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A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The quotient field is then obviously Noetherian, but the subring $k[x_1,x_2,\dots]$ is not since there is an infinite ascending chain of ideals which never stabilizes.

Is there an instance of a finitely generated Noetherian ring over some ground ring $R$, that has an intermediate ring which is not finitely generated over $R$, and hence not Noetherian either?

user26857
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Emilia
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  • This is answered by [Mariano's answer here](http://math.stackexchange.com/questions/105164/are-intermediate-rings-of-finite-ring-extensions-also-finitely-generated/105218#105218). That question did not have a Noetherian condition, but it's clearly satisfied in the example. – Dylan Moreland Feb 05 '12 at 22:23
  • Oh yeah, Hilbert's basis theorem shows the Noetherian condition. Should I delete this somehow? – Emilia Feb 05 '12 at 22:32
  • I actually don't know what the interface for asking questions is like, but if you want to then that seems fine: no one has written an answer, or anything. – Dylan Moreland Feb 05 '12 at 22:46
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    In last sentence, I think that "hence not Noetherian either " doesn't make sense ,,because Noetherian property should be checked over itself ring. (not over $R$) – hew Mar 11 '20 at 13:40
  • By Eakin - Nagata theorem, converse is true. – hew Mar 11 '20 at 13:58

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To get this question off the unanswered list, I copy Mariano's example from here.

The polynomial ring $k[x,y]$ is Noetherian (by Hilbert's basis theorem), but the subring generated by $\{xy^i:i\geq0\}$ is not finitely generated over $k$.

Community
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mdp
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