Let $I=\langle x-a,y-b \rangle$ and $J=\langle x-c,y-d \rangle$, where $a,b,c,d \in \mathbb{C}$. ($I$ and $J$ are maximal ideals of $\mathbb{C}[x,y]$).
Is it true that $I \cap J = \langle (x-a)(x-c),(y-b)(y-d) \rangle$?
($I \cap J$ is a radical ideal).
I see that $\langle (x-a)(x-c),(y-b)(y-d) \rangle \subseteq I \cap J$, but I am not sure about the other direction.
My proof:
$\langle (x-a)(x-c),(y-b)(y-d) \rangle \subseteq I \cap J$: An element $W$ on the left side is of the form $W= U(x-a)(x-c)+V(y-b)(y-d)$, where $U,V \in \mathbb{C}[x,y]$. Let $U_1=U(x-a), U_2=U(x-c), V_1=V(y-b), V_2=V(y-d)$. Then, $W=U_1(x-c)+V_1(y-d) \in J$ and $W=U_2(x-a)+V_2(y-b) \in I$, so $W \in I \cap J$.
$I \cap J \subseteq \langle (x-a)(x-c),(y-b)(y-d) \rangle$: An element $W$ on the left side is of the form $W=S_1(x-a)+T_1(y-b)$, where $S_1,T_1 \in \mathbb{C}[x,y]$ (since it belongs to $I$), and $W=S_2(x-c)+T_2(y-d)$ (since it belongs to $J$). Therefore, $W=S_1(x-a)+T_1(y-b)=S_2(x-c)+T_2(y-d)$; why should $x-c$ divide $S_1$ and $y-d$ divide $T_1$?
A proof (or a counterexample) is welcome. Thank you very much!