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

(1) LetFand be two reduced cubic forms and such that . Then and since and are reduced, it follow from Lemma 5.2 that and thatMis an automorphism of . We now have only a finite number of possibilities (exactly 16) to test forM. Let .First, if , we have hence the conditions

a>0onFand implyM=I.If

P=Rand , we have hence the conditions onFand implya=|d|, hence the additional conditions|b|<|c|onFand give a contradiction.In the case

P=RandQ=0and , then we have , and the conditions onFand implya=|d|and the additional conditions|b|<|c|also lead to a contradiction. Note that an easy calculation shows that in this case .If

P=Qand , we have and hence the conditions|b|<|3a-b|onFand give a contradiction.If

Q=0and , we have and hence the conditionsb>0onFand give a contradiction.Finally assume that

P=Q=RandMis one of the six matrices given by Lemma 5.2. Then an easy computation shows thatFis of the form . The reducedness ofFis equivalent toa>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 witha>0imply , and the matrices and as automorphisms ofF. Thus in this case, and only in this case, the automorphism group ofFis 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

Fby equivalent forms, which we will still denote by abuse of notation, until we can conclude. Set . Assume by contradiction thatFis reducible. Then by transformingFby a suitable element of we may assume thata=0and hence since . By changing if necessaryFinto-F, we may assume that . Changing into for a suitablek, we may assume that and finally by changing if necessary into , we may assume . We thus have , and .Let

pbe an odd prime dividingb. 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 unlessp=3by Proposition 3.3 (3). But ifp=3, we havea=0hence and so Proposition 3.3 (4) implies that , contradiction again.There remains the case of

p=2dividingb. 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 hencec=0orc=1. Hence if we callGthe final cubic form that we have obtained, we have for a certain , and withc=0or 1. Hence or . Since these are reduced quadratic forms equivalent to which is reduced, we have . Now, sinced<0(otherwise and is impossible), the only automorphisms of are , and since we must havea>0, this impliesM=IhenceF=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

-1are allowed), we can find such that is reduced, hence by changingFinto 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

Firreducible, otherwise the result would be false. More precisely, it is enough to check that the forms witha=0, orb=0whenQ=0, or|b|=|3a-b|whenP=Qora=|d|and|b|=|c|whenP=Rare all reducible. This finishes the proof of the proposition.