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本帖最后由 hbghlyj 于 2024-5-9 22:06 编辑 $\sin\frac{2\pi}n$的极小多项式的次数
cs.umd.edu/~gasarch/BLOGPAPERS/gov.pdf
Now that we understand how to apply Lehmer's Theorem, we prove it. Following the stated goal of this paper, we use Galois theory. A proof that appears more concrete but is really the same can be found in Section 3.4 of [11]. We write $\theta$ for $360 k / n$.
Put
\[
z:=e^{2 \pi i k / n}=\cos \theta+i \sin \theta,
\]
a primitive $n$th root of unity in the complex numbers. Then
\[
\cos \theta=\frac{z+\bar{z}}{2},
\]
where $\bar{z}$ denotes the complex conjugate of $z$. Since $k$ and $n$ are relatively prime, there is some natural number $\ell$ such that $k \ell$ is congruent to $n-k(\bmod n)$. Therefore,
\[
\bar{z}=e^{-2 \pi i k / n}=\left(e^{2 \pi i k / n}\right)^{\ell}=z^{\ell} .
\]
In particular, $\bar{z}$ belongs to $\mathbb{Q}(z)$. (Alternatively, we can remember the fact that $\mathbb{Q}(z)$ is a Galois extension of $\mathbb{Q}$.) Now $z$ satisfies the equation
\[
z^2-2 z \cos \theta+1=0
\]
because $z \bar{z}=1$, so the dimension of $\mathbb{Q}(z)$ over $\mathbb{Q}(\cos \theta)$ is 1 or 2. On the other hand, these two field extensions are not the same: $z$ cannot belong to $\mathbb{Q}(\cos \theta)$ because $z$ is not a real number (because $n$ is at least 3 ) and $\mathbb{Q}(\cos \theta)$ consists of real numbers. That is, the dimension $[\mathbb{Q}(z): \mathbb{Q}(\cos \theta)]$ is 2. Since the dimension of $\mathbb{Q}(z)$ over $\mathbb{Q}$ is $\phi(n)$ —see e.g. [2, §13.6, Cor. 42]—we obtain:
\[
[\mathbb{Q}(\cos \theta): \mathbb{Q}]=\frac{[\mathbb{Q}(z): \mathbb{Q}]}{[\mathbb{Q}(z): \mathbb{Q}(\cos \theta)]}=\frac{\phi(n)}{2} .
\]
This proves the part of Lehmer's Theorem regarding cosine. |
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