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Let $R$ be a commutative ring with a multiplicative identity such that there is a finitely generated $R$-algebra that is Noetherian. Is $R$ Noetherian then?

I tried to prove this using the fact that the homomorphic image of a Noetherian ring is Noetherian but I can not find an ideal the quotient by which is isomorphic to $R$ for general finitely generated algebras. In some special cases, that can be done (e.g. for the free polynomial algebra take the ideal $(X_1, \dots, X_n)$).

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    This is not true. Try thinking of a counterexample. What is true is that if $R$ is a subring of a Noetherian ring finite over $R$, then $R$ is also Noetherian. – Parthiv Basu Aug 08 '19 at 13:39

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Let $R$ be any non-zero ring. If you are willing to accept the existence of maximal ideals, then there's always a finitely generated non-zero $R$-algebra $S$ which is noetherian.

Just take $\mathfrak{m}$ a maximal ideal of $R$ and then $R/\mathfrak{m}$ is, being a field, noetherian.

Therefore every non-noetherian ring will still have finite non-zero morphisms to noetherian rings

Note that noetherianity won't descend down finitely presented morphisms either. Continuing with the reasoning from maximal ideals, there are many non-noetherian rings which have finitely generated maximal ideals (e.g. construct an appropriate rank > 1 valuation domain).

Badam Baplan
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  • Or, just use the zero ring. – Eric Wofsey Aug 08 '19 at 15:38
  • does there exist a non-Noetherian (commutative unital) ring such that there is no finitely presented algebra over it that is Noetherian? –  Aug 08 '19 at 15:41
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    @hello no, as Eric Wofsey just said, there's always the zero ring, which is finitely presented over every ring. – Badam Baplan Aug 08 '19 at 15:42
  • I was just trying to edit the comment to make it less trivial. Does there exist a ring $R$ such that the quotient of $R$ by any proper finitely-generated ideal is not Noetherian and such that the quotient of $R[x_1,\dots, x_d]$ for any positive integer $d$ by a proper finitely-generated ideal is not Noetherian? –  Aug 08 '19 at 15:46
  • @How about a polynomial ring in infinitely many variables? – Badam Baplan Aug 08 '19 at 15:52