Why is a Zariski closed set compact under the Zariski topology?

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.

Solution 1:

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!