# Fixed Field of rs for dihedral group of order 8

Let $f=X^4-2\in\mathbb{Q}[X]$ and $\alpha_{1,2}=\pm\sqrt{2},\ \alpha_{3,4}=\pm i\sqrt{2}$. Then $Gal(f)\cong D_4$. As for the rotation I set

$r: \alpha_1\mapsto \alpha_3\mapsto \alpha_2\mapsto \alpha_4$

and for the reflection I set

$s: \alpha_3\mapsto \alpha_4,\ \alpha _1 \mapsto\alpha_1,\ \alpha _2\mapsto\alpha_2$

Question: What is the fixed Field of $\langle rs\rangle$ and $\langle r^3 s\rangle$?

It is clear that both are of degree 4 over $\mathbb Q$. By applying $rs$ to the 4-gon declared above it is also clear that

$rs: \alpha_1\leftrightarrow\alpha_3,\ \alpha_2\leftrightarrow\alpha_4$

Is it correct to say that it follows $\mathcal{F}(\langle rs\rangle)=\mathbb{Q}(\alpha_1+\alpha_3)=\mathbb{Q}(\alpha_2+\alpha_4)$? I'm not sure if it is correct to say $rs(\alpha_1+\alpha_3)=\alpha_3+\alpha_1$.

By the same statement it would follow that $\mathcal{F}(\langle r^3 s\rangle)=\mathbb{Q}(\alpha_2+\alpha_3)=\mathbb{Q}(\alpha_1+\alpha_4)$.

• The elements of your field are of the form $\sum_{m=0}^3 c_m \alpha^m, \quad c_m\in \mathbb{Q}(i)$ and $\sigma(\sum_{m=0}^3 c_m \alpha^m) = \sum_{m=0}^3 \sigma(c_m) \sigma(\alpha)^m$ – reuns Jun 12 '17 at 14:55
• It is correct to say that $rs(\alpha_1+\alpha_3)=\alpha_3+\alpha_1$. However, only knowing that $\alpha_1+\alpha_3$ is in the fixed field of $rs$, one cannot conclude $\mathcal{F}(\langle rs\rangle)=\mathbb{Q}(\alpha_1+\alpha_3)$, although this is indeed true in this case, by considering the degree of extensions. – pisco Jun 12 '17 at 14:56

As you numbered, let $r=(1 3 2 4), s=(34)$, then $rs = (13)(24)$.
Denote the fixed field of $\langle rs \rangle$ be $F$, Then $\alpha_1\alpha_3 = i\sqrt{2}, \alpha_1+\alpha_3 = i\sqrt{2}+\sqrt{2} \in F$, however, note that $(\alpha_1+\alpha_3)^2 = 2\alpha_1\alpha_3$. Since $F$ has degree 4 over $\mathbb{Q}$, $F=\mathbb{Q}(i\sqrt{2}+\sqrt{2})$.
Similarly for the fixed field of $\langle r^3s \rangle$, you can verify it is equal to $\mathbb{Q}(i\sqrt{2}-\sqrt{2})$.
• What purpose does it serve showing $(\alpha_1+\alpha_3)^2=2\alpha_1\alpha_3$? As you said, by knowing the degree it is clear that $F=\mathbb{Q}(\alpha_1+\alpha_3)$ after observing $\alpha_1+\alpha_3\in F$. The question is why $\alpha_1+\alpha_3\in F$. Since $rs(\alpha_1+\alpha_3)\neq rs(\alpha_1)+rs(\alpha_3)$ this would need a more detailed proof. – user424862 Jun 12 '17 at 15:06
• The permutation $rs$ swaps 1 and 3, therefore it maps $\alpha_1$ to $\alpha_3$, and maps $\alpha_3$ to $\alpha_1$. This gives $$rs(\alpha_1 + \alpha_3) = \alpha_3 + \alpha_1 = \alpha_1+\alpha_3$$ therefore $\alpha_1+\alpha_3$ is in the field $F$ – pisco Jun 12 '17 at 15:09
• From a intuitive standpoint it's clear that this happens. But you can't swap 1 and 3 if you don't apply it to 1 or 3. We are applying rs to the sum of 1 and 3 and can't extend $rs(\alpha_1+\alpha_3)=rs(\alpha_1)+rs(\alpha_3)$. – user424862 Jun 12 '17 at 15:12
• We can write $rs(\alpha_1+\alpha_3)=rs(\alpha_1)+rs(\alpha_3)$ because $rs$ is an automorphism (on the splitting field of $x^4-2$). The property $f(a+b) = f(a)+f(b)$ always hold for a homomorphism. – pisco Jun 12 '17 at 15:16