Question on Noetherian/Artinian properties of a graded ring 
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!
 A: 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
A: 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.
