The algebra of invariants of a binary cubic form

Question

How to find the algebra of invariants of a binary cubic form

${a_{\overset{\,}{0}}}x^3+3{a_{\overset{\,}{1}}}x^2y+3{a_{\overset{\,}{2}}}xy^2+{a_{\overset{\,}{3}}}y^3$

The algebra of invariants

$I=4\left({a_{\overset{\,}{0}}}{a_{\overset{\,}{2}}}-{a_{\overset{\,}{1}}}\!^2\right)\left({a_{\overset{\,}{1}}}{a_{\overset{\,}{3}}}-{a_{\overset{\,}{2}}}\!^2\right)-\left({a_{\overset{\,}{0}}}{a_{\overset{\,}{3}}}-{a_{\overset{\,}{1}}}{a_{\overset{\,}{2}}}\right)^2$

Answer 1

Cubic equation ${\alpha_{\overset{\,}{0}}}t^3+{\alpha_{\overset{\,}{1}}}t^2+{\alpha_{\overset{\,}{2}}}t+{\alpha_{\overset{\,}{3}}}=0$ and its Discriminant:

$$\Delta_{(3)}= \dfrac{4}{3}\left(3{\alpha_{\overset{\,}{0}}}{\alpha_{\overset{\,}{2}}}-{\alpha_{\overset{\,}{1}}}\!^2\right)\left(3{\alpha_{\overset{\,}{1}}}{\alpha_{\overset{\,}{3}}}-{\alpha_{\overset{\,}{2}}}\!^2\right)-\dfrac{1}{3}\left(9{\alpha_{\overset{\,}{0}}}{\alpha_{\overset{\,}{3}}}-{\alpha_{\overset{\,}{1}}}{\alpha_{\overset{\,}{2}}}\right)^2$$

Take \begin{align*} \left\{ \begin{split} \alpha_{\overset{\,}{0}}&=a_{\overset{\,}{0}}\\ \alpha_{\overset{\,}{1}}&=3a_{\overset{\,}{1}}\\ \alpha_{\overset{\,}{2}}&=3a_{\overset{\,}{2}}\\ \alpha_{\overset{\,}{3}}&=a_{\overset{\,}{3}}\\ \end{split} \right. \end{align*} Then you can get your answer.

(For the Discriminant of cubic equation, you can refer to some textbooks on polynomial)