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Let $R$ be a non-negatively graded Noetherian ring such that $R_{0}$ is Artinian and $R_{+}$ is a nilpotent ideal. Prove that $R$ is Artinian. Give an example to show that this is false if the Noetherian property is removed.

This is an exercise from a note that I saw on the Internet. I can not solve it. Please help me. Thanks!

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What have you tried so far? Of course, you should use the statement in your other question math.stackexchange.com/questions/147466 –  Martin Brandenburg May 20 '12 at 19:54
    
Sorry, how can I use it ? I am not good at algebra. –  variete May 21 '12 at 1:54

2 Answers 2

By your assumption, $R_+$ is nilpotent, then of course the krull dimension of $R$ is the same as $R/R_+=R_0$ which is zero. A commutative ring is Artinian if and only if it is Noetherian and of dimension zero. So $R$ is Noetherian and of dimension zero, thus it is Artinian. We are done!

Counter-example: omit

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Thank you very much wxu! However, why do you know that if $R_+$ is nilpotent then the Krull dimension of $R$ is equal to the dimension of $R/R_+$? –  variete May 23 '12 at 11:45
    
Nilpotent ideal is contained in any prime ideal. –  wxu May 23 '12 at 12:21
    
Dear @wxu, I am sorry for my stupid question here, but can you show me a proof? –  variete May 23 '12 at 15:57
    
@variete You should be able to get this on your own: Suppose $P$ is prime and $x$ is nilpotent. Then for some $n$, $x^n=0\in P$ implies ____?____. –  rschwieb May 23 '12 at 16:54

Here is how the problem unfolded for me. ($rad(R)$ denotes the Jacobson radical, for me.)

The first thing I noticed is that $R_0\cong R/R^+$ is an Artinian ring, and since $R^+$ was given to be nilpotent it is contained in $rad(R)$, and so I had some hope that $R^+=rad(R)$ so that I could use the Hopkins-Levitzki Theorem.

Does the grading imply $R^+=rad(R)$? This will hold if $rad(R/R^+)=\{0\}$, and for Artinian rings one only needs to check that $R/R^+$ has no nilpotent nonzero ideals. If $K\supsetneq R^+$ were nilpotent, then there would have to be some $x\in K\setminus R^+$ that was nilpotent: but this would imply that there is a grade 0 element $x$ such that $x^n\in R^+$, which is an absurdity. So $R/R^+$ is semisimple artinian, and therefore $R^+=rad(R)$.

Invoking the Hopkins-Levitzki theorem, a ring such that $R/rad(R)$ is Artinian and $rad(R)$ is nilpotent is right Artinian iff right Noetherian. Since we are given that $R$ is Noetherian on both sides, $R$ is Artinian on both sides.

I'd like to encourage you though to find your own solution, possibly one that doesn't have to invoke the H-L Theorem :)

Let me give you a hint for the required counterexample: start with a polynomial ring in countably many indeterminates $\mathbb{F}[x_1,x_2,\dots]$ and see if you can make a graded ring whose set $R^+$ is nilpotent.

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Dear @rschwieb : Does rad(R) mean the Jaconson radical ? –  variete May 23 '12 at 11:43
    
@variete Correct, I'll put in a note about it. –  rschwieb May 23 '12 at 11:44
    
@variete Did you have any luck with the counterexample hint? –  rschwieb May 23 '12 at 12:59
    
Sorry, I have not found it yet –  variete May 23 '12 at 14:35
    
@variete Think of a candidate for grading that you might use for polynomials in many variables :) And let me know if more hints are necessary. –  rschwieb May 23 '12 at 15:27

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