@KarlKronenfeld Yes, it should be $\mathbb{C}$. We have this homomorphism and then we take a polynomial of $\mathbb{C}[x, y, z]$ and we want to show that if $p \in ker \phi$ then $p \in \langle y-x^2, z-x^3\rangle$. My question is : Is the step :
If $p(x, y, z) \in ker \phi$, $$\phi(p(x, y, z))=0 \Rightarrow \phi(g(x, y, z)(y-x^2)+(z-x^3)a(x, z)+b(x))=0 \Rightarrow \phi(g(x, y, z)) \phi((y-x^2))+\phi((z-x^3))\phi(a(x, z))+\phi(b(x))=0\Rightarrow \phi(b(x))=0 \Rightarrow b(\phi(x))=0 \Rightarrow b(x)=0$$
Does it work like that? $$\int_{0}^{\infty}\int_{0}^{\infty} \sin(x^{2}+y^{2}) \,dx\ dy=\Im\left\{\int_{0}^{\infty}\int_{0}^{\infty} e^{i(x^2+y^2)} \ dx \ dy\right\}=\Im\left\{\int_{0}^{\infty}e^{i x^2} \ dx\int_{0}^{\infty} e^{i y^2} \ dy\right\}=\frac{\pi}{4}$$
