Find the degree of the splitting field of $x^6+1$ over $\mathbb{Q}$
First I tried to find the roots of $x^6+1$ over $\mathbb{C}$ so I can create the splitting field:
$$x^6+1 = (x^3-i)(x^3+i)$$
but this strategy showed some complications like in Find the degree of the splitting field of $x^4 + 1$ over $\mathbb{Q}$
So I guess the goal is to find the roots in terms of $e^{\mbox{something}}$. It's easy to see that $i$ is a root of this equation, so we just convert $i$ to polar form involving $e$ like this: $i = e^{\frac{\pi}{2}}$ by thinking in the complex plane and where the $i$ vector points. Now, to find the other $5$ roots, we just add $\frac{n2\pi}{6}$ for $n=1,2,3,4,5$, or we just say that the $6$ roots are: $e^{i\left(\frac{\pi}{2} + n\frac{2\pi}{6}\right)}, n=0,1,2,3,4,5$.
We know that each root has a negative version, so they form pairs. We can just take the first $3$ representants, that is: $e^{i\left(\frac{\pi}{2} + n\frac{2\pi}{6}\right)}, n=0,1,2$, the others are the negative versions of it.
We must find $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}, \pm e^{i\frac{5\pi}{6}}, \pm e^{i\frac{7\pi}{6}}\right):\mathbb{Q}\right]$. The idea is to find $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}, \pm e^{i\frac{5\pi}{6}},\pm e^{i\frac{7\pi}{6}} \right):\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}, \pm e^{i\frac{5\pi}{6}}\right)\right]$, $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}, \pm e^{i\frac{5\pi}{6}}\right):\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}\right)\right]$ and $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}\right):\mathbb{Q}\right]$
So we can multiply and use the theorem of multiplicity of degrees to find the main degree.
In order to find the degree of each of those extensions, I must verify the basis of each one of the bigger extension over the smaller. For example, in $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}\right):\mathbb{Q}\right]$, the elements are going to be of the form $a + be^{i\frac{\pi}{2}} + ce^{i\frac{-\pi}{2}}$ for $a,b,c\in\mathbb{Q}$ (the $c$ coefficient is the inverse of $e^{i\frac{\pi}{2}}$). I should now see if repeated multiplication of $e^{i\frac{\pi}{2}}$ gives me its inverse, which doesn't, so the basis is $a + be^{i\frac{\pi}{2}} + ce^{i\frac{-\pi}{2}}$ and $\left[\mathbb{Q}\left(\pm e^{i\frac{\pi}{2}}\right):\mathbb{Q}\right]=3$? By using this reasoning I guess the other degress would be greater than $1$ and the main degree would be over $6$, what I think is impossible.
So… am I doing something wrong or am I solving it in a hard way?
Even if you have a better way to solve it, like proving $x^6+1$ is the irreducible polynomial that contains all the $6$ roots, could you tell what is wrong with my reasoning? Thank you!
Best Answer
First of all one of the mistakes you make is that you assume $\mathbb{Q}(e^{\frac{\pi i}{2}},e^{\frac{5\pi i}{3}})$ doesn't contain $e^{\frac{7\pi i}{3}}$. In fact we have that $\mathbb{Q}(e^{\frac{\pi i}{2}},e^{\frac{5\pi i}{3}}) = \mathbb{Q}(e^{\frac{\pi i}{2}},e^{\frac{\pi i}{3}})$ and it's not hard to see that this field contains all the roots of $x^6 + 1$. To see that it's a splitting field note that none of it's subfield contain all root. This can be observed by the fact that the splitting field must contain $e^{\frac{\pi i}{2}}$ and $e^{\frac{\pi i}{3}}$
Another mistake you make is to assume that elements of $\mathbb{Q}(e^{\frac{\pi i}{2}})$ are of the form $a + be^{\frac{\pi i}{2}} + ce^{-{\frac{\pi i}{2}}}$ when in fact they are of the form $a + be^{\frac{\pi i}{2}}$. In fact for any field $Q(\alpha)$ the elements are given by $\sum a_i\alpha^i$. In you case you have $\left(e^{\frac{\pi i}{2}}\right)^2 = e^{i\pi} = -1$. This will lead you to conclude that $[\mathbb{Q}(e^{\frac{\pi i}{2}}):\mathbb{Q}]=2$. In fact the minimal polynomial of $e^{\frac{\pi i}{2}}=i$ is $x^2+1$ of degree 2.
Now the minimal polynomial of $e^{\frac{\pi i}{3}}$ over $\mathbb{Q}(e^{\frac{\pi i}{2}})$ is the cyclotomatic polynomial of sixth order, i.e. $x^2 - x + 1$, hence $[\mathbb{Q}(e^{\frac{\pi i}{2}},e^{\frac{\pi i}{3}}):\mathbb{Q}]=4$ and the splitting field has degree $4$.