Let be an integral basis of a cubic number fieldKas above. Forxandyelements of , set

- For
xandyin , we have- The function is the restriction to of a binary cubic form (again denoted by ) with rational coefficients.
- .
- The form is an integral primitive irreducible cubic form.
- The class of in is independent of the integral basis that we have chosen (hence by abuse of notation we will denote this class by ).
- Let the number field
Kbe defined by a root of the polynomial withp,q,rin , such that there exists an integral basis of the form witht,u,fin and -- this is always possible. Then, when we choose and , we have explicitly:

(1) follows directly from the definitions, after noting that is the determinant of the matrix Note that if we take one of the expressions on the RHS of (1) for definition of there is a sign ambiguity. However, this will not matter in the sequel since , so that and are -equivalent.(2) The definition of shows that is the restriction to of a cubic form with coefficients in . Furthermore Galois theory shows that the coefficients of are invariant under the Galois group of over , so that in fact is a rational cubic form. Note that the rationality of when

xandyare rational also follows from the remark made before the proposition.(3) The roots of (with any choice for the square root) are by assumption the for , and this respects the convention for the choice of roots made at the beginning. Hence one checks that

So that since, by the explicit formula for , we have .

(4) This is the longest part of the proof. Let with

a,b,c,din by (2). First, we note that ifxandyare in then is in , hence by the remark made above, where so . Applying this to and we deduce that and . Then applying this to and we deduce that and . It follows thatbandcbelong to with . Now, by (3) we know that is an integer. By the explicit formula for and using thatbandcbelong to we obtain that . Since , we cannot have , hence , thus proving that is integral.Recall that by (1), if

xandyare in , we have , where for , . Now let be the GCD of the coefficients of . We thus have for eachx,yin , . But since for any and since is an integral basis, it follows that foranywe have .This means by definition that is an inessential discriminant divisor, and according to a theorem of Dedekind, for a cubic field this implies that either , or else and

2is totally split inK. Assume the latter. We thus have for an integral cubic formG, and by (3). However since2is totally split inK, it is in particular unramified, hence , which is absurd since this implies that the integral cubic formGhas non-integral discriminant. Thus we have hence is primitive.Finally, let us show that is irreducible. Since we are in the cubic case, this means that has no linear factor in . Assume the contrary. By definition of , we may assume that such a linear factor is proportional to . It follows that there exist integers

randsnot both zero such that , so that . Taking conjugates, we see that we have , so that the conjugates of are equal, hence by Galois theory , and sincerandsare not both zero, this is in contradiction with being an integral basis. It follows that is irreducible.(5) Let be another integral basis. This means that there exists integers

with , since and must generate the same lattice. It follows thatA,B,C,D,E,Fsuch that

where the second equality follows from for all and . Since , it follows from (1) that with , so and are equivalent.

(6) This follows from a straightforward but tedious computation.