Find all homomorphisms from $\mathbb{Q}(\sqrt[5]2)$ to $\mathbb{Q}(\sqrt[5]2, \omega)$

Question

We need to determine all $\mathbb{Q}$-homomorphisms from $\mathbb{Q}(\sqrt[5]2)$ to $\mathbb{Q}(\sqrt[5]2, \omega)$, where $\omega=e^{2\pi i/5}$.

I can figure it out that there are 5 injective homomorphisms because there are $5$ roots of $x^5-2$ in $\mathbb{Q}(\sqrt[5]2, \omega)$.

But how do I proceed after this to find all of them?

Answer 1

First, notice that $\mathbb{Q}(\sqrt[5]{2},\omega)$ is the splitting field over $\mathbb{Q}$ of the polynomial $f(X)=X^5-2$.

In general, it's a well known fact that the set of $\mathbb{Q}$-homomorphisms from a simple, algebraic extension $\mathbb{Q}(\alpha)$ to $\mathbb{C}$ is in bijection with the set of complex roots of $f_{\alpha,\mathbb{Q}}$ minimal polynomial of $\alpha$ over $\mathbb{Q}$, and the bijection is given by sending $\sigma\space{\in}\space{Hom_{\mathbb{Q}}(\mathbb{Q}(\alpha),\mathbb{C})}$ to $\sigma(\alpha)$.

Making this explicitly : you have a bijection from $Hom_{\mathbb{Q}}(\space\mathbb{Q}(\sqrt[5]{2}),\space \mathbb{Q}(\sqrt[5]{2},\omega)\space)$ to $\{\sqrt[5]{2},\omega\sqrt[5]{2},\omega^2\sqrt[5]{2},\omega^3\sqrt[5]{2},\omega^4\sqrt[5]{2}\}$ roots of $f$ over $\mathbb{C}$. Thus, you get five homomorphisms, each of them determined by the image of $\sqrt[5]{2}$, generator of $\mathbb{Q}(\sqrt[5]{2})$.

Still, you'd get the same result by requesting only homomorphism of field from $\mathbb{Q}(\sqrt[5]{2})$ to $\mathbb{Q}(\sqrt[5]{2},\omega)$. In fact, for every homorphism $\sigma:\mathbb{Q}(\sqrt[5]{2})\longrightarrow \mathbb{Q}(\sqrt[5]{2},\omega)$, you have $\sigma(1)=1$ implies $\left.\sigma\right|_\mathbb{Q}=id_\mathbb{Q}$.

Answer 2

We need $\varphi (\sqrt[5]2)^5=\varphi (\sqrt[5]2^5)=\varphi (2)=2,$ by homomorphismness.

Thus there are $5:$

$$\sqrt [5]2\to\sqrt [5]2\omega ^i,i=1,\dots, 5.$$