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Is the reduction $X_{red}$ of a flat, finite, surjective scheme $X$ over an integral base $S$ still flat?

I could possibly add that I am already aware we can assume the base $S$ to be local and complete, and we can assume $X$ is local (and henselien). So far this hasn't seemed to help me.

My intuation is that I should be trying to show that the dimension of the residues of the sheaf of nilpotents $\mathcal{N}$ in the structure sheaf $\mathcal{O}_{X}$ of $X$ is constant over $S$.

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up vote 5 down vote accepted

In analytic geometry, you can see to the Douady's example in Fischer book (p.151) where $X:=\lbrace{(x,s,t)\in {\Bbb C}^{3}: x^{3}+sx +t=0; 27t^{2}+4s^{3}=0}\rbrace$, $S:=\lbrace{(s,t): 27t^{2}+4s^{3}=0\rbrace}$ and $f:X\rightarrow S$ is a finite, surjective and flat map (induced by the canonical projection). Then, it easy to show that the restriction $f: X_{red}\rightarrow S$ is not flat !

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Thanks, Fischer's book certainly claims that this is a counter example. I can't seem to work out the details for myself though, so any help finding the nilpotents would be much appreciated! – name Apr 15 '11 at 0:46
The fibre $X_P$ over a nonzero point $P\in S$ is the union of a point and a double point (because $27t^2 + 4s^3$ is the discriminant). To find a nilpotent, you want an $f \in \Gamma(S,\mathcal{O}_X)$ s.t. its restriction to each $X_P$ is 0 on the simple point and in the maximal ideal of the double point. Fix $P =(s,t)$, and let $x^3+sx+t = (x-r_1)(x-r_2)^2$. Expanding, you find $r_1 = -2r_2$, hence $t = 2r_2^3$ and $s=-3r_2^2$. Notice that on the fibre $X_P$, $f_P = (x-r_1)(x-r_2)$ is a nilpotent. Mult. this by $r_2^2$ to get $r_2^2x^2+r_2^3x-2r_2^4$ which is then nilpotent and regular. – Jørgen Rennemo Apr 15 '11 at 6:38
Actually the counter examples of Douady (over a reduced base space Y) and of Cowsik-Nori (over smooth 2-dimensional base spaces) are finite, flat morphisms. Is it true that the restriction to the reduction X_{red} of a flat map-germ f: X -->C^2 which is not finite is still flat? – Trinh Feb 13 '14 at 14:52

An extremely partial answer: If $S$ is Dedekind, then the answer is yes (for quite general $X$). Indeed, for a reduced Noetherian scheme all embedded points are generic points of components, so if $X_{red} \to S$ is non-flat then $X_{red}$ would have a component lying over some closed point of $S$, hence so would the original scheme $X$, and so $X$ itself would not be flat over $S$.

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Thanks for your answer. I didn't really follow the argument, does it work for S normal as well? – name Apr 28 '11 at 14:13

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