(1) Let F and 
be two reduced cubic forms and 
such
that 
. Then 
and since 
and 
are
reduced, it follow from Lemma 5.2 that 
and that M is an
automorphism of 
. We now have only a finite number of possibilities
(exactly 16) to test for M. Let 
.
First, if 
, we have 
hence the conditions a>0 on
F and 
imply M=I.
If P=R and 
, we have 
hence the conditions 
on F and 
imply a=|d|, hence the
additional conditions |b|<|c| on F and 
give a contradiction.
In the case P=R and Q=0 and 
, then we
have 
, and the conditions 
on F and 
imply a=|d| and the additional conditions |b|<|c| also lead to
a contradiction. Note that an easy calculation shows that in this case
.
If P=Q and 
, we have
and hence the conditions
|b|<|3a-b| on F and 
give a contradiction.
If Q=0 and 
, we have
and hence the conditions b>0 on F and 
give a contradiction.
Finally assume that P=Q=R and M is one of the six matrices given by
Lemma 5.2. Then an easy computation shows that F is of the form
. The reducedness of F is equivalent to a>0, |b|<|3a-b|.
For the six matrices of Lemma 5.2, we have respectively
, 
and 
. The first two leads to a
contradiction because of the condition |b|<|3a-b|, and the other four
together with a>0 imply 
, and the matrices
and
as
automorphisms of F. Thus in this case, and only in this case, the
automorphism group of F is nontrivial and is cyclic of order 3 generated
by one of the above two matrices. This finishes the proof of (1).
(2) Let 
be a reducible real cubic form. We will
successively replace F by equivalent forms, which we will still denote
by abuse of notation, until we can conclude. Set 
.
Assume by contradiction that F is reducible. Then by transforming F by
a suitable element of 
we may assume that a=0 and hence 
since 
. By changing if necessary F into -F, we may assume
that 
. Changing 
into 
for a suitable k, we may
assume that 
and finally by changing if necessary 
into
, we may assume 
. We thus have 
,
and 
.
Let p be an odd prime dividing b. Since 
, we have
, but since 
, by definition this means that
plus an additional condition. In particular, by
Proposition 3.3 (2) we have 
, hence 
, so
, hence 
, and this leads to a contradiction unless
p=3 by Proposition 3.3 (3). But if p=3, we have a=0 hence 
and
so Proposition 3.3 (4) implies that 
, contradiction
again.
There remains the case of p=2 dividing b. In that case we have
hence
, so 
.
Since 
, we have 
. We conclude as before by Proposition
3.3 (2) that 
, hence 
hence 
, in
contradiction with 
coming from Proposition 3.3 (3).
We deduce from the different cases above that b=1, and hence c=0 or
c=1. Hence if we call G the final cubic form that we have obtained,
we have 
for a certain 
, and 
with c=0 or 1. Hence 
or 
. Since
these are reduced quadratic forms equivalent to 
which is reduced,
we have 
. Now, since d<0 (otherwise 
and
is impossible), the only automorphisms of 
are 
, and
since we must have a>0, this implies M=I hence F=G, contradiction.
(3) The uniqueness statement follows from (1). Let 
be an
irreducible real cubic form and 
its Hessian. By the usual theory of
reduction of quadratic forms (with the added twist that matrices of determinant
-1 are allowed), we can find 
such that 
is
reduced, hence by changing F into 
we may assume 
reduced.
In all the cases of Definition 5.3, it is easily checked that the use of the
16 matrices of Lemma 5.2 will lead to a reduced form. Note that it is
essential to assume F irreducible, otherwise the result would be false.
More precisely, it is enough to check that the forms with a=0, or
b=0 when Q=0, or |b|=|3a-b| when P=Q or a=|d| and |b|=|c| when
P=R are all reducible. This finishes the proof of the proposition.
