重點提要
定義：若某個n次單位根所有正整數乘冪中，可化為1的最小次數為n，則稱該單位根為本原單位根(primitive)。
規約：$R = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$。（式(7)，第10節）
事實：1的所有n次單位根有：$R, R^2, \cdots, R^n$。（第10節）
定理：1的n次單位根$R, R^2, \cdots, R^n$中，$R^k$為本原單位根若且唯若指數k與n互質。
習題4
1. Show that the primitive cube roots of unity are $\omega$ and $\omega^2$.
2. For $R$ given by (7), prove that the primitive nth roots of unity are
(i) for $n = 6$, $R, R^5$;
(ii) for $n = 8$, $R, R^3, R^5, R^7$;
(iii) for $n = 12$, $R, R^5, R^7, R^{11}$.
3. When n is a prime, prove that any nth root of unity, other than 1, is primitive.
4. Let $R$ be a primitive nth root (7) of unity, where n is a product of two different primes p and q. Show that $R, \cdots, R^n$ are primitive with the exception of $R^p, R^{2p}, \cdots, R^{qp}$, whose qth power are unity, and $R^q, R^{2q}, \cdots, R^{pq}$, whose pth power are unity. These two sets of exceptions have only $R^{qp}$ in common. Hence there are exactly $pq - p - q + 1$ primitive nth roots of unity.
5. Find the number of primitive nth roots of unity if n is a square of a prime p.
6. Extend Ex. 4 to the case in which n is a product of three distinct primes.
7. If $R$ is a primitive 15th root (7) of unity, verify that $R^3, R^5, R^9, R^{12}$ are the primitive fifth roots of unity, and $R^5$ and $R^{10}$ are the primitive cube roots of unity. Show that their eight products by pairs give all the primitive 15th roots of unity.
8. If $\rho$ is any primitive nth root of unity, prove that $\rho, \rho^2, \cdots, \rho^n$ are distinct and give all the nth roots of unity. Of these show that $\rho^k$ is a primitive nth root of unity if and only if k is relatively prime to n.
9. Show that the six primitive 18th roots of unity are the negative of the primitive ninth roots of unity.
習題解答
1. $n = 3, R = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}, \omega = R = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}, \omega = R$。由定理可知，本原單位根有$R^1, R^2$，亦即$\omega, \omega^2$。
2. (i) 不小於6而與6互質的數有1, 5。
(ii) 不小於8而與8互質的數有1, 3, 5, 7。
(iii) 不小於12而與12互質的數有1, 5, 7, 11。
3. 不小於質數p而與質數p互質的數有$p - 1$個。
4. #(1的$n = pq$次本原單位根)
= #(不小於pq且與pq互質的數)
= n $-$ #(不小於pq且與pq不互質的數)
= n $-$ #(不小於pq，p的倍數或q的倍數)
= n $- \left( \lfloor {\frac{pq}{p}} \rfloor + \lfloor {\frac{pq}{q}} \rfloor - 1 \right)$
= $n - (q + p - 1)$
= $pq - p - q + 1$
= $(p - 1)(q - 1)$ 。
5. #(1的$n = p^2$次本原單位根)
= #(不小於$p^2$且與$p^2$互質的數)
= $p^2 -$ #(不小於$p^2$且與$p^2$不互質的數)
= $p^2 - $ #(不小於$p^2$，p的倍數或$p^2$的倍數)
= $p^2 - \left( \lfloor \frac{p^2}{p} \rfloor + \lfloor \frac{p^2}{p^2} \rfloor - 1 \right)$
= $p^2 - p - 1 + 1$
= $p(p - 1)$
6. 命$n = pqr$。
#(1的$n = pqr$次本原單位根)
= #(不小於pqr且與pqr互質的數)
= pqr $-$ #(不小於pqr且與pqr不互質的數)
= pqr $-$ #(不小於pq，p的倍數或q的倍數或r的倍數)
= pqr $- \left( \lfloor {\frac{pqr}{p}} \rfloor + \lfloor {\frac{pqr}{q}} \rfloor + \lfloor {\frac{pqr}{r}} \rfloor - \lfloor {\frac{pqr}{pq}} \rfloor - \lfloor {\frac{pqr}{pr}} \rfloor - \lfloor {\frac{pqr}{qr}} \rfloor + \lfloor {\frac{pqr}{pqr}} \rfloor \right)$
= $pqr - (qr + pr + pq - r - q - p + 1)$
= $(p - 1)(q - 1)(r - 1)$ 。
7. $R = \cos \frac{2\pi}{15} + i \sin \frac{2\pi}{15}$。
$R^3 = \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^1$,
$R^6 = \cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^2$,
$R^9 = \cos \frac{6\pi}{5} + i \sin \frac{6\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^3$,
$R^{12} = \cos \frac{8\pi}{5} + i \sin \frac{8\pi}{5} = \left( \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5} \right)^4$.
