Before specializing to the cubic case, we consider the general case of binary forms of degree n. Let K be a field (usually , or ).
The roots of F in K are the solutions of . If for and , the point at infinity is a root of order exactly n-m (if m=n, i.e , it is of course not a root), and the other roots are the roots in K of the polynomial of degree m. In particular if K is algebraically closed, F has always exactly n roots, counted with multiplicity. Note that the point at infinity is rational over any base field, algebraically closed or not.
Denote by , with , the roots of F in . It is easily seen that we can choose representatives in so that we haveOf course the choice of representative is not unique: for each i we can change into as long as . We will always assume that the representatives of the roots are chosen in this manner. We define the discriminant of the form F by the following formula: This makes sense since if we change into with , the product is multiplied by
If is a polynomial of degree exactly equal to n ( i.e if ), then one immediately checks that .
In degrees up to 3 we have the following formulas.
If F is a form of degree n and is a matrix with coefficients in K we define the action of on F by .
Let . Then
We now restrict to the case of integral binary forms, i.e binary forms whose coefficients are all in . As a corollary of the proposition, we see that the action of preserves the discriminant of F, since is even.
We will say that the form F is irreducible if is irreducible as a polynomial in . Equivalently, F is irreducible if and the polynomial is irreducible in (or ).
We will say that an integral form F is primitive if the GCD of all its coefficients is equal to 1.
Let F be an integral form and . Then is irreducible if and only if F is, and is primitive if and only F is.