Why is $\mathbb{Q}(\sqrt[4]{2}) $ is not normal over $\mathbb{Q}$?

Question

I read somewhere that $\mathbb{Q}(\sqrt[4]{2}) $ is not normal over $\mathbb{Q}$, and the reason given is: if it were normal over $\mathbb{Q}$, then $i \in \mathbb{Q}(\sqrt[4]{2})$.

Could someone explain to me why this is the case?

Answer 1

One definition of normal extension $L/K$ is:

Every irreducible polynomial in $K[X]$ that has one root in $L$, has all of its roots in $L$.

$\sqrt[4]{2}$ is a root of $X^4-2$, which is irreducible over $\mathbb Q$.

$i\sqrt[4]{2}$ is another root of $X^4-2$, but it is not in $\mathbb Q(\sqrt[4]{2})$ because this field is contained in $\mathbb R$.

Therefore, $\mathbb Q(\sqrt[4]{2})$ does not contain all the roots of $X^4-2$ and so cannot be normal over $\mathbb Q$.