# Math Help - Application of the Fund. Theorem of Galois Theory

1. ## Application of the Fund. Theorem of Galois Theory

I'm having trouble with this question: Find the splitting field of $f(x)=x^4 - 7$ over $F= \mathbb{Q}$and find all the intermediate fields of the extension.

The splitting field is $\mathbb{Q} (\sqrt[4]{7} , \zeta) = K$ where $\zeta = e^{\frac{2i \pi}{4}}=i$ . Now this extension has degree $8$ since $2, 4\vert [K:F]$ and $[K:f] \leq 8$. By the isomorphis extension theorem there are $\sigma , \tau: K\longrightarrow K$ automorphisms such that $\sigma (\sqrt[4]{7}) = i\sqrt[4]{7}$ and $\sigma (i) =i$ and $\tau (\sqrt[4]{7}) = \sqrt[4]{7}$ $\tau (i)=-i$
from this we know that $G=Gal(\frac{K}{F}) \cong D_8$ . Now the fixed field of $< \sigma>$ is $F(i)$ and the fixed field of $< \tau>= F(\sqrt[4]{7})$, but I'm having trouble with the other subgroups $<\sigma ^2>$ , $<\sigma \tau>$, $<\sigma ^2 \tau>$, $< \sigma ^3 \tau>$. I know they must be degree $4$ extensions of $F$, but using the trace I get either trivial fixed elements or ones that I don't know how to check their degree over $F$.

Tnaks in advance

2. Originally Posted by Jose27
I'm having trouble with this question: Find the splitting field of $f(x)=x^4 - 7$ over $F= \mathbb{Q}$and find all the intermediate fields of the extension.

The splitting field is $\mathbb{Q} (\sqrt[4]{7} , \zeta) = K$ where $\zeta = e^{\frac{2i \pi}{4}}=i$ . Now this extension has degree $8$ since $2, 4\vert [K:F]$ and $[K:f] \leq 8$. By the isomorphis extension theorem there are $\sigma , \tau: K\longrightarrow K$ automorphisms such that $\sigma (\sqrt[4]{7}) = i\sqrt[4]{7}$ and $\sigma (i) =i$ and $\tau (\sqrt[4]{7}) = \sqrt[4]{7}$ $\tau (i)=-i$
from this we know that $G=Gal(\frac{K}{F}) \cong D_8$ . Now the fixed field of $< \sigma>$ is $F(i)$ and the fixed field of $< \tau>= F(\sqrt[4]{7})$, but I'm having trouble with the other subgroups $<\sigma ^2>$ , $<\sigma \tau>$, $<\sigma ^2 \tau>$, $< \sigma ^3 \tau>$. I know they must be degree $4$ extensions of $F$, but using the trace I get either trivial fixed elements or ones that I don't know how to check their degree over $F$.

Tnaks in advance
the fixed field of $<\sigma^2>$ is $\mathbb{Q}(\sqrt{7},i).$ this is quite easy to see: let $\sqrt[4]{7}=a.$ then an element of $K$ is in the form $z=r_1+r_2a+r_3a^2+r_4a^3+r_5i + r_6ia+r_7ia^2+r_8ia^3,$ where $r_j \in \mathbb{Q}.$

then since $\sigma^2(a)=-a,$ we see immediately that $\sigma^2(z)=z$ if and only if $r_2=r_4=r_6=r_8=0.$ therefore $\sigma^2(z)=z$ if and only if $z=r_1+r_3a^2+r_5i + r_7ia^2 \in \mathbb{Q}(a^2,i).$

3. Okay, I think I got it:

The fixed field of $< \sigma \tau >= F((1+i) \sqrt[4]{7})$. This is easily seen to be fixed by $\sigma \tau$ and since it cannot be a root of a degree $2$ polynomial over $F$, it's minimal polynomial must have degree $4$ (since otherwise it would generate the whole extension which cannot be).

Proceeding in almost the same way I get:

Fixed field of $< \sigma ^3 \tau> = F((1-i) \sqrt[4]{7})$ and $< \sigma ^2 \tau>= F(i \sqrt[4]{7})$.

Is this right?

4. Originally Posted by Jose27
Okay, I think I got it:

The fixed field of $< \sigma \tau >= F((1+i) \sqrt[4]{7})$. This is easily seen to be fixed by $\sigma \tau$ and since it cannot be a root of a degree $2$ polynomial over $F$, it's minimal polynomial must have degree $4$ (since otherwise it would generate the whole extension which cannot be).

Proceeding in almost the same way I get:

Fixed field of $< \sigma ^3 \tau> = F((1-i) \sqrt[4]{7})$ and $< \sigma ^2 \tau>= F(i \sqrt[4]{7})$.

Is this right?
you're right about the fixed field of $<\sigma \tau>.$ i didn't check the other ones.