Timeline for commutative algebra, diagonal morphism
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2015 at 20:18 | history | edited | darij grinberg | CC BY-SA 3.0 |
disambiguation for "algebra"
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| Jul 21, 2014 at 8:42 | vote | accept | AYK | ||
| Jul 21, 2014 at 5:01 | answer | added | tj_ | timeline score: -1 | |
| Jul 21, 2014 at 4:15 | answer | added | Julian Rosen | timeline score: 6 | |
| Jul 21, 2014 at 1:50 | comment | added | Steven Landsburg | @VivekShende: But we don't need $ker(A[x]\rightarrow B)$ to be finitely generated; we just need a certain quotient of it to be finitely generated, no? | |
| Jul 21, 2014 at 1:42 | comment | added | Vivek Shende | is finite type really enough? One has a map from the sequence $0 \to I_A \to A[\mathbf{x}] \otimes_A A[\mathbf{x}] \to A[\mathbf{x}] \to 0$ to the sequence $0 \to I \to B \otimes_A B \to B \to 0$. Then you have an exact sequence $I_A \to I \to \mathrm{coker}(I_A \to I)$ and also by the snake lemma a surjection $\mathrm{ker}(A[\mathbf{x}] \to B) \twoheadrightarrow \mathrm{coker}(I_A \to I)$. It's easy to see $I_A$ is finitely generated, but the condition that $\mathrm{ker}(A[\mathbf{x}] \to B)$ is precisely asks $B$ to be finitely presented and not just finitely generated. | |
| Jul 20, 2014 at 22:23 | comment | added | AYK | Yes, the elements $db_i\in \Omega^1$ form a generating set. But how to go from $I/I^2$ to $I$? | |
| Jul 20, 2014 at 22:10 | comment | added | მამუკა ჯიბლაძე | Maybe one could use that $I/I^2$ is Kähler differentials, it must be well known how the number of generators of $\Omega^1$ as a module relates to the number of generators of $B$ as an algebra over $A$... | |
| Jul 20, 2014 at 21:43 | review | Close votes | |||
| Jul 22, 2014 at 12:58 | |||||
| Jul 20, 2014 at 21:22 | review | First posts | |||
| Jul 20, 2014 at 21:28 | |||||
| Jul 20, 2014 at 21:19 | history | asked | AYK | CC BY-SA 3.0 |