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1. If { — ^ — -!^ j = — 1, then x' -\- mx + n is irreducible and we liave 



F(x) = (x-1-1)2 (x2 + mx + n) (mod. p) 

 and hence 



(p) = (p, e + i)2(p, e2 + me + n). 



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



fr — s)2 = — 4 (a + 21^) (mod p) and we get r and a from the con- 

 gruences 



r -f s ^m ) 



_ \ (mod. p). 

 r — s ^ k J 



We have then 



F (x) = (x + \r- (x + r) (X + s) (mod. p) 



and hence 



(p) = (p, + 1)2 (p,0 + r) (p, e^-s). 



Case II. p =fc A- 

 In this case we have the two relations 



f,(ei-l) + fjea - 1) + f3(e3 - 1) + l,i^, -1)^0. 

 fjCj + faBa + fjeg + i^Gi^A. 

 Now since A is the only prime which is div^isible by the square of a prime 

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

 ing ways : 



1. (p) = Ai . Aa . A3 . A4 where Ai,A2,A3,A^ are all of 1st degree. 



2. (p)i= Aj . A2 . Ag where A^ is of 2d degree and AjjAj of Ist de- 



gree. 



3. (p) = Ai . A2 where A^ and A 2 are both of 2d degree. 



4. (p) = Ai . A2 where Aj is of 1st degree and A.^ of 3d degree. 



5. (p) = Ai where A^ 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) . P:!(x) (mod. p). 



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



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



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



