Timeline for Non-cyclotomic abelian extensions
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2012 at 18:12 | comment | added | Franz Lemmermeyer | @unknown: you're right, I did not see the inclusion signs. There is, by the way, a beautiful article by Waterhouse (The normal closures of certain Kummer extensions, Can. Math. Bull. 37, No.1, 133-139 (1994)) dealing with problems like this one. | |
| Nov 19, 2012 at 17:36 | comment | added | John Pardon | @Franz Lemmermeyer: OK, I fixed a problem, but I don't think it's exactly what you said. The edit now ensures that $\operatorname{Gal}(L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/\mathbb Q)$ is nonabelian. | |
| Nov 19, 2012 at 17:34 | history | edited | John Pardon | CC BY-SA 3.0 |
added 164 characters in body
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| Nov 19, 2012 at 17:05 | comment | added | John Pardon | @Franz Lemmermeyer: I don't understand your objection. Do you mean that the Galois closure of $L(\sqrt c)$ is $K$, which has degree four over $L$? I think this is ok, since I said that the Galois group of $L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})$ is contained in $(\mathbb Z/2\mathbb Z)^n$, (it may be a proper subgroup). What I need is only that it is a nontrivial subgroup (hence I need that $\gamma$ is not a square) because this is what makes (2) true. | |
| Nov 19, 2012 at 5:53 | comment | added | Franz Lemmermeyer | I think this argument has a little hole. Consider an A4-extension K/Q of the rationals, let L be its cyclic cubic subfield, and F/L a quadratic extensions of F/L. If the square root of c in L generates F, then your argument predicts that c has an orbit of length 2 under the action of the cyclic Galois group of order 3. For getting rid of this problem you probably have to consider orbits in L^* modulo squares. | |
| Nov 19, 2012 at 5:43 | comment | added | Will Sawin | Yes. It is easy to compute the action by conjugation of $\operatorname{Gal}(L/\mathbb Q)$ on $(\mathbb Z/2 \mathbb Z)^n$, and it is nontrivial. | |
| Nov 19, 2012 at 5:32 | comment | added | LMN | In #2 is it really obvious that the extension is non-abelian? | |
| Nov 19, 2012 at 5:25 | vote | accept | LMN | ||
| Nov 19, 2012 at 5:18 | history | answered | John Pardon | CC BY-SA 3.0 |