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Let $K=\mathbb{Q}(\sqrt[8]{7},i)$, let $F_1 = \mathbb{Q}(\sqrt{7})$ and let $F_2=\mathbb{Q}(\sqrt{-7})$.

(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.

(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.

I first thought that $K/\mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=\mathbb{Q}(\sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $\mathbb{Q}(\sqrt{7})$ or over $\mathbb{Q}(\sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since $$\mathbb{Q}(\sqrt{7})\subseteq\mathbb{Q}(\sqrt[8]{7})\subseteq\mathbb{Q}(\sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!

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If $F$ has characteristic zero, and $K=F(\sqrt[4]a,i)$ where $a\in F$, then $K$ is Galois over $F$, being the splitting field of $x^4-a$. Your first example has $a=\sqrt7$.

The Galois group of $K/F_1$ has order $8$, and computing it is very similar to standard examples such as $\Bbb Q(\sqrt[4]2,i)/\Bbb Q$. The group is dihedral of order $8$.

I don't think $K/F_2$ is Galois.

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  • $\begingroup$ You only need $\operatorname{char}(F) \ne 2$ $\endgroup$ – Kenny Lau Jan 4 at 22:08
  • $\begingroup$ I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois? $\endgroup$ – Alex Jan 5 at 7:31
  • $\begingroup$ @Alex You asked what the Galois group of a non-Galois extension is.... $\endgroup$ – Lord Shark the Unknown Jan 5 at 7:34
  • $\begingroup$ @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem. $\endgroup$ – Alex Jan 5 at 7:38
  • $\begingroup$ Exams have misprints, just as all other types of writing. @Alex $\endgroup$ – Lord Shark the Unknown Jan 5 at 7:40

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