"un-nil-ifying" ideals via deformation

Question

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 $X = \textrm{Spec}\ k[\epsilon] / \epsilon^2$, $Y = \textrm{Spec} \ k[x, y] / (y^2 = x) \to T = \textrm{Spec}\ k[x]$.

Answer 1

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.

Answer 2

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 $X$.