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If $K$ is a number field, is it always possible to find a finite extension $L/K$ such that $L$ is the composite of fields $L_1,\ldots, L_n$, with the property that at most one prime ramifies in $L_i/\mathbb{Q}$? Equivalently, is the composite of all extensions ramified at $\leq 1$ place all of $\overline{\mathbb{Q}}$?

The first question has an affirmative answer when $K$ is abelian, but for the general case, the equivalent second question sounds too strong to be true. Any ideas?

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As I read it, the first question is equivalent to find one finite extension $L/K$ ramified at most at one place. And I don't see why this is equivalent to the second question. –  Chris Wuthrich Nov 13 '10 at 9:58
    
Sorry about that; edited for clarity. –  Kevin Ventullo Nov 13 '10 at 10:30
    
I'm still a little confused. For example, your first question could be answered affirmatively if you could find, say, a quadratic extension of K ramified at a single prime. This seems unlikely to imply your second question... –  Cam McLeman Nov 13 '10 at 11:33
    
Cam, $K$ could be heavily ramified, while the $L_i$ are required to be ramified at at most one place over $\mathbb{Q}$. –  Alex B. Nov 13 '10 at 11:41
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I would try proving that the answer is no for the field K generated by a cube root of 2. –  Franz Lemmermeyer Nov 13 '10 at 13:00

1 Answer 1

up vote 14 down vote accepted

I believe the answer is `No', and Franz Lemmermeyer's example $K=Q(2^{1/3})$ and his strategy of the proof do the trick.

Suppose this particular $K$ is contained in the compositum $F$ of $L_i$, with every $L_i$ ramified at only one prime. Assume each $L_i$ is Galois over $Q$ (otherwise replace it by its Galois closure) and that no two $L_i$ ramify at the same prime $p$ (otherwise replace this pair by their compositum). The $L_i$ are then linearly disjoint over $Q$, since their pairwise intersections would have to be unramified at all primes. So $G=Gal(F/Q)$ is the direct product of $Gal(L_i/Q)$'s.

Now the group $G$ has a 2-dimensional irreducible representation $\rho$, the one that factors through the Galois closure of $K/Q$ (an $S_3$-extension of $Q$). As $G$ is the direct product of groups, we can write $\rho=\rho_1\otimes...\otimes\rho_n$ uniquely, with $\rho_i$ irreducible representations of $Gal(L_i/Q)$. Moreover, $\rho$ is self-dual, so all the $\rho_i$ are self-dual as well. Of these $\rho_i$ one must be 2-dimensional and the others are 1-dimensional. Because 1-dimensional self-dual characters have order 2, this shows that at all primes $p$ except at most one (the one corresponding to the 2-dimensional $\rho_i$) inertia $I_p$ acts on $\rho$ through a quotient of order 2. But $I_2$ and $I_3$ act through quotients of order 3 and 6 respectively, contradiction!

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Dear Tim: Very nice. A tiny correction is that pairwise triviality of intersections does not quite suffice to justify the required linear disjointness in general (but we have more ramification information, and so have disjointness of each $L_i$ against the compositum of the others, say also proceeding by induction on the number of $L_i$ or whatever; of course, you had all of this in mind so as to not really need to worry about it). For example, in a biquadratic field the 3 quadratic subfields have pairwise trivial intersections but the triple is not a linearly disjoint collection. –  BCnrd Nov 13 '10 at 21:26
    
You're right of course, I did not put this correctly! Also, as my brother has pointed out, all this representation theory nonsense can be avoided: the group $G=G_1*...*G_n$ acts on three points (the embeddings of $K$ into $C$), and its very difficult to have pairwise commuting groups to act on 3 points. As one of them, corresponding to $p=3$, is the whole of $S_3$, all the others must act trivially, so $K$ cannot be ramified at any other prime. –  Tim Dokchitser Nov 13 '10 at 22:48
    
@Tim: Nice answer (both of them)! That second one is quite slick. –  Kevin Ventullo Nov 13 '10 at 23:04

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