Hilbert's Nullstellensatz - Proof by Localization 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 $I A_f =A_f$. 
 A: In general for  a morphism of rings $\phi:A\to B$ and its associated morphism of schemes $F=\phi^*:Spec(B)\to Spec(A)$, we have for every ideal $I\subset A$ the equality  $F^{-1}(V(I))=V(I\cdot B)$.  
So in the  case $A=k[x_1,...,x_n]\hookrightarrow B=A_f$, the morphism  $F:Spec(A_f)=Spec(A)\setminus V(f)\hookrightarrow  Spec(A)$ is the inclusion and we have : $$I\cdot A_f= A_f  \iff V(I\cdot A_f) =\emptyset     \iff      F^{-1}(V(I))=\emptyset \iff  V(I)\subset V(f)  \iff f\in \sqrt {I}$$    These equivalences are purely formal and do not require the Nullstellensatz ! 
The Nullstellensatz may be used to interpret the last condition  $f\in \sqrt {I} $:  this condition is indeed equivalent to the condition that $f$ vanish on the zero set in $\overline {k}^n$ of the ideal $I$, a condition that  one might write as $V_\overline {k}^n(I)\subset V_\overline {k}^n(f) $.
Beware that in the above I have used $V(I)$ in the scheme-theoretic sense: it denotes the set of primes in $A$ containing $I$.
