Timeline for Non-split extension of the rationals by the integers
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2012 at 7:23 | vote | accept | Mark Opitz | ||
| Aug 24, 2012 at 3:21 | answer | added | Noah Stein | timeline score: 14 | |
| Aug 23, 2012 at 19:21 | answer | added | Ralph | timeline score: 28 | |
| Aug 23, 2012 at 1:11 | comment | added | Kevin Ventullo | @algori: Uncountable, but contains a copy of $\mathbb{Q}$. | |
| Aug 22, 2012 at 22:34 | comment | added | anon | Consider the canonical exact sequence Z --> \prod_p Z_p --> M, where the product takes place over all primes p. Then M is a Q-vector space (exercise). Picking any non-zero element m in M gives an extension A of Q by Z via pullback; explicitly, A is the set of all elements in \prod_p Z_p that map to the Q-subspace generated by m in M. This extension is non-zero since Hom(Q,\prod_p Z_p) = \prod_p Hom(Q,Z_p) = 0. | |
| Aug 22, 2012 at 22:29 | comment | added | Andreas Blass | See my answer to mathoverflow.net/questions/90586/are-these-abelian-groups-free for a stronger result, due to Fuchs and Loonstra. | |
| Aug 22, 2012 at 22:29 | comment | added | Damian Rössler | See also p. 5 of math.jhu.edu/~jmb/note/torext.pdf | |
| Aug 22, 2012 at 22:02 | answer | added | Alexander Shamov | timeline score: 10 | |
| Aug 22, 2012 at 21:54 | comment | added | algori | Mark -- something uncountable. | |
| Aug 22, 2012 at 21:46 | comment | added | Mark Grant | There should in fact be uncountably many of them! journals.cambridge.org/… What do you get if you quotient the $p$-adic integers by the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Z}_p$? | |
| Aug 22, 2012 at 21:02 | history | asked | Mark Opitz | CC BY-SA 3.0 |