Timeline for Hasse principle for rational times square
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2016 at 18:22 | vote | accept | JadeSnail | ||
| Jun 14, 2016 at 11:08 | comment | added | Daniel Loughran | You can find a reference for this fact in the proof of Theorem 2.2. of arxiv.org/PS_cache/math/pdf/0612/0612528v4.pdf. | |
| Jun 14, 2016 at 10:32 | comment | added | David E Speyer | Thanks! That does do it! The point is that the only cyclic subgroup of $G$ containing $z$ is $\langle z \rangle$, so this means that every place of $\KK$ either splits in $\KK(\sqrt{a)}$, or else is split over $\QQ$. I'll rewrite to spell this out when I get the chance. So any extension with this group and with all decomposition groups cyclic works. | |
| Jun 14, 2016 at 7:12 | comment | added | Daniel Loughran | It is known that for every solvable group $G$, there exists a Galois extension $\mathbb{K}/\mathbb{Q}$ whose Galois group is $G$ and such that the decomposition group at all primes $p$ is cyclic. Does this help? | |
| Jun 14, 2016 at 2:59 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Jun 14, 2016 at 2:31 | history | answered | David E Speyer | CC BY-SA 3.0 |