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This is a somewhat vague question, but given my general ignorance in algebraic geometry, I hoped that maybe someone could give me some hint...

Suppose that I have a variety $X$ (over an algebraically closed field of characteristic 0) and I want to prove that it's normal. Assume (for simplicity) that I know that away from one point $x\in X$.

Assume now that I have an explicit resolution $\pi:Y\to X$ and I know everything about its scheme-theoretic fibers (in my particular case they are irreducible but unfortunately not reduced; however, the corresponding reduced scheme is always smooth). Is there some statement about that resolution (maybe its fibers) that would imply normality of $X$ (once again: it is clear that if the fibers were irreducible and reduced, that would be it; in my case they are not reduced, but completely describe them. Does it help?)

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2 Answers 2

Unfortunately it is not true that if $Y$ is smooth, $f$ is proper and the fibers of $f:Y \rightarrow X$ are reduced and connected that $X$ is normal. Unfortunately I am all to familiar with the following example, which has ended any optimism I have had about finding a criteria for normality using a desingularization. The question is local so I will just use the affine coordinates. Let $ R = k[x^4,x^3y,xy^3, y^4] $. This ring is the cone of a projection of the quartic normal curve, so it is not normal at the origin. One checks easily using local coordinates that the resolution is just $\bf P^1$. I learned of this example from Mohan Kumar, but it also appears in one or more of Eisenbud's books, as well as a counter example to other questions asked here. It seems to be a universal counter example to any seemingly true facts about depth etc for which one cannot find a proof.

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Thank you, that really changes my picture... –  Alexander Braverman Oct 3 '11 at 15:29

Well, the affine line normalizes the cusp $x^2=y^3$, and the fibers are about as nice as they can be conditional on not being reduced. Yet the cusp is not normal. So it seems unlikely to me that there could be a criterion of this kind.

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It is obvious that if the fibers are reduced and connected then the variety is normal. In your case the fibers are obviously not reduced. –  Alexander Braverman Sep 30 '11 at 4:53
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Alexander: Have I misundersood your question? I thought you were asking "If the fibers of the map $\pi$ are not reduced but nevertheless very nice, can I conclude that $X$ is normal?". I gave an example of a map where the fibers are not reduced but nevertheless as nice as non-reduced fibers can possibly be, yet $X$ is not normal. So if I understood your question correctly, I believe this pretty much answers it. –  Steven Landsburg Sep 30 '11 at 5:44
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I think the question is interesting only for dimension >1, since for curves normalization and desingularization coincide. –  rita Sep 30 '11 at 5:54
    
My question was not formulated precisely - what I had in mind was for example whether some careful analysis of dimensions on the non-reduced locus compared to some other parameters of the situation can help; or maybe a group action with some properties. Basically I am wondering if there are examples when one can prove normality based on some purely geometric properties of the situation. –  Alexander Braverman Sep 30 '11 at 11:31
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I don't think there can be a criterion based only on the dimension of the non-reduced locus. To expand on Steven's example, you can take the origin in affine n-space and put a "cusp", i.e., non-normal point, only at that point. The resolution will be a homeomorphism with domain equal to affine n-space, and the only non-reduced fiber will be the origin. But of course there could be some result in your situation. Can you tell us more about your situation? –  Jason Starr Sep 30 '11 at 13:01

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