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 $S$.



Is the reduction of a flat, finite, surjective scheme over an integral base still flat?
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 !


