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Is every Zariski closed set defined by a set of equations? Then the complement of a Zariski closed set is a Zariski open set. Thank you very much.

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3 Answers 3

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Given an algebraically closed field $k$, a subset $X\subseteq k^n$ is Zariski closed, by definition, if $X= V(I)$ for some ideal $I\subseteq k[x_1,\ldots,x_n]$. But because $k[x_1,\ldots,x_n]$ is noetherian (by the Hilbert basis theorem), any ideal is finitely generated, so $I=(f_1,\ldots,f_t)$ for some finite collection of $f_i\in k[x_1,\ldots,x_n]$. Thus $$X=V(I)=V(f_1,\ldots,f_t)=\{p\in k^n\mid f_i(p)=0\text{ for all }i\}.$$ Thus, we have shown that any Zariski closed subset of $k^n$ is defined by a set of equations.

Is this what you are after?

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Let $A$ be a ring, and let $X=\mathrm{Spec} \ A$. A subset $V\subset X$ is Zariski closed if and only if there is some subset $S\subset A$ such that $$V=\{x\in X : s(x) =0 \ \mathrm{in} \ k(x) \ \mathrm{for} \ \mathrm{all} \ s \in S\}.$$ Here $s(x)$ is the image of $s$ under the natural morphism $A\to k(x)$, where $k(x)$ denotes the residue field of $x$. Informally, $V$ is the zero locus of the subset $S\in A$.

If $A$ is noetherian, we can choose $S$ to be finite.

Zev Chonoles' answer treats the case $A=k[x_1,\ldots,x_n]$, i.e, $X= \mathbf{A}^n_k$ with $k$ an algebraically closed field.

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The Zariski open sets are formed by the topological basis of basic open sets $$ D_f = \left\{ x \in X \mid f(x) \ne 0 \right\} $$ and any Zariski open set is a union (maybe infinite union) of such sets.

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