I have the field $\mathbb{Q}(\sqrt[4]{3})$ and i want to find its automorphisms that leave $\mathbb{Q}$ invariant, so my attempt is: I find its minimal polynomial which is $$P(x)=x^4-3$$ Then i know that its automorphisms must send $\sqrt[4]{3}$ to every possible root of $P(x)$ The roots are $\pm \sqrt[4]{3}, \pm \sqrt[4]{3} i$ and the possible maps are the identity $id$, $$s_1 : \sqrt[4]{3} \mapsto - \sqrt[4]{3}$$ $$s_2 : \sqrt[4]{3} \mapsto \sqrt[4]{3} i$$ $$s_3 : \sqrt[4]{3} \mapsto - \sqrt[4]{3} i$$ But $ \pm \sqrt[4]{3} i$ are not elements of $\mathbb{Q}(\sqrt[4]{3})$ So the only automorphisms are $id$ and $s_1$. Am i right to exclude $s_2$ and $s_3$? Generally in such problems, should i keep the maps that map only to elements of the field?

  • $\begingroup$ Automorphisms of $\mathbb{Q}(\sqrt[4]{3})$ must send $\sqrt[4]{3}$ to an element of $\mathbb{Q}(\sqrt[4]{3})$. In particular, a real number. And yes, at the same time a root of $x^4-3$. $\endgroup$ – user553213 May 10 '18 at 13:47
  • $\begingroup$ How do i find those elements, aren't those the roots of the minimal polynomial? $\endgroup$ – Dimtsol May 10 '18 at 13:49
  • 1
    $\begingroup$ Yes, $\pm\sqrt[4]{3}$ only. $\endgroup$ – user553213 May 10 '18 at 13:50

Automorphisms of $\mathbb{Q}(\sqrt[4]{3})$ are maps $f:\mathbb{Q}(\sqrt[4]{3})\to\mathbb{Q}(\sqrt[4]{3})$ so yes, as @deyore said $f(\sqrt[4]{3})\in \mathbb{Q}(\sqrt[4]{3})\subset \mathbb{R} $


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