Given the quotient ring $R:=\mathbb{C}[x,y,z]/(xy+z^2,x^2y+z^4-xy^3-1)$, we set $f_1=xy+z^2$ and $f_2=x^2y+z^4-xy^3-1$. We define $f_3$ by $f_3:=f_2+(xy-z^2)f_1=x^2y-xy^3+x^2y^2-1 \in \mathbb{C}[x,y,z] $. Then we get $(f_1,f_2)=(f_1,f_3)$. Now, $R'$ is the $\mathbb{C}$-subalgebra of $R$ generated by $x$ and $y$. My question is:
Why can I write $f_3=0$ in $R'$?
How is this possible? I don't see it.
Well, you have $R={\Bbb C}[x,y,z]/\langle f_1,f_2\rangle$ and $$f_3= f_2+(xy-z^2)f_1 = x^2y-xy^3+x^2y^2-1.$$ Then $f_3=0$ in $R$ and so also in the subalgebra $R' = \mathbb C[\bar x,\bar y]$.