0

Can somebody give me example of ring $R$ such that $R/I$ is Noetherian but $R$ is not Noetherian ring? $I$ is finitely generated ideal of $R$.

Also please search example such that $I$ is not nilpotent because if $I$ is nilpotent then R become Notherian ring.

rschwieb
  • 160,592
maths student
  • 1,535
  • 6
    $R$ is any non-Noetherian ring, and $I = R$ works... – darij grinberg Sep 24 '18 at 15:42
  • I has to be finitely generated. – maths student Sep 24 '18 at 15:44
  • 1
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 24 '18 at 15:55
  • 5
    $R$ is finitely generated as an iedal. – darij grinberg Sep 24 '18 at 15:55
  • 1
    A non-Noetherian ring $R$ with finitely generated maximal ideals (and $I$ any of those) works. For instance, this and these. –  Sep 24 '18 at 15:56
  • 1
    @darijgrinberg While for modules it is fine to consider the submodule $R<R$, using it for rings stretches quite thin the definitions, because often your interest lies in the category of commutative rings with unity, where you are not allowed to have the zero ring. –  Sep 24 '18 at 15:59
  • @Ninjahatori I edited the question to correct the punctuation, formatting and title to be closer to our standards to show you how easy it is. It doesn't have to be perfect, but it can't have problems in every sentence either. I'm sure that if you persisted in posting with the quality you originally used, I and others would quickly lose patience and just vote to close until you fixed it. So please take Shaun's advice above about post quality to heart. – rschwieb Sep 24 '18 at 16:17
  • 3
    @SaucyO'Path: The zero ring has a unity. It's just zero :) – darij grinberg Sep 24 '18 at 16:21
  • @SaucyO'Path Sir, can you explain me 2ℤ()×ℚ why this is not -notherian and why ideal of this is principal? I guess non noetherian beacause since it is not finitely generated but how to write it properly I don't know? – – maths student Sep 24 '18 at 17:25

1 Answers1

3

$R=\prod_{i\in \mathbb N}F_2$, with $I=(0,1,1,\ldots)R$.

$R/I$ is the field $F_2$ and $I$ is principal.

rschwieb
  • 160,592