與5互質的數：1, 2, 3 ,4。
$R^5 = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} = \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^1$,
$R^{10} = \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} = \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^2$.
與3互質的數：1, 2。
$R^3 \times R^5 = R^8$,
$R^3 \times R^{10} = R^{13}$,
$R^6 \times R^5 = R^{11}$,
$R^6 \times R^{10} = R^{16} = R^1$,
$R^9 \times R^5 = R^{14}$,
$R^9 \times R^{10} = R^{19} = R^4$,
$R^{12} \times R^5 = R^{17} = R^2$,
$R^{12} \times R^{10} = R^{22} = R^7$,
與15互質的數：1, 2, 4, 7, 8, 11, 13, 14。
8. 因為$\rho$是本原單位根，所以存在正整數a滿足$1 \le a \le n$且$\gcd (a, n) = 1$使得$\rho = R^a$，其中$R = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$。
於是$\{ \rho, \rho^2, \cdots, \rho^n \} = \{ R^a, R^{2a}, \cdots, R^{na} \}$。任取兩相異正整數$k, j$滿足$1 \le k < j \le n$。若$\rho^j = \rho^k$，即$R^{ja} = R^{ka}$，則有$\cos \frac{2ja\pi}{n} + i \sin \frac{2ja\pi}{n} = \cos \frac{2ka\pi}{n} + i \sin \frac{2ka\pi}{n}$。因此$\frac{2ja\pi}{n}$與$\frac{2ka\pi}{n}$為同界角，得$j - k$為n的倍數。然而$1 \le j-k \le n$，所以$j - k$不可能是n的倍數。這意味著$R^a, R^{2a}, \cdots, R^{na}$完全相異。
因$\left( \rho^k \right)^n = \left( \rho^n \right)^k = 1^k = 1$，所以$\{ \rho, \rho^2, \cdots, \rho^n \}$是1的n次單位根。
假定$\rho^k$是本原根，由前文$\rho$與$R$的關係，得$R^{ak}$是本原根。再由定理可知$\gcd (ak, n) = 1$。因為$\gcd (a, n) = 1$，所以$\gcd (k, n) = 1$。
假定k與n互質，則$\gcd (ak, n) = 1$，故$R^{ak}$是本原根，即$\rho^k$是本原根。
9. $n = 18$，與18互質的數有1, 5, 7, 11, 13, 17。取$R_{18} = \cos \frac{2\pi}{18} + i \sin \frac{2\pi}{18}$。
$R_{18}^1 = \cos \frac{2\pi}{18} + i \sin \frac{2\pi}{18} = \cos \frac{\pi}{9} + i \sin \frac{\pi}{9}$,
$R_{18}^5 = \cos \frac{5\pi}{9} + i \sin \frac{5\pi}{9}$,
$R_{18}^7 = \cos \frac{7\pi}{9} + i \sin \frac{7\pi}{9}$,
$R_{18}^{11} = \cos \frac{11\pi}{9} + i \sin \frac{11\pi}{9}$,
$R_{18}^{13} = \cos \frac{13\pi}{9} + i \sin \frac{13\pi}{9}$,
$R_{18}^{17} = \cos \frac{17\pi}{9} + i \sin \frac{17\pi}{9}$.
$n = 9$，與9互質的數有1, 2, 4, 5, 7, 8。取$R_{9} = \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}$。
$R_{9}^1 = \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}$,
$R_{9}^2 = \cos \frac{4\pi}{9} + i \sin \frac{4\pi}{9}$,
$R_{9}^4 = \cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9}$,
$R_{9}^5 = \cos \frac{10\pi}{9} + i \sin \frac{10\pi}{9}$,
$R_{9}^7 = \cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9}$,
$R_{9}^8 = \cos \frac{16\pi}{9} + i \sin \frac{16pi}{9}$.
觀察幅角，有
$\frac{10\pi}{9} - \frac{\pi}{9} = \pi$,
$\frac{14\pi}{9} - \frac{5\pi}{9} = \pi$,
$\frac{16\pi}{9} - \frac{7\pi}{9} = \pi$,
$\frac{11\pi}{9} - \frac{2\pi}{9} = \pi$,
$\frac{13\pi}{9} - \frac{4\pi}{9} = \pi$,
$\frac{17\pi}{9} - \frac{8\pi}{9} = \pi$.
所以互為相反數。
