Timeline for Are the abelian absolute Galois groups of these local fields isomorphic?
Current License: CC BY-SA 3.0
12 events
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| Mar 13, 2016 at 16:11 | history | edited | znt | CC BY-SA 3.0 |
added 54 characters in body
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| Mar 12, 2016 at 21:16 | comment | added | znt | I think it must be positive. There is no odd order torsion, and for 2-power order we see that both fields contain $\pm1$ and if either one contained $i$ then it would contain $1+i$ which is an $8$th root of 16, meaning that the fields themselves would be isomorphic (this is Lubin's observation). Hence either the fields aren't isomorphic and the only torsion in their units is $\pm1$ or they're isomorphic. I think we're done. | |
| Mar 12, 2016 at 19:51 | comment | added | Pablo | @znt so the answer is positive, right? | |
| Mar 12, 2016 at 19:43 | comment | added | znt | pari-gp says [type "idealfactor(nfinit(x^8-48),2)" into it ] that Q_2(48^{1/8}) is also totally ramified of degree 8. | |
| Mar 12, 2016 at 18:26 | comment | added | Lubin | It seems to turn out that $\Bbb Q_2(3^{1/8},\sqrt2\,)$, which contains both fields, is totally ramified (of degree $16$). The argument that I gave above therefore applies to both fields: neither has any more $2$-power torsion in the principal units than $\{\pm1\}$, and the residue field is $\Bbb F_2$ for both as well. My argument involves Higher Ramification, and right now I’m so sick of ramified extensions of $\Bbb Q_2$ that it’ll have to wait till tonight for me to write the whole mess up. | |
| Mar 12, 2016 at 14:27 | history | edited | znt | CC BY-SA 3.0 |
attempt to deal with GH's (valid) objection about me using Weil groups by accident.
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| Mar 12, 2016 at 14:18 | comment | added | znt | But you're right, I should edit. | |
| Mar 12, 2016 at 14:12 | comment | added | znt | The exact sequence splits and everything is finitely-generated. Am I missing something? | |
| Mar 12, 2016 at 11:58 | comment | added | GH from MO | Why is the question equivalent to asking if $K_1^\times$ and $K_2^\times$ are isomorphic? By class field theory I only see that there is an exact sequence $1\to\mathcal{O}_K^\times\to \mathrm{Gal}(K_{\mathrm{ab}}/K)\to\hat{\mathbb{Z}}\to 1$ for $K=K_1$ and for $K=K_2$. I agree that if the torsion subgroups of $\mathcal{O}_K^\times$ are not the same for $K=K_1$ and $K=K_2$, then $\mathrm{Gal}(K_{\mathrm{ab}}/K)$ are not the same for $K=K_1$ and $K=K_2$. | |
| Mar 12, 2016 at 4:41 | comment | added | Lubin | It’s clear that the only torsion in $\Bbb Q_2(3^{1/8})$ is $\{\pm1\}$, ’cause if $i$ were there, $\sqrt{-3}$ would be there too, inducing an unramified quadratic subfield. But $\Bbb Q_2(3^{1/8})$ is certainly totally ramified of degree $8$, the Eisenstein polynomial being $(X+1)^8-3$. I have no insight about $\Bbb Q_2(48^{1/8})$, though. It would help to know a prime element of that field. | |
| Mar 11, 2016 at 23:05 | review | First posts | |||
| Mar 11, 2016 at 23:09 | |||||
| Mar 11, 2016 at 23:03 | history | answered | znt | CC BY-SA 3.0 |