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If $R$ is a commutative ring unital ring satisfying ascending chain condition on finitely generated ideals then is $R$ Noetherian ?

So we would have to prove a.c.c. for any ideals or by Cohen's theorem we would have to show every prime ideal of $R$ is finitely generated, but I cannot think of anything in either direction. Please help. Thanks in advance.

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  • $\begingroup$ Note it is well-known a commutative ring is noetherian if its prime ideals are finitely generated. $\endgroup$
    – Bernard
    Commented Aug 4, 2016 at 15:38
  • $\begingroup$ Somewhat interestingly, a ring satisfying the d.c.c. on finitely generated right ideals is not necessarily Artinian. If I remember right, I think it is equivalent to right perfectness. $\endgroup$
    – rschwieb
    Commented Aug 5, 2016 at 0:36
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    $\begingroup$ This question is Exercise 1, page 52, from Kaplansky, Commutative Rings. $\endgroup$
    – user26857
    Commented Aug 20, 2016 at 18:52

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A noetherian ring $R$ can also be characterized by the fact that every ideal of $R$ is finitely generated.

Suppose that there exists an ideal $I$ in $R$ which is not finitely generated. Let $x_1\in I$. The ideal generated by $x_1$, $(x_1)\neq I$. There exists $x_2\in I, x_2$ is not in $(x_1)=I_1$. Suppose defined $I_n$ such that $I_{n-1}\subset I_n\subset I$, $I_n$ is finitely generated. There exists $x_{n+1}\in I$, $x_{n+1}$ is not in $I_{n}$, write $I_{n+1}=I_n+(x_{n+1})$. The sequence $I_n$ does not stabilize, contradiction.

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  • $\begingroup$ oh so simple it was , thanks ( b.t.w. you didn't have to write $gr$ in each case to mean generating ... I would understand just $(I,x)$ as well :) ) $\endgroup$
    – user228169
    Commented Aug 4, 2016 at 15:29

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