Bourbaki

Lemma 5

Tag 00IU, permanent. Cite it as [Bourbaki, Tag 00IU].

Algebra, VIII, p. 300, § 16. Other Descriptions of the Brauer Group

If cc is a 2-coboundary, then so is c~s\widetilde{c}_s.

Let t:H→Ft: H\rightarrow F be a mapping such that c=∂tc=\partial t. Let t~s:G→F\widetilde{t}_s: G\rightarrow F be the mapping defined by

t~s(g)=∏x∈H\Gs(x)−1t(s(x)gs(x⋅g)−1) \widetilde{t}_s(g) =\prod_{x\in H\backslash G}^{s(x)^{-1}}t(s(x)gs(x\cdot g)^{-1})

for g∈Gg\in G. It suffices to prove that c~s=∂t~s\widetilde{c}_s=\partial \widetilde{t}_s, which follows from the relations c~s(g1,g2)=∏x∈H\Gs(x)−1s(x)g1s(x⋅g1)−1t(s(x⋅g1)g2s(x⋅g1g2)−1)\widetilde{c}_s(g_1, g_2) =\prod_{x\in H\backslash G}^{s(x)^{-1}s(x)g_1s(x\cdot g_1)^{-1}}t(s(x\cdot g_1)g_2s(x\cdot g_1g_2)^{-1})

×∏x∈H\Gs(x)−1(t(s(x)g1g2s(x⋅g1g2)−1)−1t(s(x)g1s(x⋅g1)−1)) \times \prod_{x\in H\backslash G}^{s(x)^{-1}}(t(s(x)g_1g_2s(x\cdot g_1g_2)^{-1})^{-1}t(s(x)g_1s(x\cdot g_1)^{-1}))
=∂t~s(g1,g2) =\partial \widetilde{t}_s(g_1, g_2)

for g1,g2∈Gg_1, g_2\in G.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

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Cites

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.