40 CHAPTER III. 



It will belong to the included field &F[5 3 ] if, and only if, <J 125 = (5. 

 Applying # 7 ^?1, we have (mod 5) 



5 5 



1 25 = 



* / = 1 



Applying 32), this becomes 



(b- c i) - V + fe- <i)a 8 + (c 4 - Oa 8 + (ft,- 

 The conditions that this shall be identical with 6 are 



c = 0, e 2 = c 6 , C 3 = C 4 (mod 5). 

 Hence the 5 3 marks of the 6r.F[5 3 ] are given by 



33) C Q + c. 2 (x* + x*) + c s (x* + x*) fa, c 2 ,c s = 0, 1, 2, 3, 4]. 

 Since (x 2 + x^ =X B + x*, 



we infer that % = x* + % 5 defines the GrF\5F\. In fact, we find 



T 5 = X 3 + X*, T= X + X*, T 80 EE X 5 -f X* + # 3 + ^ 2 ? 



and finally that T 31 = 1. Hence A = 2(a? 2 + a? 5 ) belongs to the exponent 

 4 31 and is therefore a primitive root in the 6r.F[5 3 ]. We derive 

 at once the P/$[3 ? 5] satisfied by A 7 viz., 



2A 3 -A 2 + A + 1 (mod 5). 



We next verify that x 2 belongs to the exponent 2 3 - 3 2 - 31, 

 so that Q = a? (re 2) belongs to the exponent 



5 6 - lEE2 3 -3 2 - 7-31, 

 so that Q is a primitive root in the 6r.F[5 6 ]. We have 



(a; - 2) 126 - (x 6 - 2)(x - 2) = - 2(ff -f ^ 6 ) = - 2r 25 . 



But T 25 belongs to the exponent 31. Hence the exponent of x 2 

 contains the factor 31 and, moreover, the factor 2 3 , since 



(* - 2)^ (5 '- I) -(* - 2) 126 '"=(- 2) M = - 1 (mod 5). 

 We next prove that the power 2 3 - 3 2 - 31 of x 2 gives unity. Indeed, 



(x - 2) 15 = (x 5 -2) 3 =2x*-x*+x + 2 (mod 5), 

 and, by a slight calculation, 



(x - 2) 18 = 2x 5 + a* + a 8 + 2# 2 + 4. 

 This being of the form 33), we have 



(x - 2)*' ' 3 *- [(x - 2)i*] m = 1 (mod 5). 

 For the same reason , 

 ( X - 2)2 3 -3-3i= [( X 



