Timeline for Are epimorphisms from a division ring isomorphisms ?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2013 at 11:42 | comment | added | tj_ | Are you sure it has been downvoted ? The last time I looked in, your answer had 1pt - now it has 3pts. | |
| Feb 7, 2013 at 10:57 | comment | added | Martin Brandenburg | Hm, I wonder why my answer has been downvoted. It contains a correct proof, right? | |
| Feb 7, 2013 at 2:31 | comment | added | Tom Goodwillie | Oh, I didn't understand. | |
| Feb 6, 2013 at 13:57 | comment | added | Martin Brandenburg | @Tom: Sure, but the condition follows from the criterion mentioned by TJ. I've added the proof. | |
| Feb 6, 2013 at 13:56 | comment | added | Torsten Schoeneberg | One surely has to be careful as a priori the maps $i_j$ are not homomorphisms, but under the given criterion they are; maybe one also has to extend $1, s$ once to a left and once to a right basis, and then check Bourbaki, Algebra ch. II §3 no. 7 cor. 1 and following remarks to see that what Martin writes in his first comment is true after making a left-/right-distinction. | |
| Feb 6, 2013 at 13:56 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
added 539 characters in body; deleted 5 characters in body; deleted 44 characters in body
|
| Feb 6, 2013 at 1:52 | comment | added | Tom Goodwillie | Martin, pushouts in Ring are not given by tensor product as in CommRing. | |
| Feb 6, 2013 at 1:41 | comment | added | Martin Brandenburg | I use the following elementary fact: If $M$ is a free $R$-module and $m,n$ are part of a basis such that $m \otimes n = n \otimes m$ in $M \otimes_R N$, then $m=n$. | |
| Feb 6, 2013 at 0:48 | comment | added | tj_ | I don't understand which contradiction you obtain. Could you please give some more details. Thanks. | |
| Feb 6, 2013 at 0:34 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
deleted 10 characters in body
|
| Feb 6, 2013 at 0:24 | history | answered | Martin Brandenburg | CC BY-SA 3.0 |