Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Properties of the map I

[Math Amateur]

I am reading Dummit and Foote Ch 15, Commutative Rings and Algebraic Geometry. In Section 15.1 Noetherian Rings and Affine Algebraic Sets, the set $$\mathcal{I} (A)$$ is defined in the following text on page 660: (see attachment)

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"While the ideal whose locus determines a particular algebraic set V is not unique, there is a unique largest ideal that determines V, given by the set of all polynomials that vanish on V.

In general, for any subset A of $$\mathbb{A}^n$$ define

$$\mathcal{I}(A) = \{ f \in k[x_1, x_2, ... \ ... \ , x_n ] \ | \ f ( a_1, a_2, ... \ ... \ , a_n) = 0$$ for all $$( a_1, a_2, ... \ ... \ , a_n) \in A \}$$"

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Then at the top of page 661 D&F write: (see attachment)

"The following properties of the map $$\mathcal{I}$$ are very easy exercises ...

Among these easy exercises is $$\mathcal{I}( \mathbb{A}^n ) = 0$$

Despite this being an easy exercise, I cannot see exactly why $$\mathcal{I}( \mathbb{A}^n ) = 0$$ :(

Can someone please help?

Peter

[This has also been posted on MHF]

 

[jvicens]

Dear Peter,

Thank you for bringing up this question. It is indeed an interesting and important property of the ideal map \mathcal{I} . Let me try to explain why \mathcal{I}( \mathbb{A}^n ) = 0 .

Firstly, it is important to note that the set \mathbb{A}^n represents the entire affine space, which means that it includes all possible points in \mathbb{A}^n , including points with coordinates that are not algebraic numbers. This means that for any polynomial f in \mathcal{I}( \mathbb{A}^n ) , its value at these points would be undefined, and therefore it cannot be a member of \mathcal{I}( \mathbb{A}^n ) .

Secondly, the definition of \mathcal{I}(A) states that it contains all polynomials that vanish on the set A . Since \mathbb{A}^n contains all possible points, any polynomial that vanishes on all points in \mathbb{A}^n must be identically zero. This means that the only polynomial in \mathcal{I}( \mathbb{A}^n ) is the zero polynomial, which makes \mathcal{I}( \mathbb{A}^n ) = 0 .

I hope this helps to clarify your doubts. If you have any further questions, please do not hesitate to ask. Keep up the good work with your studies!