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1. If (- j^ — 1, then x--|- mx + n is irreducible and we have 



F(x) = (x-|-l)2(x2 + mx + n) (mod. p) 



and hence 



(P) = (P, e4-l)-(P, e2 + me + n). 



2. If f ~^^"^ -) = + 1, then let r — s = k be a solution of 



(r — s)- = — 4 (a +21-) (mod p) and we get r and s from the con- 

 gruences 



r+s=m| 



_ \ (mod. p). 

 r — s = k J 



We have then 



F(x) = (x + 1)2 (x + r) (x + s) (mod. p) 

 and hence 



(p) = (p, + l)'(p, + r) (p, e+s). 



Case //. p rh A- 

 In this case we have the two relations 



f:(ei- 1) + f2(e, - 1) + f,,(e., - 1) -f f,(e, - 1) = 0. 



Now since A is the only prime which is divisible by the square of a prime 

 ideal, the relations given above show that p can be factored in one of the follow- 

 ing ways : 



A4 where AijAjjAjjA^ are all of 1st degree. 



where Aj is of 2d degree and A 2, A3 of Ist de- 

 gree, 

 where Aj and Aj are both of 2d degree, 

 where Ai is of 1st degree and Aj of 3d degree, 

 where Aj is of 4th degree, in which case (p) is 

 a prime ideal. 

 Hence F(x) can be factored in one of the following ways : 



1. F(x) = Pi(x) . P2(x) . P3(x) . P4(x) (mod. p). 



2. F(x) = Pi(x) . P2(x) . P3(x) (mod. p). 



3. F(x) = Pi(x) . P2(x) (mod. p). 



4. F(x) = Pi(x) . P2(x) (mod. p). 



5. F^x) = Pl(x) (mod. p). 



where Pi(x) is a prime function of the same degree as the corresponding A . 



