Timeline for Can one prove vanishing of higher direct images fiber-wise?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2012 at 21:21 | comment | added | Rami | OK, I think that I got it finely. Is there still some hope to part 4.b?. Thank you very much again | |
| Apr 20, 2012 at 16:55 | comment | added | Rami | Thank you very much, I need to think about it in order to understand it | |
| Apr 19, 2012 at 6:37 | comment | added | Mohan | Yes, Rami, that is what I mean. This can be done generally, but to convince oneself, take $R=k[x^4,x^3y,xy^3,y^4]$, blow up the maximal ideal and call it $X$. The one can explicitly check that $X$ is non-singular, the fibers over $R$ are either a point (outside the vertex) and the projective line over the vertex, all scheme-theoretically. Of course $R$ is not normal. If you would like the more general argument, I could write it up and post. | |
| Apr 18, 2012 at 20:32 | comment | added | Rami | Dear Mohan, I'll be very happy to see more details about your above answer. I'm confused. Does your answer means that Sasha is wrong? Do you claim that one can have a resolution of singularities of a non-normal variety s.t. all the scheme theoretic fibers are reduced and connected? | |
| Apr 13, 2012 at 7:36 | comment | added | Mohan | Sandor, this has been a common mistake. There is no reason for the scheme theoretic inverse image of a fat point to be non-reduced under a non-flat map. (Most recently, Vermiere and others have stated this as a lemma in one of their published papers and used it, and I pointed out the error). This would be a method of choice in proving some variety is normal, if that were true and I first encountered it in some one's attempt at proving Schubert varieties are normal. Though I can give a complete proof of what I stated above, the space is limited. If you desire, I can spell it out as an answer. | |
| Apr 11, 2012 at 8:04 | comment | added | Sándor Kovács | Are you sure that the scheme-theoretic pre-image of that point is the reduced exceptional divisor? $\pi$ would have to factor through the normalization of $Y$ and I would expect that the pre-image of the non-normal point on the normalization is not a single reduced point, but something fatter. If that's true, then the pre-image on $X$ is unlikely to be reduced. | |
| Apr 11, 2012 at 4:02 | comment | added | Mohan | I am sorry, I meant an $\mathbb{A}^1$-bundle over $\mathbb{P}^1$. | |
| Apr 11, 2012 at 3:59 | history | answered | Mohan | CC BY-SA 3.0 |