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

Case 1: p $ \not\!$| XYZ

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

3X $\displaystyle \cong$ X + Y + Z $\displaystyle \cong$ Xp + Yp + Zp $\displaystyle \cong$ 0(mod p)

Now, we are assuming that p $ \geq$ 5 and so we obtain X $ \cong$ 0(mod p); this contradicts our hypothesis for Case 1. Secondly, we see that (X + $ \omega^{j}_{}$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 + $ \omega^{j}_{}$Y, X + $ \omega^{k}_{}$Y). Then this ideal P contains (1 - $ \omega^{j-k}_{}$)Y. Now from the factorisation

(- Z)p = Xp + Yp = (X + Y)(X + $\displaystyle \omega$Y) ... (X + $\displaystyle \omega^{p-1}_{}$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 - $ \omega^{l}_{}$) for some 0 $ \leq$ l $ \leq$ p - 1. By the description of prime ideals in R as in section 1 we see that P = $ \lambda$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 + $ \omega^{j}_{}$Y)R. Assume I1 is principal; then we have an equation

(X + $\displaystyle \omega$Y) = u . $\displaystyle \alpha^{p}_{}$

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

(X + $\displaystyle \omega^{-1}_{}$Y) = $\displaystyle \overline{u}$ . $\displaystyle \overline{\alpha}^{p}_{}$

By the results mentioned in section 1 we have $ \omega^{r}_{}$$ \overline{u}$ = u for some r. Moreover, $ \alpha^{p}_{}$ is congruent to an integer modulo pR and hence is congruent to its own complex conjugate. Thus we obtain an equation

X + $\displaystyle \omega$Y - $\displaystyle \omega^{r}_{}$X - $\displaystyle \omega^{r-1}_{}$Y $\displaystyle \cong$ 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,$ \omega$,...,$ \omega^{p-1}_{}$}. From this and the fact that X and Y are prime to p it follows that r = 1 and X $ \cong$ 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 + $ \omega$Z). Assuming J1 is principal we see by an argument like the one above that X $ \cong$ Z(mod p). But as seen above the two congruences

X $\displaystyle \cong$ Y(mod p) andX $\displaystyle \cong$ 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 up previous
Next: Case 2: p| XYZ Up: 2 Construction of cyclic Previous: 2 Construction of cyclic
Kapil Hari Paranjape 2002-11-22