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I was going through Atiyah and Macdonald's exercises and found his 17th exercise in chapter 5 (integral dependence and valuation) a bit dicey.

He is basically trying to prove that for any affine variety $X$, if $I(X)\ne (1)$, then $X\ne \emptyset$. And he then asks the reader to prove the weak form of the Nullstellensatz.

My problem is the following. If $I(X)\ne (1)$, then, there exists and $f\in k[t_1,...,t_n]$ and $x\in X$ such that $f(x)\ne 0$. This immediately implies that $X\ne \emptyset$. I do not understand why he uses Noether's normalisation for this. Secondly, I do not see how one can prove the weak form of Nullstellensatz just using this theorem because we still do not know that $I(V(m))=m$ for a maximal ideal $m\subset k[t_1,...,t_n]$ which is the strong version of the Nullstellensatz. Or is there a way to prove that specifically for maximal ideals without using the strong form the Nullstellensatz?

Sure, we can use Zariski's theorem, which is Corollary 5.24 in this book to prove it (which is what I believe is happening in this post), but that would not be in line with the problem. Moreover, I am not really worried about the proof of the Nullstellensatz. What bothers me the usage of Noether's normalisation for something that does not need it. I hope I have been able to explain my question clearly.

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  • $\begingroup$ Your post is missing the text of this exercise which makes it less likely that you'll get helpful responses. Do consider making it easier for others to help you by adding the excerpt that's causing you problems to your post. $\endgroup$
    – KReiser
    Aug 11, 2020 at 20:46
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    $\begingroup$ It seems to me that you are correct, we do not need Noether's Normalization Lemma to show that $I(X) \neq (1) \implies X \neq \varnothing$. However, we do need NNL to prove the claim about maximal ideals. $\endgroup$
    – Brian Shin
    Aug 11, 2020 at 21:25

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This is a fleshing-out of my comment.

You're observation is correct: we do not need Noether's Normalization Lemma to prove the implication $I(X) \neq (1) \implies X \neq \varnothing$. However, we do need the normalization lemma to prove that maximal ideals of $A = k[t_1, \ldots, t_n]$ are of the form $(t_1 - a_1, \ldots, t_n - a_n)$.

Indeed, let $\mathfrak{m}$ be a maximal ideal of $A$. By Noether Normalization, we have an injective finite map $k[X_1, \ldots, X_m] \to A/\mathfrak{m}$. (In particular, this map is an integral extension.) However, we know the following fact: if $R \to S$ is an integral extension, then $R$ is a field if and only $S$ is a field. This implies that $m=0$, and so $A/\mathfrak{m} = k$, by the assumption that $k$ is algebraically closed. (This is assumed Atiyah-Macdonald.) We conclude by observing where the $t_i$ map to under the projection $A\to A/\mathfrak{m}$.

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