Need to show that $\textbf{V}(y-x^2,xz-y^2)=\textbf{V}(y-x^2,xz-x^4)$ in $\mathbb{R}^3$.
I was trying to use the fact that $\textbf{V}(f,g)=\textbf{V}(f,g_1)\cup\textbf{V}(f,g_2)$ when $g=g_1g_2$. That led to the right side being $\textbf{V}(y-x^2,x)\cup\textbf{V}(y-x^2,z-x^3)$. But how is it equal to the left side?
Or is giving a geometric interpretation the only way to prove this?
Just show a two sided inclusion. If a point $(x,y)$ belongs to $V(y-x^2, xz-y^2)$ then we have $y=x^2$ and $xz=y^2$. Putting the first equation into the second one gives us $xz=y^2=x^4$. Hence $(x,y)\in V(y-x^2, xz-x^4)$. The reverse inclusion is similar.