# Closure of difference of varieties is ideal quotient

For ideals $$J_1,J_2$$ of a commutative ring $$R$$, the ideal quotient is defined by $$(J_1:J_2)=\{f\in R\mid fJ_2\subset J_1\}$$. Suppose $$X$$ is an affine variety. I am trying to understand why if $$J_1$$ and $$J_2$$ are radical ideal in the coordinate ring $$A(X)=k[x_1,...,x_n]/I(X)$$ (where $$k$$ is an algebraically closed field), then $$\overline{V(J_1)-V(J_2)}=V(J_1:J_2)$$.

The only tools I can think of to use are the fact that $$\overline{S}=V(S)$$ and the Nullstellensatz. But I still can't see why it's true. What am I missing?

Suppose $$f(x)=0$$ for all $$x\in W:=V(J_1)-V(J_2)$$. Then for all $$g\in J_2$$ we have $$fg(x)=0$$ for all $$x\in V(J_1)$$, so $$fg\in J_1$$, and hence $$f\in(J_1:J_2)$$. Thus $$I(W)\subseteq(J_1:J_2)$$, and so $$\overline W\supseteq V(J_1:J_2)$$.
Conversely, if $$f\in(J_1:J_2)$$ and $$x\in W$$, then $$fg(x)=0$$ for any $$g\in J_2$$. In particular, since $$x\not\in V(J_2)$$, we know that there exists $$g\in J_2$$ with $$g(x)\neq0$$, and hence $$f(x)=0$$. Thus $$W\subseteq V(J_1:J_2)$$, so $$\overline W\subseteq V(J_1:J_2)$$.