# 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.

1. 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?

2. 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 $n$?

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Look up (complex) roots of unity, cyclotomic polynomials, and show that $K=\mathbb{Q}(e^{\pi i/13})$. – Jyrki Lahtonen Aug 13 '12 at 17:55
What do you mean by "we only include powers that are coprime to $n$"? – tomasz Aug 13 '12 at 18:29

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.

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I downvoted ’cause it looks incomplete to me. – Lubin Aug 13 '12 at 21:51
@Lubin If it were a complete solution that would hardly be a hint, now would it. Not that I care very much about your vote, but I would recommend you go read some guidelines on downvoting. I don't think the reason you gave falls under any of them. – rschwieb Aug 13 '12 at 23:23

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.

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I would have replaced $f(x)=x^13+1$ by $F(x)=-f(-x)$ to get a polynomial whose roots are the $13$-th roots of $1$, avoiding some complications. But otherwise, this is just the way I would hve done it. +1. – Lubin Aug 13 '12 at 21:49