Question

Let k be a commutative ring and G be any group. If kG is left noetherian, show that kG is right noetherian.

Answer 1

Define a map $\varepsilon : kG \to kG$ by $\varepsilon\left(\sum a_{g} g\right) = \sum a_{g} g^{-1}$ . Since $(gh)^{-1} = h^{-1} g^{-1}$ , and $k$ is commutative, we can show that $\varepsilon(\alpha \beta) = \varepsilon(\beta) \varepsilon(\alpha)$ . Of course $\varepsilon$ is one-one, onto, and an additive homomorphism. Since we also have $\varepsilon^{2} = 1$ , $\varepsilon$ is an involution on $kG$ . If $I_{1} \subset I_{2} \subset \cdots$ was an ascending chain of right ideals in $kG$ , $\varepsilon(I_{1}) \subset \varepsilon(I_{2}) \subset \cdots$ would have given an ascending chain of left ideals in $kG$ . This gives the desired conclusion in the noetherian case.