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How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois?

At first I thought it was the splitting field of $x^4-7$, but I was only able to prove that it was a subfield of the splitting field. Any ideas?

I'm trying to find all the intermediate fields in terms of their generators, but I don't understand how. I am trying to imitate Dummit and Foote on this. I am looking at the subgroups of the Galois group in terms of $\tau$ where $\tau$ is the automorphism that takes $\sqrt[4]{7}$ to itself and $i$ to $-i$, and $\sigma$ that takes $\sqrt[4]{7}$ to $i\sqrt[4]{7}$ and $i$ to itself. How do I, for example, find the subfield corresponding to $\{1, \tau\sigma^3\}$? I know I am supposed to find four elements of the galois group that $\tau\sigma^3$ fixes, but so far i can only find $-\sqrt[4]{7}^3$.

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  • $\begingroup$ If you can show that $\mathbb{Q}(\sqrt[4]{7},i)$ is a subfield of the splitting field, then show that every root of $x^4-7$ is in $\mathbb{Q}(\sqrt[4]{7})$ and you'll be done. $\endgroup$
    – Arturo Magidin
    Dec 4, 2011 at 6:55
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    $\begingroup$ @MathMastersStudent: You are on the right track with only a small step left, what went wrong? $\endgroup$
    – Eric Naslund
    Dec 4, 2011 at 6:58
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    $\begingroup$ You have asked 9 questions, and you haven't accepted any answers. Do you know how to accept answers? Do you understand the importance to this website of accepting answers? $\endgroup$
    – Gerry Myerson
    Dec 4, 2011 at 7:00
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    $\begingroup$ Gerry, so sorry! I didn't know there was such a thing. I will check the meta-section and accept ASAP. This ite is wonderful. $\endgroup$
    – MathTeacher
    Dec 4, 2011 at 7:24

1 Answer 1

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The splitting field of your polynomial is generated by the $a=\sqrt 7$, $-a$, $ia$ and $-ia$, its roots. It therefore contains $a$ and $i=\frac{ia}{a}$. Obviously, the field generated by $a$ and $i$ also contains the four roots.

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  • $\begingroup$ Thanks Mariano. I am also asked to find explicit generators of the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[4]{7},i)$. Should I just first look at the subgroups of the Galois group, or would this be doing down the wrong path? $\endgroup$
    – MathTeacher
    Dec 4, 2011 at 9:28
  • $\begingroup$ You should :) ${}$ $\endgroup$
    – Mariano Suárez-Álvarez
    Dec 4, 2011 at 13:08

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