extension is trivial. For $K^S$ then to be infinite, there (at the very least) needs to exist an extension $L/K$ unramified outside $S$ whose Galois group $G$ is simple and non-abelian. Heuristics concerning the existence of such extensions suggest that the probability of this occurring is very small... $\endgroup$ – Infinity Mar 18 '18 at 5:41
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    $\begingroup$ ... indeed sufficiently small so that there is a non-zero probability that there are no such extensions for any such $G$, and then $K^S = K$. Of course, we don't even know a single example with $S = \emptyset$ such that $K^S = K$ and the root discriminant of $K$ exceeds the Odlyzko bounds. There are some contexts (e.g. projecteuclid.org/euclid.ant/1510842310) which suggest the heuristics above need to be taken with a grain of salt, but these examples only involve $S$ containing all primes above $2$ and so are somewhat orthogonal to the question you are posing. $\endgroup$ – Infinity Mar 18 '18 at 5:41
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@David Does your question mean (even in vague terms) : "Given $K$, what is the "smallest" $S$ such that $K^S/K$ is infinite ?". Because the examination of miscellaneous known cases seems to suggest the contrary of a uniform answer. Throughout, fix a prime number $p$ (assumed to be odd to avoid petty trouble) and shift the problem from $K^S$ to $K^S (p)$ = the maximal pro-$p$-extension of $K$ unramified outside $S$, with Galois group $ {G^S_K}(p)= Gal(K^S (p)/K)$ .

1) Take $S$ containing all the $p$-places of $K$. It is well known by CFT that ${G^S_K}(p)^{ab}$ is a $\mathbf Z_p$-module of rank $1+r_2+ \delta$, where $r_2$ is the number of complex places of $K$ and $\delta \in \mathbf N$ is conjecturally null. So the point is whether $S$ can be downsized or not.

2) Let us first remove some $p$-places from $S$. Consider an imaginary quadratic field $k$ s.t. $p$ splits completely in $k$, say $(p)=P.P^*$, and an elliptic curve $E$ defined over a number field $F$, with complex multiplication by the ring of integers of $k$. Suppose moreover that $E$ has good ordinary reduction at any $p$-place. Then the field $F(E_{P^{\infty}})$ obtained by adding all the $P^n$-torsion points of $E$ is a $\mathbf Z_p$-extension of $K=F(E_P)$. Thus $K^S/K$ is infinite, with $S$={$P$}.

3) Let us be go to extremities and cut $S$ down to the empty set, so that $K^S (p)$ is the so-called "$p$-class field tower". The Golod-Safarevic theorem says that there exists a function $\gamma (n)$ s.t., for all number fields $k$ of degree $n$ with finite $p$-class field tower, $\gamma(n)$ > the minimal number of generators of the $p$-class group of $k$. One can even take $\gamma(n)=2+2\sqrt(n+1)$. Moreover, for a finite $p$-group $G$, it is known by cohomological considerations that $r(G) > \frac 14 d(G)^2$ , where $d(G)$ (resp. $r(G)$) dotes the minimal number of generators (resp. relations) of $G$. The combination of these two properties produces examples of finite as well as of infinite $p$-class field towers.

4) Another approach lies in the "tame" situation, in which $S$ contains no $p$-place, hence $K^S (p)$ contains no $\mathbf Z_p$-extension, hence ${G^S_K}(p)^{ab}$ is finite (this property is called FAB). At the beginning of this century, somewhat to the general surprise, J. Labute produced a family of Galois groups $G^S_{\mathbf Q}(p)$ called mild, of cohomological dimension $2$, hence torsion free, hence infinite. Roughly speaking, the initial terms of the relations of a mild pro-$p$-group $G$ must satisfy special combinatorial properties in order that the graded algebra built on the lower $p$-central series of $G$ has a nice convenient description in terms of the corresponding free graded algebra. Subsequent developments by A. Schmidt et. al. provide an infinite supply of examples over arbitrary number fields. One should also remark the relevance of arithmetical FAB groups with regard to the Fontaine-Mazur conjecture on $p$-adic representations ./.

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  • $\begingroup$ First, $K^S(p)$ is a very different type of extension from $K^S$ and is much more closely related to class field theory (it also seems to assume that $S$ is a subset of primes dividing some fixed $p$). Second, it seems clear from the original question (and the mention of G-S) that the OP is not expecting a "uniform" answer, just asking if there are any non-trivial general results which guarantee that $K^S$ is infinite. $\endgroup$ – Infinity Mar 18 '18 at 10:55
  • $\begingroup$ @Infinity I don't understand your comment. If $K^S (p)/K$ is infinite, then so is $K^S/K$. Besides, in the "tame" case, $S$ is coprime to $p$. So I did actually give "non trivial general results" implying the unfiniteness of $K^S/K$. $\endgroup$ – nguyen quang do Mar 18 '18 at 11:42
  • $\begingroup$ What you write is very interesting, but it appears to be aiming at a slightly different question than the one I actually asked! As @infinity points out, it might (potentially) happen that $K^S(p) / K$ might be finite for every $p$, for some fixed $K$ and $S$, but $K^S / K$ infinite. $\endgroup$ – David Loeffler Mar 18 '18 at 13:02
  • $\begingroup$ Put as this, I agree, although the point did not appear very clearly in your original question. That was why my answer only dealt with examples of infinite $K^S(p)/K$ ($p$ arbitrary). $\endgroup$ – nguyen quang do Mar 18 '18 at 13:32

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