## Transformation of a cubic form

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How can I change an integral binary form $ax^3+bx^2y+cxy^2+dy^3$ with the usual discriminant $D =b^2c^2-27a^2d^2+18abcd-4ac^3-4b^3d$ into a form $ax^3+dy^3$ which has a simple discriminant $-27a^2d^2$, which matrix (with $\operatorname{det} =1,-1$) can transforms it ? thanks...

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That is not always possible. – Felipe Voloch Oct 10 2011 at 16:23

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Under a linear change of variables (x,y) |--> (Ax+By,Cx+Dy), the discriminant of your cubic form becomes (AD-BC)^6*disc(original). If the cubic form can be diagonalized then this formula has to be -27 times a square, as you wrote, so the discriminant of the original cubic form has to be -3 times a perfect square. Lots of examples are not like that, e.g., x^3 + x^2y + y^3 has discriminant -31. Thus this cubic form can't be diagonalized over Q. That you can diagonalize quadratic forms in characteristic 0, or more generally outside characteristic 2, is something special about degree 2. – KConrad Oct 10 2011 at 17:23

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It is diagonalizable over $\mathbb{C}$ (if the discriminant is not zero, anyway). The space of binary cubic forms is "prehomogeneous", which means that $GL_2(\mathbb{C})$ acts transitively on the set of cubic forms with discriminant not zero. To prove this, and to find explicit matrices: Factor your cubic form over $\mathbb{C}$. $GL_2$ acts on the three roots of a binary cubic form, and this action is triply transitive, so figure out the action sending your roots to the roots of $x^3 + y^3$. Then, you can scale as appropriate to get a matrix in $SL_2$. – Frank Thorne Oct 10 2011 at 18:01

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Please use TeX on this site. – GH Oct 10 2011 at 19:05
let us assume that the matrix of transformation is in GL2(Z). indeed i like to proof the relation for a cubic form F with H hessian form and J jacobian form in this form : J^2+27D.F^2-4H^3=0 when i can transform F into ax^3+dy^3 then it should be very easy ! – ahmad Oct 11 2011 at 11:43
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## 1 Answer

One can solve the problem explicitly over $\mathbb{C}$ then try to work out the extra constraints due to working over the integers. For the complex case, and continuing on the approach in the comment by Frank Thorne, you can also use the so-called canonizant. Let $C(x,y)$ be your cubic. The canonizant here is the same as the Hessian which classically normalized is: $$H(x,y)=\frac{1}{36} \left(\frac{\partial^2}{\partial x \partial v}-\frac{\partial^2}{\partial y\partial u}\right)^2\ C(x,y) C(u,v)\ |_{u:=x, v:=y}$$ meaning: take the derivatives then set $u=x$ and $v=y$. If one can write $C=L_1^3+L_2^3$ where $L_1$, $L_2$ are linear forms in $x,y$, then $$H(x,y)=2 \Delta(L_1,L_2)^2\ L_1(x,y)\ L_2(x,y)$$ where $\Delta(L_1,L_2)$ is the determinant formed with the coefficients of the two linear forms. The matrix you are looking for essentially is that which sends $(x,y)$ to $(L_1(x,y),L_2(x,y))$. So to find it you need to compute the Hessian and then factor it. This means solving a second degree equation instead of a cubic equation as suggested in Frank's comment.

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