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This is perhaps a naïve question; given an irreducible scheme $X$, is there a general procedure to find a flat family $Y \to T$ such that over some point $t_0$ we have $Y \times_T {t_0} \cong X$ but for every other point $t \in T$ the scheme $Y \times_T t$ is a disjoint union of reduced schemes?

The example I have in mind is $X = \textrm{Spec}\ k[\epsilon] / \epsilon^2$, $Y = \textrm{Spec} \ k[x, y] / (y^2 = x) \to T = \textrm{Spec}\ k[x]$.

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So, you don't really have a scheme to start with, but a reduced scheme and you want a family for which that reduced scheme is the support of the special fiber with whatever scheme structure, right? Otherwise, for instance if your $X$ is reduced, then this is not going to happen. In the case when you don't care about the scheme structure, take two trivial families of your reduced scheme over the same base and glue them at $t_0$ like in your example. –  Sándor Kovács Dec 5 '10 at 0:35
    
Sandor, are you sure that you obtain a flat family in this way? If so, where is the mistake in my example of double projective line? –  Francesco Polizzi Dec 5 '10 at 0:49
    
I'm a bit confused. The double projective line cannot be the central element of a flat family whose general element is the union of two disjoint lines (as your example seems to suggest), because the arithmetic genus is different (0 instead of -1). It seems to me that this argument does not involve the irreducibility of the total space. Or am I missing something? –  Francesco Polizzi Dec 5 '10 at 1:56
    
Francesco, if the base is a smooth curve, then flatness only requires that all associated points (in this case irreducible components) dominate the base. I think this is OK. –  Sándor Kovács Dec 5 '10 at 1:57
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Sandor, I think one of the problems is, as you said, the scheme structure on "$2L$", and this obviously depends on the ambient space where you see $L$. Results on rigidity on Hartshorne's book apply only to $L$ a line on a smooth cubic surface (the arithmetic genus is $-2$). In this case, it is sure that no possible smoothing exist, neither over the dual numbers nor over a curve. Probably your construction provides another nilpotent structure, which is smoothable. The arithmatic genus must be $-1$, so one possibility is the scheme "$2L$" where $L$ is a fibre of a ruled surface... –  Francesco Polizzi Dec 5 '10 at 2:50
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2 Answers

The answer is no, in fact there exist examples of non-reduced projective curves which are non smoothable.

Perhaps the easiest example is the double line, i.e. the scheme $X=2L$, where $L$ is a line on a smooth cubic surface in $\mathbb{P}^3$. In fact, one can show that

  1. $H^0(X, T^1_X)=0$, so every deformation of $X$ over the dual numbers is locally trivial;
  2. $H^1(X, T_X)=0$, so every locally trivial deformation of $X$ is actually trivial.

It follows that $X$ is a rigid scheme. Since $X$ is projective, rigidity implies that for any flat family $Y \to T$ whose central fibre is isomorphic to $X$, nearby fibres are isomorphic to $X$ too (see [Hartshorne, Deformation Theory, Section 5, in particular Exercise 5.10]).

So $X$ provides a counterexamples to your question.

Notice that Sandor's answer and comments provide a different nilpotent structure on $\mathbb{P}^1$, which is instead smoothable.

I do not know whether it is possible to give a counterexample with $X$ affine.

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WOuldn't $p_n(z)$ need to be $z^n$? –  Mariano Suárez-Alvarez Dec 4 '10 at 23:35
    
In any case, your "For the sake of simplicity" simplified the problem waaaay too much! :) There are many more interesting ways in which a scheme can be non-reduced than the way the fiber $q^{-1}(0)$. –  Mariano Suárez-Alvarez Dec 4 '10 at 23:36
    
Mariano, you are completely right and my first answer was not correct. I edited it, now it should be ok (at least, I hope :) ) –  Francesco Polizzi Dec 5 '10 at 0:36
    
I am not familiar with this notation $2L$... –  Kevin H. Lin Dec 5 '10 at 0:38
    
It is just a line "counted twice". –  Francesco Polizzi Dec 5 '10 at 0:41
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Let $B={\rm Spec}\\, k[x,y]/(xy)$, i.e., the union of two lines. There is an obvious flat morphism to the line $p:B\to A={\rm Spec}\\, k[x]$. Now let $X$ be a reduced scheme. $Z=X\times B$, and $f:Z\to A$ the composition of the projection to $B$ with $p$. The projection is flat, and hence so is $f$.

Now the fiber over $(x)\in A$ is "$2X$", a non-reduced scheme with support equal to $X$.

As I mentioned in my comment above, one can definitely not prescribe the scheme structure on $X$.

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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 $0$). –  Francesco Polizzi Dec 5 '10 at 3:16
    
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. –  Sándor Kovács Dec 5 '10 at 6:17
    
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. –  Sándor Kovács Dec 5 '10 at 6:17
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