Let $F \subseteq E$ be an arbitrary algebraic extension. I want to show that there exists $L \supseteq E$ such that $L$ is a normal closure for $E$ over $F$ and this extension is unique up to $E$-isomorphism.
First, I have to give the definition of algebraic closure. $L$ is a normal closure for $E$ over $F$, if $E$ is normal over $F$ and there is no field $K$ ($F\subseteq E \subseteq K \subset L$) such thah $K$ is normal over $F$.
I have a hint: I should work inside an algebraic closure $\bar{E}$ of $E$ for existance and uniqueness.
How can I construct the desired extension in algebraic closure?