Fields and Galois Theory: an Example

Apr 3—2025

I have learned from class that if $E/F$ is a finite normal extension, then given $\alpha$ , $\beta$ $\in E$ with the same minimal polynomial, there exists $\Psi \in Aut_{F}(E)$ such that $\Psi(\alpha) = \beta$ .

Sketch Proof:

Since $\alpha$ , $\beta$ have the same minimal polynomial, it follows that $F(\alpha) \cong F[x]/<p(x)> \cong F(\beta)$ . Hence, we can find an isomorphism $\theta: F(\alpha) \rightarrow F(\beta)$ such that $\theta|_{F} = id$ . BECAUSE of the fact that E is a finite normal extension of F, it is the splitting field of some $f(x) \in F[x]$ . We can in turn also think of E as the splitting field of $f(x) \in F(\alpha)$ and also $F(\beta)$ . Thus, we can find an automorphism $\Psi: E \rightarrow E$ that extends $\theta$ . Hence, we have what we want.

Now, the question is: what goes wrong when $E/F$ is NOT a finite normal extension?

Can we find a field extension $E$ such that there exist $\alpha, \beta \in E$ that share the SAME minimal polynomial, but we cannot find $\Psi \in Aut_F(E)$ such that $\Psi(\alpha) = \beta$ ?

Yes, take $E = Q(\sqrt[4]{2})$ . The minimal polynomial of $\sqrt[4]{2}$ is $f(x) = x^4 -2 = (x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$ . Hence, $Q(\sqrt[4]{2})$ is not normal.

Take $\sqrt{2} \in Q(\sqrt[4]{2})$ . Notice that the minimal polynomial of $\sqrt{2}$ is $g(x) = x^2 - 2 = (x-\sqrt{2})(x+\sqrt{2})$ . And so, $\sqrt{2}, -\sqrt{2}$ are conjugate. But we claim there does not exist $\Psi \in Aut_Q(Q(\sqrt[4]{2}))$ such that $\Psi(\sqrt{2}) = -\sqrt{2}$ . Suppose not. Then for such $\Psi$ , we have $\Psi(\sqrt[4]{2})^2 = \Psi(\sqrt{2}) = -\sqrt{2}$ . This means $\Psi(\sqrt[4]{2}) = \sqrt[4]{-2}$ , but this is not in $Q(\sqrt[4]{2})$ , a contradiction.