Timeline for finite generation of a certain type of subring
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2015 at 18:42 | vote | accept | Neil Epstein | ||
| Jun 1, 2015 at 18:41 | comment | added | Neil Epstein | That's exactly what I was looking for but was not aware of (i.e. the Artin-Tate lemma). Thank you! | |
| Jun 1, 2015 at 14:46 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
rewritten entirely
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| May 31, 2015 at 17:39 | comment | added | Wilberd van der Kallen | @Neil Epstein. It is not true that $R$ is module-generated over $A$ by any lifting of any $k$-basis of $R/I$. But if 1 is in the lifting then $R$ is indeed module generated by the lifting and the finite generation of $A$ follows from the wikipedia Artin-Tate lemma. | |
| May 29, 2015 at 17:25 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
more detail
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| May 29, 2015 at 6:51 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
clarification
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| May 28, 2015 at 19:09 | comment | added | Neil Epstein | Also, I don't understand the explicit construction you give afterwards. How do you know you can pick $v_i$s and $x_j$s that satisfy the conditions (under the clauses starting with "so that" in the second sentence of the second paragraph) which you specify? | |
| May 28, 2015 at 18:47 | comment | added | Neil Epstein | I'm confused. If you are claiming that whenever the associated graded ring of a ring $B$ is finitely generated over $k$, so is $B$ itself, this is wrong (e.g. $B=k[\![x]\!]$). | |
| May 28, 2015 at 10:50 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
corrected mistakes
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| May 28, 2015 at 9:57 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
added 692 characters in body
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| May 28, 2015 at 9:09 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
added 11 characters in body
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| May 28, 2015 at 8:13 | history | answered | Wilberd van der Kallen | CC BY-SA 3.0 |