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

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.

• An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted? – Joel Pereira Nov 19 '18 at 5:11
• @JoelPereira only $3$ ways to permute $\sqrt{13},$ $-\sqrt{13}$ and $\sqrt[3]{7}$ as other roots are not in the extension? – Idonknow Nov 19 '18 at 5:20
• ${\bf Q}(\sqrt{13}),\root3\of7)$ doesn't make sense. Do you mean ${\bf Q}(\sqrt{13},\root3\of7)$? – Gerry Myerson Nov 19 '18 at 5:31
• @GerryMyerson Yes. Edited. – Idonknow Nov 19 '18 at 5:33
• "The group is not Galois." I think you mean, "the extension is not Galois." – Gerry Myerson Nov 19 '18 at 5:34

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 $$Aut(\mathbb{Q}(\sqrt{13},\sqrt[3]{7})/\mathbb{Q})$$
• Exactlu two: one that maps $\sqrt{13}$ to itself, and one which maps $\sqrt{13}$ to $-\sqrt{13}$. Hope it's clear. – Bilo Nov 23 '18 at 15:06