Galois group of splitting field over $\mathbb{Q}$

Question

Describe the structure of the Galois group of the splitting field $L$ of the polynomial $X^4-7$ over $\mathbb{Q}$

I believe that $L=\mathbb{Q}(\sqrt[4]{7}, i)$ is the splitting field of the polynomial

Now I need to describe the structure of $Gal(L, \mathbb{Q})$

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$H=Gal(L, \mathbb{Q}(i))$ is a cyclic group of order $2$ generated by $\tau_1$ such that:

$\tau_1(\sqrt[4]{7})=i\sqrt[4]{7}$

$Gal(L, \mathbb{Q}(\sqrt[4]{7}))$ is a cyclic group generated by $\tau_2$ such that:

$\tau_2(\sqrt[4]{7})=\sqrt[4]{7}$ and $\tau_2(i)=-i$

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I believe that $Gal(L, \mathbb{Q})$ consists of the identity element $e$ and some multiplied combinations of $\tau_1$ and $\tau_2$ but could do with some help filling in the gaps and improving my understanding

Answer 1

We know that the minimal polynomial of $\sqrt[4] 7$ has degree $4$ and the minimal polynomial of $i$ has degree $2$, so $[\Bbb{Q}(\sqrt[4] 7, i) : \Bbb{Q}]=[\Bbb{Q}(\sqrt[4] 7, i) : \Bbb{Q}(i)][\Bbb{Q}(i) : \Bbb{Q}]=4*2=8$. Thus, the Galois group has order $8$ because of the Fundamental Theorem of Galois Theory.

Now, this is the splitting field of $x^4-7$, meaning all of the roots of $x^4-7$ must permute with each other. Thus, the Galois group is a subgroup of $S_4$. However, according to this really helpful table, the only subgroup of $S_4$ of order $8$ is $D_8$, so this is the structure of the Galois group.

This can also be shown with your work: $\tau_1$ represents a rotation of 90 degrees in the complex plane while $\tau_2$ represents a reflection over the real number line in the complex plane. You are rotating and reflecting $4$ elements (the roots of $x^4-7$) without just permuting them at free will, which means the group has a structure of $D_8$.

Answer 2

There's a general tool to deal with this type of extensions. Suppose you have a diamond of extensions $F \subseteq L,L' \subseteq E$ (make the drawing) where $L\cap L'=F,LL'=E$.

Suppose moreover that $L/F$ is normal, and $E/F$ is Galois. Let $G={\rm Gal}(E/F)$, $H={\rm Gal}(E/L)$ and $K={\rm Gal}(E/L')$. Then show that $HK=G, H\cap K=1$ and $K\lhd G$. All this means that $G$ is the semidirect product of $H$ and $K$, in particular every element $\sigma$ of $G$ is written uniquely as $\tau\rho$ with $\tau\in H,\rho\in K$. In your case one gets $G$ is a semidirect product $C_2\rtimes C_4$ which is the dihedral group of order four. But one gets more than this, the above gives an explicit "decomposition" of the Galois group in terms of the subextensions $L,L'$.

To be more precise, take $L=\Bbb Q(i),L'=\Bbb Q(7^{1/4}),F=\Bbb Q$ and $E=LL'$ (your extension). Then $L/\Bbb Q$ is normal and $L\cap L'=\Bbb Q$ (this is not hard to prove, since $L$ is obtained by adjoining $i$, and its a degree two extension). One can see that $E/L$ is of degree four and $E/L'$ is of degree two.

The technique above is very useful when one adjoins more than two elements, since one can iterate the decomposition. For example, one can explicitly calculate the Galois group of $(X^4-3)(X^3-5)$ over $\Bbb Q$ as a semidirect product $(C_3\rtimes C_4)\rtimes C_2$ explicitly: it is generated by $\omega,\eta,\psi$ such that $\omega^3=\eta^4=\psi^2=1$

1. $\eta(3^{1/4})=3^{1/4}i$ and $\eta$ fixes $i,5^{1/3}$
2. $\psi(i)=-i$ and $\psi$ fixes $3^{1/4},5^{1/4}$,
3. $\omega(5^{1/3})=5^{1/3}\xi$ and $\omega$ fixes $i,3^{1/4}$.

Here $\xi$ a primitive cuberoot of $1$, and every element of $G$ is written uniquely as $\omega^i\eta^j\psi^k$ with $i=0,1,2,j=0,1,2,3,k=0,1$. Moreover $\psi\omega\psi =\omega^{-1},\eta\omega\eta^{-1}=\omega^{-1}$. A bit more calculations can give you a presentation for $G$, which has order $24$.