I was wondering how to find the normal closure of Q(∜2) in K, where K is the algebraic closure of the set of rationals.
I think the normal closure should be Q(∜2,i),based on the fact that it is the splitting field of the polynomial f(x)=x^4+2 over Q, but the task ahead is if I am right, how to show it is the smallest normal extension?
Thanks for your time!
So your answer is correct that is the normal closure. To show this, we have to assume that there is another field extension $E$ of $\mathbb{Q}(\sqrt[4]{2})$ that is the normal closure. The minimal polynomial of $\sqrt[4]{2}$ is $x^4-2$ (why?), and that has roots, $\pm \sqrt[4]{2}$ and $\pm i\sqrt[4]{2}$. So all this $4$ elements are in $E$, and thus, the quotient $(i\sqrt[4]{2})/(\sqrt[4]{2})$ is also in $E$ cause we are in a field, so $i$ is in $E$, so $E$ contains $\mathbb{Q}(\sqrt[4]{2},i)$ and that means that this is the smallest one.