Arbitrary Subset of A^n - Is it necessarily an affine algebraic set?

[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 definition of an affine algebraic set ... ...

D&F explain and define an affine algebraic set in the following text ... ...View attachment 4750
View attachment 4751My question regarding the above definition is this:

Does any arbitrary subset (that is any subset) of the affine plane $$\mathbb{A}^n$$ qualify as an affine algebraic set ...

I was thinking maybe that it does because ... ...

If we have a set $$A_1$$, say that has no non-zero functions with zeros at each of its points, then could we regard the function corresponding to the zero polynomial as having zeros at each of the points of $$A_1$$ ... ... so that the set of functions {$$z$$} where $$z$$ is the zero function is zero on all points of $$A_1$$ ... and hence $$A_1$$ qualifies as an affine algebraic set ... ... I suspect that something is wrong with my analysis above ... but what exactly ...

Can someone clarify this issue for me please ... ...

Peter

 

[mathbalarka]

No, it's completely false that any subset of $\Bbb A^n$ appears as an affine algebraic set (also called affine variety).

First, let's review the definition : For some subset $S \in k[x_1,\cdots, x_n]$, the affine variety corresponding to it is the common zero locus of all polynomials in $S$. That is, $V = \{(a_1, \cdots, a_n) \in \Bbb A^n : f(a_1, \cdots, a_n) = 0\}$. If $S$ is a singleton, we get zero locus of a single polynomial in $\Bbb A^n$. This is called a hypersurface.

Thus, you can visualize an affine variety as intersection of (not necessarily finitely many) hypersurfaces in $\Bbb A^n$. As a side-note, it's a fact that any affine variety can be realized as intersection of finitely many hypersurfaces - this is Hilbert's basis theorem.

OK, so now that we are done reviewing the definitions, let's get back to your question. Why should it be true that any subset of $\Bbb A^n$ is an affine variety? Let $U \subset \Bbb A^n$ be an arbitrary subset of $\Bbb A^n$. You can of course consider the set of all functions vanishing on $U$. Call the set $S \in k[x_1, \cdots, x_n]$. However, what you don't know is whether functions in $S$ vanishes on places other than $U$. By definition, however, functions in $S$ on the affine variety $\mathcal{Z}(S)$ doesn't vanish anywhere else. So $U$ might not necessarily be an affine variety.

So, there's no reason for what you say to be true. To get an explicit example, the subset $\Bbb A^n - \{(0, 0, \cdots, 0)\}$ of $\Bbb A^n$ is not an affine variety. The proofs are quite involved, and require much more than basic commutative algebra. As a side-note, it is sometimes desired that locally closed subsets of affine varieties be varieties. Thus, we introduce the notion of quasi-affine varieties.

If we have a set A1, say that has no non-zero functions with zeros at each of its points, then could we regard the function corresponding to the zero polynomial as having zeros at each of the points of A1 ... ... so that the set of functions {z} where z is the zero function is zero on all points of A1 ... and hence A1 qualifies as an affine algebraic set ... ..

Ah, but note that your "zero function" vanishes not only on $A_1$, but in all of $\Bbb A^n$. To prove that $A_1$ is affine, you have to find a set of functions $S$ which vanish only on $A_1$ - that is $A_1$ is the whole zero locus of the functions in $S$.

 

[Math Amateur]

K

No, it's completely false that any subset of $\Bbb A^n$ appears as an affine algebraic set (also called affine variety).

First, let's review the definition : For some subset $S \in k[x_1,\cdots, x_n]$, the affine variety corresponding to it is the common zero locus of all polynomials in $S$. That is, $V = \{(a_1, \cdots, a_n) \in \Bbb A^n : f(a_1, \cdots, a_n) = 0\}$. If $S$ is a singleton, we get zero locus of a single polynomial in $\Bbb A^n$. This is called a hypersurface.

Thus, you can visualize an affine variety as intersection of (not necessarily finitely many) hypersurfaces in $\Bbb A^n$. As a side-note, it's a fact that any affine variety can be realized as intersection of finitely many hypersurfaces - this is Hilbert's basis theorem.

OK, so now that we are done reviewing the definitions, let's get back to your question. Why should it be true that any subset of $\Bbb A^n$ is an affine variety? Let $U \subset \Bbb A^n$ be an arbitrary subset of $\Bbb A^n$. You can of course consider the set of all functions vanishing on $U$. Call the set $S \in k[x_1, \cdots, x_n]$. However, what you don't know is whether functions in $S$ vanishes on places other than $U$. By definition, however, functions in $S$ on the affine variety $\mathcal{Z}(S)$ doesn't vanish anywhere else. So $U$ might not necessarily be an affine variety.

So, there's no reason for what you say to be true. To get an explicit example, the subset $\Bbb A^n - \{(0, 0, \cdots, 0)\}$ of $\Bbb A^n$ is not an affine variety. The proofs are quite involved, and require much more than basic commutative algebra. As a side-note, it is sometimes desired that locally closed subsets of affine varieties be varieties. Thus, we introduce the notion of quasi-affine varieties.
Ah, but note that your "zero function" vanishes not only on $A_1$, but in all of $\Bbb A^n$. To prove that $A_1$ is affine, you have to find a set of functions $S$ which vanish only on $A_1$ - that is $A_1$ is the whole zero locus of the functions in $S$.

Mathbalarka,

Thanks for your very interesting and helpful post ...

I am now thinking and reflecting over what you have written ...

Thanks again,

Peter***EDIT***Can I just confirm something that I think you have said ...

I think you a saying that there are some subsets of affine space that are not algebraic sets ... is that correct?If it is correct can you specify some examples ...?

 

[mathbalarka]

I think you a saying that there are some subsets of affine space that are not algebraic sets ... is that correct?

If it is correct can you specify some examples ...?

Yes, that is correct. I have given an example in my answer :

So, there's no reason for what you say to be true. To get an explicit example, the subset $\Bbb A^n - \{(0, 0, \cdots, 0)\}$ of $\Bbb A^n$ is not an affine variety. The proofs are quite involved, and require much more than basic commutative algebra. As a side-note, it is sometimes desired that locally closed subsets of affine varieties be varieties. Thus, we introduce the notion of quasi-affine varieties.

 

[Math Amateur]

Yes, that is correct. I have given an example in my answer :

Hi Mathbalarka,

Indeed you did give such an example ... Thanks ...

Peter