Existential second-order logic (ESO) formulas have the form $$\Phi = \exists R_1 ... \exists R_k. \phi$$ where $R_1...R_k$ are relation symbols and $\phi$ is a FO formula, which can use the relation symbols $R_1...R_K$ as well as other relation symbols. My claim is that
$\Phi$ is satisfiable if and only if $\phi$ is satisfiable.
Indeed, satisfiability of a FO formula means finding a universe and an interpretation of all relation symbols. Therefore, we have an implicit quantification $\exists R_1 ... \exists R_k$ in front of the FO formula $\phi$, when considering satisfiability. (For validity, the claim does not hold.)
But the claim must be wrong since people study satisfiability of ESO separately from that of FO. What do I miss?