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I read a message about the ring of complex entire functions, that is neither Artinian nor Noetherian (see here).

Can you show me other examples of rings that are neither Artinian nor Noetherian?

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up vote 7 down vote accepted

For commutative rings (edit: I required commutative rings to have a $1$), the condition that $R$ is Artinian is equivalent to the condition that $R$ is Noetherian and has Krull dimension $0$. Thus any commutative ring which is not Noetherian is not Artinian either. One common example is the ring $A$ of algebraic integers. It has an infinite ascending chain of ideals as follows: $$(2)\subset (\sqrt{2})\subset (\sqrt[3]{2})\subset\cdots$$ and an infinite descending chain of ideals: $$(2)\supset (2^2)\supset(2^3)\supset\cdots$$ Another example is the ring $R$ of polynomials over the field $k$ in infinitely many variables, which has an infinite ascending chain of ideals: $$(x_1)\subset(x_1,x_2)\subset(x_1,x_2,x_3)\subset\cdots$$ and an infinite descending chain chain of ideals: $$(x_1)\supset(x_1^2)\supset(x_1^3)\supset\cdots$$ Note that these are just examples of infinite ascending and descending chains of ideals in these rings; many more exist.

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I think you've got your inclusions the wrong way round in your second and fourth chains –  Daniel Freedman Mar 16 '12 at 21:51
    
@DanielFreedman Yes, thanks. –  Alex Becker Mar 16 '12 at 21:53
    
I don't think the inclusions in your first chain work as stated - the roots need to be 1st, 2nd, 4th, 8th etc –  Mark Bennet Mar 16 '12 at 22:09
    
@MarkBennet $\sqrt{2}=\sqrt[6]{2}\sqrt[3]{2}$ and $\sqrt[6]{2}\in A$. –  Alex Becker Mar 16 '12 at 22:14
    
@AlexBecker - thanks, brain on hold –  Mark Bennet Mar 16 '12 at 22:26
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For any field $k$ consider the ring of polynomials in infnite many variables $k[\{x_n:n\in \mathbb N\}]$. It is not noetherian and hence is not artinian.

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