Timeline for A question on $SK_1$ of rings
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 15, 2020 at 4:43 | comment | added | nikhilesh dasgupta | Thank you sir very much for the clarification. | |
| Sep 14, 2020 at 8:30 | comment | added | Wilberd van der Kallen | You can multiply with a diagonal matrix that is congruent to the identity and get the determinant equal to one. | |
| Sep 14, 2020 at 7:04 | vote | accept | nikhilesh dasgupta | ||
| Sep 14, 2020 at 7:04 | comment | added | nikhilesh dasgupta | Thank you sir for the response. It is clear to me that the map $SK_1(B) \rightarrow SK_1(B_{red})$ is injective. But why that map is {\bf surjective } ? I understand the preimage of a determinant 1 matrix is invertible. But why the preimage has determinant 1 is not what is clear to me. | |
| Sep 14, 2020 at 4:14 | history | edited | Steven Landsburg | CC BY-SA 4.0 |
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| Sep 13, 2020 at 17:07 | history | edited | Steven Landsburg | CC BY-SA 4.0 |
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| Sep 13, 2020 at 14:48 | history | answered | Steven Landsburg | CC BY-SA 4.0 |