Splitting field of $x^{13}+1$ over $\mathbb{Q}$ I want to know how to calculate the degree of the field extension $[K:Q]$ where $K$ is the splitting field of $x^{13}+1$ over $\mathbb{Q}$.
I'm new to this area and this is not really covered in my course properly. So please don't assume I'm familiar to much when answering. 


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*Since $-1$ is a root should I conclude that all roots are $-1w^{n}$, where $w\in\mathbb{C}$ and $w^{13}=1$ or am I searching for the solutions to $x^{13}=-1 \in\mathbb{C}$, or is this just the same thing since $-1\in\mathbb{Q}$ already? 

*How do I go about finding solutions to these equations in $\mathbb{C}$? After finding solutions how do I know which are the minimal polynomials satisfying these? 
A lot of questions at the same time, but I don't really have anyone else to ask. Btw this is not a school assignment!
Are the roots $-w^{n}$ where $1\leq$n$\leq12$ and $w=\mathbb{e}^{\frac{2\pi}{13}i}$? And if we are searching for the $n$'th roots of unity when $n$ is composite why do we only include powers that are coprime to $n$?
 A: Let's first factor $f(x) = x^{13} + 1$ into irreducible factors. It's clear that $-1$ is a zero, and so the linear factor $x + 1$ divides $x^{13} + 1$.
Dividing by $x + 1$ gives $f(x) = x^{13} + 1 = (x + 1)(x^{12} - x^{11} + x^{10} - \dots + 1)$. 
Now take 
\begin{align*}
g(x) &= (x - 1)^{13} + 1 \\
&= \sum_{k = 0}^{13} \binom{13}{k} (-1)^{13 - k}x^k + 1 \\
&= x \sum_{k = 0}^{12} \binom{13}{k + 1} (-1)^{13 - (k + 1)}x^k
\end{align*}
Now when $k = 12$, then $\binom{13}{k + 1} = 1$, and for $0 \leq k < 12$, $13 \mid \binom{13}{k + 1}$, but $13^2 \nmid \binom{13}{1}$, and so by Eisenstein's criterion, the polynomial represented by the sum is irreducible, and so we have the factorization $g = xp$, with $p$ irreducible. Now since $f(x) = g(x + 1)$, we know that $f$ factors into $f = (x + 1)q$, with $q$ irreducible, and as we have already seen $q$ must equal
$(x^{12} - x^{11} + x^{10} - \dots + 1)$, which is thus irreducible.
Now take any zero of $\zeta$ of $q$, and consider the field $\mathbb{Q}(\zeta)$. Since $q$ is irreducible, the elements of the form $1,\zeta,\zeta^2,\dots,\zeta^{11}$ are linearly independent (for otherwise $\zeta$ would be a zero of a polynomial of smaller degree, contradicting the minimality of $q$), and since $\zeta^{12} = \zeta^{11} - \zeta^{10} + \dots - 1$, any higher power of $\zeta$ can be written as a linear combination in $1,\zeta,\zeta^2,\dots,\zeta^{11}$. This proves that $\mathbb{Q}(\zeta)$ has degree 12. Now for each $0 < k \leq 12$, take the elements
$$
\zeta_k = 
\begin{cases}
  \zeta^k & \text{when $k$ is odd} \\
  -\zeta^k & \text{when $k$ is even}
\end{cases}
$$
Clearly $\zeta_k^{13} = -1$, and these $\zeta_k$ must all be different (for otherwise, we must either have $\zeta_{12} = \zeta_0 = 1$, but we know that's not true, or otherwise, if $\zeta_i = \zeta_j$, with $i \neq j$, then $\zeta$ would be a zero of a polynomial of degree lower than 12, which contradicts the minimality of $q$). Since we now have 13 distinct zeros, we know that $\mathbb{Q}(\zeta)$ is the splitting field of $x^{13} + 1$, and as we have already seen the degree $[\mathbb{Q}(\zeta) : \mathbb{Q}]$ is 12.
A: Hint: given a primitive $26$th root of unity $\omega$, show that we would have $\omega^{13}=-1$. Then also $(\omega^3)^{13}=(-1)^3=-1$...
Can you see how to get the rest of the roots of $x^{13}+1$ this way? Do you see why being primitive is important?
If you can see that, then you can see why $\mathbb{Q}[\omega]$ splits $x^{13}+1$.
To see the degree of the extension, you have to determine the degree of the minimal polynomial for $\omega$ over $\mathbb{Q}$. Clearly the minimal polynomial divides $x^{13}+1$, so it's got degree less or equal to 13. In fact, since x+1 factors out of that, it has degree 12 or less.
