algebraic sets and polynomial idealsAlgebraicSetsAndPolynomialIdeals2002-10-08 14:44:462007-05-09 22:03:09Definitionzero setalgebraic setideal of an algebraic setaffine algebraic set% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\A}{\mathbb{A}}Suppose $k$ is a field. Let $\A^n_k$ denote affine $n$-space over $k$.\\
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For $S \subseteq k[x_1,\ldots,x_n]$, define $V(S)$, the {\em zero set of $S$}, by
\[ V(S) = \{(a_1,\ldots,a_n) \in k^n \mid f(a_1,\ldots,a_n)=0 \text{ for all } f \in S\}\]
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We say that $Y \subseteq \A^n_k$ is an (affine) {\em 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 $\A^n_k$.\\
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For $Y \subseteq \A^n_k$, define the {\em {\PMlinkescapetext ideal} of $Y$ in $k[x_1,\ldots,x_n]$} by \[ I(Y)=\{f \in k[x_1,\ldots,x_n] \mid f(P)=0 \text{ for all } P \in Y\}. \]
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It is easily shown that $I(Y)$ is an ideal of $k[x_1,\ldots,x_n]$.\\
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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 $k[x_1,\ldots,x_n]$.\\
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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:
\begin{enumerate}
\item $S_1 \subseteq S_2 \subseteq k[x_1,\ldots,x_n]$ implies
$V(S_1) \supseteq V(S_2)$.
\item $Y_1 \subseteq Y_2 \subseteq \A_k^n$ implies
$I(Y_1) \supseteq I(Y_2)$.
\item For any ideal $\mathfrak{a} \subset k[x_1,\ldots,x_n]$,
$I(V(\mathfrak{a}))=\operatorname{Rad}(\mathfrak{a})$.
\item For any $Y \subset \A^n_k$, $V(I(Y))=\overline{Y}$, the closure
of $Y$ in the Zariski topology.
\end{enumerate}
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.