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Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring.

I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki).

I suppose that $2\mathbb{Z}$ is principal ideal ring:
Let (2), (4), (6), ... are the ideals, therefore all elements are generated by one ideal, so $2\mathbb{Z}$ is principal ideal ring. And we conclude that $2\mathbb{Z}$ is a noetherian ring. Why can't use Hilbert's basis theorem for $R[x]$?

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  • $\begingroup$ Does the definition of "ring" in the wikipedia article require the existence of $1$? And are you certain that $R[x]$ isn't Noetherian? $\endgroup$ – Arthur Mar 11 '18 at 13:40
  • $\begingroup$ I think is because $2\mathbb{Z}$ has not unity. $\endgroup$ – Gödel Mar 11 '18 at 13:42
  • $\begingroup$ @Arthur I believe Hilbert's theorem holds for non unity rings $\endgroup$ – krirkrirk Mar 11 '18 at 13:43
  • $\begingroup$ @Arthur I think, that by definition of Noetherian ring, Hilbert's theorem should be require the existence of 1 (or I'm wrong?). $\endgroup$ – Pennywise Mar 11 '18 at 13:54
  • $\begingroup$ I don’t know if the Hilbert basis theorem works in rings without identity or not, but it still seems like $R[x]$ is noetherian. A bigger question is why $R[x]$ looks like when $R$ doesn’t have identity. Where do you see the claim? $\endgroup$ – rschwieb Mar 11 '18 at 14:18
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The Hilbert basis theorem only applies to unital rings (this is often not stated explicitly since unital is often included in the definition of "ring"). Since $2\mathbb{Z}$ is not unital, the Hilbert basis theorem does not apply in this case.

An example of a non-finitely generated ideal in $2\mathbb{Z}[x]$ is the entire ring $2\mathbb{Z}[x]$ itself. (Note that for this to be true, $2\mathbb{Z}[x]$ must be defined as the set of polynomials all of whose coefficients are in $2\mathbb{Z}$, rather than the ring obtained from $2\mathbb{Z}$ by freely adjoining a central element $x$. This distinction does not make a difference for unital rings.)

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