Let $I$ be an ideal of $A=k[x_1,\cdots,x_n]$ and $f \in A$ such that $f$ vanishes at all zeros of $I$ in the algebraic closure $\bar{k}$ of $k$. I can see that Hilbert's Nullstellensatz is equivalent to saying that $I A_f = A_f$ where $A_f$ is the localized ring with respect to the multiplicative set $1,f,f^2,\cdots$ and $I A_f$ is the ideal generated by the image of $I$ in $A_f$. My question is: how can we prove that $I A_f = A_f$? (obviously without using Hilbert's Nullstellensatz)
(Edited) Idea: Let $f, I$ be defined as above. It is enough to show that $IA_f$ is an ideal that does not have any zeros in $\bar{k}$. Then we can use the result that if an ideal does not have any zeros, then it must contain $1$ and so $I A_f = A_f$ follows. How do we define the zeros of $I A_f$ in $\bar{k}$? Intuitively, these would consist of $n$-tuples over $\bar{k}$ such that $f$ does not vanish. (How can we make this more rigorous?) Then clearly $I A_f$ does not have any zeros and so $I A_f =A_f$.