If I have two polynomials in $K(t_1,t_2)[x]$ where $t_i$ are indeterminates, and each of them has coefficients in the rational functions of only one of the indeterminates, i.e. $p_1\in K(t_1)[x]$ and $p_2\in K(t_2)[x]$, do the splitting fields considered as subfields of some bigger field necessarily intersect only on $K(t_1,t_2)$?
I think furthermore we'd need for the polynomials to have $t_i$ present in their coefficients, because otherwise we could have $p_1=p_2\in K[x]$. Which would be a counter example.
I am thinking of using the Galois groups of the splitting fields $p_1,p_2$ over both $K(t_1,t_2)$ and $K(t_1),K(t_2)$.
It is certainly possible for the two splitting fields to contain elements outside of $K(t_1,t_2)$ but algebraic over $K$.
Consider the case $K=\Bbb{Q}$, $p_1(x)=x^3-t_1$, $p_2(x)=x^3-t_2$. To get the splitting field of either $p_1(x)$ or $p_2(x)$ you need to include the third roots of unity, $\omega$ and $\omega^2$. And $K(t_1,t_2,\omega)$ will then be contained in the intersection of the splitting fields.
I think that if $K$ is algebraically closed, then this cannot happen, and the intersection of the two splitting fields will be just $K(t_1,t_2)$. I'm afraid I don't have a clean argument in mind though.