Suppose we have a proper morphism $f:X\rightarrow Y$ and $0\in Y$. If the fiber $f^{-1}(0)$ is irreducible and reduced, is the set $\{y\in Y|f^{-1}(y) \text{ is irreducible and reduced}\}$ open?
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1$\begingroup$ mathoverflow.net/questions/57802/… $\endgroup$– Daniel LoughranJan 15, 2016 at 14:50
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1$\begingroup$ That is false. Let $Y$ be $\mathbb{A}^1_k = \text{Spec}\ k[t]$. Let $\mathbb{P}^1_k$ denote $\text{Proj}\ k[u,v]$. Let $X$ be the closed subscheme of $\mathbb{A}^1_k\times_k \mathbb{P}^1_k$ with defining equation $tuv=0$. Let $f:X\to Y$ be the projection. The fiber of $f$ over $t=0$ is $\mathbb{P}^1$, which is irreducible and reduced. Every other fiber is reducible, consisting of two closed points, $[u,v] =[1,0]$ and $[u,v]=[0,1]$. $\endgroup$– Jason StarrJan 15, 2016 at 14:54
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$\begingroup$ Hey, Jason, i just saw your answer here mathoverflow.net/questions/217488/… $\endgroup$– user42804Jan 15, 2016 at 16:32
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$\begingroup$ Hey, Jason, i just saw your answer here mathoverflow.net/questions/217488/… and asked the following question: if $f:X\rightarrow Y$ is proper and flat, can we always do normalization such that we can asssume X and Y are normal and f is still proper flat. I am a little confused by this, since when we normalize $Y$ to &Y^\prime&, then base change $X$ to $X^\prime:=X \times_{Y}Y^\prime$, $X^\prime$ is not nessasary normal, if we normalized $X^\prime$, then the flatness can not keep. $\endgroup$– user42804Jan 15, 2016 at 16:38
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I am just posting my comment as an answer so that this question will not continue unanswered. Without further hypothesis, the result is false. If $f$ is open, then there is a positive answer in the link Daniel Loughran provides, What properties define open loci in families? Without the flatness hypothesis, there are many counterexamples. One counterexample has $Y= \mathbb{A}^1_k = \text{Spec}\ k[t]$, and has $X= \text{Zero}(tuv) \subset \mathbb{A}^1_k \times \mathbb{P}^1_k$, where $\mathbb{P}^1_k$ denote $\text{Proj}\ k[u,v]$.