Why is the normalization of a general fiber the general fiber of the normalization?

Question

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.

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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.

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.