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Let $w$ be primitive nth root of unity over $\mathbb{Q}$. Find the minimal polynomial of $w+w^{-1}$ over $\mathbb{Q}$

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Note that $w+w^{-1} = 2 \cos(2\pi/n)$. A minimal polynomial can be recursively found using trigonometric identities: $1=\cos(2\pi) = \cos(n \cdot 2\pi/n) = \dotsc$ –  Martin Brandenburg Oct 28 '13 at 14:42

2 Answers 2

What happens here is similar to what happens with the polynomial $M_w$ of $w$ : there is no general formula, the answers depends on the prime factorization of $n$ (for example, its degree is $\phi(n)$). In general, we only know that $M_w$ divides $X^{n-1}+X^{n-2}+\ldots +1$, and $M_w$ coincides with this polynomial when $n$ is prime.

Similarly, if $n$ is odd say $n=2m+1$, we may define

$$ T_m=\Bigg(\sum_{j=0}^{\lfloor \frac{m}{2} \rfloor} \binom{m-j}{j} (-1)^j x^{m-2j}\Bigg) + \Bigg(\sum_{j=0}^{\lfloor \frac{m-1}{2} \rfloor} \binom{m-1-j}{j} (-1)^j x^{m-1-2j}\Bigg) \tag{1} $$

Then, we have the algebraic identity

$$ T_m\bigg(w+\frac{1}{w}\bigg)=\frac{1}{w^m}\sum_{i=0}^{2m} w^i \tag{2} $$

which shows that $T_m$ annihilates $v=w+\frac{1}{w}$, so the minimal polynomial $M_v$ of $v$ divides $T_m$. When $n$ is an odd prime we have $M_v=T_m$.

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furthermore, how can i calculate the degree of following extension [$\mathbb{Q}(w+w^{-1}):\mathbb{Q}$] –  Le Van Tu Oct 28 '13 at 15:07
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@LeVanTu This degree is $\frac{\phi(n)}{2}$ since the degree $[{\mathbb Q}(w):{\mathbb Q}(v)]$ is equal to $2$ ; indeed $w$ satisfies $w^2-vw+1=0$ –  Ewan Delanoy Oct 28 '13 at 15:19
    
thank you.But i wonder how can you determine polynomial $T_m$? –  Le Van Tu Oct 28 '13 at 15:34

The minimum polynomial satisfied by $w$ is palindromic. Making the substitution $z=w+w^{-1}$ is a standard method of treating such polynomials, since it divides the degree of the polynomial by half. Look up palindromic polynomials.

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