Next: Case 2: p| XYZ Up: 2 Construction of cyclic Previous: 2 Construction of cyclic

## Case 1: p | XYZ

First of all we see easily that (X, Y, Z) are not all congruent modulo p. If not, we have

3X X + Y + Z Xp + Yp + Zp 0(mod p)

Now, we are assuming that p 5 and so we obtain X 0(mod p); this contradicts our hypothesis for Case 1. Secondly, we see that (X + Y) are mutually co-prime in R as j runs over 0,..., p - 1. If not, then we have a prime ideal P in R containing (X + Y, X + Y). Then this ideal P contains (1 - )Y. Now from the factorisation

(- Z)p = Xp + Yp = (X + Y)(X + Y) ... (X + Y)

we see that P contains Z. Hence, by the assumption that (X, Y, Z) are mutually co-prime we see that P contains (1 - ) for some 0 l p - 1. By the description of prime ideals in R as in section 1 we see that P = R. But then Z is a multiple of p which contradicts our hypothesis in Case 1. By the above paragraph and unique factorization of ideals we see that we have ideals Ij of R such that Ijp = (X + Y)R. Assume I1 is principal; then we have an equation

(X + Y) = u .

for some R and u a unit in R. Applying complex conjugation we obtain

(X + Y) = .

By the results mentioned in section 1 we have = u for some r. Moreover, is congruent to an integer modulo pR and hence is congruent to its own complex conjugate. Thus we obtain an equation

X + Y - X - Y 0(mod p)

Now it follows from the description of R given in Section 1 that it is a free abelian group with basis consisting of any (p - 1) elements of the set {1,,...,}. From this and the fact that X and Y are prime to p it follows that r = 1 and X Y(mod p). By similar reasoning interchanging the roles of Y and Z we can conclude that there is an ideal J1 such that J1p = (X + Z). Assuming J1 is principal we see by an argument like the one above that X Z(mod p). But as seen above the two congruences

X Y(mod p) andX Z(mod p)

contradict the hypothesis of Case 1. Hence, either I1 or J1 must be non-principal. But then by the principal result of Class Field theory as mentioned in section 1 we have required cyclic extension of K.

Next: Case 2: p| XYZ Up: 2 Construction of cyclic Previous: 2 Construction of cyclic
Kapil Hari Paranjape 2002-11-22