Timeline for Proof of Cartan's solvability criterion
Current License: CC BY-SA 3.0
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| Aug 27, 2013 at 10:24 | comment | added | David MJC | Yes, that's the standard argument I had in mind: it uses that $ad(y)$ is a polynomial in $ad(x)$ without constant term, so that $y\in m$. Clearly I confused this property with $y$ being a polynomial in $x$! Thanks for setting me straight. | |
| Aug 27, 2013 at 9:17 | history | answered | Dietrich Burde | CC BY-SA 3.0 |