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I am looking for an answer to why one has to assume commutativity of a ring $R$ in proving some results about Noetherian rings. For example, Let $R$ be a commutative ring; look at the proof(s) of the theorem "if $H$ and $R/H$ are Noetherian, then so is $R$". Where does commutativity comes in?

EDIT

For example, consider the following sketch of a proof. Consider a generic ascending chain of ideals $\{H_i\}_i$ in $R$. We know (by hypothesis) that $\{H_i \cap H\}_i$ is stationary so is $\{(H_i + H)/H\}_i$. Pick the max index that makes both chains stationary (say $m$), then using Dedekind's law for groups starting from $H_i = H_i\cap(H_i+H)$ (note that it is $H \cap H_i = H \cap H_m$ and $H+H_i =H+H_m$ from $m$ on), we finally get that $\{H_i\}_i$ is stationary.

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What book are you studying from? It's possible that the book places emphasis on commutative rings for some reason. Which proof do you want us to "look at"? –  Arturo Magidin Apr 2 '12 at 15:20
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The implication you want holds in noncommutative rings; but if your ring has emphasis on commutative rings, it may not want to bother dealing with noncommutative rings (in which you have the distinct notions of "left-noetherian", "right-noetherian", and "noetherian"). –  Arturo Magidin Apr 2 '12 at 15:54
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You should pick a book on noncommutative rings to answer this yourself. The book by Lam or the bible by McConner and Robson will paint a pretty good picture of what the lack of commutativity causes and how it is handled. That particular theorem you mention does not depend on commutativity but if the book you are reading is a commutative algebra book, then it is unsurprising if it will not mention the more general result —very often that would introduce big complications or simply be impossible; already the elementary study of localizations is considerably more complicated for non-comm. rings. –  Mariano Suárez-Alvarez Apr 2 '12 at 15:58
    
Before that thanks Can you help me about the Noetherian ring –  user31622 May 17 '12 at 13:04
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@asefa If your remark refers to a new question then please post it separately. –  Bill Dubuque May 17 '12 at 13:47

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