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feat(representation_theory/group_cohomology_resolution): show k[G^(n …
…+ 1)] is free over k[G] (#15501) Defines an isomorphism $k[G^{n + 1}] \cong k[G] \otimes_k k[G^n].$ Also shows that given a $k$-algebra $R$ and a $k$-basis for a module $M,$ we get an $R$-basis of $R \otimes_k M.$ Then, using that, we show $k[G^{n + 1}]$ is free.
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Amelia Livingston
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Aug 11, 2022
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