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In his first Eilenberg Lecture at Columbia, Benedict Gross says that only recently have we been able to give examples of finite galoisian extensions $K$ of ${\bf Q}$ which are ramified only at $2$ (respectively $3$) and for which the group Gal$(K|{\bf Q})$ is not solvable. He seemed to suggest that this was made possible by recent progress in the Langlands Programme.

Question. Which such number fields have been discovered recently, and which bits of the Langlands Programme are needed to construct them ?

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These must be things roughly like 2-adic or 3-adic reps associated to the $\Delta$ function. I guess these are not too recent. But it's curious that there aren't suppose to be elementary constructions of finite extensions of the right sort. Is this really true? – Minhyong Kim Oct 18 '11 at 6:22
By the way, one really striking recent application of a similar flavor is this paper of Clozel and Chenevier:… – Minhyong Kim Oct 18 '11 at 6:27
OK, forget all my silly comments. $GL2(F_3)$ is still solvable, of course. But anyways, I guess we can figure out what needs to be done and that we will need automorphic forms on bigger groups. That's where recent work comes in, I suppose. As usual, I will leave my comments up, so others can benefit from my stupidity. – Minhyong Kim Oct 18 '11 at 6:39
One could try to adapt the argument by looking at Artin reps with image $\mathrm{GL}_2(F)$ where $F$ is a finite field of char. $p \in \{2,3\}$. But Serre and Tate showed that every such representation has to be ramified outside $p$ (this is one of the first steps of the proof of Serre's conjecture). So I guess one has to look at other kinds of automorphic forms. – François Brunault Oct 18 '11 at 6:48
@Francois/Minhyong: exactly! One needs bigger groups, so one has to wait until we can compute Hilbert modular forms better, and that's what happened, thanks mostly to Dembele. – Kevin Buzzard Oct 18 '11 at 6:55
up vote 12 down vote accepted

Minhyong's comments indicate the issue here. If I want to come up with an extension unramified outside $p$ then why not look at the 2-dimensional mod $p$ representation attached to the $\Delta$ function? This works for all but a very small set of $p$, where either the mod $p$ representation is degenerate, or $p$ is so small that $GL(2,p)$ is solvable anyway. For example the semisimple mod 2 representation attached to the $\Delta$ function is trivial. Gross' question was how to deal with this small set of primes.

For these small primes one can try other level 1 forms, of course. For example $p=691$ is a funny case where the representation attached to Delta is reducible, but for $p=691$ you can just use the level 1 weight 16 form instead. The problem with the smaller primes is harder to deal with, because e.g. a modular mod 2 representation unramified outside 2 must be reducible by an old theorem of Tate (look at bounds on discriminants -- the argument is delicate). So one has to try and look elsewhere. The trick with $p=2$ is due to Lassina Dembélé and you can get the paper at his website

Classical modular forms don't cut the mustard, so one seeks to try the same trick with Hilbert modular forms defined over a totally real field ramified only at 2. Such totally real fields are not hard to find, so the issue is now the following computational one -- how to compute the level 1 forms? Dembélé did this, and found an explicit example which gave a Galois representation into $GL(2,k)$ with $k$ of size 8 if I remember correctly (I have to get the kids out of bed -1 minutes ago so can't check any more details).

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So it is "recent progress in the Langlands programme" in the following sense. The key theoretical result is Carayol's theorem from 1986 beefed up by Blasius-Rogawski and Taylor in 1991 or so, attaching Galois representations to Hilbert modular forms. The recent progress is in our ability to compute examples of such things on a computer. – Kevin Buzzard Oct 18 '11 at 6:54
Many thanks. The paper in question is, with and appendix by Serre. – Chandan Singh Dalawat Oct 18 '11 at 6:56
The case $p=3$ seems also to be done in – François Brunault Oct 18 '11 at 6:58
For the record, the case $p=5$ is treated in, and the case $p=7$ in (as I learnt from this answer by Kevin Ventullo:…). – Chandan Singh Dalawat Oct 11 '12 at 14:32

The construction of such number fields is not limited to the Langlands program. For instance, to what extent are there number fields with few ramified primes and Galois group S_n? You won't be able to cook these out of automorphic forms. In this connection there is a remarkable recent example of David Roberts: a number field whose Galois group is the symmetric group on 15875 letters, and whose discriminant is $-2^{130729}5^{63437}$!

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See also, where David Roberts writes down an explicit polynomial of degree $25$ whose splitting field $K$ is ramified only at $5$ and the group $\mathrm{Gal}(K|\mathbf{Q})$ is not solvable. – Chandan Singh Dalawat May 23 '12 at 10:10

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