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

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

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

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

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

$\tau_1(\sqrt{7})=i\sqrt{7}$

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

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

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

• How would I verify that roots of unity are part of splitting field? – amiz9 Apr 19 '16 at 1:08
• Sorry, I read the problem wrong! I thought it was $x^7-4$! You are completely correct; again, I am so sorry. – Noble Mushtak Apr 19 '16 at 1:10
• I will consider accepting your apology one day. It will take time though – amiz9 Apr 19 '16 at 1:31
• @amiz9 I guess you're being funny here? – Pedro Tamaroff Apr 19 '16 at 2:19
• @PedroTamaroff yes – amiz9 Apr 19 '16 at 4:10

## 2 Answers

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 $24$.

• I am a bit confused here. If you calculate the Galois group of $(x^4-3)(x^3-5)$ over $\Bbb{Q}$, then how does $i$ come into the picture? All of the roots of this polynomial are real. – Noble Mushtak Apr 19 '16 at 1:46
• @NobleMushtak No, the roots of $X^4-3$ are not all real. There are two that are real, and two that are complex conjugates. The roots of $X^3-5$ are not all real either, in fact only one is. – Pedro Tamaroff Apr 19 '16 at 1:49
• Oh, OK. I see now. However, now I feel like there should be another permutation of order $3$ because you have an extension of $x^4-3$, $x^3-5$, $x^2+1$, and $x^3+1$, which is $4*3*2*3=72$. Sorry if I am understanding this wrong; I'd just like to get a better understanding of what is happening. Is the tower $\Bbb{Q}(\zeta, \sqrt 5, i, \sqrt 3) \supset \Bbb{Q}(\sqrt 5, i, \sqrt 3) \supset \Bbb{Q}(i, \sqrt 3) \supset \Bbb{Q}(\sqrt 3) \supset \Bbb{Q}$ or is this wrong? – Noble Mushtak Apr 19 '16 at 1:58
• @NobleMushtak You can write $\zeta$ in terms of $i$ and $3^{1/4}$: it is $\dfrac{-1+\sqrt 3 i}2$. Roots interact, and you cannot forget this. – Pedro Tamaroff Apr 19 '16 at 2:08
• OK, that really helps me understand. Thank you for all of this explaining! I feel like I learned more now. – Noble Mushtak Apr 19 '16 at 2:10

We know that the minimal polynomial of $\sqrt 7$ has degree $4$ and the minimal polynomial of $i$ has degree $2$, so $[\Bbb{Q}(\sqrt 7, i) : \Bbb{Q}]=[\Bbb{Q}(\sqrt 7, i) : \Bbb{Q}(i)][\Bbb{Q}(i) : \Bbb{Q}]=4*2=8$. 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 $D_8$.

• The first sentence is not correct: in the example in my answer one could be mistakenly led to believe the Galois group has order $3\times 4=12$; which is not the case. You could observe that $\Bbb Q(7^{1/4},i)\supseteq \Bbb Q(7^{1/4})\supseteq Q$ is a tower with degrees $4$ and $2$; however, and appeal to the fact degrees are multiplicative in towers. – Pedro Tamaroff Apr 19 '16 at 1:53
• @PedroTamaroff In your case, you also have $x^2+1$ because of $i$, so it is $3*4*2=24$. – Noble Mushtak Apr 19 '16 at 1:54
• You're not explaining your logic. Why are you also adding $i$? What's the rule? (Sure, there's one, but not making it explicit is misleading). – Pedro Tamaroff Apr 19 '16 at 1:56
• @PedroTamaroff I'm going off of the work that the original questioner already did. – Noble Mushtak Apr 19 '16 at 1:56