I'm having trouble with the following problem:
Let $E/F$ be a finite extension where $\operatorname{char}(F)=0$, and A be an algebraically closed field containing $E$. Prove that if $\alpha\in E$ such that $\sigma(\alpha)=\alpha$ for all embedding $\sigma:E\to A$ such that $\sigma|_F=\operatorname{id}_F$, then $\alpha\in F$.
My idea is to apply the primitive element theorem (due to the given conditions), say $E=F(\gamma)$ for some $\gamma\in E$. Then, $\alpha$ can be written in terms of $\gamma$. I'm not sure if this will be useful. I think the main issue is that I don't how to use the assumption that $A$ is algebraically closed.
Any help would be appreciated.
Over $A$, the minimal polynomial of $\alpha$ splits as $\prod_\sigma (x-\sigma(\alpha))=(x-\alpha)^n=x - n \alpha x^{n-1} + \cdots$. Since this polynomial has coefficients in $F$, we have $n \alpha \in F$. Since $\operatorname{char}(F)=0$, we have $n\ne0$ and so $\alpha \in F$.