In some recent computation I came across certain cubic forms and was wondering about analogue of following result for quadratic forms.


If $k^\*/(k^\*)^2$ is finite then there are only finitely many quadratic forms up to equivalence. This can be easily seen because of diagonalisation.


I have been wondering if we can say following: given $k^\*/(k^\*)^3$ is finite is it true that there are only finitely many cubic forms up to equivalence. Or can I say something about being isotropic.



cubic forms and finiteness of $k^*/(k^*)^3$
To expand a bit on the comments. For simplicity, let's assume $k$ is characteristic $0$ (or at least, not characteristic 2 or 3). Let $f(X,Y)=aX^3+bX^2Y+cXY^2+dY^3$ be a cubic form. The case where $f$ is reducible comes down to classifying quadratic forms and linear forms, so let's assume that $f$ is irreducible. More generally, assume that $f(X,1)$ has distinct roots $e\_1,e\_2,e\_3\in\overline k$. Let $\lambda=(e\_1-e\_2)/(e\_1-e\_3)$. Then the quantity $j(f)=(\lambda^2-\lambda+1)^3/\lambda^2(\lambda-1)^2$ is an invariant of $f$. And conversely, if $j(f)=j(f')$, then there is a transformation defined over $\overline k$ taking $f$ to $f'$. So what you need to do is fix a value $j\_0\in k$ and then look at all (separable) $f\in k[X,Y]$ satisfying $j(f)=j\_0$. Fixing one such $f\_0$, you should then be able to classify the $\overline k/k$-twists of $f\_0$ using Galois cohomology in the usual way.


