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Let $\pi:\mathcal{X} \to B$ be a flat family of projective schemes, $B$ is irreducible. Let $\mathrm{Spec} K$ be a generic point on $B$. Denote by $\mathcal{X}_K$, the pull-back of $\mathcal{X}$. This is flat, projective on $K$. It will then follow that $\mathcal{X}_{K_{\mathrm{red}}}$, the associated reduced scheme, is also flat, projective scheme over $K$. Assume that there exists a subfamily of $\pi$, say $\pi':\mathcal{X}' \to B$, a flat family of projective schemes satisfying $\mathcal{X}'_b \subset \mathcal{X}_b$ for all $b \in B$ and $\mathcal{X}'_K = \mathcal{X}_{K_{\mathrm{red}}}$. Does this mean that for a general closed point $b \in B$, the fiber $\mathcal{X}'_b=\mathcal{X}_{b_{\mathrm{red}}}$, the associated reduced scheme? If not true in general, is there any known condition on $B$ or $\pi$ under which this could hold true?

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This is true if you assume moreover that $(\mathcal{X}_K)_{red}$ is geometrically reduced -- in particular, in characteristic $0$. First of all, note that set-theoretically $\mathcal{X}'_b=\mathcal{X}_b$ for all $b$ in $B$ : since $\pi $ open, $\pi (\mathcal{X}-\mathcal{X}')$ is an open subset of $B$ which does not contain the generic point, therefore it is empty. Now the subset $U\subset B$ of points $b$ such that $\mathcal{X}'_b$ is geometrically reduced is open in $B$ (EGA IV, Thm. 12.2.4), so for $b\in U\ $ $\mathcal{X}'_b$ is reduced, and therefore equal to $(\mathcal{X}_b)_{red}$.

I am not an expert in characteristic $p$ but I suspect there might be counter-examples if one does not assume that $(\mathcal{X}_K)_{red}$ is geometrically reduced.

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  • $\begingroup$ Thank you very much. I am interested only in the characteristic $0$ case. $\endgroup$ – user46578 Jul 26 '14 at 16:32
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    $\begingroup$ Here is a counterexample as abx suggests. Let $X$ be smooth connected of dimension $d>0$ over a perfect field $k$ of characteristic $p>0$ and let $\pi$ be the relative Frobenius morphism $X\rightarrow X^{(p)} =: B$ over $k$ (e.g., $t\mapsto t^p$ on the affine $t$-line over $k$). Let $X'=X$. The map $\pi$ is finite flat of degree $p^d$ with generic fiber that is reduced (corresponds to $K \hookrightarrow K^{1/p}$ since $k$ is perfect, as one sees by expressing a dense open in $X$ as etale over an affine space over $k$) but the fiber over every closed point of $B$ is non-reduced since $d>0$. $\endgroup$ – user27920 Jul 26 '14 at 17:33
  • $\begingroup$ If one replaces "reduced" with "irreducible" but omits the "geometric" aspect then the same failure of spreading-out occurs in any characteristic; e.g., $t \mapsto t^n$ on the affine line over an algebraically closed field of char. not dividing $n$ (irreducible generic fiber but reducible fibers over all closed points). So it is genuinely valuable to understand the proof of EGA IV$_3$, 12.2.4 so as to appreciate the origin of the "geometric fiber" requirement for spreading-out to work. This is not a matter of pathology-avoidance in char. $p$, but rather of the power of the Nullstellensatz. $\endgroup$ – user27920 Jul 26 '14 at 17:43
  • $\begingroup$ @abx: This is most probably, a dumb question. Is it obvious that if every fiber of $\pi'$ is non-reduced then the generic fiber $\mathcal{X}'_K$ is non-reduced? In other words, why is the $U$ mentioned in your answer non-empty? $\endgroup$ – user46578 Jul 26 '14 at 18:39
  • $\begingroup$ Because it contains the generic point by your hypothesis. $\endgroup$ – abx Jul 26 '14 at 18:43

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