I want to prove that
$I(X_1 \cap X_2) = \sqrt{I(X_1)+I(X_2)}$
for algebraic sets $X_1=Z(G_1)$ and $X_2=Z(G_2)$, with $G_1,G_2 \subseteq \mathbb{K}[X_1,\ldots,X_n]$.
Remark: Unfortunately I named the indeterminates $X_1,X_2,\ldots$ but obviously the first ones $X_1,X_2$ are algebraic sets and not indeterminates.
I've already proved $\supseteq$ but I'm struggling with proving $\subseteq$.
The problem is equivalent to prove that, if $g \in I(X_1 \cap X_2)$, then $g^n \in I(X_1)+I(X_2)$ for some $n \in \mathbb{N}$, but the $+$ is bothering me... some hint can definitely put myself in the right way.
Also, I'd like to:
Interpret geometrically what it means to have $I(X_1 \cap X_2) \neq I(X_1)+I(X_2)$,
thing that I don't see at all.
Since both $I(X_1\cap X_2)$ and $\sqrt{I(X_1)+I(X_2)}$ are radical ideals, by the Nullstellensatz it is enough to prove that $Z(I(X_1\cap X_2))=Z(\sqrt{I(X_1)+I(X_2)})$. We have $Z(I(X_1\cap X_2))=X_1\cap X_2$ and $Z(\sqrt{I(X_1)+I(X_2)})=Z(I(X_1)+I(X_2))=Z(I(X_1))\cap Z(I(X_2))=X_1\cap X_2$.
The inequality $I(X_1\cap X_2)\neq I(X_1)+ I(X_2)$ is because an algebraic reason: the sum of two radical ideals is not in general a radical ideal.
Edit: In the proof I assumed that the field is algebraically closed. The equality does not hold without this hypothesis. For example, take $\mathbb{K}=\mathbb{R}$, $X_1:\ x^2+y^2=1$, and $X_2:\ x=2$. Then $X_1\cap X_2=\emptyset$, so that $I(X_1\cap X_2)=\mathbb{R}[x,y]$, whereas the ideal $I(X_1)+I(X_2)=(x^2+y^2-1)+(x-2)$ (and hence also its radical) is a proper ideal.