Let $E/F$ be an algebraic field extension. Let us take the definition of normal closure to be the normal extension $N$ of $E$ over $F$ such that $N/F$ is normal and $N$ is minimal with this property containing $E$.
There is a theorem in many textbooks which states that if $E/F$ is finite then $N$ is unique up to isomorphism. My confusion is how can we have distinct $N$ in this case?
For example, take $\alpha\in\mathbb{R},\beta,\bar{\beta}\in\mathbb{C}$ to be the roots of $x^3-2$, let $F=\mathbb{Q},E=\mathbb{Q}(\alpha)$. Then one choice of $N$ is $\mathbb{Q}(\beta)$, which contains $E$ as a subfield. But how to construct another $N$ (maybe even all $N$s) in this case?
The other normal closures are not subfields of $\Bbb{C}$. In your example case we can also use $N'=E[x]/(x^2+x+1)$, effectively adjoining a non-complex third root of unity $\omega=x+(x^2+x+1)$ to the field $E$. It is easy to show that $E(\omega)$ is a splitting field of $x^3-2$. Of course, it is isomorphic to the field $N=E(\beta)=E(e^{2\pi i/3})$ that is a subfield of $\Bbb{C}$.
To make this argument more precise I should first give a definition of $\Bbb{C}$, but I will skip that.
But, it is not difficult to show that if $N$ and $N'$ are both isomorphic Galois extensions of $\Bbb{Q}$, and both are subfields of $\Bbb{C}$, then $N=N'$ as sets. Hint: