square-detecting ideal

Question

Let $\phi : k[a_0,a_1,a_2,a_3,a_4] \to k[C_1,C_2,D_1,D_2]$ be a morphism of rings defined by $$ a_0 \mapsto D_1^2D_2^2,\quad a_1 \mapsto 2D_1D_2(C_2D_1+C_1D_2),\quad a_2 \mapsto4C_1C_2D_1D_2 + C_2^2D_1^2+C_1^2D_2^2$$ $$ a_3 \mapsto 2C_1C_2(C_2D_1+C_1D_2), \quad a_4 \mapsto C_1^2C_2^2.$$ Using the kernel subroutine of Macaulay2 I found that the kernel is $$I = (8a_0^2a_3-4a_0a_1a_2+a_1^3,16a_0^2a_4 + 2a_0 a_1 a_3-4a_0a_2^2 + a_1^2 a_2, 8a_0a_1a_4-4a_0a_2a_3+a_1^2a_3, $$ $$ a_0a_3^2-a_1^2a_4,8a_0a_3a_4-4a_1a_2a_4+a_1a_3^2,16a_0a_4^2+2a_1a_3a_4-4a_2^2a_4+a_2a_3^2,$$ $$8a_1a_4^2-4a_2a_3a_4+a_3^3).$$ But how could I arrive at this answer myself? It is easy to show that $I \subset \ker \phi$ by just substituting the variables in the generators of $I$, but how can I show $\ker \phi \subset I$?

Answer 1

Let $I'=\langle a_0-\phi(a_0), \ldots, a_4-\phi(a_4)\rangle$ in the big ring $k[a_0,a_1,a_2,a_3,a_4,C_1,C_2,D_1,D_2]$. Then: $$\ker \phi = I' \cap k[a_0,a_1,a_2,a_3,a_4].$$ There is an algorithm to compute this:

1. Calculate a Gröbner basis $G$ for $I'$ with respect to an elimination ordering, where the $C_1,C_2,D_1,D_2$ variables are greater than the $a_0,a_1,a_2,a_3,a_4$ variables.

2. The polynomials in $G$ without any $C_1,C_2,D_1,D_2$ variables form a Gröbner basis for $I' \cap k[a_0,a_1,a_2,a_3,a_4] =\ker \phi$.