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I'm familiar with this formulation of the weak form of Hilbert's Nullstellensatz.

If $k$ is an algebraically closed field, $I\in k[x_1,\dots,x_n]$ is an ideal. Then $V(I)=\varnothing$ if and only if $1\in\sqrt{I}$.

In Atiyah-Macdonald "Introduction to Commutative Algebra" the theorem is stated in a (probably) more general form (Corollary $7.9$):

Let $k$ be a field, $A$ a finitely generated $k$-algebra. Let $\mathfrak{M}$ be a maximal ideal of $A$. Then the field $A/\mathfrak{M}$ is a finite algebraic extension of $k$. In particular, if $k$ is algebraically closed then $A/\mathfrak{M}\simeq k$.

I would like to understand the connection between the two statements.

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The second form is indeed, as you said, a more general statement. First, we prove that, if $\mathbb{K}$ is algebraically closed, every maximal ideal of $R = \mathbb{K}[x_1, x_2, \dots, x_n]$ is of the form $\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$. Obviously any such ideal is maximal. To prove the converse, consider a maximal ideal $\mathfrak{n}$ and the projection $\varphi: R\to R/\mathfrak{n}$. As you said, $R/\mathfrak{n} \simeq \mathbb{K}$ by the Nullstellensatz. Call $a_i$ the image of $x_i$. Then we easily see that $\mathfrak{m} = (x_1-a_1, \dots, x_n-a_n) \subset \ker(\varphi)$. By maximality of $\mathfrak{m}$, it must coincide with the kernel $\mathfrak{n}$.

We now pass on to $V(I)$. Note that, if $(a_1,\dots, a_n) \in V(I)$, by considering the evaluation morphism in $(a_1, \dots, a_n)$, we get that $I \subseteq M = (x_1 -a_1, \dots, x_n - a_n)$: indeed, $M$ is in the kernel, and by assumption $I$ is in the kernel, too. If, on the other hand, $I$ is proper, then it is contained in some maximal ideal and we also know, by above, that every maximal ideal of $R$ is in a one-to-one correspondence with $n$-tuples $(a_1, \dots, a_n)$ and that it vanishes when evaluated on such tuple. So, considering again the evaluation morphism in the $n$-tuple corresponding to one of the maximal ideals containing $I$, we get that $(a_1, \dots, a_n) \in V(I)$ since $I$ is contained in the kernel of such morphism. We have proven and can now formulate the following: $$V(I) \neq \varnothing \iff I \text{ is proper}$$ Your statement is the contrapositive, since $1 \in \sqrt{I} \implies 1 \in I$, which in turn that implies $I$ is not proper.

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  • $\begingroup$ Thank you. You've been very clear. I've only one question, $\varphi$ is defined on $R$(instead of $\mathbb{K}$), am I right? $\endgroup$
    – Luca
    Sep 1, 2019 at 11:01
  • $\begingroup$ Yes, you're correct. I've edited the answer. $\endgroup$ Sep 1, 2019 at 19:08
  • $\begingroup$ Thank you for the answer. "Obviously any such ideal is maximal” - I cannot see this, can anyone clear this? $\endgroup$ Jul 2, 2021 at 9:48
  • $\begingroup$ @DanishA.Alvi just take the quotient of $R$ by $\mathfrak{m}$. In the quotient the image of $x_i$ will be $a_i$, for all $i$'s, hence you end up with $K$. Is it clearer? $\endgroup$ Jul 2, 2021 at 16:51
  • $\begingroup$ I’m sorry, I still cannot see why $\mathfrak{m}$ will be maximal. But i think you’ve done your best, I will figure it out! $\endgroup$ Jul 2, 2021 at 18:08

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