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The book "Invitation to Algebraic Geometry" says the following:

A Zariski closed set in $\Bbb{A}^n$ is compact in the Zariski topology.

Why is this this the case? According to the Hilbert Basis Theorem, I can understand that any open cover of the complement of a variety will have a finite subcover. But I don't understand why an open cover of a variety will have a finite subcover.

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  • $\begingroup$ It is quasi-compact. $\endgroup$ – Bernard Sep 23 '16 at 18:40
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Let $Z\subset\mathbb{A}^n=X$ a closed subset; by definition $$ \exists f_1,\dots,f_r\in\mathbb{K}[x_1,\dots,x_n]=R\mid Z=V(f_1,\dots,f_r)=\bigcap_{i=1}^rV(f_r), $$ withous loss of the generality we can assume $r=1$ and we put $f_1=f$ so that $Z=V(f)$; because $R$ is an U.F.D., then \begin{gather*} \exists g_1,\dots,g_s\in R\mid f=g_1\cdot\dots\cdot g_s,\,g_j\,\text{is a prime polynomial}\\ Z=V(g_1\cdot\dots\cdot g_s)=\bigcup_{j=1}^sV(g_j), \end{gather*} without loss of generality we can assume $s=1$, in other words $f$ is a prime polynomial and $Z$ is an irreducible closed subset of $X$.

Let $\{U_i\}_{i\in I}$ an open covering of $Z$, for exact: $$ Z\subseteq\bigcup_{i\in I}U_i, $$ by previous reasoning, we can assume (without loss of the generality) that $U_i$'s are irreducible; by definition: \begin{gather*} \forall i\in I,\exists f_i\in R\mid U_i=D(f_i)=X\setminus V(f_i),\\ \bigcup_{i\in I}U_i=\dots=X\setminus\bigcap_{i\in I}V(f_i)=X\setminus W; \end{gather*} we know that $W$ is a closed subset of $X$, let $I(W)$ be the associated ideal of $W$, by Hilbert's Base theorem, it is finitely generated; by this statement \begin{gather*} \exists I_F\subseteq I\,\text{finite,}\,\{f_i\in R\}_{i\in I_F}\mid I(W)=(f_i\mid i\in I_F)\Rightarrow\\ \Rightarrow Z\subseteq\bigcup_{i\in I}U_i=X\setminus V(I(W))=X\setminus\bigcap_{i\in I_F}V(f_i)=\bigcup_{i\in I_F}U_i \end{gather*} and the claims follows. (Q.E.D.) $\Box$

Remark: I had use only the hypothesys that $X$ is an affine space over a field, independently from its characteristic and other algebraic properties!

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  • $\begingroup$ What is $D(f_i)$? $\endgroup$ – user67803 Sep 27 '16 at 4:27
  • $\begingroup$ Sorry me, I wrote $U_i=D(f_i)=X\setminus V(f_i)$; or in other words, by definition $D(f_i)$ is equal to $X\setminus V(f_i)$! $\endgroup$ – Armando j18eos Sep 27 '16 at 18:46
  • $\begingroup$ How do you gurantee that $W$ will be generated by a finite subset of $\{f_i\}$? It will be generated by a finite collection for sure by Hilbert basis theorem, but why that will be a finite subset of $\{f_i\}$? $\endgroup$ – Ri-Li Apr 22 '19 at 1:39
  • $\begingroup$ $I(W)$ is an ideal of $\mathbb{K}[x_1,\dots,x_n]$, so it is a finitely generated ideal by Hilbert's Base Theorem; because $W$ is closed then $W=\overline{W}=V(I(W))$. Is it more clear? $\endgroup$ – Armando j18eos Apr 22 '19 at 11:01
  • $\begingroup$ I have mentioned in my comment that it will be generated by a finite collection by Hilbert base theorem(this part is clear). My question is why that finite collection has to be a finite subset of $\{f_i\}_{i \in I}$? It can be any other finite subcollection, right! say $\{g_1,...,g_n\} \nsubseteq \{f_i\}_{i \in I}$. $\endgroup$ – Ri-Li Apr 22 '19 at 22:39
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Hint: If $V$ is closed and additionally irreducible and covered by $\bigcup _{i\in I}U_i$, pick $x\in V$ and then $i\in I$ with $x\in U_i$. Then $V\setminus U_i$ is closed and of lower dimension.

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  • $\begingroup$ I'm no sure I understand what "of lower dimension" means. Are you using a form of induction on a cover that could possibly be infinite to begin with? $\endgroup$ – user67803 Sep 23 '16 at 18:00
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    $\begingroup$ Oh okay. I guess you're using the fact that a closed set in he Zariski topology is Noetherian. $\endgroup$ – user67803 Sep 23 '16 at 18:01
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A Zariski closed set in $A^n$ is a noetherian space. Thus it is compact.

https://en.m.wikipedia.org/wiki/Noetherian_topological_space

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