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Aug 12, 2014 at 16:16 comment added Jeremy Rouse No. If $K$ is any infinite field and $P(x,y) \in K[x,y]$ is non-zero, then there exist $a_{1}$ and $a_{2}$ in $K$ so that $P(a_{1},a_{2}) \ne 0$. See Lemma 1 of this file.
Aug 12, 2014 at 15:28 comment added Mehdi Note that it may for all $t_1, t_2$ in the countable number field $K$ we have $P(t_1,t_2)=0$ but $P(t_1, t_2)\neq 0$ over the complex number field $\mathbb{C}$.
Aug 12, 2014 at 15:12 comment added Jeremy Rouse You say "I want to prove there exist two parameters $t_{1}$ and $t_{2}$ in $K$ for which the corresponding polynomials $Q_{1}$ and $Q_{2}$ have no common root." Doesn't the fact that $P(t_{1},t_{2})$ is not identically zero do that?
Aug 12, 2014 at 15:01 comment added Mehdi You can see that there exists a monomial e.g. $t_1^{12}$ in the expansion of resultant with a nonzero coefficient. So this is not identically $0$.
Aug 12, 2014 at 14:17 comment added Jeremy Rouse If you're trying to prove that there exist $t_{1}$ and $t_{2}$ in $K$ so that $P(t_{1},t_{2})$ is not zero, don't you just need to know that $P(x,y)$ is not identically zero?
Aug 12, 2014 at 11:51 comment added Mehdi One Raw Idea: If somehow we could construct a morphism from the curve $C$ to the curve $E'$, i.e. $\phi : C: \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula applied to the morphism $\phi$ we have \begin{align*} 2g(C)-2 \ge (2g(E') -2) \deg ( \phi) + \sum_{p \in C}(e_p-1), \end{align*} and we can easily see that the right side is greater than $1$ since $g(E') =9$, so $2g(C)-2 >1$ and we have $g(C) > \frac{3}{2}>1$.
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