106 



and also 



f i^'i + h''2 + f;f 3 + f4*4 = 4. 

 From these two relations we see, remembering that /\ is divisible by the 

 square of a prime ideal (2), that the re(iuired factorization of p is either 



(p)=Ai-.A2.A3 

 where Ai, As, A;; are prime ideals of first degree, or 



(p) = Al2A2 



where Ai is of first degree and As of second degree. 

 Hence the factors of F(x) are either 



F(x)= |Pi(x)|' P^Cx). Pafx) (mod. p), 



where Pi(x), P2(x), P^Cx) are prime functions of first degree, or 



F(x) = {Pi(x)|' P,(x) 



where Pifx) is of first degree and P^(x) of second degree. 



In order to find the prime ideal factors of p we have thus to resolve F(x) 

 into its prime factors with re.-pect to the modulus p. To do this we set 

 X^ + ax2 + bx + c # (x + 1)2 (x^ + mr + n) (mod. p) 



= xi + (m + 21)x» + (n4-12 + 2ml)x2-f (ml2-f 

 21n)x+nl'. (mod. p.) 

 Hence, for determining 1, m, n we have the congruences 

 m -f 21 = o 1 



n-r2ml + l^ = a I 



> (mod. p). 

 mP + 21n = b i ^ ^ 



(r — s)2 = — 4 (a + 212) (mod. p). 



(1^ Hilbert, p. 201. (21 Hilbert, p. 195. 



