- Two equivalent reduced complex cubic forms are equal.
- A reduced complex cubic form belonging to
Uis irreducible.- Any irreducible complex cubic form is equivalent to a unique reduced form.

(1) Let whereFand are reduced and . Then by the formula proved above, there exists such that . As before, we deduce from the inequalities|Q|<P<Rand that , and hence thatMis 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

Fbelonging toUmust 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 eitherQ=0orQ=Pfor the formF, 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 ofF. We must have otherwiseFis reducible, and sinceP=a, we havea>0. Changing into if necessary (which changesQinto-Qand leavePandRunchanged), we may also assume that . Finally, ifb=0, changing again into of necessary we may assumed>0since .