Subsets in the Affine Plane and ideals of K[A^n]

[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 confirm some thoughts on the correspondence between D&F's definition of \(\displaystyle \mathcal{I}\) and the ideals of \(\displaystyle k[ \mathbb{A}^n ]\) ...

On page 660 (in Section 15.1) of D&F we find the following text:View attachment 4737In the above text we find the following text:

" ... ... It is immediate that \(\displaystyle \mathcal{I} (A)\) is an ideal, and is the unique largest ideal of functions that are identically zero on \(\displaystyle A\). This defines a correspondence

\(\displaystyle \mathcal{I} \ : \ \{ \text{ subsets in } k[ \mathbb{A}^n ] \} \ \rightarrow \ \{ \text{ ideals of } k[ \mathbb{A}^n ] \}\). ... ... "I am thinking about the above correspondence and how we can take any arbitrary set of points \(\displaystyle A\) in \(\displaystyle \mathbb{A}^n \)and find a set of functions that are identically zero on \(\displaystyle A\) ... ... I thought this may be impossible for many sets, maybe infinitely many sets \(\displaystyle A\) ... but then realized that in these cases the set of functions would be the zero function and the corresponding ideal would be \(\displaystyle \{ z \}\) where \(\displaystyle z\) is the zero function ...

Can someone please confirm that my thinking/reflections above are on the right track ... I am also worrying a bit about why \(\displaystyle \mathcal{I} (A)\) is the unique and largest such ideal ... ...

Hope someone can help with these issues ... simple though they may be ...

Peter

 

[mathbalarka]

I am thinking about the above correspondence and how we can take any arbitrary set of points A in A^n and find a set of functions that are identically zero on A ... ... I thought this may be impossible for many sets, maybe infinitely many sets A

I have no idea what you mean by that. Why should this be impossible? You go through the polynomials in $k[x_1, \cdots, x_n]$ which vanish in all of $A$ identically, and throw out the ones that doesn't. The remaining polynomials is your set.

but then realized that in these cases the set of functions would be the zero function and the corresponding ideal would be {z} where z is the zero function ...

Sorry, that's not valid. If you have a subset $A$ of $\Bbb A^n$, then $\mathcal{I}(A)$ is the set of _all_ polynomials in $k[x_1, \cdots, x_n]$ which vanish identically on $A$. The zero function is an element of $\mathcal{I}$, but there maybe more functions in there.

I am also worrying a bit about why I(A) is the unique and largest such ideal ... ...

This follows by definition. $\mathcal{I}(A)$ is _by definition_ the unique and largest ideal of functions vanishing on $A$.