Let $K$ be a number field. $K$ is a normal extension of $\mathbb{Q}$ iff $\exists f(x)\in\mathbb{Q}[x]: K$ is the splitting field for $f(x)$. A field extension is Galois iff it is normal and separable, but since all number fields are separable extensions of $\mathbb{Q}$, normal and Galois coincide in this case. I also recall a theorem saying Galois number fields must have either all totally real or all complex infinite places (embeddings).
A common example is $K=\mathbb{Q}(\sqrt[3]{2})$. I understand why this is not normal: it lacks the complex cube roots of 2. But what if we instead extend $\mathbb{Q}$ by one of the complex roots?
Consider the case where $K=\mathbb{Q}(\zeta_3\sqrt[3]{2})$, where $\zeta_3$ is the primitive cube root of unity. This is a degree 3 extension of $\mathbb{Q}$, with 2 of those degrees corresponding to the identity/conjugation complex place. Since complex places contribute multiples of 2 to the degree, the remaining 1 must correspond to a real place (in particular, having as image the field in the previous paragraph), meaning $K$ cannot be normal. But $K$ is the splitting field for $x^3-2$, [Note: $K$ is not the splitting field - that was the nature of the confusion here. -JQ]. which is the minimal polynomial of $\zeta_3\sqrt[3]{2}$, is it not?
In general I'd like to understand the case of number fields with exactly 1 complex place, and at least 1 real place. I'm able to produce lots of examples, but I don't get why they are not considered normal.
