Timeline for Expressing a number field as a composite of extensions ramified at one place
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2010 at 23:05 | vote | accept | Kevin Ventullo | ||
| Nov 13, 2010 at 19:40 | answer | added | Tim Dokchitser | timeline score: 15 | |
| Nov 13, 2010 at 17:38 | comment | added | BCnrd | Dear Franz: Ah, whoops, that was dumb of me. So the mystery of the structure of the $G_i$'s seems even worse than I was imagining. | |
| Nov 13, 2010 at 17:32 | comment | added | Franz Lemmermeyer | @BCnrd: if L_1 and L_2 are ramified at the same prime, replace them by their compositum. | |
| Nov 13, 2010 at 17:25 | comment | added | BCnrd | Dear Franz: Some of the $L_i$'s may be ramified over the same rational prime, so this collection of fields is not necessarily mutually linearly disjoint over $\mathbf{Q}$ (i.e., the tensor product of the $L_i$'s may not be a field). So their composite field has Galois group that is merely contained in the direct product, and hence the Galois group of the Galois closure of $K$ over $\mathbf{Q}$ may not be (naturally) a quotient of the direct product of the Galois groups of the $L_i$'s over $\mathbf{Q}$. | |
| Nov 13, 2010 at 17:04 | comment | added | Franz Lemmermeyer | Since the Galois closure of an extension ramified at a single prime has the same property, we may assume that the $L_i$ are normal with Galois groups $G_i$. Their direct product must have a quotient isomorphic to $S_3$, the Galois group of the normal closure of $Q(2^{1/3})$. Can we extract information on the structure of the $G_i$ from that? | |
| Nov 13, 2010 at 13:22 | comment | added | Tim Dokchitser | I agree with Franz, the field K generated by the cube root of 2 looks like a good starting point, but it might be tricky to prove that the answer is no for such fields. E.g., Lassina Dembele constructs a non-solvable Galois extension F_2/Q ramified only at 2 with Galois group 2.SL(2,F_8)^2, and it might account for the correct tame ramification at 2 (at least its degree is a multiple of 3). So K might well be contained in the compositum of F_2 and some non-solvable F_3/Q ramified only at 3. These exist as well by Dembele- Greenberg-Voight, so it looks really hard... Nice question! | |
| Nov 13, 2010 at 13:00 | comment | added | Franz Lemmermeyer | I would try proving that the answer is no for the field K generated by a cube root of 2. | |
| Nov 13, 2010 at 11:41 | comment | added | Alex B. | Cam, $K$ could be heavily ramified, while the $L_i$ are required to be ramified at at most one place over $\mathbb{Q}$. | |
| Nov 13, 2010 at 11:33 | comment | added | Cam McLeman | 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... | |
| Nov 13, 2010 at 10:30 | comment | added | Kevin Ventullo | Sorry about that; edited for clarity. | |
| Nov 13, 2010 at 10:29 | history | edited | Kevin Ventullo | CC BY-SA 2.5 |
Edited for clarity
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| Nov 13, 2010 at 9:58 | comment | added | Chris Wuthrich | 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. | |
| Nov 13, 2010 at 9:25 | history | asked | Kevin Ventullo | CC BY-SA 2.5 |