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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!

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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 $4$.

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  • $\begingroup$ Why did you use $e^{i\frac{5\pi}{3}}$ instead of $e^{i\frac{5\pi}{6}}$? $\endgroup$ – Guerlando OCs Sep 22 '17 at 2:28
  • $\begingroup$ $e^{i\frac{7\pi}{6}} = (e^{i\frac{5\pi}{6}})^2\cdot e^{-i\frac{3\pi}{6}}$, 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? $\endgroup$ – Guerlando OCs Sep 22 '17 at 2:47
  • $\begingroup$ I think I might have misunderstood what a splitting field is. I thought it was just a field with the roots adjoined. Does my reasoning and your answer work for the case where the degree to be calculated is not of the splitting field but instead is of the field with roots adjoined? $\endgroup$ – Guerlando OCs Sep 22 '17 at 3:12
  • $\begingroup$ @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. $\endgroup$ – Stefan4024 Sep 22 '17 at 12:05
  • $\begingroup$ @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. $\endgroup$ – Stefan4024 Sep 22 '17 at 12:09

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