Timeline for A ring such that all projectives are stably free but not all projectives are free?
Current License: CC BY-SA 2.5
15 events
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Jun 8, 2010 at 5:46 | comment | added | algori | Hailong -- yes, in the algebraic case it's the stable triviality that's causing problems. But my guess is there is an example along these lines. | |
Jun 8, 2010 at 5:42 | comment | added | Hailong Dao | @algori: R_5 is the ring of polynomial functions on 5-spheres. It then injects into the ring of continuous functions, unless I am missing something. One still needs to show projectives are stably free though. | |
Jun 8, 2010 at 5:31 | comment | added | algori | Hailong -- your comment says $R_5$ is a subring of $R$. But how is it defined exactly? | |
Jun 8, 2010 at 5:24 | comment | added | Hailong Dao | @algori: $R_5$ is Noetherian. My comment is meant to use Tyler example to show that $R_5$ works. | |
Jun 8, 2010 at 5:20 | comment | added | algori | Hailong -- if you are happy with non-Noetherian rings, then Tyler's example works as advertized (and there are many more). | |
Jun 8, 2010 at 5:17 | comment | added | Hailong Dao | @algori: R is the ring of continuous function in Tyler's answer. | |
Jun 8, 2010 at 5:16 | comment | added | algori | Hailong -- and what is $R$? | |
Jun 8, 2010 at 5:15 | comment | added | Hailong Dao | algori: $R_n$ is defined in my question. | |
Jun 8, 2010 at 5:13 | comment | added | Hailong Dao | Namely, let T be the $R_5$ module defined as kernel of the map defined by the column with all the $x_i$s. $R_5$ is a subring of your ring R, and T tensoring with R gives the tangent bundle which shows T is not free. | |
Jun 8, 2010 at 5:12 | comment | added | algori | Hailong -- what is $R_5$? My bet would be the easiest example is an algebraic vector bundle. | |
Jun 8, 2010 at 5:00 | comment | added | Hailong Dao | But it may still be true that $R_5$ works. | |
Jun 8, 2010 at 4:45 | history | edited | Tyler Lawson | CC BY-SA 2.5 |
warning to the unwary
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Jun 8, 2010 at 4:43 | comment | added | algori | Tyler -- the ring of continuous functions on the 5-sphere is definitely not Noetherian. | |
Jun 8, 2010 at 4:36 | comment | added | Hailong Dao | Hy Tyler, very interesting! For completeness and benefit of other viewers, can you provide some references? | |
Jun 8, 2010 at 4:27 | history | answered | Tyler Lawson | CC BY-SA 2.5 |