[Merged by Bors] - feat(representation_theory/group_cohomology_resolution): show k[G^(n + 1)] is free over k[G] #15501
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| @[simp] lemma prod_unique_apply {α β : Type*} [unique β] (x : α × β) : | ||
| prod_unique α β x = x.1 := rfl | ||
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| @[simp] lemma prod_unique_symm_apply {α β : Type*} [unique β] (x : α) : |
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Similarly here:
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| @[simp] lemma prod_unique_symm_apply {α β : Type*} [unique β] (x : α) : | |
| @[simp] lemma coe_prod_unique_symm {α β : Type*} [unique β] : | |
| ⇑(prod_unique α β).symm = pi.prod (id : α → α) default := rfl | |
| lemma prod_unique_symm_apply {α β : Type*} [unique β] (x : α) : |
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It seems logic/equiv/basic doesn't import data/pi/algebra where pi.prod is defined, and vice versa :(
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Is there some kind of tool which finds the first file which imports both logic/equiv/basic and data/pi/algebra?
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…+ 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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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.