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Here is a problem from the textbook Algebra by Hungerford, which I seem to be stuck on for quite some time now:

Let S be a multiplicative subset of a commutative Noetherian ring R with identity. Then the ring $S^{-1}R$ is Noetherian.

I assume $S^{-1}$ means ${\{ s^{-1} \ | \ s \in S }\}$ so $S^{-1}R = {\{ s^{-1}r \ | \ s \in S, r \in R }\}$. If both R is Noetherian so is S and S^{-1}. An ideal of S^{-1}R is an ideal of S^{-1} since RI=IR=I in a commutative ring. So any sequence of increasing ideals in S^{-1}R stops from some point on. How to make this rigorous, if the approach is on the right track?

user26857
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Algebra
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    Perhaps you should check what $S^{-1}R$ actually means. It is a ring of fractions, not a subring of $R$ by any means. Not sure why are you even interested in this question if you don't know it. – Mark Feb 07 '22 at 15:41
  • hi Algebra; I'd like to second the comment of Mark – you need to know what $S^{-1}R$ (the "localization of $R$ at $S$") is before trying to prove any of its properties! I'm not familiar with Hungerford's book, but he almost surely defines localization before assigning this exercise; I would recommend working through that section – Atticus Stonestrom Feb 07 '22 at 16:19

1 Answers1

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Assume the contrary that the following ascending chain of ideals in $S^{-1}R$ does not stop.

$$S^{-1}I_1 \subseteq S^{-1}I_2 \subseteq S^{-1}I_3 \subseteq S^{-1}I_4 \subseteq \cdots . $$

So there is not any integer $n$ such that $S^{-1}I_{n}=S^{-1}I_{n+1}$. Hence we find the following ascending chain of ideals in $R$ which does not stop.

$$I_1\subseteq I_2\subseteq I_3\subseteq I_4\cdots. $$

This is a contradiction.

Saeed Yazdani
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    You should mention the fact that all ideals of $S^{-1}R$ have that form. And also explain why we get an increasing sequence of ideals in $R$. In general, $S^{-1}I\subseteq S^{-1}J$ doesn't imply $I\subseteq J$. – Mark Feb 07 '22 at 16:20
  • In fact ideals of $S^{-1}R$ have the form ${S^{-1}I : I ,\text{is an ideal of R and },S\cap I=\varnothing }$. So if $S^{-1}I$ is an ideal of $S^{-1}R$, $I$ is an ideal of $R$. – Saeed Yazdani Feb 07 '22 at 18:41
  • I know, but it's not like this is absolutely trivial. You should mention this in the answer. (not prove, but mention). And as I said $S^{-1}I\subseteq S^{-1}J$ doesn't necessary imply $I\subseteq J$. So in order to get an ascending chain in $R$ you need to choose the ideals correctly. Thing is, every ideal of $S^{-1}R$ satisfies $J=S^{-1}(f^{-1}(J))$ where $f:R\to S^{-1}R$ is the natural homomorphism. So given an ascending chain $J_1\subseteq J_2\subseteq...$ in $S^{-1}R$ you can look at the (definitely ascending) chain $f^{-1}(J_1)\subseteq f^{-1}(J_2)\subseteq$ in $R$. – Mark Feb 07 '22 at 20:07