Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $M-I$. – Alex B. Nov 30 '11 at 7:27
@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. – Kiera Nov 30 '11 at 7:56
You are welcome, Kiera! – Alex B. Nov 30 '11 at 15:56
up vote 3 down vote accepted

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 $\mathbb{Q}\bigl((1+i)\sqrt[4]{7}\bigr)$.

share|cite|improve this answer
My book had mentioned these symmetrizing functions, but the explanation wasn't as clear as yours here. Cheers! – Kiera Nov 30 '11 at 8:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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