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本帖最后由 hbghlyj 于 2024-5-9 22:06 编辑 $\cos\frac{2\pi}n$的极小多项式的次数
cs.umd.edu/~gasarch/BLOGPAPERS/gov.pdf
The result for sine can be derived from the result for cosine using the complementary angle formula
\[
\sin \left(360 \frac{k}{n}\right)=\cos \left(360 \frac{k}{n}-90\right)=\cos \left(360 \frac{4 k-n}{4 n}\right) .
\]
This part of the proof is identical to the one on p. 38 of Niven's book, so we just sketch it. First, observe that since $n$ is not 4 , the fraction $(4 k-n) /(4 n)$, when put in lowest terms, has denominator at least 3, so we may indeed apply the cosine result. Next, divide the proof into cases depending on the highest power of 2—say $2^e$—dividing $n$. For example, if $e=1$, then $(4 k-n) /(4 n)$ in lowest terms has denominator $2 n$, and $\sin (\theta)$ has degree $\phi(2 n) / 2$. But $n=2 m$ for some $m$ odd, so
\[
\frac{\phi(2 n)}{2}=\frac{\phi(4 m)}{2}=\frac{\phi(4) \phi(m)}{2}=\phi(m)=\phi(2) \phi(m)=\phi(n) .
\]
The remaining cases are left to the reader. This completes the proof of Lehmer's Theorem.
For later use, we note that $\mathbb{Q}(z)$ is Galois over $\mathbb{Q}$ with abelian Galois group, as we learned in our Galois theory course. Consequently, every subfield of $\mathbb{Q}(z)$ is also Galois over $\mathbb{Q}$ with abelian Galois group, including $\mathbb{Q}(\cos \theta)$ and any subfield of it.
Remark. The proof of the cosine part of Lehmer's Theorem amounts to the observation that $\mathbb{Q}(\cos \theta)$ is the maximal real subfield of $\mathbb{Q}(z)$, i.e., the intersection of $\mathbb{Q}(z)$ with the real numbers. |
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