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.

  • $\begingroup$ An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted? $\endgroup$ – Joel Pereira Nov 19 '18 at 5:11
  • $\begingroup$ @JoelPereira only $3$ ways to permute $\sqrt{13},$ $-\sqrt{13}$ and $\sqrt[3]{7}$ as other roots are not in the extension? $\endgroup$ – Idonknow Nov 19 '18 at 5:20
  • $\begingroup$ ${\bf Q}(\sqrt{13}),\root3\of7)$ doesn't make sense. Do you mean ${\bf Q}(\sqrt{13},\root3\of7)$? $\endgroup$ – Gerry Myerson Nov 19 '18 at 5:31
  • 1
    $\begingroup$ @GerryMyerson Yes. Edited. $\endgroup$ – Idonknow Nov 19 '18 at 5:33
  • $\begingroup$ "The group is not Galois." I think you mean, "the extension is not Galois." $\endgroup$ – 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 $Aut(\mathbb{Q}(\sqrt{13},\sqrt[3]{7})/\mathbb{Q})$

  • $\begingroup$ Just a quick question. How many element does the automorphism group have? $\endgroup$ – Idonknow Nov 23 '18 at 13:48
  • $\begingroup$ Exactlu two: one that maps $\sqrt{13}$ to itself, and one which maps $\sqrt{13}$ to $-\sqrt{13}$. Hope it's clear. $\endgroup$ – Bilo Nov 23 '18 at 15:06

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