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.
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 $k(a,b)$.