Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is an idle question. Is there an example of a number field $K$ for which the maximal everywhere-unramified extension $M|K$ is of infinite degree, and yet the group $\mathrm{Gal}(M|K)$ has been completely determined ?

share|improve this question
    
@EE John: We say that an unknown group $G$ has been completely determined if we have shown it to be isomorphic to a known group $G'$. For example, when $L$ is the splitting field over $\mathbf{Q}$ of the polynomial $T^3-2$, the group $G=Gal(L|\mathbf{Q})$ is completely known once we show that it is isomorphic to $G'=\mathfrak{S}_3$. –  Chandan Singh Dalawat Mar 10 '11 at 11:47
    
To complement above : more generally than an isomorphism with a "basic" group, completely determining a group can mean one has it as a product of "basic" groups, as extension ; that it has a local description ; that one knows what its normal subgroups are, etc (and that is a big caetera!) –  Julien Puydt Mar 10 '11 at 17:01
    
Is there an example of an infinite unramified extension of a number field (not necessarily maximal) and known Galois group? –  Franz Lemmermeyer Mar 10 '11 at 18:29
    
@Franz: Nope, don't think so. –  Cam McLeman Mar 10 '11 at 19:02
2  
Tim, two Russians (Golod-Shafarevich) did it first. There is a somewhat dated write-up by Roquette in Cassels-Fröhlich. –  Chandan Singh Dalawat Mar 11 '11 at 11:53

4 Answers 4

up vote 7 down vote accepted

No, I'm pretty sure not.

In general, the theory is much more developed for the maximal pro-p-quotient of the groups you're asking about, and even in this more explored setting, not a single explicit presentation of an infinite such group is known (to me, for sure, but I think to anyone -- Edit: Nigel Boston appears to concur in his survey paper Galois $p$-groups unramified at $p$). In fact, this is true even if we generalize to ask about the Galois groups $G_{S,p}(K)$ of maximal $p$-extensions unramified outside of a finite set of primes $S$ (with some tameness conditions on $S$ -- obviously taking $K=\mathbf{Q}(\zeta_p)$ and $S$ as the set of primes above $p$ gives a counterexample to my claims.)

To elaborate, what we do have are certain approximations to presentations for such groups. If you consider a pro-$p$ presentation of $G_{S,p}(K)$: \begin{align*} 1\to R\to F\to G_{S,p}(K)\to 1, \end{align*} where $F$ is a free pro-$p$-group, then in some cases (for example, $K=\mathbb{Q}$) you can find an approximation to a minimal generating set for the relation module $R$ in the sense that you can give an explicit description of these generators in some quotient of $F$, e.g., modulo the third step in the lower central series of $F$. (Actually, it's the "Zassenhaus filtration" for which the results are sharpest.) A lot of authors have written on this idea, which roughly originated with Koch -- I'd recommend NSW's "Cohomology of Number Fields" for an overview, and then work of Morishita, Vogel, or possibly myself for more details. In brief, these approximations are determined by the arithmetic of $K$ and $S$ (e.g., $p$-th power residue symbols or other class-field-theoretic symbols evaluated at the primes in $S$). This was a fairly resounding triumph of the theory -- it would be outright revolutionary to lift these congruences of relations mod $F_3$ to literal equalities in $F$.

Let me finish by bringing the general case back toward your original unramified setting. The relevance of the more general case is as follows: Say, for example, that you have a quadratic extension $K$ of $\mathbb{Q}$ and would like to know its maximal unramified 2-extension. If we let $S$ be the set of primes dividing the discriminant of $K$, then the maximal unramified 2-extension of $K$ corresponds to an index-2 subgroup of $G_{S,2}(\mathbb{Q})$, which is, as above, difficult to get ones hands on (if infinite). Finally, I should also mention work of Boston conjecturing explicit (non-approximate) presentations for $G_{S,p}(K)$, though even here, the unramified case is less concrete.

share|improve this answer

It seems the question is not so idle after all, for the Fontaine-Mazur conjecture has something to say about it, as I discovered today in Neukirch-Schmidt-Wingberg (the NSW of Cam's answer). To paraphrase their (10.8.13), every representation

$$ \operatorname{Gal}(M|K)\to \operatorname{GL}_n(\mathbf{Q}_p) $$

of the group of automorphisms of the maximal unramified extension $M$ of a number field $K$ has finite image, even if the degree $[M:K]$ is infinite. This is certainly a very strong restriction.

share|improve this answer

Here are a few complementary remarks. The first nontrivial result on class field towers before Golod and Shafarevich was Scholz's proof that for every positive integer $n$ there exist class field towers of length $n$. His proof was completely constructive: given a prime $\ell$, he chose $n+1$ prime numbers $p_j \equiv 1 \bmod \ell$ satisfying certain congruence conditions and showed that the compositum of the cyclic $\ell$-extensions inside the fields of $p_j$-th roots of unity has an unramified extension whose Galois group has stage $n$. I guess that it is possible to extract the Galois group from his proof; note, however, that his unramified extension certainly is not maximal in general.

Artin later found a short proof of Scholz's result: given any finite group $G$, let $L/K$ be a normal extension with Galois group $G$ (this is easy). Then choose an extension $F/K$ disjoint from $L/K$ and generated by a root of a polynomial $p$-adically so close to that of $L$ that the whole ramification vanishes when the extension is lifted to $F$; then $LF/F$ is unramified with Galois group $G$.

This construction is, unfortunately, pretty useless for infinite groups.

share|improve this answer
    
Franz, I have seen references to Scholz but not to Artin. Could you please indicate the place where this argument is carried out ? –  Chandan Singh Dalawat Mar 12 '11 at 2:43
    
Fröhlich (On non-ramified extensions with prescribed Galois group, Mathematika, 9 (1962) 133-134) seems to adopt an approach similar to what you describe. –  Chandan Singh Dalawat Mar 12 '11 at 7:57
    
Chandan: it should be somewhere in the correspondence between Artin and Hasse, available here: goedoc.uni-goettingen.de/goescholar/handle/1/5143 (or google for artin hasse correspondence if the link does not work). –  Franz Lemmermeyer Mar 12 '11 at 10:52
    
Franz, Peter Roquette has kindly given me a copy. But it is good to know that the volume is available online. –  Chandan Singh Dalawat Mar 13 '11 at 4:39

What is the meaning of "Gal(M/K) has been completely determined?"

share|improve this answer
    
Please use the comment box. –  Chandan Singh Dalawat Mar 10 '11 at 11:49
1  
It's not possible to do that with only 11 points of reputation, unfortunately. –  Konstantin Ardakov Mar 11 '11 at 13:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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