14 CHAPTER II. 



the equation W n (x) = completely decomposes in the F'[jp n ~\. Any 

 one of its roots w' is a primitive root in the F'|j9 w ]. Indeed, by its 

 definition, W n (x) does not divide x e x in the F\jp] for e <p n . The 

 powers of w' therefore give all the marks of the F 1 [$> n '\. Hence 

 F[p n ~\ and F 1 \j) n ~\ are abstract forms of the same Galois Field. These 

 results, first proven by Moore (loc. cit), may be stated as follows: 

 Theorem. Every existent field of finite order s may be represented 

 as a Galois Field of order s=p n . The 6r-F[j} n ] is defined uniquely 

 by its order; in particular, it is independent of the special irreducible 

 congruence used in its construction. 



* CHAPTEE tt 



PROOF OF THE EXISTENCE OF THE GF[p*] FOR EVERY 

 PRIME p AND INTEGER m. 



19. The next step is to prove the existence, for every prime 

 number p and positive integer w, of a congruence of degree m irre- 

 ducible modulo p 7 from which will follow the existence of the GF[p m ~\. 

 We will, however, make a more general investigation, taking as our 

 basis a fixed GF[p n ~] (in its abstract form), whose existence is supposed 

 known. We will prove that functions belonging to and irreducible 

 in the GF\j)*~\ exist for every integer m and will determine their 

 number. Since the GF[p\, the field of integers taken modulo p, is 

 known to exist, we shall have proven (taking n = 1) the existence, 

 for every value of m, of functions belonging to and irreducible in 

 the GF[p\, i. e., irreducible modulo p. 



At the same time, we shall have deduced some important pro- 

 perties of the GF[p* m ~\ with respect to the included field, the GF[jp*]. 



20. Theorem. - - If two functions F(x) and P(x) belonging to 

 the G-F[p n '\ have in the field no common divisor containing x, we can 

 determine two functions F'(x) and P\x), belonging to the G-F[p n ~] 

 such that F? ^ . F( ^ _ p ,^ t p ^ _ L 



The proof is quite analogous to that of 7. 



21. Theorem. - - J/J in the GF\p n ~\, P\x] lias no factor invol- 

 ving x in common with F(x) but divides the product F(x) F(x), then P(x) 

 divides E(x) in the G-F^]. 



Indeed, by multiplying the given equation 

 E(x) F(x) = P(x) - S(x) 

 by F'(x), determined as in 20, we find 



E(x) - P(x)[8(x) - F'(x) - E(x) - P'(xj]. 



