Let $\mathbb{Q}(2^\frac{1}{3})$ be the extension field of $\mathbb{Q}$.
Find the number of homomorphisms $\mathbb{Q}(2^\frac{1}{3}) \to \mathbb{Q}(2^\frac{1}{3})$
My attempt:
Denote $E:=\mathbb{Q}(2^\frac{1}{3})$.
$[E:\mathbb{Q}]=3$ ,since $[E:\mathbb{Q}]$ is finite $ \text{# number of homomorphism} \leq 3$
Any idea how to continue ?
Let $\phi: \Bbb Q(2^{1/3}) \to \Bbb Q(2^{1/3})$ be a field homomorphism. Any such homomorphism $\phi$ fixes $\Bbb Q$ pointwise, since $\phi(1) = 1$. Furthermore, $\phi$ is determined completely by the image of $2^{1/3}$ under $\phi$. We have $$\phi(2^{1/3})^3 = \phi((2^{1/3})^3) = \phi(2) = 2$$ allowing three possibilities for $\phi(2^{1/3})$, of which two are not real. As $\Bbb Q(2^{1/3})\subset \Bbb R$, we must have $\phi(2^{1/3}) = 2^{1/3}$. Therefore, $$\operatorname{Hom}(\Bbb Q(2^{1/3}), \Bbb Q(2^{1/3})) = \{ \operatorname{Id}_{\Bbb Q(2^{1/3})} \}$$ and $|\operatorname{Hom}(\Bbb Q(2^{1/3}), \Bbb Q(2^{1/3}))| = 1$.