Timeline for Abelianization of general linear group of a polynomial ring
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2020 at 20:59 | comment | added | François Brunault | This is related to algebraic $K$-theory: given a ring $R$, the Whitehead group $K_1(R)$ is defined as the abelianization of $\mathrm{GL}_\infty(R) = \bigcup_{n \geq 1} \mathrm{GL}_n(R)$. If $R$ is an Euclidean domain then $K_1(R) \cong R^\times$. You can read about these things in Milnor's Introduction to algebraic $K$-theory. | |
| Apr 17, 2020 at 20:30 | answer | added | Luc Guyot | timeline score: 6 | |
| Apr 17, 2020 at 14:48 | comment | added | YCor | Indeed I should have quoted Nagao (I referred to the Serre proof). J-P. Serre duly quotes Nagao (J. Poly. Ozaka Univ. 1959) who originally proved the amalgamation decomposition. Serre (Astérisque 1977) provided a new proof using action on trees (now known as Bass-Serre theory). | |
| Apr 17, 2020 at 14:45 | vote | accept | qqqqqqw | ||
| Apr 17, 2020 at 9:23 | history | became hot network question | |||
| Apr 17, 2020 at 8:30 | comment | added | Wilberd van der Kallen | One may use Nagao's Theorem. Especially when the field has only two elements, as the easy way is then blocked. | |
| Apr 17, 2020 at 8:24 | answer | added | YCor | timeline score: 21 | |
| Apr 17, 2020 at 7:57 | comment | added | YCor | "Determining the relations", which is a sloppy way to mean describe defining relators with respect to given (not yet here) generators, is much harder than describing the abelianization. | |
| Apr 17, 2020 at 1:31 | comment | added | qqqqqqw | Is it easy to determine the relations? | |
| Apr 17, 2020 at 1:22 | comment | added | user6976 | So it is the group of 2 by 2 nonsingular matrices ring of polynomials in one variable? If you know the generators (elementary matrices, ...) you should be able to find the abelianization yourself. | |
| Apr 17, 2020 at 1:18 | comment | added | qqqqqqw | X is a polynomial variable. K[X] is not a field extension. | |
| Apr 17, 2020 at 1:17 | comment | added | user6976 | It depends on what $X$ is. | |
| Apr 17, 2020 at 1:09 | history | asked | qqqqqqw | CC BY-SA 4.0 |