[proofplan] We prove the equivalence by passing through the ascending-chain formulation of the Noetherian condition. First, from the finite-generation characterisation of Noetherian rings, every ascending chain of ideals stabilises. If a nonempty family of ideals had no maximal member, one could recursively choose a strictly increasing sequence inside it, contradicting this stabilisation. Conversely, the maximal condition applied to the set of terms in any ascending chain forces that chain to stabilise, and the finite-generation characterisation then identifies this with Noetherianity. [/proofplan]

[step:Derive stabilisation of ascending chains from finite generation of ideals]Assume that $R$ is Noetherian. By [citetheorem:9999], every ideal of $R$ is finitely generated. Let $(I_n)_{n \in \mathbb{N}}$ be an ascending chain of ideals of $R$, so $I_n \trianglelefteq R$ and $I_n \subset I_{n+1}$ for every $n \in \mathbb{N}$. Define $I := \bigcup_{n=1}^{\infty} I_n$. Then $I \trianglelefteq R$. Indeed, if $a,b \in I$, then there exist $m,n \in \mathbb{N}$ with $a \in I_m$ and $b \in I_n$; for $N := \max\{m,n\}$, the ascending-chain property gives $a,b \in I_N$, hence $a-b \in I_N \subset I$. If $r \in R$ and $a \in I$, then $a \in I_m$ for some $m \in \mathbb{N}$, so $ra \in I_m \subset I$. Since every ideal of $R$ is finitely generated, choose elements $a_1,\ldots,a_k \in I$ such that $I = (a_1,\ldots,a_k)$. For each index $j \in \{1,\ldots,k\}$, choose $n_j \in \mathbb{N}$ such that $a_j \in I_{n_j}$. Let $N := \max\{n_1,\ldots,n_k\}$. Then $a_j \in I_N$ for every $j \in \{1,\ldots,k\}$, so $I = (a_1,\ldots,a_k) \subset I_N$. Since also $I_N \subset I$, we have $I = I_N$. Therefore, for every $n \geq N$, $I_N \subset I_n \subset I = I_N$, and hence $I_n = I_N$. Thus every ascending chain of ideals of $R$ stabilises.[/step]

[guided]The goal of this step is to convert finite generation into a chain condition. Assume $R$ is Noetherian. By [citetheorem:9999], this means that every ideal of $R$ is finitely generated. Let $(I_n)_{n \in \mathbb{N}}$ be an ascending chain of ideals: $I_1 \subset I_2 \subset I_3 \subset \cdots$. We want to prove that the chain eventually stops changing. The standard way to do this is to collect all terms of the chain into one ideal. Define $I := \bigcup_{n=1}^{\infty} I_n$. We first verify that $I$ is an ideal, because finite generation can only be applied to ideals. If $a,b \in I$, then $a \in I_m$ and $b \in I_n$ for some $m,n \in \mathbb{N}$. Taking $N := \max\{m,n\}$, the inclusions in the chain imply $a,b \in I_N$. Since $I_N$ is an ideal, $a-b \in I_N$, hence $a-b \in I$. Similarly, if $r \in R$ and $a \in I$, then $a \in I_m$ for some $m \in \mathbb{N}$, and the ideal property of $I_m$ gives $ra \in I_m \subset I$. Thus $I \trianglelefteq R$. Now apply finite generation to $I$. There exist elements $a_1,\ldots,a_k \in I$ such that $I = (a_1,\ldots,a_k)$. Each generator $a_j$ belongs to the union $I$, so each generator occurs in some stage of the chain. For every $j \in \{1,\ldots,k\}$, choose $n_j \in \mathbb{N}$ with $a_j \in I_{n_j}$, and define $N := \max\{n_1,\ldots,n_k\}$. Because the chain is ascending, every generator $a_j$ lies in $I_N$. Therefore the ideal generated by these elements is contained in $I_N$: $I = (a_1,\ldots,a_k) \subset I_N$. The reverse inclusion $I_N \subset I$ holds by definition of $I$ as the union of the chain. Hence $I = I_N$. Finally, for every $n \geq N$, the chain gives $I_N \subset I_n \subset I = I_N$. Therefore $I_n = I_N$ for every $n \geq N$. This proves that every ascending chain of ideals stabilises.[/guided]

[step:Use stabilisation to find a maximal ideal in every nonempty family] Let $\mathcal{F}$ be a nonempty collection of ideals of $R$. Suppose, for contradiction, that $\mathcal{F}$ has no maximal element with respect to inclusion. Choose $I_1 \in \mathcal{F}$. Since $I_1$ is not maximal in $\mathcal{F}$, there exists $I_2 \in \mathcal{F}$ such that $I_1 \subsetneq I_2$. Recursively, after $I_n \in \mathcal{F}$ has been chosen, the failure of maximality gives an ideal $I_{n+1} \in \mathcal{F}$ such that $I_n \subsetneq I_{n+1}$. This produces a strictly ascending chain of ideals of $R$, contradicting the stabilisation proved in the previous step. Hence $\mathcal{F}$ contains a maximal element. [/step]

[step:Show the maximal condition forces every ascending chain to stabilise] Assume that every nonempty collection of ideals of $R$ has a maximal element with respect to inclusion. Let $(I_n)_{n \in \mathbb{N}}$ be an ascending chain of ideals of $R$. Define the nonempty collection $\mathcal{F} := \{I_n : n \in \mathbb{N}\}$. By hypothesis, there exists $I_N \in \mathcal{F}$ that is maximal in $\mathcal{F}$ with respect to inclusion. For every $n \geq N$, the ascending-chain property gives $I_N \subset I_n$. Since $I_n \in \mathcal{F}$ and $I_N$ is maximal in $\mathcal{F}$, this inclusion forces $I_n = I_N$. Therefore the chain stabilises at $N$. [/step]

[step:Recover finite generation of every ideal from chain stabilisation] Assume every ascending chain of ideals of $R$ stabilises. Let $I \trianglelefteq R$ be an ideal. Suppose, for contradiction, that $I$ is not finitely generated. Choose $a_1 \in I$. Since $(a_1) \neq I$, choose $a_2 \in I \setminus (a_1)$. Recursively, after $a_1,\ldots,a_n \in I$ have been chosen, the ideal $J_n := (a_1,\ldots,a_n)$ is a finitely generated subideal of $I$, hence $J_n \neq I$. Choose $a_{n+1} \in I \setminus J_n$. Then $J_n \subsetneq J_{n+1}$, where $J_{n+1} := (a_1,\ldots,a_n,a_{n+1})$. Thus $(J_n)_{n \in \mathbb{N}}$ is a strictly ascending chain of ideals of $R$, contradicting stabilisation. Hence every ideal of $R$ is finitely generated. By [citetheorem:9999], $R$ is Noetherian. [/step]

[step:Combine the two implications] The first two steps show that if $R$ is Noetherian, then every nonempty collection of ideals of $R$ has a maximal element with respect to inclusion. The last two steps show that this maximal condition forces every ascending chain of ideals to stabilise, hence every ideal is finitely generated, and therefore $R$ is Noetherian by [citetheorem:9999]. This proves the equivalence. [/step]