PlanetMath: algebraic sets and polynomial ideals

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algebraic sets and polynomial ideals

(Definition)

Suppose is a field. Let $\mathbb{A}^n_k$ denote affine -space over .

For $S \subseteq k[x_1,\ldots,x_n]$ , define , the zero set of , by

$\displaystyle V(S) = \{(a_1,\ldots,a_n) \in k^n \mid f(a_1,\ldots,a_n)=0$ for all $\displaystyle f \in S\}$

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

For $Y \subseteq \mathbb{A}^n_k$ , define the ideal of in $k[x_1,\ldots,x_n]$ by

$\displaystyle I(Y)=\{f \in k[x_1,\ldots,x_n] \mid f(P)=0$ for all $\displaystyle P \in Y\}.$

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

Thus we have defined a function mapping from subsets of $k[x_1,\ldots,x_n]$ to algebraic sets in $\mathbb{A}^n_k$ , and a function mapping from subsets of $\mathbb{A}^n$ 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:

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

From the above, we see that there is a 1-1 correspondence between algebraic sets in $\mathbb{A}^n_k$ and radical ideals of $k[x_1,\ldots,x_n]$ . Furthermore, an algebraic set $Y \subseteq \mathbb{A}^n_k$ is an affine variety if and only if is a prime ideal. As an example of how things can go wrong, the radical ideals and 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.