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Nov 26, 2021 at 11:27 history closed YCor
user44191
abx
Ben McKay
Friedrich Knop
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Nov 24, 2021 at 16:42 history became hot network question
Nov 24, 2021 at 11:51 answer added Ben C timeline score: 6
Nov 24, 2021 at 10:33 review Close votes
Nov 26, 2021 at 11:27
Nov 24, 2021 at 10:11 comment added user11090426 @Lao-tzu Thank you for providing me this result.
Nov 24, 2021 at 9:52 comment added Lao-tzu Alg(A,C)⊂Spec(A) is just the set of closed points in Spec(A).
Nov 24, 2021 at 9:04 history edited Ben McKay CC BY-SA 4.0
improved the grammar a little
Nov 24, 2021 at 9:00 comment added user11090426 @LeoAlonso Thanks
Nov 24, 2021 at 8:56 comment added Leo Alonso If $A$ is a finite type algebra this is essentially the content of Hilbert's Nullstellensatz.
Nov 24, 2021 at 8:53 comment added user11090426 @YCor I want to know for what algebra A, the two sets are equal.
Nov 24, 2021 at 8:51 comment added YCor What is the question (the last sentence does't make sense)?
Nov 24, 2021 at 8:50 comment added YCor So you have an inclusion Alg(A,C)$\subset$Spec(A). The first correspond to quotients reduced to C. The second corresponds to quotients that are integral C-algebras.
Nov 24, 2021 at 8:49 history edited YCor CC BY-SA 4.0
fixed typo and completed definition
Nov 24, 2021 at 8:45 comment added user11090426 @YCor I have added it in the question, sorry.
Nov 24, 2021 at 8:44 history edited user11090426 CC BY-SA 4.0
added 17 characters in body
Nov 24, 2021 at 8:43 comment added YCor What do you mean by "algebra"? the same as $\mathbf{Z}$-algebra? as $\mathbf{C}$-algebra? something else?
Nov 24, 2021 at 8:43 comment added user11090426 Does some theorems about this question? We know that if f $\in Alg(A, \mathbb{C})$, then $ker f$ is a maximal ideal. What about else?
Nov 24, 2021 at 8:43 history edited YCor CC BY-SA 4.0
formatting
Nov 24, 2021 at 8:39 history asked user11090426 CC BY-SA 4.0