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$R$ is a commutative artinian ring. $G$ is a finite group. Is it true that $R[G]$ is artinian?

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A decreasing chain of ideals in $R[G]$ is a decreasing chain of $R$-submodules of the finitely generated $R$-module $R[G]$, no? – Dylan Moreland Dec 15 '11 at 19:14

Yes. Let $A_1\supset A_2 \supset ...$ be a descending chain of right ideals in $R[G]$. For each $g$ in $G,$ let $\pi_g:R[G] \rightarrow R$ be the projection map which sends an element $x$ to it's $g$-th coefficient. Note $\pi_g$ is a right $R$-module homomorphism. It follows

$$\pi_g(A_1) \supset \pi_g(A_2) \supset ...$$

is a descending chain of right ideals in $R.$ Choose an index $N_g$ such that $\pi_g(A_{N_g}) = \pi_g(A_{N_g + k})$ for all $k>N_g.$ Setting $N = \displaystyle\max_{g\in G}\{N_g\},$ it follows $$\displaystyle\bigoplus_{g\in G}\mbox{ } \pi_g(A_N) =\displaystyle\bigoplus_{g\in G} \mbox{ }\pi(A_{N+k})$$ for all $k>N.$ Therefore, as $$\displaystyle\bigoplus_{g\in G}:R[G] \rightarrow R^G$$ is an isomorphism of left $R$-modules, $A_N = A_{N+k}$ for all $k>N$ and $R[G]$ is artinian.

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