Find the degree of the splitting field of $x^6+1$ over $\mathbb{Q}$

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!

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 $4$.
• Why did you use $e^{i\frac{5\pi}{3}}$ instead of $e^{i\frac{5\pi}{6}}$? – Guerlando OCs Sep 22 '17 at 2:28
• $e^{i\frac{7\pi}{6}} = (e^{i\frac{5\pi}{6}})^2\cdot e^{-i\frac{3\pi}{6}}$, where $e^{-i\frac{3\pi}{6}} = e^{-i\frac{\pi}{2}}$, that's what I did wrong in the first part. In the second, I forgot to see that $(e^{\frac{\pi}{2}})^3 = e^{\frac{-\pi}{2}}$. Thank you :). But I didn't get the argument of the cyclotomatic polynomial. Can't I calculate the degree by finding a basis? Where did you find this polynomial? – Guerlando OCs Sep 22 '17 at 2:47
• @GuerlandoOCs For the first question note that the roots of unity are given by $e^{\frac{2k\pi i}{n}}$, when you plug in $n=6$ you get the wanted form. – Stefan4024 Sep 22 '17 at 12:05
• @GuerlandoOCs For the second one of course you can find the basis, as you have that $\left(e^{\frac{\pi i}{3}}\right)^2 = -1 + e^{\frac{\pi i}{3}}$, but this is effectivelly same as finding the minimal polynomial. In fact you have to prove that $\left(e^{\frac{\pi i}{3}}\right)^2$ can be written as a linear combination of $1$ and $e^{\frac{\pi i}{3}}$, otherwise you have to include it into the basis. For the cyclotomatic polynomial note that $e^{\frac{\pi i}{3}}$ is the sixth root of unity and the cyclotomatic polynomial of order $6$ is the minimal polynomial for it. – Stefan4024 Sep 22 '17 at 12:09