2
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thank you very much. I am interested only in the characteristic $0$ case. $\endgroup$
    – user46578
    Jul 26, 2014 at 16:32
  • 1
    $\begingroup$ 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, 2014 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, 2014 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, 2014 at 18:39
  • $\begingroup$ Because it contains the generic point by your hypothesis. $\endgroup$
    – abx
    Jul 26, 2014 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.