# 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 $I A_f =A_f$.

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Perhaps I'm missing something, but what is $A$? – Alex Becker Dec 26 '12 at 19:44
@AlexBecker: Thanks, i fixed this. – Manos Dec 26 '12 at 19:57

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 $I$.
Dear Georges, your answer is as always remarkably instructive both in terms of insight and notation. Let me make it perfect: you are missing a foolstop at the end :) There is a point in my question, which remains unclearified, please let me know if it merits a new question: is there a meaning in talking about the zero-set of the ideal $I A_f$ in $\bar{k}$? (analogously in talking about the zero set of the ideal $I$ in $\bar{k}$) If yes, how is this defined formally? – Manos Dec 26 '12 at 23:27
Dear Manos, it certainly makes sense to talk about the zero-set in $\overline k ^n$ of the ideal $IA_f\subset A_f$: you will simply obtain the set-theoretic difference $V_\overline {k}^n(I)\setminus V_\overline {k}^n(f) \subset \overline {k}^n$. (You might look at the example $n=3, I=(x_1,x_2), f=x_3$.) Also, I have taken your remark into account: so you made a fool stop omitting a full stop :-) – Georges Elencwajg Dec 27 '12 at 0:30