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Here is one exercise from some notes on graded rings. I tried but I got no idea to solve it. Please help me. Thanks.

Let $R$ be a graded ring. Prove that $R$ is Noetherian (Artinian) if and only if $R$ satisfies the ascending (descending) condition on homogeneous ideals.

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What have you tried so far? The proof imitates the usual proof of the Hilbert Basis Theorem. This is already a direct hint to the solution ... –  Martin Brandenburg May 20 '12 at 19:55
Can you post the solution? Sorry I am not good at algebra.... –  variete May 21 '12 at 3:13
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Proof that $R$ is Noetherian: Suppose $R$ satisfying ACC on homogeneous ideals. Then $R_0$ is Noetherian. And $R_+$ is generated by homogeneous elements, by acc, we may find finitely many homogeneous elements, $x_1,\ldots,x_n$, such that $R_+=(x_1,\ldots,x_n)$, we claim that $R=R_0[x_1,\ldots,x_n]$. For any homogeneous $f\in R_+$, $f=x_1g_1+\ldots+x_ng_n$, where $g_i$ are homogeneous and with degree less than $f$, so by induction, $g_i\in R_0[x_1,\ldots,x_n]$, in this case, $f\in R_0[x_1,\ldots,x_n]$. Thus $R=R_0[x_1,\ldots,x_n]$. Finally, by Hilbert's basis theorem, $R$ is Noetherian.

Proof that $R$ is Artinian: First of all, $R_0$ is Artinian. By condition of dcc, the ideal $R_+$ is indeed has only finitely many term, i.e., $R_+=R_1\oplus R_2\ldots \oplus R_n$. So $R_+$ is a nipotent ideal, $(R_+)^{n+1}=0$. The subring $R_0$ is Artinian, it has finitely many maximal ideals, say $\mathfrak{m}_1,\ldots,\mathfrak{m}_k$, then there exists an $l$ such that $\mathfrak{m}_1^l\cdots\mathfrak{m}_k^l=0$. It is clear that maximal ideals of $R$ are exactly $(\mathfrak{m}_i,R_+),i=1,\ldots,k$ and the ideals $\mathfrak{m}_i^lR$ pairwise are comaximal! So by chinese remainder theorem, we only need to show $R/\mathfrak{m}_i^lR$ is Artinian for each $i$. Our dcc conditons on homogeneous ideals still hold. Hence we may assume $R_0$ is a Artinian local ring with unique maximal $\mathfrak{m}_0$ and $R$ satisfys dcc conditions on homogeneous ideals. Ok, let us considet the ring $R/\mathfrak{m}_0R$, it must be finite over $R_0/\mathfrak{m}_0$, otherwise, the image of $R_+$ is generated by the image of homogeneous of $R_+$ which is of infinite dimension over $R_0/\mathfrak{m}_0$ and hence we can find an strictly decended infinite chain of homogeneous ideals of $R$. This is a contradition. Thus $R/\mathfrak{m}_0R$ is finite over $R_0/\mathfrak{m}_0$. Let us pick $f_1,\ldots,f_k$ be homogenous of $R$ such that the image of $R_0+\sum R_0f_i$ is $R/\mathfrak{m}_0R$. Then we claim that $R=R_0+\sum R_0f_i$. Denote $S=R_0+\sum R_0f_i$. Then $R=S+\mathfrak{m}_0R$, thus $R/S=\mathfrak{m}_0R/S$, notice that $\mathfrak{m}_0$ is nilpotent! It follows imediately that $R/S=0$. Eventually, we have proved that $R$ is finite over $R_0$ and thus $R$ is Artinian.

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Dear @wxu, why $R^{+}$ is nilpotent ? –  variete May 23 '12 at 1:30
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