[proofplan]
We use the ascending-chain formulation of the Noetherian condition: a ring is Noetherian when every ascending chain of ideals stabilises. First, assuming this chain condition, we prove that an ideal which were not finitely generated would allow us to choose elements producing a strictly increasing chain of finitely generated subideals. Conversely, if every ideal is finitely generated, the union of any ascending chain of ideals is itself an ideal, and finite generation of that union forces all generators to appear at one finite stage, so the chain stabilises.
[/proofplan]
[step:Derive finite generation of every ideal from the ascending chain condition]
Assume that $R$ is Noetherian, meaning that every ascending chain of ideals of $R$ stabilises. Let $I \trianglelefteq R$ be an ideal.
Suppose, for contradiction, that $I$ is not finitely generated. Since the zero ideal is generated by $0_R$, the ideal $I$ is not the zero ideal. Choose $a_1 \in I$. Having chosen elements $a_1,\ldots,a_n \in I$ for some $n \in \mathbb{N}$, define the finitely generated ideal
\begin{align*}
J_n := (a_1,\ldots,a_n) \trianglelefteq R.
\end{align*}
Because $I$ is not finitely generated, $J_n \neq I$. Since $J_n \subset I$, there exists $a_{n+1} \in I \setminus J_n$. This recursive construction gives a sequence $(a_n)_{n \in \mathbb{N}}$ of elements of $I$ such that, for every $n \in \mathbb{N}$,
\begin{align*}
J_n \subsetneq J_{n+1}.
\end{align*}
Indeed, $J_n \subset J_{n+1}$ because the generators of $J_n$ are among the generators of $J_{n+1}$, and the inclusion is proper because $a_{n+1} \in J_{n+1}$ while $a_{n+1} \notin J_n$. Thus
\begin{align*}
J_1 \subsetneq J_2 \subsetneq J_3 \subsetneq \cdots
\end{align*}
is an ascending chain of ideals which does not stabilise. This contradicts the Noetherian hypothesis. Therefore $I$ is finitely generated.
[/step]
[step:Show that finite generation of all ideals forces every ascending chain to stabilise]
Assume that every ideal of $R$ is finitely generated. Let $(I_n)_{n \in \mathbb{N}}$ be an ascending chain of ideals of $R$, so
\begin{align*}
I_1 \subset I_2 \subset I_3 \subset \cdots.
\end{align*}
Define the subset
\begin{align*}
I := \bigcup_{n=1}^{\infty} I_n \subset R.
\end{align*}
We first check that $I$ is an ideal. If $x,y \in I$, then there exist $m,n \in \mathbb{N}$ such that $x \in I_m$ and $y \in I_n$. Let $N := \max\{m,n\}$. Since the chain is ascending, $x,y \in I_N$, and hence $x+y \in I_N \subset I$. If $r \in R$ and $x \in I$, choose $m \in \mathbb{N}$ with $x \in I_m$. Since $I_m$ is an ideal, $rx \in I_m \subset I$. Also $0_R \in I_1 \subset I$. Hence $I \trianglelefteq R$.
By hypothesis, $I$ is finitely generated. Choose generators $a_1,\ldots,a_k \in I$, where $k \in \mathbb{N}$, such that
\begin{align*}
I = (a_1,\ldots,a_k).
\end{align*}
For each index $j \in \{1,\ldots,k\}$, since $a_j \in I = \bigcup_{n=1}^{\infty} I_n$, there exists $n_j \in \mathbb{N}$ such that $a_j \in I_{n_j}$. Define
\begin{align*}
N := \max\{n_1,\ldots,n_k\}.
\end{align*}
Because the chain is ascending, every generator $a_j$ lies in $I_N$. Therefore the ideal generated by these elements satisfies
\begin{align*}
I = (a_1,\ldots,a_k) \subset I_N.
\end{align*}
The reverse inclusion $I_N \subset I$ follows from the definition of $I$ as the union of the chain. Hence $I = I_N$. For every $m \geq N$, we have
\begin{align*}
I_N \subset I_m \subset I = I_N.
\end{align*}
Thus $I_m = I_N$ for every $m \geq N$, so the chain stabilises.
[guided]
The goal is to prove the ascending chain condition. So we begin with an arbitrary ascending chain of ideals
\begin{align*}
I_1 \subset I_2 \subset I_3 \subset \cdots.
\end{align*}
The natural object attached to this chain is its union. Define
\begin{align*}
I := \bigcup_{n=1}^{\infty} I_n \subset R.
\end{align*}
Before using the hypothesis that every ideal is finitely generated, we must verify that this union is actually an ideal. The ascending condition is exactly what makes this true. If $x,y \in I$, then $x \in I_m$ and $y \in I_n$ for some $m,n \in \mathbb{N}$. Set $N := \max\{m,n\}$. Since the chain is ascending, both $I_m$ and $I_n$ are contained in $I_N$, so $x,y \in I_N$. Because $I_N$ is an ideal, $x+y \in I_N$, and therefore $x+y \in I$. If $r \in R$ and $x \in I$, choose $m \in \mathbb{N}$ with $x \in I_m$. Since $I_m$ is an ideal, $rx \in I_m$, hence $rx \in I$. Finally, $0_R \in I_1 \subset I$. Thus $I \trianglelefteq R$.
Now the finite-generation hypothesis applies to this ideal $I$. Choose elements $a_1,\ldots,a_k \in I$, for some $k \in \mathbb{N}$, such that
\begin{align*}
I = (a_1,\ldots,a_k).
\end{align*}
Each generator $a_j$ belongs to the union $I$, so each generator appears at some finite stage of the chain. For every $j \in \{1,\ldots,k\}$, choose $n_j \in \mathbb{N}$ such that $a_j \in I_{n_j}$. Define
\begin{align*}
N := \max\{n_1,\ldots,n_k\}.
\end{align*}
Because the chain is ascending, $I_{n_j} \subset I_N$ for every $j$, so every generator $a_j$ lies in $I_N$. Since $I_N$ is an ideal containing all the generators of $I$, it contains the ideal they generate:
\begin{align*}
I = (a_1,\ldots,a_k) \subset I_N.
\end{align*}
The containment $I_N \subset I$ follows directly from $I = \bigcup_{n=1}^{\infty} I_n$. Therefore $I = I_N$. Now if $m \geq N$, the chain gives $I_N \subset I_m$, while the definition of $I$ gives $I_m \subset I = I_N$. Hence
\begin{align*}
I_m = I_N.
\end{align*}
So the chain is constant from stage $N$ onward. Since the original chain was arbitrary, every ascending chain of ideals stabilises.
[/guided]
[/step]
[step:Conclude the equivalence]
The first step proves that the Noetherian ascending chain condition implies finite generation of every ideal. The second step proves that finite generation of every ideal implies the ascending chain condition. Therefore $R$ is Noetherian if and only if every ideal $I \trianglelefteq R$ is finitely generated.
[/step]
