Fibered surfaces with sections and generic fibers iso to projective space. This is an exercise from Liu's book on Algebraic Geometry, exercise 8.3.5c)
Let $f:X \rightarrow S$ be a fibered surface over a Dedekind scheme of dimension $1$, with generic fiber $X_{K(S)} \cong \mathbb{P}^1_{K(S)}$. Let $s \in S$ be such that $X_s$ is geometrically integral. 
I am trying to show that there exists an open neighborhood $V$ of $s$ such that $f^{-1}(V) \cong \mathbb{P}^1_V$.
In a) one showed that if $S$ is local, then $X_s \cong \mathbb{P}^1_{k(s)}$ .
In b) one showed that if $D$ is the Cartier divisor corresponding to a section of $X \rightarrow S$, then under the assumption that $S$ is local, $\mathcal{O}_X(D)$ is very ample. 
I believe I can do a proof of $c)$ using very general properties that are not all too illuminating, and I suspect Liu wanted one to use $a$ and $b)$ in conjunction to show $c)$, but I can't seem to get it to work. I suppose one wants to show that there is a non-empty open subset $V$ of $S$ such that $\mathcal{O}_X(D)$ restricted to $V$ is very ample. Further, a trivial thing to note is to note that we have an isomorphism on the generic point, and we want to extend this somehow. 
Any help would be greatly appreciated!
 A: Let $f:X\to S$ be as in the question, and consider $E = f_\ast \mathcal \omega_{X/S}^{\vee}$, where $\omega_{X/S}$ is the relative dualizing sheaf of $X\to S$. (What you want is local on $S$, so you can replace $S$ from the start by a dense open such that $f$ is smooth over this open.) 
The morphism $f$ factors via $\mathbb P(E)\to S$. Note that $E$ is a vector bundle on $S$ of rank $3$:
$$\mathrm{rk} f_\ast \omega_{X/S}^\vee = h^0(\mathbb P^1_k, \omega_{\mathbb P^1_k}^\vee) = 3.$$ Choose $U = \mathrm{Spec} \ A$ dense open affine in $X$ such that $E|_U$ is trivial and $U$ contains $s$. Then $f:f^{-1}U\to U$ is still as in the question and factors through $\mathbb P(E|_U) = \mathbb P^2_U$.  (I'm not going to write $f|_{f^{-1}U}$.)
Since the degree of $f_\eta :X_\eta \to \mathrm{Spec} \ K(U)$ is two (as it is the $2$-uple embedding of $\mathbb P^1$ in $\mathbb P^2$), we see that $f$ is a relative conic in $\mathbb P^2_S$ with a smooth fibre over $s$. (Explanation: the embedding of $X$ in $\mathbb P^2_A$ is given by a polynomial which has degree two regarded as a polynomial with coefficients in the function field of $A$. So then it has degree two to start with.) Choose $V\subset U$ such that $f$ is smooth over $V$ (unnecessary step now). Then $f$ is a smooth conic in $\mathbb P^2_V$ with a section. Therefore, $f$ is isomorphic to $\mathbb P^1_V$ over $V$. (Explanation:  any Brauer-Severi variety with a section is split.)
Edit: added some more details (and rewrote the answer). Everyone is allowed to add more details of course.
