Consider the projective plane $\mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ over the algebraic closure of the function field $\mathbb{C}(t)$.


Take the point $p\_0 = [0:1:0]\in \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ and eight more general points $p\_1,\dots,p\_8\in \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ which are not defined over $\mathbb{C}(t)$. Assume that there is a cubic curve $C \subset \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$, defined over $\mathbb{C}(t)$, passing through $p\_0,p\_1,\dots,p\_8$. We can write $C$ as
$$C = \{ \alpha\_0(t) x^3 + \alpha\_1(t)x^2y + \alpha\_2(t)x^2z + \alpha\_3(t)xy^2 + \alpha\_4(t)xyz + \alpha\_5(t)xz^2 + \alpha\_6(t)y^3 + \alpha\_7(t)y^2z + \alpha\_8(t)yz^2 + \alpha\_9(t)z^3 = 0 \}$$
where $\alpha\_0,\dots,\alpha\_9$ are polynomials in $t$, and extend $C$ to a fibration on $\mathbb{P}^1$ as
$$C' = \{A\_0(s,t) x^3 + A\_1(s,t)x^2y + A\_2(s,t)x^2z + A\_3(s,t)xy^2 + A\_4(s,t)xyz + A\_5(s,t)xz^2 + A\_6(s,t)y^3 + A\_7(s,t)y^2z + A\_8(s,t)yz^2 + A\_9(s,t)z^3 = 0 \}$$
where now $A\_0,\dots,A\_9$ are homogeneous polynomials on $\mathbb{P}^1$.


Can we find such a curve $C'$ such that the polynomials $A\_0,\dots,A\_9$ appearing in the expression of $C'$ are of degree one or at least $A\_9$ is of degree one?



Coefficients of elliptic curves over function fields
Are you asking if, for all tuples $p\_1,\dots, p\_8$, there exists such a $C'$ with $A\_9$ of degree one? This is false, assuming $A\_9=0$ does not count as degree one.


We can for example choose one of the $p\_i$ to equal $[0: 0 : 1]$ for $t=1$ and and one to equal $[0,0,1]$ for $t=2$. Then regardless of which $C'$ we take, $A\_9$ will equal $0$ for $t=1$ and $t=2$ and thus cannot have degree one.


