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I was looking at the Galois group of the splitting field of $x^4-7$ over $\mathbb{Q}$. I found it to be $\mathbb{Q}(\sqrt[4]{7},i)$, and the Galois group to be the dihedral group of order $8$. Now $D_8=\langle \sigma,\rho\mid \sigma^4=\rho^2=1, \sigma\rho=\rho\sigma^3\rangle$.

I know $D_8$ has $5$ subgroups of index $4$, there should be $5$ subfields of $\mathbb{Q}(\sqrt[4]{7},i)$ of degree $4$ over $\mathbb{Q}$ by the fundamental theorem of Galois theory.

I found $3$ of them to be $\mathbb{Q}(i\sqrt[4]{7})$, $\mathbb{Q}(\sqrt[4]{7})$, and $\mathbb{Q}(\sqrt{7},i)$.

I can't figure out what the other $2$ are, but I think they're the fixed fields of the subgroups $\{1,\rho\sigma\}$ and $\{1,\rho\sigma^3\}$, but I can't determine those. Does anyone have any idea what the other two subfields of degree $4$ over $\mathbb{Q}$ are? Many thanks.

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    $\begingroup$ This is just linear algebra: write down the action of $\rho\sigma$ on the splitting field as a matrix $M$ with respect to the $\mathbb{Q}$-basis $\{\sqrt[4]{7}^ni^m\}$ and compute the kernel of $M-I$. $\endgroup$ – Alex B. Nov 30 '11 at 7:27
  • $\begingroup$ @AlexB. The book I was using gave some ad hoc examples only. I wish I had thought of this method before, it works like a charm! Thank you very much. $\endgroup$ – Kiera Nov 30 '11 at 7:56
  • $\begingroup$ You are welcome, Kiera! $\endgroup$ – Alex B. Nov 30 '11 at 15:56
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Let $a=\sqrt[4]{7}$, $b=i\sqrt[4]{7}$, $c=-\sqrt[4]{7}$, and $d=-i\sqrt[4]{7}$ be the four roots of $x^4-7$. Then the automorphisms $\rho$ and $\sigma$ permute these roots, with $$ \rho \,=\, (b\;\;d) \qquad\text{and}\qquad \sigma \,=\, (a\;\;b\;\;c\;\;d) $$ Observe then that: $$ \rho\sigma \,=\, (a\;\;d)(b\;\;c), \quad \rho\sigma^2 \,=\, (a\;\;c), \quad \rho\sigma^3\,=\, (a\;\;b)(c\;\;d), \quad\text{and}\quad \sigma^2 \,=\, (a\;\;c)(b\;\;d) $$ As you suggest, the intermediate fields $\mathbb{Q}(\sqrt[4]{7})$, $\mathbb{Q}(i\sqrt[4]{7})$, and $\mathbb{Q}(i,\sqrt{7})$ correspond to three of the subgroups of order two: $$ \mathbb{Q}(\sqrt[4]{7}) \leftrightarrow \{1,\rho\},\qquad \mathbb{Q}(i\sqrt[4]{7})\leftrightarrow \{1,\rho\sigma^2\}, \qquad \mathbb{Q}(i,\sqrt{7})\leftrightarrow\{1,\sigma^2\} $$ The main trick for figuring out the fixed field of $\{1,\rho\sigma\}$ is to consider the symmetrizing function $1+\rho\sigma$. Because of symmetry, anything in the image of $1+\rho\sigma$ must be fixed under $\rho\sigma$: $$ \rho\sigma(1+\rho\sigma)(x) \,=\, (\rho\sigma + (\rho\sigma)^2)(x) = (\rho\sigma + 1)(x) = (1+\rho\sigma)(x) $$ Now, observe that $$ (1+\rho\sigma)(a) \,=\, a+d = (1-i)\sqrt[4]{7} $$ and $$ (1+\rho\sigma)(b) \,=\, b+c = (i-1)\sqrt[4]{7} $$ Therefore, the fixed field of $\{1,\rho\sigma\}$ is $\mathbb{Q}\bigl((1-i)\sqrt[4]{7}\bigr)$. Using a similar calculation, we can see that the fixed field of $\{1,\rho\sigma^3\}$ is $\mathbb{Q}\bigl((1+i)\sqrt[4]{7}\bigr)$.

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  • $\begingroup$ My book had mentioned these symmetrizing functions, but the explanation wasn't as clear as yours here. Cheers! $\endgroup$ – Kiera Nov 30 '11 at 8:03
  • $\begingroup$ In the first part of your explanation, mustn't $\left\{1,\sigma^2\right\}$ correspond with $\mathbb{Q}(i)$? If $\sigma^2 = (a c) (b d)$, then the only fixed element under $\sigma^2$ is $i$, isn't it? Also there is a minor typo in both your answer and the question - it seems you have copied this and thus made the same mistake as OP: you both typed $\sqrt{7}$ instead of $\sqrt[4]{7}$ for the subfield corresponding to $\left\{1,\sigma^2\right\}$. $\endgroup$ – Kasper Cools Dec 14 '16 at 22:23
  • $\begingroup$ @KasperCools The subgroup $\{1,\sigma^2\}$ has index four, so it must correspond to a degree four extension of $\mathbb{Q}$. Since $\mathbb{Q}(i)$ only has degree two, there must be more in the fixed field of $\{1,\sigma^2\}$ than just $\mathbb{Q}(i)$. Indeed, since $i = b/a = d/c$ and $\sqrt{7} = a^2 = c^2$, the permutation $\sigma^2 = (ac)(bd)$ fixes both $i$ and $\sqrt{7}$. (The field $\mathbb{Q}(i)$ actually corresponds to the subgroup $\{1,\sigma,\sigma^2,\sigma^3\}$.) I also don't agree that there's a typo. $\endgroup$ – Jim Belk Dec 15 '16 at 4:13

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