# Find all field homomorphisms of $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} e^{\frac{2 \pi i}{3}})$

Let $\xi := e^{\frac{2 \pi i}{3}}$ and $\overline{\mathbb{Q}}$ the set of all algebraic numbers (i.e. $\alpha \in \overline{\mathbb{Q}} \Rightarrow \exists f \in \mathbb{Q}[X]: f(\alpha) = 0$ ).

I need to find all field homomorphisms between..

(1) .. $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\sqrt[3]{2})$

(2) .. $\mathbb{Q}(\sqrt[3]{2})$ and $\overline{\mathbb{Q}}$

(3) .. $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{C}$

(4) .. $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)$ and $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)$

(5) .. $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)$ and $\overline{\mathbb{Q}}$

(6) .. $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)$ and $\mathbb{C}$.

So far i was (?) able to solve (1), (2) and (3). I do already now (and can proof) that any field homomorphism is injective and that any of the above homomorphisms fix $\mathbb{Q}$ (i.e. $\varphi_{| \mathbb{Q}} = \operatorname{Id}_{\mathbb{Q}}$ for any of the above field homomorphisms $\varphi$).

(1) Let $\varphi: \mathbb{Q}(\sqrt[3]{2}) \rightarrow \mathbb{Q}(\sqrt[3]{2})$, then $\varphi(\sqrt[3]{2})^3 = \varphi(\sqrt[3]{2}^3) = \varphi(2) = 2$ and therefore $\varphi(\sqrt[3]{2}) \in \{ \sqrt[3]{2}, \sqrt[3]{2} \xi, \sqrt[3]{2} \xi^2 \}$ (i.e. the roots of $X^3 -2 \in \mathbb{Q}[X]$). Since $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R} \Rightarrow \xi \notin \mathbb{Q}(\sqrt[3]{2})$, we see that $\varphi(\sqrt[3]{2}) = \sqrt[3]{2}$ and $\varphi$ must be the identity. So $\operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\sqrt[3]{2})) = \{ \operatorname{Id}_{\mathbb{Q}(\sqrt[3]{2})} \}$.

(2) We know that $\sqrt[3]{2}, \sqrt[3]{2}\xi, \sqrt[3]{2}\xi^2 \in \overline{\mathbb{Q}}$ are algebraic, since they are the roots of $X^3 -2 \in \mathbb{Q}[X]$. Therefore (freely written) $\operatorname{Id}_{\mathbb{Q}(\sqrt[3]{2})}: \sqrt[3]{2} \mapsto \sqrt[3]{2}$, $\varphi_1: \sqrt[3]{2} \mapsto \sqrt[3]{2} \xi$ and $\varphi_2: \sqrt[3]{2} \mapsto \sqrt[3]{2} \xi^2$. So $\operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}), \overline{\mathbb{Q}}) = \{ \operatorname{Id}_{\mathbb{Q}(\sqrt[3]{2})}, \varphi_1, \varphi_2 \}$.

(3) It has to be $\operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}), \overline{\mathbb{Q}}) = \operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}), \mathbb{C})$, since we already found all possible mappings (see (1)) at (2).

Is this correct so far?

Now comes the part i can't seem to solve.

(4) So far we know that the degree of $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}]$ is $3$, since the minimal polynomial of $\sqrt[3]{2}$ is $X^3 -2 \in \mathbb{Q}[X]$. After some calculations i found out that $X^2 + \sqrt[3]{2} X + \sqrt[3]{2}^2 \in \mathbb{Q}(\sqrt[3]{2})[X]$ is the minimal polynomial of $\sqrt[3]{2} \xi$. Therefore $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi) : \mathbb{Q}(\sqrt[3]{2})] = 2$ and so $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi) : \mathbb{Q}] = 6$. So we could construct a base containing $2 \cdot 3 = 6$ elements. These elements will be either $1$ or some product of $\sqrt[3]{2}$ and $\sqrt[3]{2} \xi$. So to get all possible homomorphisms, we need to find all possible values for $\varphi(\sqrt[3]{2})$ and $\varphi(\xi)$. Similar to (1) i would argue that $\varphi(\sqrt[3]{2}) \in \{ \sqrt[3]{2}, \sqrt[3]{2} \xi, \sqrt[3]{2} \xi^2 \}$ and $\varphi(\xi) \in \{ \xi, \xi^2 \}$. By combining these two we get $6$ distinct field homomorphisms. So $\operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi), \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)) = \{ \operatorname{Id}_{\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)}, \varphi_1, \varphi_2, \varphi_3, \varphi_4, \varphi_5 \}$.

(5)/(6) Since we already found all possible mappings (see (4)), i'd finish with $\operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi), \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi)) = \operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi), \overline{\mathbb{Q}}) = \operatorname{Hom}(\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2} \xi), \mathbb{C})$.

Is this correct? If so, could it have been shown better/faster? If not, could you please give me some hints?