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8 votes
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Brauer group of complete DVR

This is more trivial (in the sense that we have hidden all the non-trivial parts among general preliminaries...) than the identification of $\text{Br}(K)$ that Alex is talking about. We have that $\te …
Torsten Ekedahl's user avatar
19 votes
Accepted

Hilbert 90 for algebras

It's actually easier to go the other way around. Finite dimensional modules over an algebra $A$ fulfils the Krull-Remak-Schmidt theorem of being isomorphic to a direct sum of indecomposable modules wi …
Torsten Ekedahl's user avatar