Timeline for Decomposition of finitely generated algebras
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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Jul 31, 2017 at 15:29 | vote | accept | Luis Turcio | ||
Jun 8, 2017 at 6:54 | review | Close votes | |||
Jun 8, 2017 at 10:20 | |||||
Jun 8, 2017 at 1:52 | answer | added | Luis Turcio | timeline score: 1 | |
Apr 24, 2017 at 21:33 | comment | added | Luis Turcio | I'll try to see the rings as global sections. Thanks for the hint Benjamin Steinberg | |
Apr 24, 2017 at 21:28 | comment | added | Luis Turcio | thanks for the commentaries, of course, the question is about f.g algebras. Thanks for the nice example with continuous functions. | |
Apr 24, 2017 at 8:07 | review | Close votes | |||
Apr 24, 2017 at 9:22 | |||||
Apr 24, 2017 at 7:49 | comment | added | YCor | anyway after reading twice, the question seems to be definitely about f.g. algebras. | |
Apr 24, 2017 at 1:05 | comment | added | Benjamin Steinberg | For general commutative rings the substitute is to view the ring as the global sections of a sheaf of indecomposable rings on its Pierce spectrum | |
Apr 23, 2017 at 23:31 | comment | added | YCor | Of course there's no decomposition for arbitrary commutative algebras: maybe the simplest example is the (Boolean) algebra of continuous functions from the Cantor set to $\mathbf{Z}/2\mathbf{Z}$. | |
Apr 23, 2017 at 23:29 | comment | added | YCor | Yes for finitely generated algebras. More generally if $A$ is an arbitrary noetherian ring, it has a finite number of minimal prime ideals. This implies that the number of idempotents is finite. This is standard material; I guess it's in most commutative algebra textbooks. | |
Apr 23, 2017 at 23:19 | history | asked | Luis Turcio | CC BY-SA 3.0 |