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I'm trying to find all the subfields of the splitting field of $x^4-2$. It's $\mathbb{Q}(\sqrt[4]{2}, i)$. I already figured out the Galois group is isomorphic to $D_4$ and I found the correspondent fixed fields for all subgroups but $\langle sr^1\rangle$ and $\langle sr^3\rangle$. I know the fixed field for $\langle sr,sr^3\rangle$ is $\mathbb{Q}(\sqrt2i)$ and so the two I'm missing should be in between $\mathbb{Q}(\sqrt[4]{2}, i)$ and $\mathbb{Q}(\sqrt2i)$, but none of them is $\mathbb{Q}(\sqrt2, i)$, for that is the fixed field of $\langle r^2\rangle$. I just can't figure out what to try with.

Mind that I'm calling the complex conjugation $s$ and $r$ to the element satisfying $r(\sqrt[4]{2})=\sqrt[4]{2}i$ and $r(i)=i$.

Besides the particular problem, any kind of advice or method to proceed with in order to find subfields corresponding to Galois subgroups is welcome.

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  • $\begingroup$ Use \langle and \rangle, not < and >, for delimiters $\endgroup$
    – Arturo Magidin
    Dec 26, 2022 at 16:20
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    $\begingroup$ What you call $r$ cannot be an element of order $4$ in $D_4$; because applying the transformation $\sqrt[4]{2}\mapsto -\sqrt[4]{2}$, $i\mapsto i$ twice gives the identity. But $r^2$ is not supposed to be the identity. So you have misidentified the automorphisms with the elements of the dihedral group. That is almost certainly the source of your error. (silently fixed in response, apparently).... $\endgroup$
    – Arturo Magidin
    Dec 26, 2022 at 16:22
  • $\begingroup$ Instead, the automorphism corresponding to $r$ should map $\sqrt[4]{2}$ to $i\sqrt[4]{2}$, and $i$ to itself. That will give you an automorphism of order $4$; and moreover, doing this and then doing complex conjugation will send $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$ and $i$ to $-i$, which is the same as doing complex conjugation and then applying this transformation three times; that is, it satisfies $sr = r^3s$., which is what you really need. $\endgroup$
    – Arturo Magidin
    Dec 26, 2022 at 16:25
  • $\begingroup$ @ArturoMagidin yep, sorry, that was a typo. I was working all the way with that $r$ but still I can't find the corresponding subfields $\endgroup$
    – Fernando
    Dec 26, 2022 at 16:27
  • $\begingroup$ Another error in the first paragraph: $s^2=1$, so the fixed field of $\langle s^2\rangle$ is the whole thing. You probably meant $\langle r^2\rangle$. $\endgroup$
    – Arturo Magidin
    Dec 26, 2022 at 16:28

1 Answer 1

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So thanks to @AndrewHubery I found the answer. As both $sr$ and $sr^3$ have order two, any element of the form $x+sr(x)$ will remain fixed by $sr$, and likewise, any $x+sr^3(x)$ will remain fixed by $sr^3$. Choosing an $x$ such that the obtained fixed element extends $\mathbb{Q}(\sqrt2i)$ we have the solution. $x=\sqrt[4]{2}$ does the trick and hence $\mathbb{Q}(\sqrt[4]{2}(1+i))$ is $sr$'s fixed field and $\mathbb{Q}(\sqrt[4]{2}(1-i))$ si $sr^3$'s.

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