Normal extension, why $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal and $\mathbb Q(\sqrt[3]2,\zeta_3)/\mathbb Q$ is normal.

Question

Normal extension, why $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal and $\mathbb Q(\sqrt[3]2,\zeta_3)/\mathbb Q$ is normal where $\zeta_3=e^{\frac{2i\pi}{3}}$. My definition of normal extension is $L/K$ is normal if $$\text{Hom}_K(L,K^{alg})=\text{Aut}_K(L),$$ but I don't hide you that I can't apply it, it's too abstract for me.

Answer 1

Building up to your definition, an extension $N/K$ is normal when the image of every $K$-homomorphism of the extension $N$ is contained in $N$ i.e. $\sigma(N) \subseteq N$ for a $K$-homomorphism $\sigma$. Since $\sigma$ is an injective ring homomorphism between the fields $N\to N$, we must have that $\sigma(N)\cong N$ since $\sigma (N)$ is a subspace of $N$ with equal dimension. Therefore $\sigma$ is an automorphism.

By properties of $K$-homomorphisms, we know that a $K$-homomorphism $\sigma$ will send an algebraic element to its conjugates over $K$. But in this case we have that $\mathbb{Q}(\sqrt[3]{2})$ does not contain the conjugates of $\sqrt[3]{2}$ over $\mathbb{Q}$, and so it's image does not lie in $\mathbb{Q}(\sqrt[3]{2})$, and therefore it is not normal.