My question arises from the following problem. Consider the field $K=\Bbb Q(\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{5}),$ and define a map $f$ via $$f(\sqrt[3]{2})=\sqrt[3]{2} e^{2i\pi/3}, f(\sqrt[3]{3})=\sqrt[3]{3}, f(\sqrt[3]{5})=\sqrt[3]{5}.$$ I want to show that this extends (uniquely) into a field morphism $f' : K \to \overline{\Bbb Q}$. The issue is that the different cubic roots must be proved to be linearly independent, and even more, we want for instance $\sqrt[3]{3} \not\in \Bbb Q(\sqrt[3]{2})$ (because otherwise the value of $f'$ on $\sqrt[3]{3}$ will be determined by its value on $\sqrt[3]{2}$).
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So more generally, assume that $K / \Bbb Q$ is a finite extension and $E \subset K$ is a finite set with $K = \Bbb Q(E)$. I want to know nice sufficient conditions such that:
- $(*)$ any map $f : E \to \Bbb C$ such that $f(a)$ has the same minimal polynomial as $a$, for any $a \in E$, extends into a field morphism $f' : K \to \Bbb C$.
My guess is that the condition
- $a \not \in \Bbb Q(E \setminus \{a\})$ for any $a \in E$
is sufficient. Question 1 : is it correct? If not, what are sufficient conditions?
(The aside question $2$ is to determine whether the set $\{\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{5}\}$ satisfies my condition just above).
This condition does not imply your desired property in general. For example, suppose that $f \in \mathbb{Q}[x]$ is an irreducible quartic polynomial with Galois group $D_4$. Then the splitting field $K$ of $f$ is generated by two roots of $f$, say $K = \mathbb{Q}(\alpha_1,\alpha_2)$. The element $\alpha_2$ has degree two over $\mathbb{Q}(\alpha_1)$, and vice versa, so your criterion is satisfied. Let's denote the other roots by $\{\alpha_3, \alpha_4\}$. Then not every injective mapping of $\{\alpha_1,\alpha_2\} \to \{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ extends to an automorphism $K \to K$, since there are $12$ such maps, but only $8$ automorphisms of $K$.
An example of such an $f$ is $x^4 - 2x^2 - 2$.