# Weak form of Hilbert's Nullstellensatz (Atiyah-Macdonald)

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.

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.
• Thank you. You've been very clear. I've only one question, $\varphi$ is defined on $R$(instead of $\mathbb{K}$), am I right?
• @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? Jul 2, 2021 at 16:51
• 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! Jul 2, 2021 at 18:08