Timeline for Rationality of algebraic groups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2017 at 1:27 | comment | added | Mikhail Borovoi | Note that any two-dimensional linear algebraic group $G$ over a field $k$ of characteristic 0 is $k$-rational. Indeed, since ${\rm char}(k)=0$, we may assume that $G$ is reductive. Since there are no semisimple groups of dimension $\le2$, our group $G$ is a torus. Finally, in Section 4.9 of Voskresenskii's book it is proved that any $k$-torus of dimension one or two is $k$-rational. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 5, 2012 at 13:20 | answer | added | George McNinch | timeline score: 6 | |
| Mar 5, 2012 at 1:17 | comment | added | Igor Rivin | @Noam: thanks, that's very interesting! | |
| Mar 4, 2012 at 5:11 | comment | added | Noam D. Elkies | The answer to this MO query mathoverflow.net/questions/76882/… cites a paper showing there are three-dimensional commutative algebraic groups over ${\bf Q}$ that are rational over $\overline{\bf Q}$ but not over ${\bf Q}$! | |
| Mar 3, 2012 at 23:15 | comment | added | Jim Humphreys | As the commentw on Ryan's answers indicate, you need to say more precisely what you mean by "algebraic group" and by "rational". There are results going back to Rosenlicht (in older algebraic geometry language) about which varieties are "rational" in the sense that the function field is a purely transcendental field. The linguistic issues tend to get in the way of clarity. | |
| Mar 3, 2012 at 23:09 | answer | added | Peter McNamara | timeline score: 2 | |
| Mar 3, 2012 at 21:32 | answer | added | Ryan Reich | timeline score: 3 | |
| Mar 3, 2012 at 19:58 | history | asked | Igor Rivin | CC BY-SA 3.0 |