# Non-unital ring $(2\mathbb{Z})[X]$ is not Noetherian

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 $$R[x]$$?

• Does the definition of "ring" in the wikipedia article require the existence of $1$? And are you certain that $R[x]$ isn't Noetherian? – Arthur Mar 11 '18 at 13:40
• I think is because $2\mathbb{Z}$ has not unity. – Gödel Mar 11 '18 at 13:42
• @Arthur I believe Hilbert's theorem holds for non unity rings – krirkrirk Mar 11 '18 at 13:43
• @Arthur I think, that by definition of Noetherian ring, Hilbert's theorem should be require the existence of 1 (or I'm wrong?). – Pennywise Mar 11 '18 at 13:54
• 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? – rschwieb Mar 11 '18 at 14:18

## 1 Answer

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.)