Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $I A_f =A_f$.

share|cite|improve this question
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
up vote 4 down vote accepted

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 $I$.

share|cite|improve this answer
Dear Manos, you are very welcome to a follow-up question, but I suggest you make it a new question, unless it is a request for clarification of some point in my answer. – Georges Elencwajg Dec 26 '12 at 23:17
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
Ooops! Wrong spelling :) – Manos Dec 27 '12 at 1:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.