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本帖最后由 hbghlyj 于 2024-10-27 08:59 编辑 TTAANN001 发表于 2022-8-25 10:13$\DeclareMathOperator\Gal{Gal}$
证明方程$z^{n-1}+z^{n-2}+\dots+z+1=0$（也就是所谓的分圆方程）对于$n$为以上所提到的数的解（如果是复数，那就是复数的实部和虚部）可以只被加、减、乘、除、开平方所表达。
Constructibility of n-gon: relation between cyclotomic polynomial and field of constructible points
To construct the regular n-gon we just need to now for which $n\geq 3$, the central angle $2\pi/n$ is constructible. The connections with the unity roots and the complex plane comes from here, since $2\cos (\frac{2\pi}{n})=\zeta_n+\zeta_n^{-1}$. We will need a few lemmas that uses your two theorems.
Lemma 1: For every $r\geq 2$ the angle $\frac{2\pi}{2^r}$ is constructible.
Hint: use induction and the bisection of the angle formed by two lines that intersect.
Lemma 2: Let $m$ and $n$ dos positive coprime integers. The the angle $\frac{2\pi}{mn}$ is constructible if and only if $\frac{2\pi}{n}$ and $\frac{2\pi}{m}$ are both constructible.
Hint: Use the fact that if $\alpha$ and $\beta$ are constructible then $s\alpha+r\beta$ are constructible for $s,r\in\mathbb{Z}$. For the reciprocal just use this fact applied to Bezout's identity.
Lemma 3: Let $p$ a prime number greater than $2$, $k$ a positive integer and $n:=p^k$. Then the angle $\frac{2\pi}{n}$ is constructible if and only if $k=1$ and $p=2^{2^r}+1$ for some $r\geq 0$.
Proof:
Lets suppose that the angle $\frac{2\pi}{n}$ is constructible, i.e the real number $c_n=\cos(2\pi/n)$ is constructible, so it implies that $[\mathbb{Q}(c_n):\mathbb{Q}]=2^{l-1}$ for some $l\geq 1$. Since we know that $[\mathbb{Q}(c_n):\mathbb{Q}]=\frac{\phi(n)}{2}$ it implies that $$p^{k-1}(p-1)=\phi(n)=2[\mathbb{Q}(c_n):\mathbb{Q}]=2^l$$
In consequence, $k=1$ and $p=2^l+1$. If $l$ is not a power of $2$ then some odd $q$ would divide $l$ and then $l=qm$ for some even $m$. Since $q$ is odd
$$ t^q+1=(t+1)\sum_{j=0}^{q-1}(-1)^jt^j$$
and it implies
$$p=2^l+1=2^{mq}+1=(2^m+1)\sum_{j=0}^{q-1}(-1)^j2^{mj}$$
but that's impossible since $p$ is prime and $1<2^m+1<p$. Therefore exists some integer $r\geq 0$ such as $l=2^r$, and $p=2^{2^r}+1$.
Reciprocally, let's suppose that $p=2^l+1$ for some integer $l\geq 1$, and we have to prove that $c:=\cos (2\pi/p)$ is constructible. Let $s:=\sin (2\pi/p)$, $i:=\sqrt{-1}$, and consider $\zeta=c+is\in\mathbb{C}$. We know that $\zeta$ is an pth-root of unity, so $\mathbb{Q}(\zeta)|\mathbb{Q}$ is a Galois extension which Galois group is isomorphic to $\mathbb{Z}_{p}^*$, which order is $\phi(p)=p-1=2^l$. Also $2c=\zeta+\zeta^{-1}\in\mathbb{Q}(\zeta)$, so we have a field tower $\mathbb{Q}\subset \mathbb{Q}(c)\subset \mathbb{Q}(\zeta)$, and since the Galois group is cyclic, the subgroups are normal, so it follow from the second fundamental theorem of Galois theory that $$\Gal(\mathbb{Q}(c):\mathbb{Q})\cong\frac{\Gal(\mathbb{Q}(\zeta):\mathbb{Q})}{\Gal(\mathbb{Q}(\zeta):\mathbb{Q}(c))}$$
In particular, $\Gal(\mathbb{Q}(c):\mathbb{Q})$ is cyclic and it's order is $ord(\frac{\Gal(\mathbb{Q}(\zeta):\mathbb{Q})}{\Gal(\mathbb{Q}(\zeta):\mathbb{Q}(c))})=\frac{\phi(n)}{2}=2^{l-1}$. It follows from the Sylow theorems and the fact that is cyclic that for every $1\geq j\geq l-1$, $G(\mathbb{Q}(c):\mathbb{Q})$ has an unique normal subgroup $H_j$ with order $2^{l-1-j}$ and $H_{j+1}\subset H_j$. So we have a field tower
$$\mathbb{Q}=K_0\subset K_1\subset \cdots \subset K_{l-1}=\mathbb{Q}(c)$$ for $K_j=Fix(H_j)$ and every extension $K_{j+1}|K_j$ has degree $2$. It implies from the definition of constructibility (your second theorem), that $c$ is constructible
Theorem Let $n\geq 3$ a positive integer. Then the angle $\frac{2\pi}{n}$ is constructible if and only if n is a power of $2$ or $n:=2^r p_1\cdots p_k$ where $r$ is non negative and the primes $p_i$ are different two by two and they are of the form $p_j=2^{2^{r_j}}+1$ where $r_j\geq q$ for $1\leq j\leq k$.
Proof
From the lemma 1 the angle $\frac{2\pi}{2^l}$ is constructible for every integer $l\geq 1$. Suppose that $n:=2^r p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ for $r\geq 0$ and $\alpha_j\geq 1$. If the angle $\frac{2\pi}{n}$ is constructible it follows from lemma 2 it follows that every angle $\frac{2\pi}{p_j^{\alpha_j}}$ is constructible. It implies from lemma 3 that $\alpha_j=1$, $p_j=2^{2^{r_j}}+1$ for some positive integer $r_j$ for $1\leq j\leq k$.
Reciprocally, if $n:=2^r p_1\cdots p_k$ and the prime numbers $p_j$ verifies the conditions of the theorem, it follow from lemma 1 and 3 that the angles $\frac{2\pi}{2^r}$ and $\frac{2\pi}{p_j}$ are constructible. Applying the lemma 2 it concludes that the angle $\frac{2\pi}{n}$ is constructible. |
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色k
发表于 2022-8-25 08:07
hbghlyj
发表于 2024-10-27 03:59
