Finite Field Problems: Solutions

Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The fundamental equation is α 2 + 1 = 0, hence α 2 = 1. So α 3 = α and α 4 = 1. Thus, α =4. Solution: (α +1) 2 = α 2 +2α +1=( 1)+2α +1=2α. (α +1) 3 =(α +1) 2 (α +1)=2α(α +1)=2α 2 +2α =2α 2. (α +1) 4 =(2α) 2 =4α 2 = 4. (α +1) 5 = 4(α +1)= 4α 4. (α +1) 6 = 4 2α = 8α. (α +1) 8 =( 4) 2 =16=5. (α +1) 10 =5 2α = α. Thus, (α +1) 20 =( α) 2 = 1, and (α +1) 40 =( 1) 2 =1,So α +1 divides 40. But we ve tested every divisor of 40 already, and of them, only 40 is an exponent for α + 1. Thus, α + 1 has order 40. d) What is the order of (α +1) 392 in F? Solution: (α +1) 392 = 40 = 40 =5. (40,392) 8 e) What is the order of α +2inF? Solution: (α +2) 2 = α 2 +4α +4=( 1)+4α +4=4α +3, so (α +2) 3 = α 3 +6α 2 +12α +8=( α)+6( 1)+12α +8 =11α +2=2 Here, we ve used that α 3 = α 2 α = α and that 11 = 0. It is easy to see that 2 has order 10 in Z 11, as2 5 =32= 1. Thus, (α +2) 30 =((α +2) 3 ) 10 =2 10 =1. Thus, the order of α + 2 divides 30. Moreover, if (α +2) 3k =1, then 1 = (α +2) 3k =2 k,sok must be divisible by 10. So

2 Finite Field Problems: Solutions the only divisors of 30 we have not checked are the divisors of 30 = 10. But 3 (α +2) 10 =(α +2) 9 (α +2)=2(α +2)=2α +4 1, so 30 is the smallest positive exponent of α+2, and hence α+2 has order 30. 2. Let f = x 2 +1 Z 19 [x] and let F = Z 19 [x]/(f), a field. Let a) What is the order of α in F? Solution: The fundamental equation is α 2 + 1 = 0, hence α 2 = 1. So α 3 = α and α 4 = 1. Thus, α =4. b) What is the order of α +1inF? Solution: Here, the order must divide 19 2 1 = 360. We have: (α +1) 2 = α 2 +2α +1=( 1)+2α +1=2α. (α +1) 3 =(α +1) 2 (α +1)=2α(α +1)=2α 2 +2α =2α 2. (α +1) 4 =(2α) 2 =4α 2 = 4. (α +1) 5 = 4(α +1)= 4α 4. (α +1) 6 = 4 2α = 8α. (α +1) 8 =( 4) 2 =16= 3. (α +1) 9 = 3(α +1)= 3α 3. (α +1) 10 =5 2α =10α. (α +1) 12 =( 8α) 2 = 64 = 12. (α +1) 15 = 12(2α 2)=24α 24=5α 5. (α +1) 18 = 12( 8α) = 96α = α. Thus, (α +1) 36 =( α) 2 = 1, and (α +1) 72 = 1, so the order of α + 1 divides 72. The only divisor of 72 we haven t checked is 24. (α +1) 24 =12 2 = 144 = 11, so α + 1 has order 72. 3. Let f = x 2 x +1 Z 5 [x] and let F = Z 5 [x]/(f), a field. Let Solution: The possible orders are the divisors of 5 2 1 = 24, or 1, 2, 3, 4, 6, 8, 12, 24.

Finite Field Problems: Solutions 3 Solution: The fundamental equation is α 2 α +1=0. so α 2 = α 1. Thus, α 3 = α(α 1) = α 2 α =(α 1) α = 1. Thus, α 6 = 1. Since we ve already tested all divisors of 6, α has order 6. Solution: We have (α +1) 2 = α 2 +2α +1=(α 1)+2α +1=3α. (α +1) 3 =3α(α +1)=3α 2 +3α =3(α 1)+3α =6α 3=α 3. (α +1) 4 =(3α) 2 =9(α 1) = α +1. (α +1) 6 =(α 3) 2 = α 2 6α +9=(α 1) α 1= 2. (α +1) 8 = 2(3α) = α. (α +1) 12 =( 2) 2 = 1. Thus, (α +1) 24 = 1, and 24 is the lowest positive exponent for α +1,soα + 1 has order 24 (and hence is a primitive element of F). 4. Let f = x 2 x 1 Z 7 [x] and let F = Z 7 [x]/(f), a field. Let Solution: The order must divide F = F 1= Z 7 deg f 1 = 48. So the possible orders are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

