[proofplan]
We prove the Noetherian property by translating a descending chain of Zariski closed subsets into an ascending chain of ideals in the [polynomial ring](/page/Polynomial%20Ring) $k[x_1,\dots,x_n]$. The inclusion reversal between subsets and vanishing ideals gives monotonicity in the correct direction. Since $k[x_1,\dots,x_n]$ is Noetherian by the [Hilbert Basis Theorem](/theorems/860), the ideal chain stabilizes. Finally, every Zariski closed subset is recovered from its vanishing ideal, so stabilization of the ideals forces stabilization of the original closed subsets.
[/proofplan]
[step:Translate the descending chain of closed sets into vanishing ideals]
Let $F_1\supseteq F_2\supseteq F_3\supseteq \cdots$ be a descending chain of Zariski closed subsets of $\mathbb A_k^n$. Define $R:=k[x_1,\dots,x_n]$. For every subset $E\subseteq R$, let $V(E)$ denote the zero set $V(E):=\{a\in \mathbb A_k^n:f(a)=0\text{ for every }f\in E\}$. For every subset $S\subseteq \mathbb A_k^n$, let $I(S):=\{f\in R:f(a)=0\text{ for every }a\in S\}$ be its vanishing ideal.
For each $i\geq 1$, the set $I(F_i)$ is an ideal of $R$: the zero polynomial vanishes on $F_i$; if $f,g\in I(F_i)$ and $a\in F_i$, then $(f+g)(a)=f(a)+g(a)=0$; and if $h\in R$, then $(hf)(a)=h(a)f(a)=0$. Since $F_{i+1}\subseteq F_i$ for every $i\geq 1$, every polynomial that vanishes on $F_i$ also vanishes on $F_{i+1}$. Hence
\begin{align*} I(F_i)\subseteq I(F_{i+1}) \end{align*}
for every $i\geq 1$. Thus
\begin{align*} I(F_1)\subseteq I(F_2)\subseteq I(F_3)\subseteq \cdots \end{align*}
is an ascending chain of ideals in $R$.
[guided]
Start with an arbitrary descending chain of Zariski closed subsets:
\begin{align*} F_1\supseteq F_2\supseteq F_3\supseteq \cdots. \end{align*}
The goal is to prove that this chain eventually stops changing. The useful algebraic object attached to a subset $S\subseteq \mathbb A_k^n$ is its vanishing ideal, so we define
\begin{align*} R:=k[x_1,\dots,x_n] \end{align*}
and
\begin{align*} I(S):=\{f\in R:f(a)=0\text{ for every }a\in S\}. \end{align*}
For every subset $E\subseteq R$, the notation $V(E)$ denotes the zero set
\begin{align*} V(E):=\{a\in \mathbb A_k^n:f(a)=0\text{ for every }f\in E\}. \end{align*}
First check that $I(F_i)$ is an ideal of $R$ for every $i\geq 1$. The zero polynomial is in $I(F_i)$ because it vanishes at every point. If $f,g\in I(F_i)$ and $a\in F_i$, then $(f+g)(a)=f(a)+g(a)=0$, so $f+g\in I(F_i)$. If $h\in R$ and $f\in I(F_i)$, then $(hf)(a)=h(a)f(a)=0$ for every $a\in F_i$, so $hf\in I(F_i)$.
Why does this construction reverse inclusions? If $F_{i+1}\subseteq F_i$, then a polynomial that vanishes on the larger set $F_i$ automatically vanishes on the smaller set $F_{i+1}$. Therefore every element of $I(F_i)$ is an element of $I(F_{i+1})$, and we have
\begin{align*} I(F_i)\subseteq I(F_{i+1}) \end{align*}
for every $i\geq 1$.
Thus the descending chain of closed subsets has been converted into an ascending chain of ideals:
\begin{align*} I(F_1)\subseteq I(F_2)\subseteq I(F_3)\subseteq \cdots. \end{align*}
This is the point of passing to vanishing ideals: Noetherianity of the polynomial ring controls ascending ideal chains.
[/guided]
[/step]
[step:Use Hilbert Basis Theorem to stabilize the ideal chain]
Since $k$ is a field, its only ideals are $(0)$ and $k$, so every ascending chain of ideals in $k$ stabilizes and $k$ is a Noetherian ring. By the Hilbert Basis Theorem applied successively to adjoining the variables $x_1,\dots,x_n$, the polynomial ring
\begin{align*} R=k[x_1,\dots,x_n] \end{align*}
is Noetherian. By the definition of a Noetherian ring, every ascending chain of ideals in $R$ stabilizes. Applied to the chain from the previous step, there exists an integer $N\geq 1$ such that
\begin{align*} I(F_i)=I(F_N) \end{align*}
for every $i\geq N$.
[/step]
[step:Recover each closed set from its vanishing ideal]
We claim that every Zariski closed subset $F\subseteq \mathbb A_k^n$ satisfies
\begin{align*}
F=V(I(F)).
\end{align*}
[claim:Closed subsets equal the zero set of their vanishing ideal]
Let $F\subseteq \mathbb A_k^n$ be Zariski closed. Then
\begin{align*}
F=V(I(F)).
\end{align*}
[/claim]
[proof]
By definition of the Zariski topology and the equivalent closed-set description in the theorem statement, there exists an ideal $J\trianglelefteq R$ such that
\begin{align*}
F=V(J).
\end{align*}
If $a\in F$ and $f\in I(F)$, then by definition of $I(F)$ one has $f(a)=0$. Therefore $a\in V(I(F))$, and hence
\begin{align*}
F\subseteq V(I(F)).
\end{align*}
For the reverse inclusion, let $a\in \mathbb A_k^n\setminus F$. Since $F=V(J)$, the point $a$ is not a common zero of all polynomials in $J$. Therefore there exists $f\in J$ such that
\begin{align*}
f(a)\neq 0.
\end{align*}
Because $F=V(J)$, every polynomial in $J$ vanishes on every point of $F$, so $f\in I(F)$. Since $f(a)\neq 0$, the point $a$ is not in $V(I(F))$. Thus
\begin{align*}
V(I(F))\subseteq F.
\end{align*}
Combining the two inclusions gives
\begin{align*}
F=V(I(F)).
\end{align*}
[/proof]
[/step]
[step:Conclude that the original closed-set chain stabilizes]
Let $N\geq 1$ be such that
\begin{align*}
I(F_i)=I(F_N)
\end{align*}
for every $i\geq N$. For every $i\geq N$, the previous step gives
\begin{align*}
F_i=V(I(F_i)).
\end{align*}
Using the equality of ideals, we obtain
\begin{align*}
F_i=V(I(F_N)).
\end{align*}
Again applying the recovery of closed sets from their vanishing ideals to $F_N$, we get
\begin{align*}
V(I(F_N))=F_N.
\end{align*}
Therefore
\begin{align*}
F_i=F_N
\end{align*}
for every $i\geq N$. Hence every descending chain of Zariski closed subsets of $\mathbb A_k^n$ stabilizes, so $\mathbb A_k^n$ is Noetherian.
[/step]
