Timeline for Rings over which every module is free
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2009 at 23:26 | comment | added | Mark Hovey | I object to your assertion that you refuse to think about non-Noetherian rings, especially for this problem. As pointed out in ohdarkdevil's response below, if every module is free, then every module is projective. This is equivalent to the ring being semisimple. So it is automatically Noetherian. | |
| Nov 23, 2009 at 19:35 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 15, 2009 at 20:23 | comment | added | Ben Webster♦ | Ok, now I fixed it. | |
| Oct 15, 2009 at 20:23 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 15, 2009 at 20:10 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 15, 2009 at 19:15 | comment | added | Benoit Jubin | To apply the Artin-Wedderburn theorem, you also need the ring to be artinian. How to prove this? | |
| Oct 15, 2009 at 12:11 | vote | accept | Benoit Jubin | ||
| Oct 14, 2009 at 22:42 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |