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I was trying to proof that any Artinian ring is a Noetherian ring, but I found on the Web that the proof is non trivial. However I have a proof I hope you can help me in finding errors.

Let $R $ be a non Noetherian ring, or there exists an ideal $I \lhd R$ such that $d_R (I)=\infty $ where $$d_R (I)=min (\{n\in \mathbb N | \exists a_1,...,a_n\in I : I=\displaystyle \sum _{1\leq i\leq n} Ra_i\} \cup \{\infty\}).$$ I want to show that any such ideal contains an ideal non finitely generated and thus I can create a non stationary descending chain of ideals, and show $R$ is not Artinian. Now let $G=\{a_i\}_{i\in J}$ be a generator system for $I$ and suppose it is minimal, or $\forall j\in J,\ I=\displaystyle \sum_{a\in G} Ra \supsetneq \displaystyle \sum _{a\in G\setminus \{a_j\}} Ra$. Thus, $J $ is infinite because $d_R (I)=\infty $. Let now $J'\subseteq J $ such that $J'=\{a_{k_i}\}_{i\in \mathbb N}$ is countable (its existence is assured by the minimality of cardinality $\aleph _0$ in infinite sets). Now let be $R\rhd I_1=\displaystyle \sum_ {a\in G\setminus \{a_{k_1}\}}\subseteq I.$ Now, if $I=I_1$, then $G $ is not minimal. Thus $I_1\subsetneq I $. Let now $G_1$ be a generator minimal system for $I_1$: $G_1$ is infinite because otherwise $I=I_1+a_{k_1} $ is finitely generated.

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You seem to be assuming that every ideal has a minimal generating set. This is not necessarily true for non-Noetherian rings, so you would need to justify why it is true for Artinian rings.

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  • $\begingroup$ Thus may I consider a non-finitely generated ideal constructed as follows: let $I$ be the ideal such that $d_R (I)=\infty $. For $k\in \mathbb N $ be $a_{k+1}\in I$ such that $a_{k+1}\in I\setminus \displaystyle \sum _ {i=1}^k Ra_i $ and $a_1\in I\setminus {0}$. Be $\mathfrak I=\displaystyle \sum _{i=1}^{\infty} Ra_i $ and $\mathfrak I_1$ constructed like before. $\endgroup$ – Blumer Feb 24 '17 at 14:13
  • $\begingroup$ @Blumer Even then, $\mathfrak{I}$ may not have a minimal generating set., as it could be that $Ra_1<Ra_2<Ra_3<\dots$. $\endgroup$ – Jeremy Rickard Feb 24 '17 at 15:25

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