5
$\begingroup$

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 $-\sqrt[4]{7}^3$.

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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 '11 at 6:55
  • 1
    $\begingroup$ @MathMastersStudent: You are on the right track with only a small step left, what went wrong? $\endgroup$ – Eric Naslund Dec 4 '11 at 6:58
  • 3
    $\begingroup$ 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 '11 at 7:00
  • 2
    $\begingroup$ 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 '11 at 7:24
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '11 at 9:28
  • $\begingroup$ You should :) ${}$ $\endgroup$ – Mariano Suárez-Álvarez Dec 4 '11 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.