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Given a definition of $R=2\Bbb Z$ prove that ring $R[x]$ is not noether.

I assume that the proof should be based on the fact that a ring is noether if and only if any ideal is finitely generated. However, I'm struggling with constructing a non-finitely generated ideal, because only ideals I can think of are "root-based" (i.e. polynomials that have certain roots), that are, of cource, finitely generated.

I would be very grateful for a description of such ideal or a piece of work that can explain the structure of polynomial ring ideals in detail.

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It's unconventional to treat rings without multiplicative identity, but here goes.

Let $R=2\Bbb Z$. There is an easy example of a non-finitely generated $R[X]$-ideal, namely $R[X]$ itself. To see an infinite ascending chain of $R[X]$-ideals, let $I_n$ be the ideal generated by $2X$, $2X^2,\ldots,2X^n$.

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  • $\begingroup$ but I think I can use the same reasoning for $R = \Bbb Z$ so $R[x]$ is noether due to Hilbert's basis theorem. What is the difference between these two cases? $\endgroup$ – youtur Mar 11 '18 at 17:11
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    $\begingroup$ As I pointed out, $R[x]$ is not finitely generated as a module over itself, and $R[x]$ has a nontrivial increasing sequence of ideals. If you study the proof of the Hilbert basis theorem carefully, you will see that it relies on $R$ being a unital ring. @youtur $\endgroup$ – Lord Shark the Unknown Mar 11 '18 at 17:14

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