This question asks the following question: In the polynomial $k$-algebra $A$, in $d$ variables over a field $k=\bar k$, is it true that any finite intersection of maximal ideals is generated by $d$ elements?
That question has a positive answer by Harm Derksen, and I wish to understand his answer.
Let us concentrate on the two-dimensional case $d=2$ and $k=\mathbb{C}$, namely $A=\mathbb{C}[x,y]$. By the Nullstellensatz, every maximal ideal is of the form $\langle x-\alpha, y-\beta \rangle$, for some $\alpha,\beta \in \mathbb{C}$.
How do we actually find these two generators for a concrete finite intersection of maximal ideals? Could one please apply the method explained in Harm Derksen's answer in the following example: $M_1=\langle x,y \rangle$ and $M_2=\langle x-1,y-2 \rangle$; I see four generators for $I=M_1 \cap M_2$: $\{ x(x-1),x(y-2),y(x-1),y(y-2) \}$.
My attempt to apply the method in Derksen's answer: $n=2$. $P_1=(0,0), P_2=(1,2)$. $P_1=(Q_1,r_1), P_2=(Q_2,r_2)$, so $Q_1=0,r_1=0,Q_2=1,r_2=2$ (here there are only two coordinates, so we can just write $q_1,q_2$ instead of $Q_1,Q_2$). Here we can take as the linear change of coordinates the identity, since $q_1,q_2$ are already distinct. By the induction hypothesis, there exists a polynomial $f_1 \in \mathbb{C}[x]$ that generates the finite intersection of which $n-1$ maximal ideals in $\mathbb{C}[x]$? The maximal ideal $\langle x-q_1 \rangle$? So $f_1(x)=x-q_1=x-0=x$? Using polynomial interpolation we can construct a polynomial $g=g(x) \in \mathbb{C}[x]$ such that $g(q_1)=r_1$ and $g(q_2)=r_2$ ($g$ is of degree $1$ and it it easy to see that it must be the line $g(x)=2x$ because $g(0)=0$ and $g(1)=2$). Now take $f_2=f_2(x,y)=y-g(x)=y-2x$. The zero set of $f_1=x,f_2=y-2x$ is the set $\{P_1=(0,0)\}$. But $\langle f_1,f_2 \rangle = \langle x,y-2x \rangle = \langle x,y \rangle$ is not the intersection of $M_1=\langle x,y \rangle$ and $M_2=\langle x-1,y-2 \rangle$...
The gap in my understanding is who are the $n-1$ maximal ideals in $\mathbb{C}[x]$? I do not inderstand why we cannot take all the $n$ points $q_1,q_2$ or in other words, $f_1$ that generates $\langle x \rangle \cap \langle x-1 \rangle$ is $f_1=f_1(x)=x(x-1)$. Then is $\langle f_1,f_2 \rangle = \langle x(x-1),y-2x \rangle$ the intersection of $M_1=\langle x,y \rangle$ and $M_2=\langle x-1,y-2 \rangle$? I see why $\langle x(x-1),y-2x \rangle \subseteq M_1 \cap M_2$. The other direction probably holds as well.
Thank you very much!