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I'm struggling with the exercise said in the title (Hartshorne, III, ex. 9.10). No problems in showing that $\mathbb{P}^1$ is rigid. In the second part, we want to show that $X_0$ being rigid does not imply that $X_0$ does not admit global deformations.

The problem asks us to do so building a flat, proper morphism $f:X\to \mathbb{A}^2$ over an algebriacally closed field $k$, with $\mathbb{P}^1$ in the central fiber, but such that for no neighbourhood $U$ of the origin one has $f^{-1}(U)\simeq U\times \mathbb{P}^1$.

I'd like to get hints or an example for this part. I know that some properties of the fibres have to be preserved, in particular dimension, degree and arithmetic genus (Hart III, Cor 9.10). I guess this also implies that we want some of the fibres to be singular (otherwise the geometric genus being equal to the arithmetic would force an isomorphism with $\mathbb{P}^1$), but I don't know how to obtain these singularities, or if the problem can be avoided considering a field different from the complex numbers.

I haven't tried to think about the third part of the problem yet, but any hints about that one would be well accepted anyway.

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  • $\begingroup$ Here's a hint. By considering $\mathbb{P}(\mathcal{E}):=\mathbf{Proj}(\mathrm{Sym}^\bullet(\mathcal{E}))$, you should be able to reduce this to a question about finding a coherent sheaf $\mathbb{A}^2$ which is not trivial on any neighborhood of the origin. $\endgroup$ Mar 11, 2015 at 8:38

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Consider a field of characteristic $\ne 2$ , and the projective subscheme $X \subset \mathbb A^2_k\times_k\mathbb P^2_k $ given by the equation $(a-1)x^2+(b-1)y^2+z^2=0$.
The projection morphism $f:X\to \mathbb A^2_k$ is smooth above $$S=D(a-1,b-1)=\operatorname {Spec}k[a,(a-1)^{-1},b,(b-1)^{-1}]\subset \mathbb A^2_k\quad (S\cong\mathbb G_m\times_k \mathbb G_m)$$ with all fibers isomorphic to $\mathbb P^1_k$ .
However the projection $X|S\to S$ is not locally trivial near any point of $S$ (in particular not locally trivial near $a=0,b=0$) because the generic fiber of $X|S\to S$ is the conic $(a-1)x^2+(b-1)y^2+z^2=0$ seen as having coefficients $a-1,b-1,1 \in k(a,b)$, and that conic has no rational point over the field $k(a,b)$.

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  • $\begingroup$ Very nice example Georges! I guess what you wrote is fairly intuitive. You have a family of conics which, fiber by fiber are of course trivial (they have a zero), but to create a global trivialization you'd have to choose a family of rational points which, as you mentioned there is none. $\endgroup$ Mar 11, 2015 at 10:49
  • $\begingroup$ Thanks for your interesting comment, @Alex. $\endgroup$ Mar 11, 2015 at 11:10
  • $\begingroup$ I'm sorry, I guess I'm not familiar with the terminology..I get how you build the family, but not why its not locally trivial. How do rational points come into play (and what are they?)? $\endgroup$
    – Franco
    Mar 12, 2015 at 0:02

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