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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?

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    $\begingroup$ Bound in terms of what? Take $C'$ with polynomials of arbitrary high degree, then choose $8$ points on it. $\endgroup$ Aug 6, 2021 at 15:55
  • $\begingroup$ You are right, my question was badly written. I modified it. $\endgroup$
    – Arty
    Aug 6, 2021 at 16:25
  • $\begingroup$ I repeat my objection: Weierstrass form is unique! So if you start with Weierstrass form with polynomials of high degrees, it cannot be reduced to a Weierstrass form with polynomials of low degrees. $\endgroup$ Aug 6, 2021 at 17:46
  • $\begingroup$ Sorry, I modified the question again. Now $C'$ has a completely general form. $\endgroup$
    – Arty
    Aug 6, 2021 at 18:01

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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.

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  • $\begingroup$ I was asking if there exists such a $C'$ for $p_1,\dots,p_8$ general not for all tuples $p_1,\dots,p_8$. Perhaps for a point $p_i$ to take the same value for two different values of $t$ is a closed condition. $\endgroup$
    – Arty
    Aug 6, 2021 at 23:58
  • $\begingroup$ @Arty It is certainly not closed. No congruence condition will be. $\endgroup$
    – Will Sawin
    Aug 7, 2021 at 0:48

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