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Timeline for Separability and smoothness

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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jul 13, 2015 at 13:29 comment added user237522 Thanks!! I will check the books you recommend. (I only know one sufficient condition for smoothness, which appears in the paper "A Jacobian criterion for smoothness" by Morris and Wang).
Jul 13, 2015 at 13:11 comment added Vinteuil The usual proofs show that a regular homomorphism satisfies some kind of Jacobian criterion, and so in the case of regular homomorphisms of finite type, they are smooth (not in general if they are not of finite type, but at least we have Popescu desingularization theorem). A non-homological version of the proof can be seen in Bourbaki, Algèbre Commutative, Section 7, n.10, Théorème 4. A homological one in André, Homologie des Algèbres Commutatives, Supplément, Théorème 30.
Jul 13, 2015 at 13:00 history edited user237522 CC BY-SA 3.0
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Jul 13, 2015 at 12:44 comment added user237522 Thanks. Also, I wish to remark that the comment originally referred to a question of mine where I assumed that $B$ is free over $A$ (hence flat). The thing that remains for me to check is why a regular homomorphism implies smoothness (maybe it's easy or well-known; I have not yet tried to think about it). @Vinteuil, if you like to write this as an answer, I would happily accept it.
Jul 13, 2015 at 12:32 comment added Vinteuil Yes, that's the definition I have in mind.
Jul 13, 2015 at 12:26 comment added user237522 Thank you very much!! Please, just to be sure, what is your definition of separability? Is it: $B$ is a projective $B \otimes_A B$-module + $B$ is flat over $A$? (if so, in the noetherian case, it's my definition of etale).
Jul 13, 2015 at 12:17 comment added Vinteuil As quoted, the definition of separability implies flatness. If you use a definition of "separable algebra" which does not assume flatness, take in mind that there exist non-flat local epimorphisms (see archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/… ).
Jul 13, 2015 at 7:56 comment added Vinteuil I would add a few words to the quote of the question (boldface is mine): "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$_{B\otimes_AB}B<∞$ for a flat algebra implies regularity ...". This sentence says that the easier thing to prove is not "separable implies smooth" but "separable implies regular", as it can be seen if you work with less hypothesis, such as in Theorem 1 in the paper you cite above.
Jul 13, 2015 at 2:14 comment added user237522 Thank you very much for your comments! As for Corollary 2, I only tried to guess what the comment talked about, so I applied Corollary 2 to show that "separable+flat implies smooth". As for your initial comment, I do not understand why it contradicts the claim "separable implies smooth", since this claim allows the existence of smooth algebras which are not separable (like $k \subseteq k[t]$); but maybe I am missing something. I agree that this is quite confusing, because we do not know what the exact definitions the comment used.
Jul 13, 2015 at 1:39 comment added grghxy @user237522: You are misreading that Corollary 2, as it involves the condition of merely finite "flat dimension", which is much much weaker than "projectivity", the latter being the definition of "separable" in the link you are giving at the end. I showed in my initial comment above that a smooth non-etale algebra is never separable. So you need to identify the correct definition of "separable algebra", and make sure it is really satisfied in the smooth case (else the question posed has no motivation); otherwise this is all quite confusing.
Jul 13, 2015 at 0:32 comment added user237522 Maybe his definition of separability is my definition of separability+flatness, and the comment meant: "separability and flatness implies smoothness", which is. if I am not wrong, Corollary 2 of wwwuser.gwdg.de/~subtypo3/gdz/pdf/PPN358147735_0065/… (which he mentions as a reference).
Jul 12, 2015 at 23:13 comment added grghxy @user237522: I suspect that the person who made the earlier comment must have had in mind another definition of "separable algebra" or was confused, as my argument shows that the definition of separability that you are using is never satisfied for smooth algebra which is not etale.
Jul 12, 2015 at 22:54 comment added user237522 Thanks. The comment is in: mathoverflow.net/questions/209379/…. However, I feel uncomfortable to attach the link, since it involves the person who wrote that comment (and if it's not true, maybe he wishes to delete it). Anyway, I can delete the link (after getting an answer).
Jul 12, 2015 at 22:42 comment added Henry.L @user237522 Where is the comment "separablity implies smoothness", can you give the link and thanks.
Jul 12, 2015 at 22:38 history edited user237522 CC BY-SA 3.0
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Jul 12, 2015 at 22:05 comment added user237522 Thanks. The motivation is a comment on MO which claims that "separablity implies smoothness"; the (beginning of the) explanation in the comment says: "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 ...".
Jul 12, 2015 at 21:53 comment added grghxy There is something confusing about the premise of this question, as follows. Separability means that the surjective map $\pi: B\otimes_A B \rightarrow B$ makes $B$ a projective $B\otimes_A B$-module, but $\pi$ is surjective, so by injectivity of flat local maps it follows that $I := \ker(\pi)$ satisfies $I_{\mathfrak{p}}=0$ for all primes $\mathfrak{p} \supset I$ of $B \otimes_A B$. Hence, $\Omega^1_{B/A}=0$, so smooth $A$-algebras with positive relative dimension (e.g., $B=A[T]$!) are never separable. Have I misunderstood something? If not, then what is the motivation for this question?
Jul 12, 2015 at 20:54 history asked user237522 CC BY-SA 3.0