I am trying to solve a nonlinear polynomial equation system using Janet basis, when they have finite many solutions. For example the solution of the system:
$$xy^2-y^3-3x^2=0,x^2+y^2+xy=0.$$
There is a hint saying that after finding the Janet Basis of the ideal $$I=\langle xy^2-y^3-3x^2,x^2+y^2+xy\rangle$$ in the polynomial ring $\mathbb{R}[x,y]$, one can find a basis for the $\mathbb{R}$-vector space $T:=\mathbb{R}[x,y]/I$, and then a matrix of the linear map $f_{x}:T\rightarrow T$ defined as $v\rightarrow xv$ and the same for $y$, and then somehow using the minimal polynomials of the matrices of these maps.
Ok, I found the Janet Basis of $I$, and it is $$J=\{p_{1}=x^2+xy+y^2, p_{2}=xy^2-y^3+3xy+3y^2, p_{3}=y^4-3y^3+6xy+6y^2\}$$ with multiplicative set $\{x,y\}$ for $p_{1}$ and multiplicative set $\{y\}$ for $p_{2},p_{3}$, and the basis of the factor ring as $\mathbb{R}$-vector space is $B= \{ 1, y, x, y^2, yx, y^3\}$.
So we can now easily find the matrices of $f_{x},f_{y}$ and also their minimal polynomials, but what can we do more to determine the solutions of the equations system, and what does this have to do with eigenvalues and diagonalization of the matrices of $f_{x},f_{y}$.
Thanks for any help or hint.
In this case all solutions can be found by elementary considerations. Multiplying the second equation by $y-2x$ and adding it to the first equation gives $$ x^2(2x+y+3)=0. $$ Then it follows immediately that $x=y=0$, or $$ (x,y)=\left( \frac{\sqrt{-3}-3}{2}, -\sqrt{-3}\right),\; \left( \frac{-\sqrt{-3}-3}{2}, \sqrt{-3}\right). $$ Over the real numbers, only $(x,y)=(0,0)$ is possible. If there is no elementary way to solve such a system, one needs to employ other techniques, such as Groebner bases, Janet bases and Janet-like Gröbner bases or similar tools.