Suppose $X \rightarrow Y$ is a map of reduced connected projective schemes of finite type over an algebraically closed field of characteristic 0, where $Y$ is a smooth connected curve. Let $Z \rightarrow X$ be the normalization of $X$. Then, for a general closed point $y \in Y$ (meaning for all but finitely many closed points of $Y$), is $Z_y$ (the fiber of $Z$ over $y$) the normalization of $X_y$?

Edit: This question was originally asked without the characteristic 0 hypothesis. Jason Starr pointed out in the comments that it fails in characteristic $p >0$, essentially because generic smoothness does not hold, as is witnessed in the case of quasi-elliptic fibrations.

Feel free to ignore the following, but in case it helps provide context, I will now explain why I want to know this is true for my research. For other reasons, I am trying to prove the following statement:

Suppose we have a proper flat map of quasi-projective reduced schemes over an algebraically closed field $X \rightarrow Y$ where $Y$ is a smooth connected curve. Assume further that the fiber over every point in $Y$ has two irreducible components, and the fiber over a particular closed point $y \in Y$ has two irreducible components with distinct Hilbert polynomials. Then, $X$ has two irreducible components.

Essentially by considering the normalization of $X$, and using Stein factorization, I have reduced the problem to the question above (showing that the normalization of the fiber is the fiber of the normalization for a general closed point of $Y$). However, I'm stuck on this detail.

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    $\begingroup$ The answer depends on what you mean by "general point". The geometric generic fiber of $Z\to Y$ need not be normal. One class of examples are the "quasi-elliptic fibrations". $\endgroup$ Dec 22, 2015 at 2:00
  • $\begingroup$ @JasonStarr Good point, I should have clarified that. I had intended just to ask about the general closed point (or equivalently geometrically closed point, since the field k is algebraically closed), and not the generic point. I have edited the question to reflect this clarification. $\endgroup$ Dec 22, 2015 at 17:01
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    $\begingroup$ For the quasi-elliptic fibrations that Jason mentions, all fibers are singular, hence not normal. $\endgroup$
    – abx
    Dec 22, 2015 at 18:04
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    $\begingroup$ Yes, use EGA IV$_3$, 12.2.4(iv) (which does not apply for positive generic characteristic since normal does not imply geometrically normal for schemes of finite type over an imperfect field). $\endgroup$
    – nfdc23
    Dec 23, 2015 at 2:29
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    $\begingroup$ @AaronLandesman: Think on your own about it, and ask others with more experience after you return to campus. These matters are not discussed in the Stacks Project (if I'm not mistaken), providing a very compelling illustration of why EGA remains an important reference with much valuable material not found elsewhere. If you follow the very thorough cross-references in EGA (not just a quick glance) you'll understand the "property P on geometric fibers" and "spreading-out" formalism. (Hint: see IV$_2$, 6.7.4(c) and 6.7.5--6.7.8 for the former, and consider "finite birational" for the latter). $\endgroup$
    – nfdc23
    Dec 24, 2015 at 2:15

1 Answer 1


This is merely a recollection of the discussion in the comments, that I wrote for for myself. I think it is still incomplete, could anyone help me fill the gap at the end?

I will denote the normalizations by $\nu: \hat X \to X$ and $\nu_y: \hat{(X_y)} \to X_y$, and we have to compare the two varieties $(\hat X)_y$ and $\hat{(X_y)}$.

It is clear that $\nu_y$ factors over $(\hat X)_y$, so both varieties are birational to $X_y$. Now we have the following Lemma:

Lemma. If $f: X \to Y$ is a finite, birational morphism, and $Y$ is normal, then $f$ is an isomorphism.

Proof Take an open affine subset $\operatorname{Spec} A \subset Y$, and let $\operatorname{Spec} B \subset X$ be its preimage. So $f$ corresponds to an injection $A \hookrightarrow B$, and $B$ is finite over $A$, i.e. an integral extension. Since $f$ is birational, $A$ and $B$ have the same quotient field $Q(A) = Q(B)$. But $Y$ is normal, so $A$ is integrally closed in $Q(A)$, and hence $A = B$.

So it is sufficient to show that $\hat{(X_y)}$ is indeed normal for general $y$. By Grothendieck, EGA IV2 (12.2.4) (iv) we know that the set of points $y$ with geometrically normal fibers is open. But why is it not empty?

I have two thoughts:

  1. In characteristic $0$, normal implies geometrically normal, so it is sufficient to find at least one normal fiber.
  2. If we suppose that $\hat X$ is indeed regular, then we can conclude the existence of a normal (even smooth) fiber by generic smoothness. This actually happens in my own application, but is not true in general.
  • $\begingroup$ Wouldn't $\hat{X}$ normal implies generic fibre being normal? $\endgroup$
    – hyyyyy
    Nov 28, 2021 at 20:52

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