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Dec 5, 2010 at 6:17 comment added Sándor Kovács I guess this is sort of the point. If the general fiber is the disjoint union of two components, then the nilpotent part of "$2X$" would have a nowhere zero section, so it would have to be trivial.
Dec 5, 2010 at 6:17 comment added Sándor Kovács Right. If $X=\mathbb P^1$, then this example gives ${\rm Proj}\, k[x,y]\times {\rm Spec}\, k[\varepsilon]/(\varepsilon^2)$ which is obviously different from ${\rm Proj}\, k[x,y,z]/(z^2)$ because the latter is not a product since the normal bundle of a line in $\mathbb P^2$ is not trivial.
Dec 5, 2010 at 4:59 history edited Allen Knutson CC BY-SA 2.5
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Dec 5, 2010 at 3:16 comment added Francesco Polizzi In fact, when you take $X=\mathbb{P}^1$ you get a non-reduced scheme with arithmetic genus $-1$, so not all nilpotent structures are allowed (for instance, neither the one in my answer, whose arithmetic genus is $-2$, nor $\textrm{Proj} k[x,y,z]/(z^2)$, whose arithmetic genus is $0$).
Dec 5, 2010 at 1:51 history answered Sándor Kovács CC BY-SA 2.5