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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