### Proposition 5.4

• Two equivalent reduced real cubic forms are equal.
• A reduced real cubic form belonging to U is irreducible.
• Any irreducible real cubic form is equivalent to a unique reduced form.

### Proof

(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.