Consequence of Hilberts Nullstellensatz - Dummit & Foote, Section 15.2

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote (D&F) Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.2 Radicals and Affine Varieties ... ...

I need help with understanding a consequence of Hilbert's Nullstellensatz which D&F mention after stating the Theorem ...

Hilbert's Nullstellensatz (D&F Section 15.2, page 675) and a particular consequence mentioned on page 676, read as follows:
https://www.physicsforums.com/attachments/4824
View attachment 4825In the above text from Dummit and Foote, we read the following:

" ... ... One particular consequence of the Nullstellensatz is that for any proper ideal \(\displaystyle I\), we have \(\displaystyle \mathcal{Z} (I) \neq \emptyset\) since \(\displaystyle \text{ rad } I \neq k[ \mathbb{A}^n ]\). ... ... "
I cannot see how Hilbert's Nullstellensatz implies the statement that \(\displaystyle \text{ rad } I \neq k[ \mathbb{A}^n ] \Longrightarrow \mathcal{Z} (I) \neq \emptyset \)Can anyone demonstrate formally and rigorously how this statement follows from Hilbert's Nullstellensatz as stated by D&F?Help will be appreciated ...

Peter***EDIT***

In order to help MHB readers to understand the context and the terminology of the above post fully, I am providing D&F's definition of \(\displaystyle \mathcal{Z} (S)\) where \(\displaystyle S\) is a set of \(\displaystyle k\)-valued functions on \(\displaystyle \mathbb{A}^n\) ... ... as follows:https://www.physicsforums.com/attachments/4826
https://www.physicsforums.com/attachments/4827
***EDIT 2*** ***EDIT 2*** ***EDIT 2***I have been reflecting on proving the "consequence" mentioned above ... here are my thoughts ... ...

We want to show that given Hilbert's Nullstellensatz that:

\(\displaystyle \text{ rad } I \neq k [ \mathbb{A}^n] \ \Longrightarrow \ \mathcal{Z} (I) \neq \emptyset \) ... ... ... (1)

But showing (1) is equivalent to showing the contrapositive, namely:

\(\displaystyle \mathcal{Z} (I) = \emptyset \ \Longrightarrow \ \text{ rad } I = k [ \mathbb{A}^n]\) ... ... ... (2)But by Hilbert's Nullstellensatz we have:

\(\displaystyle \text{ rad } I = \mathcal{I} ( \mathcal{Z} (I) )\)and so (2) becomes:

\(\displaystyle \mathcal{Z} (I) = \emptyset \ \Longrightarrow \ \mathcal{I} ( \mathcal{Z} (I) ) = k [ \mathbb{A}^n]\)Thus, essentially, we need to show that:

\(\displaystyle \mathcal{I} ( \emptyset ) = k [ \mathbb{A}^n]\)BUT ... how do we interpret \(\displaystyle \mathcal{I} ( \emptyset )\) ... ... it does not seem to have a straightforward interpretation ... ... ... ... thinking ... well ... \(\displaystyle \mathcal{I} ( A )\) is defined as follows:

\(\displaystyle \mathcal{I} ( A ) = \{ f \in k [ x_1, x_2, \ ... \ ... \ , x_n ] \ | \ f( a_1, a_2, \ ... \ ... \ , a_n) \text{ for all } ( a_1, a_2, \ ... \ ... \ , a_n) \in A \}\)

... so that we have:

\(\displaystyle \mathcal{I} ( \emptyset ) = \{ f \in k [ x_1, x_2, \ ... \ ... \ , x_n ] \ | \ f( a_1, a_2, \ ... \ ... \ , a_n) \text{ for all } ( a_1, a_2, \ ... \ ... \ , a_n) \in \emptyset \}\)But ... since there are no elements in \(\displaystyle \emptyset\), the above definition does not seem to make sense ... ... unless we regard any condition to be fulfilled as satisfied because there are no elements that have to satisfy it ...I hope someone can clarify this issue/problem for me ... ...

Peter

 

[Turgul]

In case you are still pondering this:

Your last observation that $\mathcal{I}(\emptyset) = k[x_1,\ldots,x_n]$ is correct (and thus your overall argument). Indeed, for any set $S \subseteq \mathbb{A}^n$, we have $\mathcal{I}(S)$ is the set of polynomials in $k[x_1,\ldots,x_n]$ which vanish for every point of $S$. But when $S = \emptyset$, any polynomial will evaluate to zero for every point of $S$ (at no point is this false!), hence $k[x_1,\ldots,x_n] \subseteq \mathcal{I}(\emptyset)$.

But a caution is in order: to prove the Nullstellensatz, or specifically this corollary, we absolutely require that the field $k$ be algebraically closed. Consider $k = \mathbb{R}$, then in $\mathbb{R}[x]$, we have $\mathcal{Z}(x^2+1) = \emptyset$ even though the ideal generated by $x^2+1$ is certainly proper and a radical ideal (maximal, even). To prove just this corollary, even when $n=1$, we fundamentally require algebraic closedness.

 

[Math Amateur]

In case you are still pondering this:

Your last observation that $\mathcal{I}(\emptyset) = k[x_1,\ldots,x_n]$ is correct (and thus your overall argument). Indeed, for any set $S \subseteq \mathbb{A}^n$, we have $\mathcal{I}(S)$ is the set of polynomials in $k[x_1,\ldots,x_n]$ which vanish for every point of $S$. But when $S = \emptyset$, any polynomial will evaluate to zero for every point of $S$ (at no point is this false!), hence $k[x_1,\ldots,x_n] \subseteq \mathcal{I}(\emptyset)$.

But a caution is in order: to prove the Nullstellensatz, or specifically this corollary, we absolutely require that the field $k$ be algebraically closed. Consider $k = \mathbb{R}$, then in $\mathbb{R}[x]$, we have $\mathcal{Z}(x^2+1) = \emptyset$ even though the ideal generated by $x^2+1$ is certainly proper and a radical ideal (maximal, even). To prove just this corollary, even when $n=1$, we fundamentally require algebraic closedness.

Thanks Turgul ... I was still pondering this issue, so your post is very welcome and helpful ...

Your post is much appreciated ...

Peter