$V(I(X)) \cap V(I(Y)) = \emptyset \Leftrightarrow\exists f \in K[x_1,...,x_n] : f(x)=0 \forall x \in X \text { and } f(x)=1 \forall x \in Y$

Question

For $A\subset K[x_1,...,x_n]$, let $V(A)$ be the affine variety. For $B \subset K^n$, let $I(B)$ be the vanishing ideal of $B$. I want to show the following equivalence for $X,Y \subset K^n$: $$V(I(X)) \cap V(I(Y)) = \emptyset \Leftrightarrow \exists f \in K[x_1,...,x_n] : f(x)=0 \ \forall \ x \in X \text{ and } f(x)=1 \ \forall \ x \in Y.$$

Unfolding the first statement, I got the equivalence $$\forall (\xi_1,...,\xi_n)\in K^n \ \forall f\in I(X) \ \forall h \in I(Y) : (g+h)(\xi_1,...,\xi_n)\neq 0, \\ \text{since} \ V(I(X)) \cap V(I(Y)) = V(I(X)+ I(Y)).$$ How can I go on from here?

Answer 1

We know that $V(I)\cap V(J)= V(I+J)$. Assume that $V(I+J)=\emptyset$. By Nulstellensatz we must have $I+J=(1)$. So there exists $f\in I$, $g\in J$ with $f+g=1$. Note that $f=0$ on $V(I)$ and $1$ on $V(J)$.