Suppose $k$ is a field. Let $\A^n_k$ denote affine $n$ -space over $k$ .

For $S \subseteq k[x_1,\ldots,x_n]$ , define $V(S)$ , the zero set of $S$ , by $$ V(S) = \{(a_1,\ldots,a_n) \in k^n \mid f(a_1,\ldots,a_n)=0 \text{ for all } f \in S\ $$

We say that $Y \subseteq \A^n_k$ is an (affine) algebraic set if there exists $T \subseteq k[x_1,\ldots,x_n]$ such that $Y=V(T)$ . Taking these subsets of $\A^n_k$ as a definition of the closed sets of a topology induces the Zariski topology over $\A^n_k$ .

For $Y \subseteq \A^n_k$ , define the ideal of $Y$ in $k[x_1,\ldots,x_n]$ by $$ I(Y)=\{f \in k[x_1,\ldots,x_n] \mid f(P)=0 \text{ for all } P \in Y\}. $$

It is easily shown that $I(Y)$ is an ideal of $k[x_1,\ldots,x_n]$ .

Thus we have defined a function $V$ mapping from subsets of $k[x_1,\ldots,x_n]$ to algebraic sets in $\A^n_k$ , and a function $I$ mapping from subsets of $\A^n$ to ideals of $k[x_1,\ldots,x_n]$ .

We remark that the theory of algebraic sets presented herein is most cleanly stated over an algebraically closed field. For example, over such a field, the above have the following properties:

1. $S_1 \subseteq S_2 \subseteq k[x_1,\ldots,x_n]$ implies $V(S_1) \supseteq V(S_2)$ .
2. $Y_1 \subseteq Y_2 \subseteq \A_k^n$ implies $I(Y_1) \supseteq I(Y_2)$ .
3. For any ideal $\mathfrak{a} \subset k[x_1,\ldots,x_n]$ , $I(V(\mathfrak{a}))=\operatorname{Rad}(\mathfrak{a})$ .
4. For any $Y \subset \A^n_k$ , $V(I(Y))=\overline{Y}$ , the closure of $Y$ in the Zariski topology.

From the above, we see that there is a 1-1 correspondence between algebraic sets in $\A^n_k$ and radical ideals of $k[x_1,\ldots,x_n]$ . Furthermore, an algebraic set $Y \subseteq \A^n_k$ is an affine variety if and only if $I(Y)$ is a prime ideal. As an example of how things can go wrong, the radical ideals $(1)$ and $(x^2+1)$ in $\mathbb{R}[x]$ define the same zero locus (the empty set) inside of $\mathbb{R}$ , but are not the same ideal, and hence there is no such 1-1 correspondence.