Determine the automorphism group $Aut(\mathbb{Q}(\sqrt{13}, \sqrt[3]{7})/\mathbb{Q})$

Question

Question: Determine the automorphism group $$Aut(\mathbb{Q}(\sqrt{13}, \sqrt[3]{7})/\mathbb{Q}).$$

My attempt: Since the polynomial $(x^2-13)(x^3-7)$ has roots $$\sqrt{13}, -\sqrt{13}, \sqrt[3]{7}, \sqrt[3]{7}\omega, \sqrt[3]{7}\omega^2$$ where $\omega$ is the cube root of unity. Since the extension does not contain all roots, so the extension is not Galois.

However, I do not know how to determine the automorphism group.

Any hint is appreciated.

Answer 1

Let $\sigma \in Aut(\mathbb{Q}(\sqrt{13},\sqrt[3]{7})/\mathbb{Q})$, then $\sigma$ is uniquely determine by $\sigma(\sqrt{13})$ and $\sigma(\sqrt[3]{7})$.

Since $\sigma(\alpha) \in \mathbb{Q}(\sqrt{13},\sqrt[3]{7}) \;\forall \alpha \in \mathbb{Q}(\sqrt{13},\sqrt[3]{7})$, we have that $\sigma(\sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $\mathbb{Q}(\sqrt{13},\sqrt[3]{7})$.

This implies that $\sigma(\sqrt[3]{7}) = \sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $\mathbb{Q}(\sqrt{13},\sqrt[3]{7})$.

Clearly $\sigma(\sqrt{13}) \in \lbrace \pm \sqrt{13} \rbrace$.

In conclusion, we see that $\sigma$ is completely determined by its behaviour on $\sqrt{13}$, and we can easily determine the automorphism group $Aut(\mathbb{Q}(\sqrt{13},\sqrt[3]{7})/\mathbb{Q})$