Q:

Why is the map $I$ from subsets of $\mathbb{A}_k^n$ to ideals of $k[x_1,\ldots,x_n]$ not surjective?

In algebraic geometry, we consider the map   
$$I:\{\text{subsets of }\mathbb{A}_k^n\}\longrightarrow\{\text{ideals of }k[x_1,\cdots,x_n]\},\qquad X\mapsto I(X)$$
This map is not injective, because $I(\mathbb{A_k^n})=0$. But why it is not surjective? How to find a counterexample? Thank you!

A:

Hint: Suppose that $x_1^2$ vanishes on a subset $S\subseteq \mathbb{A}_k^n$. What other polynomial can you deduce vanishes on the subset $S$ as well? Can the ideal $(x_1^2)\subset k[x_1,\ldots,x_n]$ be $I(S)$ for any $S$, then?

A:

The image consists precisely of the radical ideals.

