Let $F$ be a perfect field and $\Omega$ an algebraically closed field containing $F$. Let $\bar{a}$ be an $n-$tuple of elements of $\Omega$ and let $I(\bar{a}/F)=\{f(\bar{x}) \in F[\bar{x}] : f(\bar{a})=0\}$ be an ideal over $F[x_1,\ldots,x_n]$ . We denote by $I(\bar{a}/F) \Omega$ the ideal generated by $I(\bar{a}/F)$ (seen as a set) in the bigger polynomial ring $\Omega[x_1,\ldots,x_n]$. I want to prove that $I(\bar{a}/F) \Omega$ is a prime ideal if and only if $F(\bar{a}) \cap F^{alg} =F$.
I'm stuck from the beginning. I tried to suppose there is some $b \in F(\bar{a}) \cap F^{alg}$ with $b \notin F$ and with its minimal polynomial over $F$ I tried to build an element $gh \in I(\bar{a}/F) \Omega$ such that the primality of the ideal fails. For the converse I have even less ideas.
Any suggestion would be helpful thanks!