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In general how would i tell if the rings $\mathbb{Z}[\sqrt d]$ and $\mathbb{Z}[\frac{x}{y}]$ are noetherian?

I know that the ring $\mathbb{Z}$ is noetherian as all ideals are contained in a finite ascending chain where the ideal at the bottom of the chain is generated by a prime integer. However i am unsure how adjoining the $\sqrt2$ impacts the ACC condition.

I have attempted to show that $\mathbb{Z}[\frac{1}{2}]$ by arguing that the ideal generated by $<\frac{3}{2}>$ is maximal, as the only ring that contains this ideal is $Z[\frac{1}{2}]$

Also what is the difference between having an ideal noetherian as a $\mathbb{Z}$ module as opposed to having an ideal noetherian as a ring.

thanks for the help,.

Peter A
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  • In a Nethereian ring, every ascending chain of ideals is finite. This does not imply that every ideal is in the same chain. – ajotatxe May 11 '15 at 17:12
  • As for your last question, being noetherian as a $\mathbb{Z}$-module is stronger than being noetherian as a ring or module. For example, $\mathbb{Q}$ is noetherian as a $\mathbb{Q}$-module or ring, but not as a $\mathbb{Z}$-module, because we have $\mathbb{Z}[\frac{1}{2}] \subsetneq \mathbb{Z}[\frac{1}{4}]\subsetneq \mathbb{Z}[\frac{1}{8}]$, etc. – Andrew Dudzik May 11 '15 at 17:58

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By the Hilbert basis theorem, both $\Bbb Z[x]$ and $\Bbb Z[x,y]$ are Noetherian rings.

$\Bbb Z[\sqrt{n}]$ is just a quotient of the first ring, and since the quotient of a Noetherian ring is Noetherian, you have your answer.

$\Bbb Z[x/y]$ is the same as the localization of the second one at the multiplicative set $\{y^n\mid n\in 0,1,2,\ldots\}$. Since the localization of a commutative Noetherian ring is Noetherian, this ring is Noetherian.

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rschwieb
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