Deformation of a Rigid Scheme is étale Locally Trivial. This question is from ex.III.9.10, Hartshorne. All scheme are finite type over an algebraically closed field $k$. Let $f: X \rightarrow T$ be a flat projective morphism from $X$ to a nonsingular curve $T$. Assume that there is a closed point $t \in T$ s.t. $X_t \sim \mathbb{P}^1_k$.
Then we need to show that there is a nonsingular curve $T'$ and a flat morphism $g: T' \rightarrow T$ whose image contains $T$ and $X' = X \times_T T'$ is isomorphic to $\mathbb{P}^1_{T'}$.
By flatness and smoothness of the fibre of $t$, we know that $f$ is smooth in a neighbourhood of every point of $X_t$. But the singular locus is closed and its image by $f$ is closed not containing $t$. Hence we have a neighbourhood $U$ of $t$ s.t. $f^{-1}(U)$ is in the smooth locus, i.e. $f: f^{-1}(U) \rightarrow U$ is smooth. Now by flatness $f_* \mathcal{O}_X$ is torsion free hence flat and locally free. By formal function theorem we know that it is of rank one. So every fibre is smooth connected curve of arithmetic genus $0$, which must be $\mathbb{P}^1$.
Now we have reduced to the situation of a smooth family of $\mathbb{P}^1$ over a smooth curve. It remains to find the suitable local base $T'$.
Moreover, I wonder whether we can choose $g$ to be étale. It seems reasonable since étaleness is a reasonable replacement of analytically localness. And when a scheme is rigid, I don't wish that it can be trivialized only after ramified base change.
 A: The key is to find a section $s: T \rightarrow X$ after base change.
Such a section is proper, hence defines a divisor $D$ in $X$. Since it is a section, it intersects each fibre transversely at exactly one point, hence its restriction as a divisor to each fibre is $\mathcal{O}(1)$, whose linear space of global sections is of dimension $2$. By the formal function theorem, or some arguments with upper semicontinuity, $f_*\mathcal{L}(D)$ is locally free of rank $2$. We can choose $T$ small enough s.t. $f_*\mathcal{L}(D)$ is free. This means that $\mathcal{L}(D)$ has two global sections.
Now we need to carefully choose two sections to let them defines a appropiate morphism $X \rightarrow \mathbb{P}^1_T$. Global sections of $f_*\mathcal{L}(D)$ and of $\mathcal{L}(D)$ are $1$-$1$ correspondent. Since we can let $f_*\mathcal{L}(D)$ being free, we can let $s_1, s_2$ be two sections of it s.t. they are linearly independent (as two vectors in $f_*\mathcal{L}(D)_t/m_tf_*\mathcal{L}(D)_t$). Then these two sections corresponding to two sections of $\mathcal{L}(D)$ s.t. their restriction to each $X_t$ is linearly independent, which means that they have no common zero, the linear system they span is base point free and the corresponding morphism from $X_t$ to $\mathbb{P}^1_k$ is an isomorphism.
So these two sections have no intersection since their restriction to each $X_t$ doesn't. This means they are base point free and define a morphism $X \rightarrow \mathbb{P}^1_T$ which is an isomorphism restricting to each $X_t$. Now each $X_t$ and $T$ are orthogonal and $X \rightarrow \mathbb{P}^1_T$ is relative to $T$, this morphism must be smooth of degree one, hence an isomorphism (This argument must be a little vague. But if you are familiar with scheme theory, or III.10 of Hartshorne, you know what I am saying. See for example, III.10.4.).
Now it remains to find the section. Let $x \in X_t$ be a closed point. Denote the uniformiser of $\mathcal{O}_{t,T}$ also by $t$. By smoothness, $t \neq 0 \in m_x/m^2_x$. Then $t$ locally defines a smooth curve $t: T' \hookrightarrow X$ and $f\circ s: T' \rightarrow T$ is unramified at $x$. Shrinking $T'$ s.t. $f\circ s$ is étale. Then $T'$ is the desired local base since the locally closed embedding $t$ is lifted to a closed embedding which is a section after base change.
