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Let $R=\oplus_{I\geq 0}R_i$ be a positive graded ring(maybe not commutative), where $R_0$ is a commutative Noetherian ring. If $R$ is finite generated $R_0$-algebra, is $R$ Noetherian?

In here, Is every (left) graded-Noetherian graded ring (left) Noetherian?, $\mathbb Z$-graded ring is graded Noehterian iff it is Noetherian.

I found that this result is true for graded-commutative ring using Artin-Tate lemma:https://en.wikipedia.org/wiki/Artin-Tate_lemma.

Thank you in advance.

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The answer is no by Exercice 26 in the 2012 edition of Bourbaki's Algèbre VIII.1. (This seems moreover to have nothing to do with graduations.)

(Translation of the exercise: Let $K$ be a commutative field, let $A$ be the polynomial ring $K[T]$, and let $\sigma$ be the endomorphism $P(T)\mapsto P(T^2)$ of $A$. Then, the ring $A[X]_\sigma$ is not left-noetherian, although $A$ is noetherian.)

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  • $\begingroup$ What is your meaning of graduations? $\endgroup$
    – Jian
    Dec 30, 2019 at 7:07
  • $\begingroup$ The usual one, i.e. (for a ring) a given decomposition of the underlying group as a direct sum compatible with the operation of the group of degrees in the obvious way. But the point is that by the linked question it is seen that noetherianness and graded-noetherianness are the same. $\endgroup$ Dec 30, 2019 at 8:43
  • $\begingroup$ Hi Rohrer, I don't understand French in the exercise. Can you translate it in the answer? Thank you. $\endgroup$
    – Jian
    Dec 30, 2019 at 16:40
  • $\begingroup$ Done. $\mbox{ }$ $\endgroup$ Dec 30, 2019 at 18:24
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Perhaps the simplest counterexample: let $R$ be the free $R_0$-algebra on two generators $x$ and $y$. The two-sided ideal $RxR$ is not finitely generated as a left (or right) ideal.

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