MHB Affine Algebraic Sets - Properties of the map I - Dummit and Foote, page 661

[Math Amateur]

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

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need someone to help me to fully understand the reasoning/analysis following the listing of some properties of $$\mathcal{I}$$ ... in particular the reasoning pertaining to property (10) ... ...

The relevant text from D&F page 661 is as follows:View attachment 4828In the above text we see the following statement by D&F:

" ... ... The last relation shows the maps $$\mathcal{Z}$$ and $$\mathcal{I}$$ act as inverses of each other provided one restricts to the collection of affine algebraic sets $$ V = \mathcal{Z} (I) \text{ in } \mathbb{A}^n$$ and to the set of ideals in $$k [ \mathbb{A}^n ]$$ of the form $$\mathcal{I} (V)$$ ... ... "I cannot see why we have to restrict to ideals in $$k [ \mathbb{A}^n ]$$ of the form $$\mathcal{I} (V)$$ when property 10 is stated in terms of ideals $$I = \mathcal{I} (A)$$ where $$A$$ is an arbitrary subset of $$\mathbb{A}^n$$ ... and hence NOT restricted to an affine algebraic set $$V$$ ... ... ?

Can someone please help me to clarify this issue/problem ... ?

Peter
***EDIT***

To ensure MHB readers understand the notation and context of the above post I am providing D&F's definitions of $$\mathcal{Z}$$ and $$\mathcal{I}$$, as follows:View attachment 4829
View attachment 4830View attachment 4831

 

[Turgul]

Note that if I take an arbitrary subset $$A \subseteq \mathbb{A}^n$$, the zeroes of any polynomial vanishing on all of $$A$$ may be much larger than the set $$A$$. Let's look at an example: let $$k = \mathbb{C}$$ and $$A = \mathbb{Z} \subseteq \mathbb{A}^1$$. Then for a polynomial $$f \in \mathbb{C}[x]$$ to be in $$\mathcal{I}(A)$$, $f(a) = 0$ for every integer $a$. But the only polynomial in 1 variable with infinitely many zeroes is the zero polynomial. Hence $\mathcal{I}(A) = \{0\}$. We note in this case, since everything is a zero of the zero polynomial, $\mathcal{Z}(\mathcal{I}(A)) = \mathcal{Z}(\{0\}) = \mathbb{A}^1$.

In effect, $\mathcal{Z}(\mathcal{I}(A))$ fills in all of the missing parts of $A$ to give you the smallest algebraic set containing the original $A$. The closure, if you will.

By a similar token, $\mathcal{I}(\mathcal{Z}(S))$ takes an arbitrary subset $S \subseteq k[x_1,\ldots,x_n]$ and returns a certain kind of ideal of $k[x_1,\ldots,x_n]$.

Let's look at a quick example: let $S = \{x^2\} \subseteq k[x]$. Then $\mathcal{Z}(\{x^2\})$ is the set of points of $\mathbb{A}^1$ for which $x^2$ evaluates to zero. Clearly this is the single point $0$. Hence $\mathcal{Z}(\{x^2\}) = \{0\}$. Now $\mathcal{I}(\mathcal{Z}(\{x^2\})) = \mathcal{I}(\{0\})$, but any function divisible by $x$ has $0$ as a root (and these are all such functions), hence we have $\mathcal{I}(\mathcal{Z}(\{x^2\})) = xk[x] = (x)$. So we started with the single element $x^2$ and we got the entire ideal generated by $x$.
In general, both $\mathcal{Z}(\mathcal{I}(-))$ and $\mathcal{I}(\mathcal{Z}(-))$ are both processes which enlarge sets. In particular, both the maps $\mathcal{Z}(-)$ and $\mathcal{I}(-)$ are not injective in general (we've effectively given examples in the case of $\mathbb{A}^1$) so we can only hope that these maps can make a bijection between carefully chosen subsets of $\mathbb{A}^n$ and carefully chosen ideals of $k[x_1\ldots,x_n]$. The content of D&F's claim is that we do, indeed, get the bijection we're hoping for.

 

[Math Amateur]

Note that if I take an arbitrary subset $$A \subseteq \mathbb{A}^n$$, the zeroes of any polynomial vanishing on all of $$A$$ may be much larger than the set $$A$$. Let's look at an example: let $$k = \mathbb{C}$$ and $$A = \mathbb{Z} \subseteq \mathbb{A}^1$$. Then for a polynomial $$f \in \mathbb{C}[x]$$ to be in $$\mathcal{I}(A)$$, $f(a) = 0$ for every integer $a$. But the only polynomial in 1 variable with infinitely many zeroes is the zero polynomial. Hence $\mathcal{I}(A) = \{0\}$. We note in this case, since everything is a zero of the zero polynomial, $\mathcal{Z}(\mathcal{I}(A)) = \mathcal{Z}(\{0\}) = \mathbb{A}^1$.

In effect, $\mathcal{Z}(\mathcal{I}(A))$ fills in all of the missing parts of $A$ to give you the smallest algebraic set containing the original $A$. The closure, if you will.

By a similar token, $\mathcal{I}(\mathcal{Z}(S))$ takes an arbitrary subset $S \subseteq k[x_1,\ldots,x_n]$ and returns a certain kind of ideal of $k[x_1,\ldots,x_n]$.

Let's look at a quick example: let $S = \{x^2\} \subseteq k[x]$. Then $\mathcal{Z}(\{x^2\})$ is the set of points of $\mathbb{A}^1$ for which $x^2$ evaluates to zero. Clearly this is the single point $0$. Hence $\mathcal{Z}(\{x^2\}) = \{0\}$. Now $\mathcal{I}(\mathcal{Z}(\{x^2\})) = \mathcal{I}(\{0\})$, but any function divisible by $x$ has $0$ as a root (and these are all such functions), hence we have $\mathcal{I}(\mathcal{Z}(\{x^2\})) = xk[x] = (x)$. So we started with the single element $x^2$ and we got the entire ideal generated by $x$.
In general, both $\mathcal{Z}(\mathcal{I}(-))$ and $\mathcal{I}(\mathcal{Z}(-))$ are both processes which enlarge sets. In particular, both the maps $\mathcal{Z}(-)$ and $\mathcal{I}(-)$ are not injective in general (we've effectively given examples in the case of $\mathbb{A}^1$) so we can only hope that these maps can make a bijection between carefully chosen subsets of $\mathbb{A}^n$ and carefully chosen ideals of $k[x_1\ldots,x_n]$. The content of D&F's claim is that we do, indeed, get the bijection we're hoping for.

Thanks so much for the help, Turgul ... I had nearly given up on the elements of basic algebraic geometry ... much obliged for the help ...

... will be working through your post carefully, shortly

Thanks again,

Peter