4 Finite Field Problems: Solutions Solution: The fundamental equation is α 2 α 1=0,so α 2 = α +1. Thus, α 3 = α 2 + α =(α +1)+α =2α +1 α 4 =2α 2 + α =2(α +1)+α =3α +2 α 5 =3α 2 +2α =3(α +1)+2α =5α +3 α 6 =5α 2 +3α =5(α +1)+3α = α +5 α 7 = α 2 +5α =(α +1)+5α = α +1 α 8 = α 2 + α = (α +1)+α = 1 Thus, α 16 = 1. Since we ve tested all divisors of 16, α has order 16. Solution: Here, α +1=α 2,so o(α +1)= o(α) (o(α), 2) = 16 (16, 2) =8. 5. Let f = x 2 2 Z 13 [x] and let F = Z 13 [x]/(f), a field. Let Solution: The possible orders are the divisors of 13 2 1= 168=4 42=8 3 7. Solution: The fundamental equation is α 2 2=0,soα 2 =2. But it s easy to see that 2 has order 12 in Z 13: 2 4 =16=3,so 2 6 =3 2 2 =12= 1. Since any proper divisor of 12 divides either 4 or 6, 12 is the smallest positive exponent of 2. Thus, α 24 = (α 2 ) 12 = 2 12 = 1, so the order of α divides 24. Also, since α 2k = 2 k, no even, proper divisor of 24 is an exponent for α. The only odd divisor of 24 is 3, and α 3 = α 2 α =2α 1. Thus, α has order 24.

Finite Field Problems: Solutions 5 Solution: (α +1) 2 = α 2 +2α +1=2+2α +1=2α +3. (α +1) 3 =(2α + 3)(α +1)=2α 2 +5α +3=5α +7. (α +1) 4 =(2α +3) 2 =4α 2 +12α +9= α +4. (α +1) 6 =(5α +7) 2 =25α 2 +70α +49=70α +99=5α +8. (α +1) 7 =(5α + 8)(α +1)=5α 2 +13α +8=18=5. Thus, (α +1) 14 =5 2 =25= 1, and (α +1) 28 = 1. Since we ve tested all divisors of 28, α + 1 has order 28. 6. Let f = x 4 x 1 Z 3 [x] and let F = Z 3 [x]/(f), a field. Let Solution: The possible orders are the divisors of 3 4 1 = 80. b) What is the order of α +1inF? Solution: The fundamental equation is α 4 α 1=0,so α 4 = α+1. The elements of F are in one-to-one correspondence with the polynomials in α of degree 3, so the latter cannot be simplified. Since α 4 = α+1, α 5 = α 2 +α, and α 6 = α 3 +α 2. As we ve just seen, these can t be simplified further. Also, the lowest power of α+1 that could be meaningfully simplified is the 4-th power. We use the binomial theorem: (α +1) 4 = α 4 +4α 3 +6α 2 +4α +1=(α +1)+α 3 +0α 2 + α +1 = α 3 +2α +2. (α +1) 5 = α 5 +5α 4 +10α 3 +10α 2 +5α +1 =(α 2 + α) (α +1)+α 3 + α 2 α +1 = α 3 α 2 α. We now multiply out the square of this last term, obtaining (α +1) 10 = α 6 + α 5 α 4 α 3 + α 2 =(α 3 + α 2 )+(α 2 + α) (α +1) α 3 + α 2 =3α 2 1= 1.

6 Finite Field Problems: Solutions Thus, (α +1) 20 = 1, and the order divides 20. Since we ve tested all the divisors of 20 and 20 is the lowest exponent among them, α + 1 has order 20. c) What is the order of α in F? Solution: The fundamental equation gives α 4 = α+1. Thus, we just calculated the order of α 4. We know that 20 = o(α 4 )= o(α) (o(α), 4), so o(α) =20 (o(α), 4). Since 4 divides 20, 4 divides o(α). Write o(α) = 4k. Then 1=α 4k =(α 4 ) k =(α +1) k, so k must be divisible by o(α + 1) = 20. So 4k is divisible by 80, hence α has order 80, and is a primitive element of F.