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Jun 12, 2023 at 9:49 comment added Dimitri Koshelev Thank you for your idea. Is it always possible to construct in such a way an isotrivial pencil, that is, all its smooth members are of the same j-invariant?
Jun 11, 2023 at 19:21 comment added Jason Starr Fine: take the Weirrstrass model, and use a pencil of plane cubics containing the two points that includes the Weirrstrass model as one member of the pencil. The intersection matrix shows the lifts are independent.
Jun 11, 2023 at 11:51 comment added Dimitri Koshelev I require that the points $P_0(t)$, $P_1(t)$ are independent. In particular, they have to be non-torsion. Hence, your approach does not work.
Jun 11, 2023 at 11:46 comment added Jason Starr You can just choose $\mathcal{E}$ to be the "constant" family, namely $E\times_{\text{Spec}\ \mathbb{F}_q} \text{Spec}\ \mathbb{F}_q(t)$. Then every $\mathbb{F}_q$-point of $E$ lifts to a "constant" $\mathbb{F}_q(t)$-point of $\mathcal{E}$.
Jun 11, 2023 at 9:44 comment added Dimitri Koshelev Their coefficients may depend on $q$, but their heights are restricted above by a constant.
Jun 11, 2023 at 9:36 comment added Wojowu How could $P_0(t)$ etc. ever be independent of $q$ when all the data in your problem depends on it?
Jun 11, 2023 at 9:33 history edited Dimitri Koshelev CC BY-SA 4.0
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Jun 11, 2023 at 9:27 history asked Dimitri Koshelev CC BY-SA 4.0