The $\textit{ascending chain condition}$ on $k[x]:=k[x_1,\ldots,x_n]$ (where $k$ is a field) says that if $I_1\subset I_2\subset ...$ is an ascending chain of ideals in $k[x]$, then there exists some positive integer $N$ such that $I_N=I_{N+1}=I_{N+2}=\ldots$.

The ascending chain condition is equivalent to Hilbert's basis theorem, i.e. equivalent to the statement that every ideal in $k[x]$ is finitely generated.

### Solution:

First note that $I_\infty:=\bigcup_{j\in\mathbb{N}} I_j$ is an ideal of $k[x]$.

Therefore, given that every ideal of $k[x]$ is finitely generated, choose a finite set $F$ of generators of $I_\infty$. So there is an $N\in\mathbb{N}$ such that $F\subset I_N$. Since the chain is ascending, this means $I_N=I_{N+1}=I_{N+2}=\ldots$.

Conversely, given an ideal $I\subset k[x]$, choose polynomials $f_i\in I$ inductively in the following way: Start with an arbitrary $f_1\in I$. Define $I_j=\langle f_1,\ldots,f_j\rangle$. Then choose $f_{j+1}\in I\setminus I_j$. This ascending chain terminates by assumption and we have $I_{\infty}=I_N$ for some $N$. This implies we cannot find $f_{N+1}\in I$ and $f_{N+1}\notin I_N$. Thus we have exhausted all of [[$I$ and we have $I=I_{\infty}=I_N$. Therefore $I$ is finitely generated.