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=cos2πn+isin2πn。（式(7)，第10節）
事實：1的所有n次單位根有：R,R2,⋯,Rn。（第10節）
定理：1的n次單位根R,R2,⋯,Rn中，Rk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 ω and ω2.
2. For R given by (7), prove that the primitive nth roots of unity are
(i) for n=6, R,R5;
(ii) for n=8, R,R3,R5,R7;
(iii) for n=12, R,R5,R7,R11.
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,⋯,Rn are primitive with the exception of Rp,R2p,⋯,Rqp, whose qth power are unity, and Rq,R2q,⋯,Rpq, whose pth power are unity. These two sets of exceptions have only Rqp 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 R3,R5,R9,R12 are the primitive fifth roots of unity, and R5 and R10 are the primitive cube roots of unity. Show that their eight products by pairs give all the primitive 15th roots of unity.
8. If ρ is any primitive nth root of unity, prove that ρ,ρ2,⋯,ρn are distinct and give all the nth roots of unity. Of these show that ρk 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=cos2π3+isin2π3,ω=R=cos2π3+isin2π3,ω=R。由定理可知，本原單位根有R1,R2，亦即ω,ω2。
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 −(⌊pqp⌋+⌊pqq⌋−1)
= n−(q+p−1)
= pq−p−q+1
= (p−1)(q−1) 。
5. #(1的n=p2次本原單位根)
= #(不小於p2且與p2互質的數)
= p2− #(不小於p2且與p2不互質的數)
= p2− #(不小於p2，p的倍數或p2的倍數)
= p2−(⌊p2p⌋+⌊p2p2⌋−1)
= p2−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 −(⌊pqrp⌋+⌊pqrq⌋+⌊pqrr⌋−⌊pqrpq⌋−⌊pqrpr⌋−⌊pqrqr⌋+⌊pqrpqr⌋)
= pqr−(qr+pr+pq−r−q−p+1)
= (p−1)(q−1)(r−1) 。
7. R=cos2π15+isin2π15。
R3=cos2π5+isin2π5=(cos2π5+isin2π5)1,
R6=cos4π5+isin4π5=(cos2π5+isin2π5)2,
R9=cos6π5+isin6π5=(cos2π5+isin2π5)3,
R12=cos8π5+isin8π5=(cos2π5+isin2π5)4.
與5互質的數：1, 2, 3 ,4。
R5=cos2π3+isin2π3=(cos2π3+isin2π3)1,
R10=cos4π3+isin4π3=(cos2π3+isin2π3)2.
與3互質的數：1, 2。
R3×R5=R8,
R3×R10=R13,
R6×R5=R11,
R6×R10=R16=R1,
R9×R5=R14,
R9×R10=R19=R4,
R12×R5=R17=R2,
R12×R10=R22=R7,
與15互質的數：1, 2, 4, 7, 8, 11, 13, 14。
8. 因為ρ是本原單位根，所以存在正整數a滿足1≤a≤n且gcd(a,n)=1使得ρ=Ra，其中R=cos2πn+isin2πn。
於是{ρ,ρ2,⋯,ρn}={Ra,R2a,⋯,Rna}。任取兩相異正整數k,j滿足1≤k<j≤n。若ρj=ρk，即Rja=Rka，則有cos2jaπn+isin2jaπn=cos2kaπn+isin2kaπn。因此2jaπn與2kaπn為同界角，得j−k為n的倍數。然而1≤j−k≤n，所以j−k不可能是n的倍數。這意味著Ra,R2a,⋯,Rna完全相異。
因(ρk)n=(ρn)k=1k=1，所以{ρ,ρ2,⋯,ρn}是1的n次單位根。
假定ρk是本原根，由前文ρ與R的關係，得Rak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Rak是本原根，即ρk是本原根。
9. n=18，與18互質的數有1, 5, 7, 11, 13, 17。取R18=cos2π18+isin2π18。
R118=cos2π18+isin2π18=cosπ9+isinπ9,
R518=cos5π9+isin5π9,
R718=cos7π9+isin7π9,
R1118=cos11π9+isin11π9,
R1318=cos13π9+isin13π9,
R1718=cos17π9+isin17π9.
n=9，與9互質的數有1, 2, 4, 5, 7, 8。取R9=cos2π9+isin2π9。
R19=cos2π9+isin2π9,
R29=cos4π9+isin4π9,
R49=cos8π9+isin8π9,
R59=cos10π9+isin10π9,
R79=cos14π9+isin14π9,
R89=cos16π9+isin16pi9.
觀察幅角，有
10π9−π9=π,
14π9−5π9=π,
16π9−7π9=π,
11π9−2π9=π,
13π9−4π9=π,
17π9−8π9=π.
所以互為相反數。
