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Let $\pi:X\to Y$ be a proper map of algebraic varieties (over $\mathbb C$) which is a bi-rational equivalence.
are the following statements equivalent?

  1. The derived direct image of $O_X$ is $O_Y$.

  2. For any $y \in Y$ we have $H^*(O_{\pi^{-1}(y)})= \mathbb C$


  1. I do not assume that $X$ or $Y$ are smooth.
  2. The fiber is considered scheme theoretically
  3. By point one can mean a scheme theoretic point or a closed point or a geometric point. I think it is does not matter.
  4. I also have 2 variations of the question:

    a) What happens we consider conditions (1) and (2) only on the level of the zero cohomology?

    b) What happens we consider conditions (1) and (2) only on the level of higher cohomologies?

  5. Variation (b) make sense without the conditions on $\pi$. I need the answer only for the case I described (since I'm interested in rational singularities) but I'll be happy to know what happens in general

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Do you want to prove it twice? – Fernando Muro Mar 8 '12 at 23:15
Sorry, I fixed it – Rami Mar 8 '12 at 23:26
I believe that this is not true, but I don't know an example off the top of my head. – Karl Schwede Mar 9 '12 at 17:48

Hi Rami,

What about using the Grothendieck complex? Thus, there is a bounded complex $K$ of coherent locally free sheaves on Y, such that $K_y$, the application of "fiber at y" functor elementwise to $K$, computes cohomologies of fibers.Then backwards induction seems to give that if (2) holds, the direct image of $O_X$ is concentrated in degree 0.

Maybe I am wrong? I did not check carefully. Also, I did not check if we can extract information about image being exactly $O_Y$.


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What is the definition of K? What is the reference? If $Y$ is not smooth than as far as I understand one can not take $K$ to be $\pi_*(O_X)$. Are there other candidates? Thank you, Rami – Rami Apr 24 '12 at 22:04
I am sorry, I forgot to mention that one has $K$ if the map $X \to Y$ is flat.. – Sasha Aug 3 '12 at 15:03

Karl is correct. Here is an example which at least sheds some light, I hope. Embed the projective line with a large degree line bundle in $n$-space, so that the embedding is not linearly normal. Let $Y$ be the cone (polynomial ring in $n+1$ variables modulo the ideal of forms vanishing on the curve), so that $Y$ is not normal. Let $X$ be the blow up of the irrelevant maximal ideal. Then one can easily check that $X$ is smooth (a $\mathbb{P}^1$ bundle over $\mathbb{P}^1$), $\pi:X\to Y$ is birational with only one exceptional point, namely the vertex of $Y$ and the scheme theoretic inverse image of this point is just a $\mathbb{P}^1$. Thus condition b) is satisfied. But, $\pi_*O_X$ is the normalization of $O_Y$.

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I am sorry, I meant an $\mathbb{A}^1$-bundle over $\mathbb{P}^1$. – Mohan Apr 11 '12 at 4:02
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. – Sándor Kovács Apr 11 '12 at 8:04
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. – Mohan Apr 13 '12 at 7:36
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? – Rami Apr 18 '12 at 20:32
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. – Mohan Apr 19 '12 at 6:37

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