Proposition 6.3


(1) Let where F and are reduced and . Then by the formula proved above, there exists such that . As before, we deduce from the inequalities |Q|<P<R and that , and hence that M is an automorphism of . The proof then terminates as in the real case, except that there are no special cases to consider since the forms are irreducible.

(2) As in the real case, a complex reducible form F belonging to U must be equivalent to or to with . If and with , then for some . Hence the reduced form is equivalent to a multiple of with or respectively which are also reduced, and hence it is equal to that multiple, hence we have either Q=0 or Q=P for the form F, which are both excluded from the definition of a reduced form in the complex case.

(3) is clear: we can always reduce first by an element of so that it satisfies 0<|Q|<P<R, the strict inequalities being guaranteed by the irreducibility of F. We must have otherwise F is reducible, and since P=a, we have a>0. Changing into if necessary (which changes Q into -Q and leave P and R unchanged), we may also assume that . Finally, if b=0, changing again into of necessary we may assume d>0 since .