** Next:** Case 2: p| XYZ
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First of all we see easily that (*X*, *Y*, *Z*) are not all congruent
modulo *p*. If not, we have
3

*X* *X* +

*Y* +

*Z* *X*^{p} +

*Y*^{p} +

*Z*^{p} 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} =

*X*^{p} +

*Y*^{p} = (

*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 *I*_{j} of *R* such that
*I*_{j}^{p} = (*X* + *Y*)*R*. Assume *I*_{1} is principal; then we have an equation
(

*X* +

*Y*) =

*u*^{ . }
for some
*R* and *u* a unit in *R*. Applying complex
conjugation we obtain
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
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 *J*_{1} such that
*J*_{1}^{p} = (*X* + *Z*).
Assuming *J*_{1} 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*) and

*X* *Z*(mod

*p*)

contradict the hypothesis of Case 1. Hence, either *I*_{1} or *J*_{1} 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