Timeline for Separability of a simple ring extension
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jul 12, 2015 at 20:03 | comment | added | user237522 | @Vinteuil Sorry to bother you again. Can you please read and comment on the following?math.stackexchange.com/questions/1358096/…. Are we talking about the same things, or your definitions differ from mine? Thank you very much. | |
| Jun 18, 2015 at 23:01 | comment | added | user237522 | @Vinteuil Thank you very much for your explanation! It is very helpful to me. | |
| Jun 18, 2015 at 7:41 | comment | added | Vinteuil | The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general pd$_{B\otimes_AB}B<\infty$ for a flat algebra implies regularity. This result was obtained in gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002067722 . The separable case was known before, but I do not know any reference, except Cartan-Eilenberg when $A$ is a field (theorem 7.10 in chapter IX, page 179). Cartan-Eilenberg was written before Serre introduced flatness, so maybe it is easy to modify $A$ field by $A \to B$ flat in their proof. | |
| Jun 18, 2015 at 3:19 | comment | added | user237522 | @Vinteuil Please can you tell me where can I find a proof of your claim "separable implies smooth"? (Sorry if this implication may be considered trivial). Perhaps I should try Matsumura's book "Commutative Algebra"? | |
| Jun 17, 2015 at 2:41 | comment | added | user237522 | @Vinteuil Thank you very much! What I have missed is "separable implies smooth", and then the example in math.stackexchange cannot be separable, since separability would imply (after some considerations you have made) that $B$ is regular but it is not, as Brevik explained. (I have now noticed that I am not sure I know how to prove that $B=A[w]$, with $w^2=x^2y$ is not integrally closed; I guess it's not difficult to prove this). | |
| Jun 16, 2015 at 8:43 | comment | added | Vinteuil | The first answer in math.stackexchange should suffice since separable implies smooth (take in mind that if $B$ is smooth over $A$, then it is smooth over $K$ and so a regular ring). For more generality, see propositions 2.3 and 2.5 in sciencedirect.com/science/article/pii/0021869392900057 | |
| Jun 15, 2015 at 15:39 | history | asked | user237522 | CC BY-SA 3.0 |