Galois group and the fixed field of $L(\sqrt[6]{2})|L$ for $L=\mathbb{Q}(e^{2πi/3} )$

Question

I have my Algebra I exam coming up soon and I want to make sure I have understood the concepts well. I would therefore like you to check my answer to the following exercise:

Let $L = \mathbb{Q}(e^{2πi/3} )$ be an extension of $K=\mathbb{Q}$.

(a) Calculate the Galois group $G = Gal(L, K)$ and the fixed field $L^G$$. Is the extension Galois?

(b) Calculate the Galois group and the fixed field of $L(\sqrt[6]{2})|L$. Is it Galois?

For a) we have that the minimal polynomial of $e^{2πi/3}$ is the third cyclotomic polynomial $f(x)=x^2+x+1$. Since we have a field extension over $ \mathbb{Q}$ and $ \mathbb{Q}$ has $char( \mathbb{Q})=0$ the field extension is separable.

Since the two roots of $f$ are $ \xi=e^{2πi/3}$ and $\xi^2 \in L$ we have a normal field extension. Since the field extension is normal and separable we have a galois field extension.

We have only $2$ Automorphisms, namely: $\phi_1: \xi \rightarrow \xi$ and $\phi_2: \xi \rightarrow \xi^2$.

Since $\phi_2^2=\phi_1$ $Gal(L/K)\simeq \mathbb{Z}_2$

I don't know a direct method to find the fixed fields (so if you have one let me know), but going according to the definition $L^G=\{ a \in L| \forall g \in G, g(a)=a \}$ we have only $\mathbb{Q}=L^G$, right?

For the second problem b) I have some difficulty finding the minimal polynomial over $L$. Can someone help me?

Answer 1

Let us observe that if $K$ is of characteristic $0$ then any extension $L/K$ is separable and hence we only need to check normality of the extensions. However a much simpler approach to verify whether an extension is Galois is to show that $L$ is a splitting field of some non-constant polynomial $f(x) \in K[x] $. And further when the extension $L/K$ is Galois then the fixed field $L^{G} =K$ where $G=\text{Gal} (L, K) $. Another ingredient which helps a bit in finding Galois groups is that the order of Galois group $G$ is same as the degree of extension ie $[L:K] $ if the extension is Galois.

With these ideas in mind one can handle the first problem easily by noting that $L$ is the splitting field of $f(x) =x^2+x+1\in K[x] $ and the extension is therefore Galois with the Galois group $G$ being of order $2$. Your solution presented in the question is fine. And since the extension is Galois the fixed field $L^G=K$.

Next for the second problem we need to deal with extension $L(a) /L$ where $a=\sqrt [6]{2}$. Another answer here gives us the degree of this extension $[L(a) :L] =6$. Now $a$ is a root of $g(x)=x^6-2\in L[x] $. The complete set of the roots of this polynomial is $$\pm a, \pm a\omega, \pm a\omega^2$$ where $\omega =e^{2\pi i/3}$. Since $\omega \in L$ it follows that $L(a) $ contains all roots of $g(x) $ and it is the smallest such field. Thus $L(a) $ is the splitting field of $g(x) $ over $L$ and the extension $L(a) /L$ is Galois and the order of the Galois group $H=\text{Gal}(L(a),L) $ is $6$ and the fixed field $L(a)^H=L$.

To find the Galois group we can observe that an automorphism $\sigma$ of $L(a) $ is entirely determined by its action on $a$ and further the image of $a$ ie $\sigma(a) $ must be a root of $g(x) $. So we clearly have two choices here and thus two obvious automorphisms $\sigma, \tau$ given by $$\sigma(a) =-a, \tau(a) =a\omega$$ and the group $H$ is generated by $ \sigma \tau$ and is the cyclic group of order $6 $.

Answer 2

Small disclaimer: This answer only deals with the minimal polynomial portion, it's more of a comment that couldn't quite fit.

Gerry mentions the Eisenstein criterion but you'd have to choose something other than $L$ itself since that's a field, which means it does not have any prime elements to check Eisenstein with.
Eisenstein could still work but only if you apply it to $\mathbb{Z}(e^{i2\pi/3})$ I'm fairly sure, at which point you'd probably have to know what prime elements look like in there... Though my algebra and Galois theory are a bit rusty so maybe someone will correct me on that.

Instead, here is a way to do it that only uses elementary properties of the degrees of field extensions.
Let me set $j := e^{i2\pi/3}$ (standard name for that number). We have: $$\begin{cases}\left[\mathbb{Q}(j,\sqrt[6]{2}) : \mathbb{Q}\right] = \left[\mathbb{Q}(\sqrt[6]{2})(j) : \mathbb{Q}(\sqrt[6]{2})\right] \cdot \left[\mathbb{Q}(\sqrt[6]{2}) : \mathbb{Q}\right]\\ \left[\mathbb{Q}(j,\sqrt[6]{2}) : \mathbb{Q}\right] = \left[\mathbb{Q}(j)(\sqrt[6]{2}) : \mathbb{Q}(j)\right] \cdot \left[\mathbb{Q}(j) : \mathbb{Q}\right]\end{cases}$$ The degrees $\left[\mathbb{Q}(\sqrt[6]{2})(j) : \mathbb{Q}(\sqrt[6]{2})\right]$, $\left[\mathbb{Q}(\sqrt[6]{2}) : \mathbb{Q}\right]$ and $ \left[\mathbb{Q}(j) : \mathbb{Q}\right]$ are easy to calculate, since $j$ is a root of $X^2 + X + 1$ but doesn't belong to $\mathbb{R} \supset \mathbb{Q}(\sqrt[6]{2})$ and we can use Eisenstein on $\mathbb{Z}$ to get that $\sqrt[6]{2}$'s minimal polynomial over $\mathbb{Q}$ is $X^6 - 2$ of degree $6$. Hence: $$\begin{cases}\left[\mathbb{Q}(j,\sqrt[6]{2}) : \mathbb{Q}\right] = 2 \cdot 6 = 12\\ 12 = \left[\mathbb{Q}(j)(\sqrt[6]{2}) : \mathbb{Q}(j)\right] \cdot 2\end{cases}$$ thus we obtain $\left[\mathbb{Q}(j)(\sqrt[6]{2}) : \mathbb{Q}(j)\right] = 6$, but $\sqrt[6]{2}$ is a root of $X^6 - 2 \in L[X]$ which is of degree $6$, therefore $X^6 - 2$ is the minimal polynomial of $\sqrt[6]{2}$ over $L$.